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Vol. XXXV

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MATHEMATICS TODAY | OCTOBER‘17

7

Sequences and Series This column is aimed at Class XI students so that they can prepare for competitive exams such as JEE Main/Advanced, etc. and be also in command of what is being covered in their school as part of NCERT syllabus. The problems here are a happy blend of the straight and the twisted, the simple and the difﬁcult and the easy and the challenging. *ALOK KUMAR, B.Tech, IIT Kanpur z

A progression is a sequence whose terms follow a certain pattern i.e., the terms are arranged under a definite rule.

ARITHMETIC PROGRESSION A sequence of numbers < tn > is said to be in arithmetic progression (A.P.) when the difference tn – tn – 1 is a constant for all n N. This constant is called the common difference of the A.P. and is usually denoted by the letter d. If ‘a’ is the first term and ‘d’ the common difference, then an A.P. can be represented as a, a + d, a + 2d, a + 3d, .... General Term of an A.P. z nth term of an A.P i.e., Tn = a + (n – 1)d. z pth term from the end is (n – p + 1)th term from the beginning = T(n – p + 1) = a + (n – p)d. Also, if last term of an A.P. is l then term from end is l – (p – 1)d. Sum of ‘n’ Terms of an A.P. The sum of n terms of the series a + (a + d) + (a + 2d) + ... {a + (n – 1)d} is given by n n Sn = [2a + (n − 1) d] or Sn = (a + l ) , 2 2 where l = last term = a + (n – 1)d ARITHMETIC MEAN Single A.M. between a and b : If a and b are two real a+b . numbers then single A.M. between a and b is 2 z n A.M.’s between a and b : If A1, A2, A3, ..., An are n A.M.’s between a and b, then z

8

MATHEMATICS TODAY | OCTOBER ‘17

A1 = a + d = a +

b−a ⎛b − a⎞ , A2 = a + 2d = a + 2 ⎜⎝ ⎟ n +1 ⎠ n +1

⎛b − a⎞ An = a + nd = a + n ⎜ ⎝ n + 1 ⎟⎠ Properties of A.P. : Let a1, a2, a3, .... are in A.P. whose common difference is d. z For a fixed non-zero number k R a1 ± k , a2 ± k , a3 ± k ,..... ; ka1, ka2, ka3, .... ; a1 a2 a3 , , .... will be in A.P., whose common k k k difference will be d, kd, d/k respectively. z The sum of terms of an A.P. equidistant from the beginning and the end is constant and is equal to sum of first and last term. i.e. a1 + an = a2 + an−1 = a3 + an−2 and so on z If number of terms of any A.P. is odd, then sum of the terms is equal to product of middle term and number of terms. z If number of terms of any A.P. is even then A.M. of middle two terms is A.M. of first and last term. z If the number of terms of an A.P. is odd then its middle term is A.M. of first and last term. z If a1, a2, .... , an and b1, b2, .... , bn are the two A.P.’s. Then a1 ± b1 , a2 ± b2 ,.... an ± bn are also A.P.’s with common difference d1 ± d2, where d1 and d2 are the common difference of the given A.P.’s. z Three numbers a, b, c are in A.P. iff 2b = a + c.

GEOMETRIC PROGRESSION (G.P.) A progression is called a G.P. if the ratio of its each term to its previous term is always constant. If 'a' be the first term and 'r' be the common ratio then, a, ar, ar2 , ....., arn – 1 is a sequence of G.P. General Term of a G.P. z nth term of a G.P. i.e., Tn = arn – 1. T T where common ratio, r = 2 = 3 = ...... T1 T2 z

If G.P. consists of ‘n’ terms, then pth term from the end = (n – p + 1)th term from the beginning = arn – p. Also, the pth term from the end of a G.P. with last

z

z

z

z

n−1

⎛1⎞ term l and common ratio r is l ⎜ ⎟ ⎝r ⎠ Sum of First ‘n’ Terms of G.P. Sum of first n terms of a G.P. is given by z

z z

. z

z

a(1 − r n ) a − lr Sn = , (when | r | < 1) and Sn = 1− r 1− r a(r n − 1) lr − a Sn = , (when | r | > 1) and Sn = r −1 r −1 Sn = na, (when r = 1)

If each term of a G.P. with common ratio r be raised to the same power k, the resulting sequence also forms a G.P. with common ratio rk. In a finite G.P., the product of terms equidistant from the beginning and the end is always the same and is equal to the product of the first and last term. i.e., if a1, a2, a3, .... , an be in G.P., Then, a1 an = a2 an−1 = a3 an−2 = a4 an−3 = .... = ar an−r +1 If the terms of a given G.P. are chosen at regular intervals, then the new sequence so formed also forms a G.P. If a1, a2, a3, .... , an , .... is a G.P. of non-zero, nonnegative terms, then loga1, loga2, loga3, .... , logan, .... is an A.P. and vice-versa. Three non-zero numbers a, b, c are in G.P., iff b2 = ac. If first term of a G.P. of n terms is a and last term is l, then the product of all terms of the G.P. is (al)n/2.

HARMONIC PROGRESSION (H.P.) z A progression is called a harmonic progression (H.P.) if the reciprocals of its terms are in A.P. General Term of an H.P.

Sum of Infinite Terms of G.P. z

a When |r| < 1, (or –1 < r < 1) ; S∞ = 1− r

z

If r t 1, then Sf doesn’t exist.

General term Tn of H.P. is

GEOMETRIC MEAN Single G.M. between a and b : If a and b are two real numbers then single G.M. between a and b = ab . z n G.M.’s between a and b : If G1, G2, G3, ..., Gn are n G.M.’s between a and b, then z

1

2

⎛ b ⎞ n+1 ⎛ b ⎞ n+1 G1 = ar = a ⎜ ⎟ , G2 = ar 2 = a ⎜ ⎟ ⎝a⎠ ⎝a⎠ 3 ⎛ b ⎞ n+1

G3 = ar 3 = a ⎜ ⎟ ⎝a⎠

z

H.M. (H) of n non-zero numbers a1, a2, a3, ..., an 1 1 1 + + ..... + an 1 a1 a2 is given by = H n

z

Let a, b be two given numbers and n harmonic means H1, H2, .... , Hn are inserted between a and b Now, a, H1, H2, .... , Hn, b are in H.P.

n ⎛ b ⎞ n+1

Properties of G.P. If all the terms of a G.P. be multiplied or divided by the same non-zero constant, then it remains a G.P., with the same common ratio. z The reciprocal of the terms of a given G.P. form a G.P. with common ratio as reciprocal of the common ratio of the original G.P. 10

HARMONIC MEAN If three or more numbers are in H.P., then the numbers lying between the first and last are called harmonic means (H.M.’s) between them. For example 1, 1/3, 1/5, 1/7, 1/9 are in H.P. So 1/3, 1/5 and 1/7 are three H.M.’s between 1 and 1/9. z Insertion of harmonic means 2ab z Single H.M. between a and b = a +b

, ...., Gn = ar n = a ⎜ ⎟ ⎝a⎠

z

MATHEMATICS TODAY | OCTOBER ‘17

1 a + (n − 1) d

⇒

1 1 1 1 1 , , ,...., , are in A.P. a H1 H 2 Hn b

MATHEMATICS TODAY | OCTOBER‘17

11

Let D be the common difference of this A.P. 1 Then, = (n + 2)th term = Tn+2 b 1 1 a −b = + (n + 1) D ⇒ D = (n + 1) ab b a Properties of H.P. z No term of H.P. can be zero. z If H is the H.M. between a and b, then z

1 1 1 1 = + + H −a H −b a b

z

H2

z

(H – 2a)(H – 2b) = H +a H +b + =2 H −a H −b

ARITHMETICO–GEOMETRIC PROGRESSION(A.G.P) The combination of arithmetic and geometric progression is called arithmetico-geometric progression. If a1, a2, a3, ..., an is an A.P. and b1, b2, b3, ..., bn is a G.P., then the sequence a1b1, a2b2, a3b3, ..., anbn is said to be an arithmetico-geometric sequence. General Term of A.G.P. The general form of A.G.P is a,(a + d ) r ,(a + 2d ) r 2 ,(a + 3d ) r 3 ,..... ? Tn = [a + (n – 1)d]rn – 1. Sum of A.G.P. z Sum of n terms : The sum of n terms of an arithmetico-geometric sequence a, (a + d)r, (a + 2d)r2, (a + 3d)r3, .... is given by n ⎧ a (1 − r n−1 ) {a + (n − 1)d }r ⎪ − + dr , when r ≠ 1 ⎪ 1− r (1 − r )2 Sn = ⎨1 − r ⎪n ⎪⎩ 2 [2a + (n − 1)d], when r = 1 z

SPECIAL SERIES n

z

MATHEMATICS TODAY | OCTOBER ‘17

1 + 2 + 3 + ....... + n = ∑ r =

n (n + 1)

r =1

z

2

n

n (n + 1)(2n + 1)

r =1

6

12 + 22 + 32 + .... + n2 = ∑ r 2 =

z

n ⎡ n (n + 1) ⎤ 13 + 23 + 33 + 43 + .... + n3 = ∑ r 3 = ⎢ ⎥ ⎣ 2 ⎦ r =1

z

1 + 3 + 5 + ..... + (2n − 1) = ∑ (2r − 1) = n2

2

n

r =1 n

z

2 + 4 + 6 + ...... + 2n = ∑ 2r = n (n + 1) r =1

Properties of Arithmetic, Geometric, Harmonic Means between Two Given Numbers Let A, G and H be arithmetic, geometric and harmonic means of two numbers a and b. a +b 2ab , G = ab and H = Then, A = 2 a +b z A≥G ≥H ' A −G =

a+b ( a − b )2 − ab = ≥0 2 2

AtG G − H = ab −

Sum of infinite sequence : a dr ,| r | < 1 . S∞ = + 1 − r (1 − r )2

Method for Finding Sum This method is applicable for both sum of n terms and sum of infinite number of terms. First suppose that sum of the series is S, then multiply it by common ratio of the G.P. and subtract. In this way, we shall get a G.P., whose sum can be easily obtained. 12

Method of Difference If the differences of the successive terms of a series are in A.P. or G.P., we can find nth term of the series by the following steps : z Denote the nth term by Tn and the sum of the series upto n terms by Sn. z Rewrite the given series with each term shifted by one place to the right. z By subtracting the later series from the former, find Tn. z From Tn, Sn can be found by appropriate summation.

z

⎛ a + b − 2 ab ⎞ 2ab = ab ⎜ ⎟ a +b a +b ⎝ ⎠

ab ( a − b )2 ≥ 0 a +b GtH ...(ii) From (i) and (ii), we get A ≥ G ≥ H Note that the equality holds only when a = b. A, G, H form a G.P., i.e., G2 = AH The equation having a and b as its roots is =

z

…(i)

x2 – (a + b)x + ab = 0 x2 – 2Ax + G2 = 0 a+b ⎡ ⎤ ⎢' A = 2 and G = ab ⎥ ⎣ ⎦ The roots a, b are given by A ± A2 − G 2 . The equation having a, b, c as its roots is x 3 − (a + b + c)x 2 + (ab + bc + ca)x − abc = 0 1 1 1 + + 1 a +b+c a b c 1/3 = , G = (abc) and Here A = H 3 3 a + b + c = 3A, abc = G3 and ⇒ x 3 − 3 Ax 2 +

3G3 x − G3 = 0 H

3G 3 = ab + bc + ca H

RELATION BETWEEN A.P., G.P. AND H.P. If A, G, H be A.M., G.M., H.M. between a and b, ⎧ A when n = 0 an+1 + bn+1 ⎪ then = ⎨G when n = −1 / 2 an + bn ⎪H when n = −1 ⎩

z

z

If A1, A2 be two A.M.’s; G1, G2 be two G.M.’s and H1, H2 be two H.M.’s between two numbers a and b, then G1G2 A + A2 = 1 H1H 2 H1 + H 2

z

Recognization of A.P., G.P., H.P. : If a, b, c are three successive terms of a sequence. a −b a If = , then a, b, c are in A.P. b−c a a −b a = , then a, b, c are in G.P. b−c b a −b a If, = , then a, b, c are in H.P. b−c c If number of terms of any A.P./G.P./H.P. is odd, then A.M./G.M./H.M. of first and last terms is middle term of series. If number of terms of any A.P./G.P./H.P. is even, then A.M./G.M./H.M. of middle two terms is A.M./G.M./ H.M. of first and last terms respectively. If pth, qth and rth terms of a G.P. are in G.P. Then p, q, r are in A.P. If a, b, c are in A.P. as well as in G.P. then a = b = c. If a, b, c are in A.P., then xa, xb, xc will be in G.P. (x z ±1) If,

z

z

z

z z

IMPORTANT POINTS z If Tk and Tp of any A.P. are given, then formula for T − T Tp − Tk . obtaining Tn is n k = n−k p−k z

If pTp = qTq of an A.P., then Tp + q = 0.

z

If pth term of an A.P. is q and the qth term is p, then Tp + q = 0 and Tn = p + q – n. 1 1 If the pth term of an A.P. is and the qth term is q p 1 then Tpq = 1 and S pq = ( pq + 1) 2 The common difference of an A.P is given by d = S2 – 2S1 where S2 is the sum of first two terms and S1 is the sum of first term or the first term. If sum of n terms Sn is given then general term Tn = Sn – Sn – 1, where Sn – 1 is sum of (n – 1) terms of A.P. If for an A.P. sum of p terms is q and sum of q terms is p, then sum of (p + q) terms is {–(p + q)}. If for an A.P., sum of p terms is equal to sum of q terms, then sum of (p + q) terms is zero. Sum of n A.M.’s between a and b is equal to n times the single A.M. between a and b. ⎛a +b⎞ i.e., A1 + A2 + A3 + ...... + An = n ⎜ ⎝ 2 ⎟⎠

z

z

z

z

z

z

z

z z

If A1 and A2 are two A.M.’s between two numbers 1 1 a and b, then A1 = (2a + b), A2 = (a + 2b) . 3 3 Sum of m A.M.’s m Between two numbers, = Sum of n A.M.’s n If Tk and Tp of any G.P. are given, then formula for 1 ⎛ Tn ⎞ n− k

obtaining Tn is ⎜ ⎟ ⎝ Tk ⎠ z

z

z

1

⎛ Tp ⎞ p − k . =⎜ ⎟ ⎝ Tk ⎠

If a, b, c are in G.P. then b c a +b b+c ⇒ = ⇒ = a b a −b b−c a −b a a +b a or or = = b−c b b+c b Let the first term of a G.P be positive, then if r > 1, then it is an increasing G.P., but if r is positive and less than 1, i.e. 0 < r < 1, then it is a decreasing G.P. Let the first term of a G.P. be negative, then if r > 1, then it is a decreasing G.P., but if 0 < r < 1, then it is an increasing G.P. MATHEMATICS TODAY | OCTOBER‘17

13

z

Product of n G.M.’s between a and b is equal to nth power of single geometric mean between a and b, i.e., G1 G2 G3 ...... Gn = ( ab )n

z z

z

G.M. of a1 a2 a3 ... an is (a1 a2 a3 ... an)1/n. If G1 and G2 are two G.M.’s between two numbers a and b is G1 = (a2b)1/3 , G2 = (ab2 )1/3 . If H1 and H2 are two H.M.’s between a and b, then 3ab 3ab and H 2 = . H1 = 2a + b a + 2b PROBLEMS Single Correct Answer Type

1.

The number of terms in the series 101 + 99 + 97 + ... + 47 is (a) 25 (b) 28 (c) 30

(d) 20

2.

If tannT = tanmT, then the different values of Twill be in (a) A.P. (b) G.P. (c) H.P. (d) none of these

3.

The sum of integers from 1 to 100 that are divisible by 2 or 5 is (a) 3000 (b) 3050 (c) 4050 (d) none of these

4.

If the sum of n terms of an A.P. is nA + n2B, where A, B are constants, then its common difference will be (a) A – B (b) A + B (c) 2A (d) 2B

5.

If the 9th term of an A.P. is 35 and 19th is 75, then its 20th terms will be (a) 78 (b) 79 (c) 80 (d) 81

6.

7.

2 6 The 9th term of the series 27 + 9 + 5 + 3 + ..... 5 7 will be 10 10 16 17 (b) (c) (d) (a) 1 17 17 27 27 If a, b, c are in A.P., then (a) 1

8.

9.

14

(b) 2

(a − c)2

(b2 − ac) (c) 3

= (d) 4

If the pth, qth and rth term of an arithmetic sequence are a , b and c respectively, then the value of [a(q – r) + b(r – p) + c(p – q)] = (a) 1 (b) –1 (c) 0 (d) 1/2 If p times the pth term of an A.P. is equal to q times the qth term of an A.P., then (p + q)th term is (a) 0 (b) 1 (c) 2 (d) 3 MATHEMATICS TODAY | OCTOBER ‘17

10. The sums of n terms of two arithmetic series are in the ratio 2n + 3 : 6n + 5, then the ratio of their 13th terms is (a) 53 : 155 (b) 27 : 77 (c) 29 : 83 (d) 31 : 89 11. Let Tr be the rth term of an A.P. for r = 1, 2, 3, ..... 1 If for some positive integers m, n we have Tm = n 1 and Tn = , then Tmn equals m 1 1 1 + (c) 1 (a) (b) (d) 0 m n mn 12. If 1, log 9 (31− x + 2), log 3 (4.3x − 1) are in A.P. then x equals (a) log34 (b) 1 – log34 (c) 1 – log43 (d) log43 13. If the ratio of the sum of n terms of two A.P.'s be (7n + 1) : (4n + 27) , then the ratio of their 11th terms will be (a) 2 : 3 (b) 3 : 4 (c) 4 : 3 (d) 5 : 6 14. The interior angles of a polygon are in A.P. If the smallest angle be 120° and the common difference be 5°, then the number of sides is (a) 8 (b) 10 (c) 9 (d) 6 15. If the pth term of an A.P. be 1/q and qth term be 1/p, then the sum of its pqth terms will be pq −1 1 − pq (a) (b) 2 2 pq +1 pq + 1 (c) (d) − 2 2 16. If the first, second and last terms of an A.P. be a, b, 2a respectively, then its sum will be ab ab (a) (b) 2(b − a) b−a 3ab 3ab (c) (d) 2(b − a) 4(b − a) 17. If a1, a2, ..., an are in A.P. with common difference d, then the sum of the following series is sin d(cosec a1 ⋅ cosec a2 + cosec a2 ⋅ cosec a3 + ........ + cosec an−1cosec an ) (a) seca1 – secan (c) tana1 – tanan

(b) cota1 – cotan (d) coseca1 – cosecan

18. If the sum of the series 2 + 5 + 8 + 11 + .... is 60100, then the number of terms are (a) 100 (b) 200 (c) 150 (d) 250

19. The sum of all natural numbers between 1 and 100 which are multiples of 3 is (a) 1680 (b) 1683 (c) 1681 (d) 1682 20. If the sum of n terms of an A.P. is 2n2 + 5n, then the nth term will be (a) 4n + 3 (b) 4n + 5 (c) 4n + 6 (d) 4n + 7 nth

term of an A.P. is 3n – 1.Choose from the 21. The following the sum of its first five terms (a) 14 (b) 35 (c) 80 (d) 40 22. The sum of the numbers between 100 and 1000 which is divisible by 9 will be (a) 55350 (b) 57228 (c) 97015 (d) 62140 23. The ratio of sum of m and n terms of an A.P. is m2 : n2, then the ratio of mth and nth term will be n −1 2m − 1 2n − 1 m −1 (b) (c) (d) (a) m −1 2n − 1 2m − 1 n −1 24. The solution of log 3 x + log 4 x + log 6 x + ..... + log16 x = 36 is 3

(a) x = 3 (c) x = 9

3

3

(b) x = 4 3 (d) x = 3

⎛n⎞ 25. A series whose nth term is ⎜ ⎟ + y , then the sum ⎝x⎠ of r terms will be ⎧ r (r + 1) ⎫ (a) ⎨ ⎬ + ry ⎩ 2x ⎭ r (r − 1) ⎫ (c) ⎨⎧ ⎬ − ry ⎩ 2x ⎭

⎧ r (r − 1) ⎫ (b) ⎨ ⎬ ⎩ 2x ⎭ ⎧ r (r + 1) ⎫ (d) ⎨ ⎬ − rx ⎩ 2y ⎭

26. The sum of the integers from 1 to 100 which are not divisible by 3 or 5 is (a) 2489 (b) 4735 (c) 2317 (d) 2632 27. If a1, a2, ..., an + 1 are in A.P., then 1 1 1 is + + ..... + a1a2 a2a3 anan+1 n −1 1 (b) (a) a1an+1 a1an+1 n +1 n (d) (c) a1an+1 a1an+1 28. Let the sequence a1, a2, a3, ...., a2n form an A.P. Then a12 − a22 + a33 − .... + a22n−1 − a22n = n 2n 2 (a12 − a22n ) (b) (a) (a − a2 ) 2n − 1 n − 1 2n 1 n (c) (a2 + a22n ) (d) None of these n +1 1

29. The first term of an A.P. of consecutive integers is p2 + 1. The sum of (2p + 1) terms of this series can be expressed as (a) (p + 1)2 (b) (p + 1)3 2 (c) (2p + 1)(p + 1) (d) p3 + (p + 1)3 30. The number of terms of the A.P. 3,7,11,15...to be taken so that the sum is 406 is (a) 5 (b) 10 (c) 12 (d) 14 31. Three numbers are in A.P. such that their sum is 18 and sum of their squares is 158. The greatest number among them is (a) 10 (b) 11 (c) 12 (d) None of these 3 + 5 + 7 + ..... to n terms

= 7 , then the value of n is 5 + 8 + 11 + .... to 10 terms (a) 35 (b) 36 (c) 37 (d) 40 33. If A1, A2 be two arithmetic means between 1/3 and 1 , then their values are 24 17 5 7 5 (a) (b) , , 72 36 72 36 7 5 5 17 , , (c) (d) 36 72 72 72 34. A number is the reciprocal of the other. If the 13 , then arithmetic mean of the two numbers be 12 the numbers are 32. If

1 4 2 5 3 2 3 4 , (b) (c) (d) , , , 4 1 5 2 2 3 4 3 35. If A be an arithmetic mean between two numbers and S be the sum of n arithmetic means between the same numbers, then (a) S = nA (b) A = nS (c) A = S (d) None of these (a)

36. If f (x + y , x − y ) = xy , then the arithmetic mean of f (x, y) and f (y, x) is (a) x (b) y (c) 0 (d) 1 37. If the sides of a right angled traingle are in A.P., then the sides are proportional to (a) 1 : 2 : 3 (b) 2 : 3 : 4 (c) 3 : 4 : 5 (d) 4 : 5 : 6 38. If the sum of three numbers of a arithmetic sequence is 15 and the sum of their squares is 83, then the numbers are (a) 4, 5, 6 (b) 3, 5, 7 (c) 1, 5, 9 (d) 2, 5, 8 MATHEMATICS TODAY | OCTOBER‘17

15

39. The four arithmetic means between 3 and 23 are (a) 5, 9, 11, 13 (b) 7, 11, 15, 19 (c) 5, 11, 15, 22 (d) 7, 15, 19, 21 1 1 1 are in A.P., then , , p+q r + p q+r (a) p, q, r are in A.P. (b) p2 , q2 , r2 are in A.P.

40. If

1 1 1 , , are in A.P. p q r

(c)

(d) None of these 41. If 1, log y x , log z y , − 15 log x z are in A.P., then (a) z3 = x (b) x = y–1 –3 (c) z = y (d) x = y–1 = z3 Multiple Correct Answer Type

42. If in a 'ABC, a, b, c are in A.P. then it is necessary that 1 b 2 2 b < < (a) (b) < <2 3 c 3 3 c 2 b < <2 3 a

(c)

(d)

1 b 2 < < 3 a 3

43. Let S1, S2, ..., Sn be the sums of geometric series whose 1st terms are 1, 2, 3, ...., n and common 1 1 1 1 ratios are , , ...., respectively. Then 2 3 4 n +1 n(n + 3) (a) S1 + S2 + .... + Sn = 2 S ⋅ S ⋅ S ..... S = n + 1 (b) 1 2 3 n n −1 1 1 1 + + .... = (c) S1S2 S2 S3 Sn−1Sn 2(n + 1) 2 3 4 n+1 (d) S1 ⋅ S2 ⋅ S3 .... Sn = n

44. If

∑ r(r + 1) =

r =1

1024 3

(n + a)(n + b)(n + c) , where a < b < c, 3

then (a) 2b = c (b) a3 – 8b3 + c3 = 8abc (c) c is prime number (d) (a + b)2 = 0 (111,...1) , then n times (a) a912 is not prime (b) a951 is not prime (c) a480 is not prime (d) a91 is not prime

45. Let an =

46. If a, b, c are in H.P., then a b c (a) will be in H.P. , , b+c c +a a +b 16

MATHEMATICS TODAY | OCTOBER ‘17

(b)

a b c are in H.P. , , b+c −a c +a −b a +b−c

(c)

1 1 1 1 1 1 are in H.P. + , + , + a b+c b c +a c a +b

(d)

bc ca ab are in H.P. , , b+c c +a a +b

47. If n be odd or even, then the sum of n terms of the series 1 – 2 + 3 – 4 + 5 – 6 + .... will be n n −1 (b) (a) − 2 2 (c)

n +1 2

(d)

2n + 1 2

Comprehension Type

Paragraph for Q. No. 48 to 52 Let A1, A2, A3, ..., Am be arithmetic means between –2 and 1027 and G1, G2, G3, ..., Gn be geometric means between 1 and 1024. Product of geometric means is 245 and sum of arithmetic means is 1025 × 171. 48. The value of n is (a) 7 (c) 11

(b) 9 (d) none of these

49. The value of m is (a) 340 (b) 342

(c) 344

(d) 346

50. The value of G1 + G2 + G3 + .... + Gn is (a) 1022 (b) 2044 (c) 512 (d) none of these 51. The common difference of the progression A1, A3, A5, ..., Am – 1 is (a) 6 (b) 3 (c) 2 (d) 1 2

52. The numbers 2A171, G 5 + 1, 2A172 are in (a) A.P. (b) G.P. (c) H.P. (d) A.G.P. Paragraph for Q. No. 53 to 55 There are two sets A and B each of which consists of three numbers in A.P. whose sum is 15 and where D and d are the common differences such that D – d = 1. p 7 If = where p and q are the product of the numbers q 8 respectively and d > 0, in the two sets 53. Value of p is (a) 100 (b) 120

(c) 105

(d) 110

54. Value of q is (a) 100 (b) 120

(c) 105

(d) 110

55. Value of D + d is (a) 1 (b) 2

(c) 3

(d) 4

Matrix–Match Type

(B) The sum of the series

(A)

(B)

(C)

(D)

Column-I The sequence a, b, 10, c, d is an arithmetic progression. Then the value of a + b + c + d is The sides of right triangle form a three term geometric sequence. The shortest side has length 2. The length of the hypotenuse is of the form a + b where a, b N, then a2 + b2 equals The sum of first three consecutive numbers of an infinite G.P. is 70, if the two extremes be multipled each by 4, and the mean by 5, the products are in A.P. The first term of the G.P. is The diagonals of a parallelogram have a measure of 4 and 6 metres. They cut off forming an angle of 60°. If the perimeter of the parallelogram is 2( p + q ) where p, q N then (p + q) equals

Column-II (p) 52

(q)

20

32

(C) If the first two terms of a Harmonic (r) 1 1 Progression be and then the 2 3 Harmonic Mean of the first four terms is (D) Geometric mean of 4 and 9 (s)

1/3

12.4 2

+

11 4 2.7 2

+

17 7 2.102

+ ... is

6

Integer Answer Type

(r)

(s)

26

40

Column-I

Column-II 42 (A) Suppose that F (n + 1) = 2F (n) + 1 (p) 2 for n = 1, 2, 3, … and F(1) = 2. Then F(101) equals 1620 (B) If a1, a2, a3, .... , a21 are in A.P. (q) and a3 + a5 + a11 + a17 + a19 = 21

∑ ai is i=1

(C) 10 t h ter m of t he s equence (r) S = 1 + 5 + 13 + 29 + …., is (D) The sum of all two digit numbers (s) which are not divisible by 2 or 3 is

59. If the sum to infinity of a decreasing G.P. with the common ratio x is 6 k such that |x| < 1; x z 0. The ratio of the fourth term to the second term is 1/16 and the ratio of third term to the square of the second term is 1/9. Find the value of k. 60. The sum of the terms of an infinitely decreasing G.P. is equal to the greatest value of the function f (x) = x3 + 3x – 9 on the interval [–4, 3] and the difference between the first and second terms is f c(0). Then the value of 3 r (where r is common ratio) is SOLUTIONS 1. (b) : First term a = 101, common difference d = –2 and last term l = 47 ∴ Tl = a + (n − 1)d ⇒ 47 = 101 + (n − 1)(−2) ⇒ n = 28

57. Match the following.

10 then the value of

(q)

5

56. Match the following.

52 2045

58. Match the following. Column-I Column-II (A) The arithmetic mean of two (p) 240 positive numbers is 6 and their 77 geometric mean G and harmonic mean H satisfy the relation G2 + 3H = 48, then product of the two number is

2. (a) : We have tannT = tanmT nT = NS + (mT) Nπ ⇒ θ= , putting N = 1, 2, 3,...., we get n−m 2π 3π π , , ,...... n−m n−m n−m π ? Common difference (d ) = n−m So, different values of T will be in A.P. 3. (b) : The sum of integers from 1 to 100 that are divisible by 2 or 5 = sum of series divisible by 2 + sum of series divisible by 5 – sum of series divisible by 2 and 5. = (2 + 4 + 6 + .... + 100) + (5 + 10 + 15 .... + 100) – (10 + 20 + 30 + .... + 100) 20 50 = {2 × 2 + (50 − 1)2} + {2 × 5 + (20 − 1)5} 2 2 10 − {10 × 2 + (10 − 1)10} 2 = 2550 + 1050 – 550 = 3050 4. (d) : Given that Sn = nA + n2B Putting n = 1, 2, 3, ...., we get S1 = A + B, S2 = 2 A + 4 B, S3 = 3 A + 9 B and so on ∴ T1 = S1 = A + B, T2 = S2 − S1 = A + 3B, Hence, common difference (d) = 2B . MATHEMATICS TODAY | OCTOBER‘17

17

5. (b) : T9 = a + 8d = 35 and T19 = a + 18d = 75 On solving we get d = 4 and a = 3 Hence 20th term of an A.P. is a + 19d = 3 + 19 × 4 = 79 . 2 6 6. (a) : Given series 27 + 9 + 5 + 3 + ...... 5 7 27 27 27 27 = 27 + + + + ..... + + ...... 3 5 7 2n − 1 27 27 27 10 ? Tn = ⇒ T9 = = =1 2n − 1 (2 × 9) − 1 17 17 7. (d) : If a, b, c are in A.P. 2b = a + c So,

(a − c)2 (b2 − ac)

=

=

(a − c)2 ⎧⎪ ⎛ a + c ⎞2 ⎫⎪ − ac ⎬ ⎨⎜ ⎟ ⎝ 2 ⎠ ⎩⎪ ⎭⎪

(a − c)2 4 [a2 + c2 + 2ac − 4ac]

8. (c)

=

9.

10. (a) : We have

Sn

1

Sn

2

=

4(a − c)2 (a − c)2

=4

(a)

2n + 3 6n + 5

n [2a1 + (n − 1)d1 ] 2n + 3 = ⇒ 2 n 6n + 5 [2a2 + (n − 1)d2 ] 2 ⎛ n −1⎞ a1 + ⎜ d ⎝ 2 ⎟⎠ 1 2n + 3 ⇒ = 6n + 5 ⎛ n −1⎞ a2 + ⎜ d2 ⎟ ⎝ 2 ⎠ T a + 12d1 2(25) + 3 53 = Put n = 25 then 1 ⇒ 13 = a2 + 12d2 6(25) + 5 T13 155 2 11. (c) 12. (b) : The given numbers are in A.P. ∴ 2 log9 (31− x + 2) = log3 (4 ⋅ 3x − 1) + 1 ⇒ ⇒ ⇒ ⇒

2 log 2 (31− x + 2) = log3 (4 ⋅ 3x − 1) + log3 3 3 2 log (31− x + 2) = log3[3(4 ⋅ 3x − 1)] 2 3 31− x + 2 = 3(4 ⋅ 3x − 1) 3 + 2 = 12 y − 3, where y = 3x y

⇒ 12 y2 − 5 y − 3 = 0 −1 3 −1 3 or ⇒ 3x = or 3x = 3 4 3 4 x = log3 (3 / 4) ⇒ x = 1 − log3 4 . y=

13. (c) 18

MATHEMATICS TODAY | OCTOBER ‘17

14. (c) : Let the number of sides of the polygon be n Then, the sum of interior angles of the polygon π = (2n − 4) = (n − 2)π 2 Since the angles are in A.P. Also, a = 120o , d = 5, n ∴ Sn = [2 × 120 + (n − 1)5] = (n − 2)180 2 2 ⇒ n − 25n + 144 = 0 ⇒ (n − 9)(n − 16) = 0 ⇒ n = 9, 16 But n = 16 gives T16 = a + 15d = 120o + 15 ⋅ 5o = 195o , which is impossible as interior angle cannot be greater than 180°. Hence n = 9 . 15. (c) ...(i) 16. (c) : We have first term A = a Second term A + d = b ...(ii) and last term l = 2a ...(iii) b From (i), (ii) and (iii), d = (b − a) and n = b−a 3ab n b Then, sum (S) = [a + l] = [a + 2a] = 2(b − a) 2 2(b − a) 17. (b) : As given d = a2 − a1 = a3 − a2 = .... = an − an−1 ∴ sin d {cosec a1cosec a2 + ..... + cosec an−1cosec an } =

sin(an − an−1 ) sin(a2 − a1 ) + ...... + sin a1 ⋅ sin a2 sin an−1 sin an

= (cot a1 − cot a2 ) + (cot a2 − cot a3 ) + .... = cot a1 − cot an

+(cot an−1 − cot an )

18. (b) : Given, Series, 2 + 5 + 8 + 11 + ..... where a = 2, d = 3 and let number of terms is n n ? Sum of A.P. = {2a + (n − 1)d } 2 n ⇒ 60100 = {2 × 2 + (n − 1)3} ⇒ 120200 = n(3n + 1) 2 2 ⇒ 3n + n − 120200 = 0 ⇒ (n − 200)(3n + 601) = 0 Hence n = 200 . 19. (b) 20. (a) 21. (d) : Here Tn = 3n – 1 , putting n = 1, 2, 3, 4, 5 we get first five terms, 2, 5, 8, 11, 14 Hence sum is 2 + 5 + 8 + 11 + 14 = 40 . 22. (a) : Series 108 + 117 + .... + 999 is an A.P. where a = 108, common difference d = 9, 999 99 n= − = 111 − 11 = 100 9 9 100 Hence, required sum = (108 + 999) 2 = 50 × 1107 = 55350.

23. (c) 24. (d) 25. (a) : On putting n = 1, 2, 3, ..... , we get 1 First term of the series a = + y , x ⎛2 ⎞ ⎛1 ⎞ 1 2 Second term = + y and d = ⎜ + y ⎟ − ⎜ + y ⎟ = ⎝x ⎠ ⎝x ⎠ x x ? Sum of r terms of the series r ⎡ ⎛1 r 1⎤ 1⎤ r ⎡2 ⎞ = ⎢2 ⎜ + y ⎟ + (r − 1) ⎥ = ⎢ + 2 y + − ⎥ ⎠ x⎦ 2 ⎣x 2⎣ ⎝x x x⎦ =

⎡ r (r + 1) ⎤ r 2 − r + 2r + ry = ⎢ + ry ⎥ 2x ⎣ 2x ⎦

.

26. (d) : Let S = 1 + 2 + 3 + ..... + 100 100 (1 + 100) = 50(101) = 5050 2 Let S1 = 3 + 6 + 9 + 12 + ......... + 99 33 = 3(1 + 2 + 3 + 4 + .... + 33) = 3 ⋅ (1 + 33) = 99 × 17 = 1683 2 Let S2 = 5 + 10 + 15 + ........ + 100 20 = 5(1 + 2 + 3 + ........ + 20) = 5. (1 + 20) = 50 × 21 = 1050 2 Let S3 = 15 + 30 + 45 + ........ + 90 6 = 15(1 + 2 + 3 + ........ + 6) = 15. (1 + 6) = 45 × 7 = 315 2 ? Required sum = S – S1 – S2 + S3 = 5050 – 1683 – 1050 + 315 = 2632. 27. (d) 28. (a) : Since a1, a2, a3, ......, a2n form an A.P. ∴ a2 − a1 = a4 − a3 = ....... = a2n − a2n−1 = d =

Here a12 − a22 + a32 − a42 + ....... + a22n−1 − a22n = (a1 − a2 )(a1 + a2 ) + (a3 − a4 )(a3 + a4 ) + ........ ...... + (a2n−1 − a2n )(a2n−1 + a2n ) ⎧ 2n ⎫ = −d(a1 + a2 + ....... + a2n ) = −d ⎨ (a1 + a2n )⎬ ⎩2 ⎭ a2n − a1 Also, 2n = a1 + (2n − 1)d ⇒ d = 2n − 1 n(a1 − a2n ).(a1 + a2n ) n Sum is = (a2 − a22n ) = 2n − 1 2n − 1 1 2p +1 29. (d) : S2 p +1 = {2( p2 + 1) + (2 p + 1 − 1)1} 2 ⎛ 2p +1⎞ 2 =⎜ (2 p + 2 p + 2) = (2 p + 1)( p2 + p + 1) ⎝ 2 ⎟⎠ = p3 + ( p + 1)3

n 30. (d) : S = [2a + (n − 1)d] 2 n ⇒ 406 = [6 + (n − 1)4 ] ⇒ 812 = n[6 + 4n − 4] 2 ⇒ 812 = 2n + 4n2 ⇒ 406 = 2n2 + n ⇒ 2n2 + n − 406 = 0 −1 ± 1 + 4 ⋅ 2 ⋅ 406 −1 ± 3249 −1 ± 57 = = 2⋅2 4 4 −1 + 57 ∴ n= = 14 4 31. (b) 32. (a) ⇒ n=

1 1 33. (b) : Here, , A1 , A2 , will be in A.P., 3 24 1 1 3 ...(i) then A1 − = − A2 ⇒ A1 + A2 = 3 24 8 1 Now, A1 is a arithmetic mean of and A2 , we have 3 1 1 ...(ii) 2 A1 = + A2 ⇒ 2 A1 − A2 = 3 3 17 5 From (i) and (ii), we get, A1 = and A2 = 72 36 34. (d) : Suppose the required numbers are a and b Therefore according to the conditions, a = 1/6 13 a + b 13 = ⇒ a +b = 2 12 6 1 13 2 ⇒ a+ = ⇒ 6a − 13a + 6 = 0 a 6 3⎞⎛ 2⎞ 3 2 ⎛ ⇒ ⎜ a − ⎟ ⎜ a − ⎟ = 0 ⇒ a = and b = ⎝ 2⎠⎝ 3⎠ 2 3 2 3 or a = and b = . 3 2 35. (a) 36. (c) : Let x + y = u, x − y = v u+v u−v ⎛u+v ⎞ ⎛u−v ⎞ ⇒ x= ,y= , ∴ f (u, v ) = ⎜ . ⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ 2 2 f (x , y ) + f ( y , x ) Now 2 ⎛x+ y x− y⎞ ⎛ y+x y−x⎞ ⎜⎝ 2 . 2 ⎟⎠ + ⎜⎝ 2 . 2 ⎟⎠ = = 0. 2 37. (c) : Let the sides of the triangle be a – d, a, a + d, then hypotenuse being the greatest side i.e., a + d . and

So, (a + d )2 = a2 + (a − d )2 ⇒ a2 + d2 + 2ad = a2 + a2 − 2ad + d2 ⇒ a = 4d Therefore ratio of the side = a − d : a : a + d = ( 4 d − d ) : 4 d : ( 4 d + d ) = 3 : 4 : 5. MATHEMATICS TODAY | OCTOBER‘17

19

38. (b) : Let three numbers are a – d, a, a + d . Given a − d + a + a + d = 15 ⇒ a = 5 and (a − d )2 + a2 + (a + d )2 = 83

46. (a, b, c, d)

⇒ a2 + d2 − 2ad + a2 + a2 + d2 + 2ad = 83 ⇒ 2(a2 + d2 ) + a2 = 83 Putting a = 5 ⇒ 2(25 + d2 ) + 25 = 83 ⇒ d = 2 Thus numbers are 3, 5, 7. 39. (b) : Let four arithmetic means are A 1 , A 2 , A 3 and A4. So, 3, A1 , A2 , A3 , A4 , 23 ⇒ T6 = 23 = a + 5d ⇒ d = 4 Thus A1 = 3 + 4 = 7, A2 = 7 + 4 = 11, A3 = 11 + 4 = 15, A4 = 15 + 4 = 19 1 1 1 are in A.P. 40. (b) : Since , and r +q p+q r + p 1 1 1 1 − = − ∴ r + p p+q q+r r + p p+q−r − p r + p−q−r = ⇒ (r + p)( p + q) (q + r )(r + p) q−r p−q ⇒ = or q2 − r 2 = p2 − q2 p+q q+r Hence, p2 , q2 , r2 are in A.P. 41. (d) 42. (a, c) : Given, a, b, c are in A.P. ? a + c = 2b Also, a + b > c 3b > 2c 2 b And b + c > a 2c > b ⇒ < <2 3 c 2 b b Similarly for , we get < < 2 3 a a

?

2

⎛ 1 ⎞ ⎛ 1 ⎞ 43. (a, b, c) : Sr = r + r ⎜ +r + .... + ∞ ⎝ r + 1 ⎟⎠ ⎜⎝ r + 1 ⎟⎠ r = r +1 = 1 1− r +1 ? a, b, c are correct and d is false. n

n

n(n + 1)(2n + 1) n(n + 1) + 6 2 r =1 r =1 n(n + 1)(n + 2) 1 = n.(n + 1)[2n + 1 + 3] = 6 3 a = 0, b = 1, c = 2 45. (a, b, c, d) : As a912 and a480 are divisible by 3, none of them is prime, and for a91, we have 44. (a, b, c) : ∑ r 2 + ∑ r =

1091 − 1 1091 − 1 107 − 1 × = 10 − 1 107 − 1 10 − 1 = (1 + 107 + .... + 1084) (1 + 10 + .... + 106) a91 is not prime. Hence a,b,c,d are correct a91 =

20

MATHEMATICS TODAY | OCTOBER ‘17

48. (b) : G1G2 ....Gn =

47.

(

(a, c)

1 × 1024

)

n

= 25n

25n = 245 n = 9

49. (b) : A1 + A2 + A3 + ... + Am−1 + Am = 1025 × 171 ⎛ −2 + 1027 ⎞ ∴ m⎜ ⎟⎠ = 1025 × 171 ⎝ 2 ? m = 342 50. (a) : Since n = 9,

1 9 ratio (r ) = (1024) +1

∴ Common ? G1 = 2, r = 2

=2

2 ⋅ (29 − 1) = 1024 − 2 = 1022 2 −1 51. (a) : Common difference of sequence 1027 + 2 A1, A2, ..., Am is =3 342 + 1 Common difference of sequence A1, A3, A5, ..., An – 1 is 6 52. (a) : We have A171 + A172 = –2 + 1027 = 1025 2 A171 + 2 A172 ∴ = 1025 2 Also G5 = 1 × 25 = 32 G1 + G2 + ... + Gn =

∴ G52 = 1024 i.e. G52 + 1 = 1025 ∴ 2 A171 , G52 + 1, 2 A172 are in A.P. 53. (c) 54. (b) 55. (c) : Let numbers in set A be a – D , a, a + d and in set B be b – d, b, b + d 3a = 3b = 15 a = b = 5 Set A = {5 – D, 5, 5 + D}; Set B = {5 – d, 5, 5 + d} Where D = d + 1 p 5(25 − D 2 ) 7 = = q 5(25 − d 2 ) 8 25(8 – 7) = 8(d + 1)2 – 7d d = –17, 1 but d > 0 ?d = 1 So numbers in set A are 3, 5, 7 Numbers in set B are 4, 5, 6 Now, p = 3 × 5 × 7 = 105 q = 4 × 5 × 6 = 120 Value of D + d = 3 56. (A) o (s), (B) o(r), (C) o(s), (D) o(r) (A) b + c = a + d = 2 · 10 a + b + c + d = 40 (B) (ar2)2 = a2 + a2r2 [where a = 2] ? r4 = 1 + r2 , r4 – r2 – 1 = 0 [Let r2 = t] 2 t –t–1=0

1+ 5 1± 5 ⇒ t= 2 2 ⎛1 − 5 ⎞ rejected ⎟ gives negative values ⎜⎝ ⎠ 2

?

t=

1+ 5 2 ⎛1 + 5 ⎞ =1+ 5 = a + b ? Hypotenuse is 2 ⎜ ⎝ 2 ⎟⎠ On comparing we get ? a2 + b2 = 1 + 25 = 26 (C) Let a, ar, ar2 are in G.P. where | r | < 1 a + ar + ar2 = 70 ? 10ar = 4a + 4ar2 or 10r = 4 + 4r2 2r2 – 5r + 2 = 0 or (2r – 1)(r – 2) = 0 ? r = 1/2 (∵ |r| < 1) For r = 1/2, a(1 + r + r2) = 70 ⎛ 1 1⎞ ⇒ a ⎜1 + + ⎟ = 70 a = 40 ⎝ 2 4⎠ ? Series becomes 40, 20, 10 where first term is 40. (D) Using cosine rule a2 = 9 + 4 + 2 · 2 · 3 · = 13 + 6 = 19 ? a2 = 19 Similarly, b2 = 9 + 4 – 2 · 2 · 3 = 7 ? b2 = 7 ? P = 2( 19 + 7 ) = 2( p + q ) ? p + q = 19 + 7 = 26 57. (A) o (r), (B) o(p), (C) o(s), (D) o(q) 2F (n) + 1 1 (A) F (n + 1) = = F (n) + 2 2 ? F(1), F(2), F(3), ... is an A.P. with common difference 1/2 1 Hence F(101) = 2 + (100) = 52 2 (B) a1 + 2d + a1 + 4d + a1 + 10d + a1 + 16d + a1 + 18d = 5a1 + 50d a1 + 10d = 2 ∴

r2 =

21

Now,

∑ ai = i=1

21 [2a1 + 20d] = 21(a1 + 10d ) = 42 2

(C) S = 1 + 5 + 13 + 29 + … + S= 1 + 5 + 13 +… + On subtracting, we get t10 = 1 + 4 + 8+ 16 + .... up to 10 terms = 1 + (4 + 8 + 16 + ... up to 9 terms) = 2045 (D) Sum of all two digit numbers 90 = (10 + 99) = (45)(109) 2 Sum of all two digit numbers divisible by 2 45 = (10 + 98) = (45)(54) 2

Sum of all two digit numbers divisible by 3 30 = (12 + 99) = 15(111). 2 Sum of all two digit numbers is divisible by 6 15 = (12 + 96) = 15 (54) 2 ? Required sum = 45(109) + 15(54) – (45) (54) – 15 (111) = 1620 58. (A) o (q), (B) o(r), (C) o(p), (D) o(s) (A) Let the number be a, b : a + b = 12 and 6ab ab + = 48 (Given) a +b ab ab + = 48 ? ab = 32 2 5 11 17 (B) S = 2 2 + 2 2 + 2 2 + .... 1 ⋅4 4 ⋅ 7 7 ⋅ 10 3⋅5 3 ⋅11 3 ⋅17 ⇒ 3S = + + + ..... 2 2 2 2 2 1 ⋅4 4 ⋅7 7 ⋅102 ⇒ 3S =

42 − 12

+

72 − 42

+

102 − 72

+ .... 12 ⋅ 42 42 ⋅ 72 72 ⋅102 1 1 1 1 1 ⇒ 3S = 1 − + − + − + .... 2 2 2 2 4 4 7 7 102 1 ⇒ 3S = 1 ⇒ S = 3 4 240 1 1 1 1 (C) H.M of , , , is = 1 1 1 1 77 2 3 4 5 + + + 2 3 4 5 (D) G.M. = ( 4) × ( 9) = 6 59. (2) : Let the series be a, ax, ax2 , ax3, .... T4 ax 3 1 1 1 ⇒ x2 = = = ⇒ x=± T2 ax 16 16 4 1 But since it is a decreasing G.P. ∴ x = 4 T3 1 1 1 ax 2 ⇒ a=9 = ⇒ = Also, 2 = a 9 T2 (ax )2 9 Also,

a 9 9×4 = = = 12 1 1− r 3 1− 4 60. (2) : f (x) is increasing. So its greatest value is f (3) = 27. Let the G.P. be a, ar, ar2 , ... with, –1 < r < 1 a = 27 and a – ar = 3 1− r On solving, we get r = 2/3 S∞ =

MATHEMATICS TODAY | OCTOBER‘17

21

*3(::?0

:LYPLZ

CBSE Sequences and Series ? nth term from the end = am

Arithemetic Progression (A.P.) X

Common difference (d) = an + 1 – an n N

X

General term, an = nth term = a + (n – 1) d

X

nth term from the end consisting of m terms is (m – n + 1)th term from the beginning ? nth term from the end = am – n + 1

⎛1⎞ =l⎜ ⎟ ⎝r ⎠

X

X

X X

n n Sn = [2a + (n − 1)d] = [a + l] 2 2 where l is last term an = Sn – Sn–1 A=

Arithmetic mean = A =

a1 + a2 + ...... + an n

Geometric Progression (G.P.) an+1 ∀n ∈ N an

X

Common ratio = r =

X

General term an = nth term = arn

X

nth term from the end consisting of m terms is (m – n + 1)th term from the beginning

22

MATHEMATICS TODAY | OCTOBER ‘17

n

n −1

where l is last term of G.P.

Sum of finite G.P. = Sn =

= X

a +b where A is A.M. of a and b 2 If a1, a2, a3, ....., an are n numbers then

= arm

Sum of n terms of the G.P.

Sum of n terms of an A.P. X

– n + 1

a(r n − 1) lr − a = (r > 1) r −1 r −1

a(1 − r n ) a − lr (r < 1) = 1− r 1− r

Sum of an infinite = S =

a (| r | < 1) 1− r

Geometric mean (G.M.) X

If a, x, b are in G.P., then x = of a and b.

X

If a, x1, x2, ....., xn, b are in G.P., then x1, x2, ....., xn are the n geometric means between a and b. 1/(n +1)

⎛b⎞ r=⎜ ⎟ ⎝a⎠

⎛b⎞ =a⎜ ⎟ ⎝a⎠

1/(n +1)

⎛b⎞ & x1 = ar = a ⎜ ⎟ ⎝a⎠

2/(n +1)

ab is the G.M.

,...................., xn =

, x2 = ar2 1/(n +1)

b ⎛a⎞ =b⎜ ⎟ ⎝b⎠ r

WORK IT OUT VERY SHORT ANSWER TYPE

1 20 7 1 + 1 + 1 + + ... 1. 2 13 9 23 2. Find the sum of first six terms of the series: Find the nth term of the series 2 1 1 1 + + + ............ 1⋅ 2 2 ⋅ 3 3 ⋅ 4 3. If x, y, z are distinct positive numbers, then show that (x + y) (y + z) (z + x) > 8xyz 4. If the 9th term of an A.P. vanishes, then find the ratio of its 29th and 19th terms. 5. Show that the product of n terms, where n is odd, of a G.P. will be equal to nth power of the middle term. SHORT ANSWER TYPE

6. If a, b, c, d are four distinct positive numbers in A.P. then show that bc > ad. 7. If D1, D2, D3, ..., Dn are in A.P. whose common difference is d, then show that sin d[sec D1 sec D2 + sec D2 sec D3 + ... + sec Dn–1 sec Dn] = tan Dn – tan D1. 8. If x, y, z are positive then find the minimum value z–log x + zlog x–log y. of xlog y–log z + ylog n 1 9. Find the sum : ∑ . ( ar + b )( ar + a + b) r =1 10. The vibrations of system are damped so that the amplitudes of the successive deflections are 12, 8, 16/3, ... . Find the amplitude of the 6th deflection and also the total deflection before the system comes to rest. LONG ANSWER TYPE - I

11 If a1, a2, a3, ...... , an are in A.P. and n ⎡ 1 1 1 1 ⎤ 1 = (a1 + an ) ⎢ k + + + ... + ⎥ ∑ ana1 ⎦ r =1 ar ⎣ a1an a2an−1 a3an−2 and a be the one A.M. and x, y be the two G.M.s between b and c, then show that x3 + y3 = kabc.

15. If a, b, c are in A.P. and x, y, z are in G.P., then show that xb–c . yc–a . za–b = 1. LONG ANSWER TYPE - II

16. Find the sum to n terms of the series: 3 + 15 + 35 + 63 + ...... 17. If a, b, c are in G.P. and log c a, log b c and log a b are in A.P. show that the common difference of A.P. is 3/2. 18. If in an A.P. the sum of m terms is equal to n and the sum of n terms is equal to m, then prove that the sum of (m + n) terms is – (m + n). Also, find the sum of first (m – n) terms (m > n). 19. Suppose x and y are two real numbers such that the rth mean between x and 2y is equal to the rth mean between 2x and y when n arithmetic means are inserted bwteen them in both the cases. Show n +1 y that − = 1. r x 20. The ratio of the sum of n terms of two A.P.cs is (7n +1) : (4n + 27). Find the ratio of their nth terms. SOLUTIONS

1. Reciprocals of the given terms of the series are: 2 13 9 23 8 13 18 23 , , , ,... or , , , ,... 5 20 10 20 20 20 20 20 8 + (n − 1) 5 5n + 3 ? nth term(an) = = 20 20 20 Hence, nthterm of the required series = 5n + 3 2. Let Tr be the rth term of the given series. Then, 1 Tr = , r = 1, 2,..., n r(r + 1) 1 1 , r = 1, 2,..., n ⇒ Tr = − r r +1 ? Required term =

6

6 1 ⎞ ⎛1 T = ∑ 6 ∑ ⎜⎝ r − r + 1 ⎟⎠ r =1 r =1

12. If pth and qth terms of G.P. are q and p respectively, show that the (p + q)th term is (qp/pq)1/(p–q).

⎛ 1⎞ ⎛1 1⎞ ⎛1 1 ⎞ ⎛1 1⎞ = ⎜1 − ⎟ + ⎜ − ⎟ + ⎜ − ⎟ + ... + ⎜ − ⎟ ⎝ 2⎠ ⎝2 3⎠ ⎝3 4⎠ ⎝6 7⎠

13. Find the sum upto n terms of the series: 1 2 3 + + + ...... 2 4 2 4 2 1 + 1 + 1 1 + 2 + 2 1 + 3 + 34

1 6 = 1− = 7 7

14. One side of an equilateral triangle is 24 cm. The midpoints of its sides are joined to form still another triangle. This process continues, indefinitely. Find the sum of the perimeters of all the triangles.

3. Using A.M. > G.M., we obtain x+y y+z z+x > xy , > yz and > zx 2 2 2 ⇒ x + y > 2 xy , y + z > 2 yz and z + x > 2 zx MATHEMATICS TODAY | OCTOBER‘17

23

( x + y ) ( y + z ) (z + x ) > 2

⇒

xy × 2 yz × 2 zx

(x + y)(y + z)(z + x) > 8xyz. 4. T9 = 0 a + 8d = 0 a = –8d T29 a + 28d −8d + 28d 20d 2 = = = = ? T19 a + 18d −8d + 18d 10d 1 5. Let the G.P. be a, ar, ar2, ar3 ... , arn–1 Here, the number of terms is odd. ? The middle term = ar((n+1)/2)–1 = ar(n–1)/2. Then, a . ar . ar2 . ar3 ..... arn–1 = anr1+2+3 + ... +(n–1) n = a r(n–1)n/2 = [ar(n–1)/2]n = (the middle term)n. 6. Given a, b, c, d are in A.P. b is the A.M. of a and c And, G.M. of a and c is ac. ∵ A.M. of a and c > G.M. of a and c ...(i) ? b > ac b2 > ac Similarly, c is the A.M. of b and d ..(ii) ? c > bd c2 > bd From (i) and (ii), we get b2c2 > (ac) (bd) bc > ad. sin(α2 − α1) 7. We have, sin d (sec D1 sec D2) = cos α1 ⋅ cos α2 sin α2 cos α1 cos α2 sin α1 − = tan D2 – tan D1 cos α1 ⋅ cos α2 cos α1 ⋅ cos α2 Similarly, sin d sec D2 sec D3 = tan D3 – tan D2 sin d sec Dn – 1 sec Dn = tan Dn – tan Dn–1 On adding, we get sin d [sec D1 sec D2 + sec D2 sec D3 + ... + sec Dn – 1 sec Dn] = tan Dn – tan D1 8. Since A.M. ≥ G.M. =

⇒ ≥3x

x

log y − log z

+y

log z − log x

+z

log x − log y

3 log y − log z

⋅y

log z − log x

(x log y–log z

⋅z

log x − log y

ylog z–log x

=1 zlog x–log y)

because log . . = (log y – log z) log x + (log z – log x) log y + (log x–log y) log z = 0 ? xlog y–log z . ylog z–log x . zlog x–log y = e0 = 1 Hence, minimum value of given expression is 1. n

9. We have,

∑

r =1 n

=∑

r =1

24

1 (ar + b)(ar + a + b)

1⎛ 1 1 ⎞ − ⎜⎝ ⎟ a (ar + b) (ar + a + b) ⎠ MATHEMATICS TODAY | OCTOBER ‘17

1 ⎡⎛ 1 1 ⎞ ⎛ 1 1 ⎞ = ⎢⎜ − +⎜ − ⎟ ⎟ + ... a ⎣ ⎝ a + b 2a + b ⎠ ⎝ 2a + b 3a + b ⎠ ⎛ 1 ⎞⎤ 1 ... + ⎜ − ⎥ ⎟ ⎝ na + b (n + 1) a + b ⎠ ⎥⎦ n 1⎧ 1 1 ⎫ . = ⎨ − ⎬= a ⎩ a + b (n + 1)a + b ⎭ (a + b) {(n + 1)a + b} 10. The amplitudes of the successive deflections are 12, 8, 16/3, ... which form the G.P. 8 2 Here, a = 12, r = = 12 3 ? Amplitude of the 6 th deflection = ar 6–1 = ar 5 5

32 128 ⎛2⎞ = 12 × ⎜ ⎟ = 12 × = ⎝3⎠ 243 81 Total deflection before the system comes to rest = Sum a 12 to infinite terms of the G.P. = = = 36 1− r 1− 2 3 ⎡ 1 1 1 1 ⎤ 11. Given, (a1 + an ) ⎢ + + + ... + ⎥ ana1 ⎦ ⎣ a1an a2an−1 a3an−2 a +a a +a a +a a +a = 1 n + 2 n−1 + 3 n−2 + ... + n 1 a1an a2an−1 a3an−2 ana1 ⎛1 1⎞ ⎛ 1 ⎛1 1⎞ 1⎞ ⎛ 1 1⎞ =⎜ + ⎟+⎜ + ⎟+⎜ + ⎟ + ... + ⎜ + ⎟ ⎝ an a1 ⎠ ⎝ an−1 a2 ⎠ ⎝ an−2 a3 ⎠ ⎝ a1 an ⎠ n ⎛1 1 1⎞ 1 = 2 ⎜ + + ... + ⎟ = 2∑ an ⎠ r =1 ar ⎝ a1 a2

? k = 2

c ⎛ c⎞ Now, x3 + y3 = b3r3(1 + r3) = b3. ⎜1 + ⎟ b ⎝ b⎠ = bc (b + c) = bc(2a) = 2abc = kabc 12. Let a be the first term and r be the common ratio of G.P. then, tp = ar p–1 = q ... (i) and tq = arq–1 = p ... (ii) Dividing (i) by (ii), we get ⇒ r p −q =

ar p −1 ar

q −1

=

q p

1/( p −q)

⎛q⎞ q ⇒r =⎜ ⎟ p ⎝ p⎠

Now, tp+q = ar p+q–1 = ar p–1 r q ⎛q⎞ =q⎜ ⎟ ⎝ p⎠

q /( p −q)

= q.

qq /( p −q) pq /( p −q) =

=

q1+q /( p −q)

q p/( p −q) pq /( p −q)

pq /( p −q)

1/( p −q)

⎛ qp ⎞ =⎜ q⎟ ⎝p ⎠

13. Let Tr be the rth term of the given series. Then, r Tr = , r = 1, 2, 3, ..., n 2 1+ r + r4 r ⇒ Tr = 2 (r + r + 1)(r 2 − r + 1) ⎫ 1⎧ 2r = ⎨ 2 ⎬ 2 2 ⎩ (r + r + 1)(r − r + 1) ⎭ ⎫ 1⎧ 1 1 ⇒ Tr = ⎨ 2 − ⎬ , r = 1, 2, ..., n 2 ⎩ (r − r + 1) (r 2 + r + 1) ⎭ n 1 ⎪⎧ n ⎛ 1 1 ⎞ ⎪⎫ ∴ Sn = ∑ Tr = ⎨∑ ⎜ 2 − 2 ⎟⎬ 2 ⎪⎩r =1 ⎝ r − r + 1 r + r + 1 ⎠ ⎪⎭ r =1

16. The difference between the successive terms are 15 – 3 = 12, 35 – 15 = 20, 63 – 35 = 28, ... . Clearly, these differences are in A.P. Let Tn be the nth term and Sn denote the sum to n terms of the given series. Then, Sn = 3 + 15 + 35 + 63 + ... + Tn–1 + Tn ...(i) Sn = 3 + 15 + 35 + ...... + Tn–1 + Tn ...(ii) Subtracting (ii) from (i), we get 0 = 3 + {12 + 20 + 28 + ... + (Tn – Tn–1)} – Tn (n − 1) Tn = 3 + {2 × 12 + (n – 1 – 1) × 8} 2 = 3 + (n – 1) (12 + 4n – 8) Tn = 3 + (n – 1) (4n + 4) = 4n2 – 1

1 ⎧⎛ 1 ⎞ ⎛ 1 1 ⎞ ⎛ 1 1 ⎞ = ⎨ ⎜1 − ⎟ + ⎜ − ⎟ + ⎜ − ⎟ 2 ⎩ ⎝ 3 ⎠ ⎝ 3 7 ⎠ ⎝ 7 13 ⎠

n2 + n 1⎧ 1 ⎫ . = ⎨1 − 2 ⎬= 2 2 ⎩ n + n + 1⎭ 2(n + n + 1) 14. Perimeter of the 1st equilateral 'ABC = 3 × side = 3 × 24 = 72 cm. Let D, E, F are the mid-points of the sides BC, CA, AB respectively.

1 1 ∴ DE = AB = × 24 = 12cm 2 2

= x b −c x =

c −a c −a 2 z 2 z a −b

2b −(a +c) (c +a)− 2b x 2 z 2

=x

b −c +

c −a c −a a −b + 2 z 2

= x 0z 0 = 1

k =1

n

∑

k =1

n

⎧ n(n + 1)(2n + 1) ⎫ k2 − ∑ 1 = 4 ⎨ ⎬−n 6 ⎩ ⎭ k =1

n = (4n2 + 6n − 1) 3 17. Let x = logc a, y = logb c, z = loga b ?xyz = 1 ...(i) Given x, y, z are in A.P. ? 2y = x + z ...(ii) Given a, b, c are in G.P. 1 ? b2 = ac 2 log b b = logb a + logb c ⇒ 2 = + y z ...(iii) ⎛ 1⎞ From (ii) and (iii), 2 ⎜ 2 − ⎟ = x + z ⎝ z⎠

From (i),

+ 36 + 18 + ..... ǈȱ . (which forms the G.P.) =

b −c c − a a −b ? xb–c yc–a za–b = x ( xz ) z

k =1

4z − 2 − z 2 2 ⇒ x =4− −z = z z

Similarly, EF = DF = 12 cm ? Perimeter of 'DEF = 3 × 12 = 36 cm Similarly we get, perimeter of 'GHI = 3 × 6 = 18 cm. Sum of perimeters of these triangles which are infinite in number = 72

15. If a, b, c are in A.P. 2b = a + c x, y, z are in G.P. y2 = xz

n

⇒ Sn = 4

1 1 ⎛ ⎞⎫ +... + ⎜ 2 − 2 ⎬ ⎟ ⎝ n − n + 1 n + n + 1 ⎠⎭

= 144 cm

n

∴ Sn = ∑ Tk = ∑ (4k 2 − 1)

72 1 1− 2 ...(i) ...(ii)

4z − 2 − z 2 ⎛ 1 ⎞ ⋅ ⎜2 − ⎟ ⋅ z = 1 ⎝ z z⎠

(4z – 2 – z2) (2z – 1) = z 8z2 – 4z – 2z3 – 4z + 2 + z2 = z –2z3 + 9z2 – 9z + 2 = 0 2z2 – 7z + 2 = 0

[∵ z – 1 ≠ 0]

1 ? Common difference of A.P. = z – y = z − 2 + z 2 z 2 − 2z + 1 2z − 4z + 2 3z = 3 . = = = 2z 2 2z z

ANSWER KEY

477*3(::?0

[Using(ii)] 1.

(c)

6. (b) 11. (c) 16. (a)

2.

(d)

3.

(c)

4.

(c)

7. (a,b) 8. (a,b,c) 9 . (b,c) 12. (b,d) 13. (d) 14. (b) 17. (9) 18. (4) 19. (5 )

5.

(a)

10. (b,d) 15. (a) 20. (4)

MATHEMATICS TODAY | OCTOBER‘17

25

18. Let a be the first term and d be the common difference of the given A.P. Then, m Sm = n ⇒ {2a + (m − 1) d} = n 2 2am + m(m – 1) d = 2n ...(i) n Also, Sn = m ⇒ {2a + (n − 1) d} = m 2 2an + n(n – 1) d = 2m ...(ii) Subtracting (ii) from (i), we get 2a(m – n) + {m(m – 1) – n(n – 1)} d = 2n – 2m 2a(m – n) + {(m2 – n2) – (m – n)} d = –2(m – n) 2a + (m + n – 1) d = –2 ... (iii) m+n {2a + (m + n − 1) d} Now, Sm+n = 2 (m + n) ⇒ Sm+n = (−2) ? Sm + n = – (m + n) 2 From (iii), we obtain 2a = –2 – (m + n – 1)d ... (iv) Substituting this value of 2a in (i), we obtain –2m – m(m + n – 1) d + m(m – 1) d = 2n ⎛m+n⎞ ⇒ d = −2 ⎜ ⎝ mn ⎟⎠ ⎛m+n⎞ in (iv), we obtain Putting d = −2 ⎜ ⎝ mn ⎟⎠ 2a = −2 +

2 (m + n − 1)(m + n) mn

m−n {2a + (m − n − 1) d} 2 2 2 m−n⎧ (m + n − 1)(m + n) − ⇒ Sm−n = ⎨−2 + 2 ⎩ mn mn

Now, Sm−n =

(m − n − 1)(m + n)} 4n m−n⎧ ⎫ 1 (m + n)⎬ = (m − n)(m + 2n) ⇒ Sm−n = ⎨−2 + 2 ⎩ mn ⎭ m 19. Let A1, A2, ... , An be n arithmetic means between x and 2y. Then, x, A1, A2, ... , An, 2y are in A.P. with 2y − x . common difference d1 given by d1 = n +1 ⎛ 2y − x ⎞ ? rth mean = Ar = x + rd1 = x + r ⎜ ⎝ n + 1 ⎟⎠ Let Ac1, Ac2, ... , Acn be n arithmetic means between 2x and y. Then, 2x, Ac1, Ac2, ... , Acn, y are in A.P. with y − 2x . common difference d2 given by d2 = n +1 ⎛ y − 2x ⎞ ? rth mean = Ar = 2x + rd2 = 2x + r ⎜ ⎝ n + 1 ⎟⎠ 26

MATHEMATICS TODAY | OCTOBER ‘17

It is given that: Ar = Acr ⎛ 2y − x ⎞ ⎛ y − 2x ⎞ = 2x + r ⎜ ⇒ x +r ⎜ ⎟ ⎝ n +1 ⎠ ⎝ n + 1 ⎟⎠ (n + 1) x + r (2y – x) = (n + 1) 2x + r(y –2x) (n + 1) x – ry = rx n +1 y ⇒ − =1 r x 20. Let a1, a2 be the first terms and d1, d2 the common differences of the two given A.P.cs respectively. Then, the sums Sn and Scnof their n terms are given by n n Sn = {2a1 + (n − 1) d1} and Sn′ = {2a2 + (n − 1) d2} 2 2 n {2a1 + (n − 1) d1} {2a + (n − 1) d } Sn 1 ∴ = 2 = 1 n { ( 2 − 1 ) d a n + ′ Sn 2} 2 {2a2 + (n − 1) d2} 2 Sn 7n + 1 = It is given that n + 27 4 Sn′ ⇒

2a1 + (n − 1) d2 7n + 1 = 2a2 + (n − 1) d2 4n + 27

(n − 1) d2 7n + 1 2 = ⇒ (n − 1) a2 + d2 4n + 27 2 a1 +

Since, the ratio to mth terms of two A.P.'s is,

...(i) a1 + (m − 1) d2

a2 + (m − 1) d2 n −1 by (m – 1) on the L.H.S of (i) we get So, replacing 2 a + (m − 1)d1 7(2m − 1) + 1 14m − 6 i.e., 1 = = . a2 + (m − 1)d2 4(2m − 1) + 27 8m + 23

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Class XI

T

his specially designed column enables students to self analyse their extent of understanding of speciﬁed chapters. Give yourself four marks for correct answer and deduct one mark for wrong answer. Self check table given at the end will help you to check your readiness.

Sequences and Series Total Marks : 80

Only One Option Correct Type 1. If a, b, c are positive reals, then least value of ⎛1 1 1⎞ (a + b + c) ⎜ + + ⎟ is ⎝a b c⎠ (a) 1 (b) 6 (c) 9 (d) none of these 2. Two A.M’s A1 and A2 ; two G.M’s G1 and G2 and two H.M’s H1 and H2 are inserted between any two numbers, then H1–1 + H2–1 equals (a) A1–1 + A2–1 (b) G1–1 + G2–1 ( A1 + A2 ) G1G2 (c) (d) G1G2 ( A1 + A2 ) 3. If a, b, c are digits, then the rational number represented by 0. cababab .... is (99c + ba) cab (a) (b) 990 990 (99c + 10a + b) (99c + 10a + b) (c) (d) 990 99 4. If H is the harmonic mean between a and b, then H +a H +b + is equal to H −a H −b 1 1 (b) − (c) 2 (d) –2 (a) 2 2 5. If the arithmetic progression whose common difference is non zero, the sum of first 3n terms is equal to the sum of the next n terms. Then the ratio of the sum of the first 2n terms to the next 2n terms is 2 1 (b) (a) 3 5 3 (c) (d) none of these. 4

Time Taken : 60 Min.

44 , then r is equal to 9 11 (c) (d) 4 17

6. If 3 + 5r + 7r2 + .... to f to (a)

17 11

(b)

1 4

One or More Than One Option(s) Correct Type 7. If logx a, ax/2 and logb x are in G.P. then x is equal to (a) loga (logb a) (b) loga (loge a) – loga (loge b) (c) –loga (loga b) (d) loga (loge b) – loga (loge a) 8. The series of natural numbers is divided into groups 1 ; 2, 3, 4 ; 5, 6, 7, 8, 9 ; .... and so on. Then the sum of the numbers in the nth group is (a) (2n – 1) (n2 – n + 1) (b) 2n3 – 3n2 + 3n – 1 (c) n3 + (n – 1)3 (d) n3 + (n + 1)3 9. Let a, b, c, d, e be five numbers such that a, b, c are in A.P., b, c, d are in G.P. and c, d, e are in H.P. If a = 2 and e = 18, then b = (a) 2 (b) – 2 (c) 4 (d) – 4 10. There are two numbers a and b whose product is 192 and the quotient of A.M. by H.M. of their greatest common divisor and least common multiple is 169 . The smaller of a and b is 48 (a) 2 (b) 4 (c) 6 (d) 12 11. Given that 0 < x < ∞

and

∑

(−1)k tan2k x = a,

k=0 ∞

then

π π π and < y < 4 4 2 ∞

∑ (−1)k cot2k y = b,

k=0

∑ (tan)2k x cot2k y is

k=0

MATHEMATICS TODAY | OCTOBER‘17

27

Q. Three numbers a, b, c are ⎡G ⎤ 2. ⎢ ⎥ = 4 where in G.P. such that ⎣L⎦ a + b + c = 70; 4a, 5b, 4c [·] denotes the are in AP. greatest integer If G = Max{a, b, c} and L function = Min {a, b, c} then

1 1 1 (b) a + b – ab + − a b ab ab 1 (d) (c) a +b −1 1 1 1 + − a b ab 12. The first three terms of a progression are 3, –1, –1. The next term is −5 5 (d) − (a) 2 (b) 3 (c) 27 9 13. Suppose a, b, c are in A.P. and a2, b2, c2 are in G.P. If 3 a < b < c and a + b + c = , then the value of a is 2 1 1 (b) (a) 2 3 2 2 1 1 1 1 (d) − (c) − 2 2 3 2 (a)

Let {an} be a sequence such that 14.

∑

r =1

4.

2

ar = n .

13 13 + 23 13 + 23 + 33 + ... upto 16 terms is + + a1 a1 + a2 a1 + a2 + a3

(a) 346 (b) 446 (c) 546 (d) 464 a1 a1 + a2 a1 + a2 + a3 + + + ... upto ∞ is 15. 1 2 3 (a) 2e

(b) 2e – 1

(c) e2

3. G – L = 7

is equal to

Comprehension Type n

1 1 1 + + + .... to ∞ 3 5 7

R. 1 +

P

Q

R

(a) 3

2

4

(b) 3

4

1

(c) 1

4

3

(d) 2

1

3

1⎛ 1⎞ e− ⎟ ⎜ 2⎝ e⎠

Integer Answer Type 17. If x1, x2, x3, ...., x2008 are in H.P. and 2007

∑ xi xi +1 = λ x1x2008 , then sum of the digits of

i =1

(d) e – 1

O is

Matrix Match Type

18. If a, b, c are in H.P. and if

16. Match the following : Column I Column II P. Three numbers a, b, c 1⎛ 1⎞ between 2 and 18 such 1. 2 ⎜⎝ e + e ⎟⎠ that a + b + c = 25; 2, a, b are consecutive terms of an A.P. ; b, c, 18 are consecutive terms of an G.P. If G = Max {a, b, c} and L = Min {a, b, c} then

⎛ a+b ⎞ ⎛ c +b ⎞ ⎜⎝ ⎟+⎜ ⎟> 2a − b ⎠ ⎝ 2c − b ⎠

λ λ λ... ∞ , then the

value of O must be 19. The number of zeroes in the end of the product of 56 × 67 × 78 × 89 × .... × 5051 must be 57O where O is 20. Value of S = 1 +

2 3 4 5 + + + + ....to ∞ is 2 4 8 16

Keys are published in this issue. Search now! -

Check your score! If your score is No. of questions attempted …… No. of questions correct …… Marks scored in percentage ……

28

> 90%

EXCELLENT WORK ! You are well prepared to take the challenge of ﬁnal exam.

90-75%

GOOD WORK !

You can score good in the ﬁnal exam.

74-60%

SATISFACTORY !

You need to score more next time.

< 60%

MATHEMATICS TODAY | OCTOBER ‘17

NOT SATISFACTORY! Revise thoroughly and strengthen your concepts.

Limits, Continuity and Differentiability This column is aimed at Class XII students so that they can prepare for competitive exams such as JEE Main/Advanced, etc. and be also in command of what is being covered in their school as part of NCERT syllabus. The problems here are a happy blend of the straight and the twisted, the simple and the difﬁcult and the easy and the challenging. *ALOK KUMAR, B.Tech, IIT Kanpur

LIMIT OF A FUNCTION Let y = f(x) be a function of x. If at x = a f(x) takes indeterminate form, then we consider the values of the function which are very near to ‘a’. If these values tend to a definite unique number as x tends to ‘a’, then the unique number so obtained is called the limit of f(x) at x = a and we write it as lim f (x ) . x →a

z

Left hand and right hand limit : Consider the values of the functions at the points which are very near to a on the left of a. If these values tend to a definite unique number as x tends to a, then the unique number so obtained is called left-hand limit of f(x) at x = a and symbolically we write it as lim f (x ) = lim f (a − h). x →a −

h→0

Similarly we can define right-hand limit of f(x) at x = a which is expressed as lim f (x ) = lim f (a + h) x →a +

z

h→0

z

z

z

z

z

z z

z

z

lim f (x ) = lim f (x ) i.e. L.H.L. = R.H.L.

z

FUNDAMENTAL THEOREMS ON LIMITS

numbers) then

x →0

If f(x) d g(x) for all x, then lim f (x ) ≤ lim g (x ) x →a

x →a

lim[ f (x )]g ( x ) = { lim f (x )}x →a

x →a

x →a

If p and q are integers, then lim ( f (x )) p/q = l p/q , x →a

lim f ( g (x )) = f ( lim g (x )) = f (m) provided ‘f ’ is

x →a

x →a

sin x sin −1 x = 1 = lim x x →0 x x →0 lim

tan x tan −1 x = 1 = lim x lim cos x = 1 x x →0 x x →0 x →0 sin(1 / x ) sin x cos x =1 lim = lim = 0 x lim (1 / x ) x →∞ x →∞ x x →∞ x lim

Logarithmic Limits

Let lim f (x ) = l and lim g (x ) = m (l and m are real x →0

x →a

Trigonometric Limits

lim f (x ) and lim f (x ) exist and x →a +

lim log{ f (x )} = log { lim f (x )}

x →a

continuous at g(x) = m.

x →a

x →a −

f (x ) l = ,m ≠ 0 x →a x →a g ( x ) m 1 If lim f (x ) = +∞ or –f, then lim =0 x →a x →a f ( x )

x lim

lim k f (x ) = k ⋅ l

provided (l)p/q is a real number.

z

x →a +

x →a

lim g ( x )

Existence of limit : lim f (x ) exists when, both x →a −

lim ( f (x ) ± g (x )) = l ± m x lim ( f (x ) ⋅ g (x )) = l ⋅ m

x →a

z

log(1 + x ) =1 x x →0 lim

x lim log e x = 1 x →e

MATHEMATICS TODAY | OCTOBER‘17

29

z

log(1 − x ) = −1 x x →0

z

log a (1 + x ) = log a e, a > 0, ≠ 1 x x →0

z

lim lim

Exponential Limits z

ex − 1 =1 x →0 x

ax − 1 = log e a x →0 x

z

e λx − 1 = λ (λ ≠ 0) x x →0

x lim

lim lim

Based on the Form 1f z

If lim f (x ) = lim g (x ) = 0 , then x →a

x →a

1/ g ( x )

lim {1 + f (x )}

x →a

lim

f (x )

= e x →a g ( x )

or when

lim f (x ) = 1 and lim g (x ) = ∞ , then x →a

x →a

lim { f (x )} g ( x ) = lim [1 + f (x ) − 1]g ( x )

x →a

x →a

lim ( f ( x )−1) g ( x )

z

z

⎛

x →0

1/ x

lim (1 + λx )

x →0

=e

1⎞

x

x lim ⎜1 + ⎟ = e x⎠ x →∞ ⎝

lim (1 + x )1/ x = e

x

⎛ λ⎞ x e lim ⎜1 + ⎟ = e λ x⎠ x →∞ ⎝

λ

⎧∞ , if a > 1 lim a x = ⎨ x →∞ ⎩ 0 , if a < 1

z z

lim f (x ) = lim g (x ) = 0

x →a

x →a

Both are continuous and differentiable at x = a. f c(x) and gc(x) are continuous at the point f (x ) f ′(x ) provided that x = a, then lim = lim ( ) g x g x →a x → a ′( x ) gc(a) z 0. The above rule is also applicable if lim f (x ) = ∞ x →a and lim g (x ) = ∞ . x →a

CONTINUITY OF A FUNCTION AT A POINT A function f(x) is said to be continuous at a point x = a of its domain if and only if lim f (x ) exists and x →a

is equal to f(a) i.e., if lim f (x ) = f (a) = lim f (x ) x →a −

30

DISCONTINUOUS FUNCTION Discontinuous function : A function ‘f ’ which is not continuous at a point x = a in its domain is said to be discontinuous there at. The point ‘a’ is called a point of discontinuity of the function. The discontinuity may arise if lim f (x ) as well x →a +

as lim f (x ) both may exist, but either of the x →a −

two or both may not be equal to f(a).

L-Hospital’s Rule If f(x) and g(x) be two functions of x such that z

PROPERTIES OF CONTINUOUS FUNCTION Let f(x) and g(x) be two continuous functions at x = a. Then z A function f(x) is said to be everywhere continuous if it is continuous on the entire real line R i.e. (–f, f). e.g., polynomial function, ex, sinx, cosx, constant, xn, |x – a| etc. z If g(x) is continuous at x = a and f(x) is continuous at x = g(a) then (fog)(x) is continuous at x = a. z If f(x) is continuous in a closed interval [a, b] then it is bounded on this interval. z If f(x) is a continuous function defined on [a, b] such that f(a) and f(b) are of opposite signs, then there is atleast one value of x for which f(x) vanishes. i.e. if f(a) > 0, f(b) < 0 ⇒ ∃ c ∈(a, b) such that f(c) = 0. z

= e x →a z

Cauchy’s definition of continuity : A function f is said to be continuous at a point a of its domain D if for every > 0 there exists G > 0 (dependent on ) such that | x − a |< δ ⇒ | f (x ) − f (a)| < ∈.

MATHEMATICS TODAY | OCTOBER ‘17

x →a +

DIFFERENTIABILITY OF A FUNCTION AT A POINT z A function f (x) is said to be differentiable (finitely) at x = a if R.H.D. at x = a [f c(a+)] = L.H.D. at x = a [f c(a–)] = finite value f (a + h) − f (a) f (a − h) − f (a) i.e. lim = lim = finite h −h h→0 h→0 and the common limit is called the derivative of f(x) at x = a, denoted by f c(a). Clearly, f (x ) − f (a) {x o a from the left as f ′(a) = lim x −a x →a well as from the right}. SOME STANDARD RESULTS ON DIFFERENTIABILITY z Every polynomial function is differentiable at each x R. z The exponential function ax, a > 0 is differentiable at each x R.

z

z

z

z

z

Every constant function is differentiable at each x R. The logarithmic function is differentiable at each point in its domain. Trigonometric and inverse trigonometric functions are differentiable in their domains. The sum, difference, product and quotient of two differentiable functions is differentiable. The composition of differentiable function is a differentiable function.

4.

(a) 5.

IMPORTANT POINTS z Any continuous function f(x), which has at least one local maximum or local minimum, is many-one. z If lim f (x ) does not exist, then we can not remove this discontinuity. So this become a non-removable discontinuity or essential discontinuity. If f is continuous at x = c and g is discontinuous at x = c, then f + g and f – g are discontinuous f·g may be continuous . If f and g are discontinuous at x = c, then f + g, f – g and fg may still be continuous. Point functions (domain and range consists one value only) is not a continuous function. If a function is differentiable at a point, then it is continuous also at that point. i.e., Differentiability Continuity, but the converse need not be true. If f(x) and g(x) both are not differentiable at x = a then the product function f(x)·g(x) and sum function f(x) + g(x) can still be differentiable at x = a. If f(x) is differentiable at x = a and g(x) is not differentiable at x = a then the sum function f(x)+ g(x) is also not differentiable at x = a.

7.

z

z

z

z

z

z

PROBLEMS Single Correct Answer Type 1.

2.

3.

1 ⎧ ⎪ x sin , x ≠ 0 If f (x ) = ⎨ , then lim f (x ) = x x →0 ⎪⎩ 0, x =0 (a) 1 (b) 0 (c) –1 (d) 2 x 3 cot x lim = x →0 1 − cos x (a) 0 (b) 1 lim

n(2n + 1)2

n→∞ (n + 2)(n2

(a) 0

(b) 1 / 2a (c) 2a

2a

(b) 2 e1/ x − 1

lim

x →0 e1/ x

(b) 1 (d) does not exist

log cos x = x x →0

If f(9) = 9, f c(9) = 4, then lim

11.

=

+ 3n − 1) (b) 2 (c) 4

(d) 1/2 f (x ) − 3

= x −3 (c) –2 (d) –4 x →9

(a) 2

10.

(c) f

(b) 1

(b) 4

e sin x − 1 = x x →0 (a) 1 (b) e lim

(c) 1/e

sin α − cos α = π α → π/ 4 α− 4 (a) 2 (b) 1 / 2 (c) 1

(d) 3

lim

(d) –1

lim tan x log sin x =

x → π /2

(a) 0 12. lim

(b) 1

cos ax − cos bx x2

x →0

(c) –1

(d) 2

= b2 − a 2 (b) 2 2 (d) b – a2

a 2 − b2 (a) 2 (c) a2 – b2 (1 + x )5 − 1

x →0 (1 + x )3

(d) –2

=

+1

(a) 0

9.

(d) –1

lim

13. lim (c) 2

(c) 0

(a) 0 (c) –1

8.

(d) 1/2a

⎧ x , when 0 ≤ x ≤ 1 If f (x ) = ⎨ , then ⎩2 − x , when 1 < x ≤ 2 (a) 1

6.

x →a

3x − a − x + a = x −a x →a lim

(a) 0

= −1 (b) 1

sin 2 x + sin 6 x = x →0 sin 5x − sin 3x (a) 1/2 (b) 1/4

(c) 5/3

(d) 3/5

(c) 2

(d) 4

14. lim (d) f

MATHEMATICS TODAY | OCTOBER‘17

31

15. lim x log(sin x ) =

x +3

x →0

(a) –1

(b) loge1

(c) 1

(d) log 2

⎧ x3 ⎫ ⎪⎪ sin x − x + ⎪⎪ 6 = 16. lim ⎨ ⎬ 5 x →0 ⎪ x ⎪ ⎪⎩ ⎪⎭ (a) 1/120 (b) –1/120 (c) 1/20 (d) –1/20

(a)

2 3

(d)

2 3

(x + y )sec(x + y ) − x sec x = y y →0 (a) secx (x tanx + 1) (b) x tanx + secx (c) x secx + tanx (d) –2/3

19. lim

20. lim

tan x − sin x

x →0

x

(a) 1/2 21. lim

3

1− x

x →1 (cos −1 x )2

(d) 4

=

(a) 1 (c) 1/4 22.

(c) 2/3

2 cos x − 1 = x → π/ 4 cot x − 1 1 1 (a) (b) 2 2

(b) 1/2 (d) none of these

lim

1 2 2

(d) 1

π

(b)

(d) 2

is given by

1 2π

(c) 1

(d) 0

⎧ x2 − 4x + 3 , for x ≠ 1 ⎪ , then 30. If f (x ) = ⎨ x 2 − 1 ⎪ 2, for x = 1 ⎩ lim f (x ) = 2

x →1+

(b)

lim f (x ) = 3

x →1−

(c) f(x) is discontinuous at x = 1 (d) none of these ⎧ x − 1, x < 0 ⎪ ⎪ 1 , x = 0 , then 31. If f (x ) = ⎨ ⎪ 4 ⎪⎩ x 2 , x > 0

(b)

lim f (x ) = 1

x →0+

lim f (x ) = 1

x →0−

(c) f(x) is discontinuous at x = 0 (d) none of these

⎡ ⎤ x 23. lim ⎢ −1 ⎥ = x →0 ⎣ tan 2 x ⎦ (a) 0 (b) 1/2

(c) 1

1 (1 − cos 2 x ) 2 = 24. lim x x →0 (a) 1 (c) 0

(b) –1 (d) does not exist

32

1

(d) 2

⎡ ⎤ lim ⎢ x + x + x − x ⎥ is equal to x →∞ ⎣ ⎦ (a) 0 (b) 1/2 (c) log2 (d) e2

(a) (c)

x +1

x →−1

(a)

=

(b) –1/2

29.

π − cos −1 x

lim

(a)

(c)

(c) –1/2

1 x 2 sin − x x is 27. The value of lim 1− | x | x →∞ (a) 0 (b) 1 (c) –1 28.

(d) e3

is equal to

x2 (b) 1/2

(a) 0

=

3a + x − 2 x 1 2 (b) 3 3 3

x →a

(c) e2

sin x + log(1 − x )

x →0

(a + h)2 sin(a + h) − a2 sin a = h h→0 (a) acosa + a2sina (b) asina + a2cosa (c) 2asina + a2cosa (d) 2acosa + a2sina

18. lim

⎛x +2⎞ lim ⎜ is ⎟ x →∞ ⎝ x + 1 ⎠ (a) 1 (b) e

26. lim

17. lim

a + 2 x − 3x

25.

MATHEMATICS TODAY | OCTOBER ‘17

(d) f

x , where [] [x] is greatest integer function, is discontinuous (a) only positive integers (b) all positive and negative integers and (0, 1) (c) all rational numbers (d) none of these

32. At which points the function f (x ) =

⎧ − x 2 , when x ≤ 0 ⎪ ⎪ 5x − 4, when0 < x ≤ 1 , then 33. If f (x ) = ⎨ 2 ⎪4 x − 3x , when 1 < x < 2 ⎪ 3x + 4, when x ≥ 2 ⎩ (a) f (x) is continuous at x = 0 (b) f (x) is continuous at x = 2 (c) f (x) is discontinuous at x = 1 (d) none of these ⎧ x−4 ⎪ | x − 4 | + a, x < 4 ⎪⎪ a + b, x = 4 . 34. Let f (x ) = ⎨ ⎪ x−4 ⎪ + b, x > 4 ⎪⎩| x − 4 | Then f (x) is continuous at x = 4 when (a) a = 0, b = 0 (b) a = 1, b = 1 (c) a = –1, b = 1 (d) a = 1, b = –1 35. The value of f (0), so that the function f (x ) =

(27 − 2 x )1/3 − 3

,(x ≠ 0) is continuous, is 9 − 3(243 + 5x )1/5 given by (a) 2/3 (b) 6 (c) 2 (d) 4 ⎧ πx ⎪1 + sin 2 , for − ∞ < x ≤ 1 ⎪ 36. If the function f (x ) = ⎨ ax + b, for 1 < x < 3 ⎪ πx ⎪ 6 tan , for 3 ≤ x < 6 ⎩ 12 is continuous in the interval (–f, 6), then the values of a and b are respectively (a) 0, 2 (b) 1, 1 (c) 2, 0 (d) 2, 1 37. The values of A and B such that the function π ⎧ ⎪ −2 sin x , x ≤ − 2 ⎪ ⎪ π π f (x ) = ⎨ A sin x + B , − < x < , is continuous 2 2 ⎪ π ⎪ , x≥ ⎪⎩ cos x 2 everywhere are (a) A = 0, B = 1 (b) A = 1, B = 1 (c) A = –1, B = 1 (d) A = –1, B = 0 1 − cos 4 x

, where x z 0 and 8x 2 f(x) = k where x = 0 is a continous function at x = 0 then the value of k will be

38. Function

f (x ) =

(a) k = 0 (c) k = –1

(b) k = 1 (d) none of these

1 ⎧ p ⎪ x sin , x ≠ 0 39. Let f (x ) = ⎨ , then f(x) is continuous x ⎪⎩0 ,x = 0 but not differential at x = 0 if (a) 0 < p d 1 (b) 1 d p < f (c) –f < p < 0 (d) p = 0 ⎧ e(1/ x ) − e (−1/ x ) ⎪ x (1/ x ) (−1/ x ) , x ≠ 0 then which of 40. If f (x ) = ⎨ e +e ⎪ 0, x = 0 ⎩ the following is true? (a) f is continuous and differentiable at every point (b) f is continuous at every point but is not differentiable (c) f is differentiable at every point (d) f is differentiable only at the origin 41. If f (x) = |x – 3|, then f is (a) discontinuous at x = 2 (b) not differentiable at x = 2 (c) differentiable at x = 3 (d) Continuous but not differentiable at x = 3 ⎧ x2 ⎪ ,x ≠ 0 42. Consider f (x ) = ⎨| x | ⎪ 0, x = 0 ⎩ (a) f(x) is discontinuous everywhere (b) f(x) is continuous everywhere (c) f c(x) exists in (–1, 1) (d) f c(x) exists in (–2, 2) 43. The function y = e–|x| is (a) continuous and differentiable at x = 0 (b) neither continuous nor differentiable at x = 0 (c) continuous but not differentiable at x = 0 (d) not continuous but differentiable at x = 0 44. The left-hand derivative of f(x) = [x] sin(Sx) at x = k, k is an integer and [x] = greatest integer d x, is (b) (–1)k – 1 (k – 1)S (a) (–1)k (k – 1)S k (c) (–1) kS (d) (–1)k – 1 kS 45. The value of m for which the function ⎧⎪mx 2 , x ≤ 1 f (x ) = ⎨ is differentiable at x = 1, is ⎪⎩ 2 x , x > 1 (a) 0 (b) 1 (c) 2 (d) does not exist 46. f(x) = || x | – 1| is not differentiable at (a) 0 (b) ±1, 0 (c) 1 (d) ±1 MATHEMATICS TODAY | OCTOBER‘17

33

Multiple Correct Answer Type 47. Let a function f : R o R satisfies the equation f(x + y) = f(x) + f(y), x, y R then (a) f is continuous for all x R if it is continuous at x = 0 (b) f(x) = x·f (1) x R, if ‘f ’ is continuous (c) f(x) = ( f(1))x x R, if ‘f ’ is continuous (d) f(x) is differentiable for all x R

⎛ x + 2 y ⎞ φ(x ) + 2φ( y ) = 48. Let φ ⎜ ∀x , y ∈ R and ⎝ 3 ⎟⎠ 3 Ic(0) = 1 and I(0) = 2 then (a) I(x) is continuous x R (b) I(x) is differentiable x R (c) I(x) is both continuous and differentiable (d) I(x) is discontinuous at x = 0 49. Consider the function 'I' defined in [0, 1] as ⎧ 2 ⎛ 1 ⎞ ; if x ≠ 0 ⎪ x ⋅ sin ⎜ ⎟ ⎝x⎠ , then φ( x ) = ⎨ ⎪ 0 ; if x = 0 ⎩ (a) I(x) has right derivate at x = 0 (b) Ic(x) is discontinuous at x = 0 (c) Ic(x) is continuous at x = 0 (d) Ic(x) is differentiable at x = 0 50. The in-circle of 'ABC touches side BC at D. Then difference between BD and CD (R is circum-radius of 'ABC) A B −C (a) 4 R sin sin 2 2 (b)

A B −C 4 R cos sin 2 2

(c) |b – c|

⎧⎡ ⎡ 1 ⎤ ⎤ 1 ⎪ ⎢| x | ⎢ ⎥ ⎥ , | x |≠ , n ∈ N , ⎪⎣ ⎣| x |⎦ ⎦ n 54. f (x ) = ⎨ then, (where 1 ⎪ 0, | x |= ⎪⎩ n [·] denotes greatest integer function) (a) f is differentiable everywhere (b) f is continuous everywhere (c) f is periodic (d) f is not an odd function 55. If f(x) = 2 + |sin–1x|, it is (a) continuous no where (b) continuous everywhere in its domain (c) differentiable no where in its domain (d) not differentiable at x = 0 56. f (x ) = cos π(| x | +[x]), then f is (where [.] denotes greatest integer function) (a) continuous at x = 1/2 (b) continuous at x = 0 (c) differentiable in (–1, 0) (d) differentiable in (0, 1) Comprehension Type Paragraph for Q. No. 57 to 59 Let p(x) be a polynomial with positive leading coefficient x

and p(0) = 0; and p( p(x )) = x ⋅ ∫ p(t )dt , ∀ x ∈ R. Then 0

57. Degree of the polynomial p(x) is (a) 4 (b) 3 (c) 5 (d)

b−c 2

⎧ x , x≠0 ⎪ , then 51. If f ( x ) = ⎨1 + e1/ x ⎪ 0, x 0 = ⎩ (b) f c(0+) = 0 (a) f c(0+) = 1 (c) f c(0–) = 1 (d) f c(0–) = 0 52. Consider the function y = f (x ) = 1 − 1 − x 2 . Then the true statements among the following is/are (a) f is continuous in its domain (b) f is differentiable in (–1, 1) (c) Rf ′ (0) = 2 and Lf ′(0) = − 2 θ cos 3π 2 then f ′ (sin θ) = (d) If π < θ < 2 2 cos θ 34

53. If f(x) = |x – a| I(x), where I(x) is a continuous function, then (a) f c(a+) = I(a) (b) f c(a–) = –I(a) + – (c) f c(a ) = f c(a ) (d) f c(a) does not exist

MATHEMATICS TODAY | OCTOBER ‘17

58.

(d) 2

p ′(x ) is discontinuous at x = |x| (a) 0 (b) 1 (c) –1 (d) none of these

59. If p(1) = 3, p(–1) = 5 and g(x) is inverse of pc(x) then gc(0) is (a) is equal to 1/4 (b) is equal to 1/8 (c) is equal to 8 (d) does not exist Paragraph for Q. No. 60 to 62 n

For x > 0, let f ( x ) = lim

log(2 + x ) + x 2 sin x

n→∞

60.

n

1 + x2

lim f (x ) is equal to

x →0+

(a) 0 (c) loge2

(b) loge3 (d) does not exist

61. At x = 1, f c(x) is (a) continuous (b) discontinuous (c) both continuous and differentiable (d) continuous but not differentiable. 62. In (0, S/2], the number of points at which f(x) vanishes is (a) 0 (b) 1 (c) 2 (d) 3 Paragraph for Q. No. 63 to 65 A function f(x) is said to be continuous at x = a if lim f (x ) = lim f (x ) = f (a) . i.e., lim f (x ) = f (a) x →a −

x →a +

x →a

When f(x) is not continuous at x = a we say that f(x) discontinuous at x = a. 63. If f (x ) = lim sin2m x , , then number of point(s) m→∞

where f(x) is discontinuous is (a) 0 (b) 1 (c) 2 (d) infinitely many

67. Match the following. Column I (A) f(x) = |x|

(p) Continuous at x=0 (q) Discontinuous (B) f(x) = xn |x|, n N at x = 0 (r) Differentiable ⎧ x l n | sin x |, x ≠ 0 (C) f ( x ) = ⎨ at x = 0 , x=0 0 ⎩ (D)

(b) exists and it equals – 2 (c) does not exists because L.H.L. z R.H.L. (d) exists and it equals 1/2 65. In order that the function f(x) = (x + 1)cotx is continuous at x = 0, f(0) must be defined as (a) 0 (b) e (c) 1/e (d) 1 Matrix – Match Type

66. Match the following. Column I (A)

⎛ 2x ⎞ f (x ) = sin −1 ⎜ is not ⎝ 1 + x 2 ⎟⎠

Column II (p) x = 1

(q) x = –1 ⎛ 2x ⎞ is not f (x ) = tan −1 ⎜ ⎝ 1 − x 2 ⎟⎠

differentiable at (C) f(x) = cos–1(4x3 – 3x) is not differentiable at (D) f(x) = sin–1(3x – 4x3) is not differentiable at

(s) Nondifferentiable at x = 0

Column II

(A)

f (x ) = sin(n[π]) (where [] denote G.I.F)

(p) Differentiable everywhere

(B)

f (x ) = sin((x − [x])π) (where [] denote G.I.F)

(q) Nondifferentiable at x = 2 (r) Nondifferentiable at –1 and 1

(C)

⎧ ⎛1⎞ ⎪ x sin ⎜ ⎟ , if x ≠ 0 ⎝x⎠ f (x ) = ⎨ ⎪ 0, if x = 0 ⎩

(D)

f (x ) =| 2 − x | + [2 + x] (where [] denote G.I.F)

(s)

Continuous at x = 0 but not differentiable at x = 0

Integer Answer Type

1⎤ ⎡ 1⎤ ⎡ 3⎤ ⎡ 69. Let f (x ) = [x] + ⎢ x + ⎥ + ⎢ x + ⎥ + ⎢ x + ⎥ . Then 4⎦ ⎣ 2⎦ ⎣ 4⎦ ⎣ no. of points of discontinuity of f(x) in [0,1] is ([·] denotes G.I.F) 70. If α ∈(−∞, − 1) ∪ (−1, 0) then the number of points where the function f (x ) = x 2 + (α − 1)| x | −α is not differentiable is

differentiable at (B)

x ≠0 x =0

Column I

1 − cos 2(x − 1) x −1 x →1 2

⎧⎪ xe1/ x , f (x ) = ⎨ ⎩⎪ 0,

68. Match the following.

64. lim

(a) exists and it equals

Column II

(r)

x = 1/2

(s)

x = –1/2

⎡1⎤ ⎧ ⎢ ⎥ ⎪ ⎣x⎦ ⎪⎪ x 2 ∑ r; x ≠0 71. Let f (x ) = ⎨ r =0 ⎪ ⎪k ; otherwise ⎪⎩ 2 ([·]denotes the greatest integer function). The value of k such that f become continuous at x = 0 is MATHEMATICS TODAY | OCTOBER‘17

35

72. Let f(x) = [x2]sinSx, x R the number of points in the interval (0, 3] at which the function is discontinuous is 73. If the function f defined by f ( x ) =

log(1 + x )1+ x 2

1 − , x

x x z 0 is continuous at x = 0, then 6 ( f(0)) = ⎧ tan[x 2 ] π ⎪ + ax 3 + b , 0 ≤ x ≤ 1 74. If f (x ) = ⎨ ax 2 is ⎪ −1 ⎩ 2 cos πx + tan x , 1 < x ≤ 2 π 26 . Find differentiable in [0, 2], then b = − k1 k2 k2 – k1 {where [] denotes greatest integer function}. SOLUTIONS

n (2n + 1)2

12. (b) : lim

+ 3n − 1)

x →0

4n3 + 4n2 + n

h→0

2 sin 4 x cos 2 x x →0 2 sin x cos 4 x

14. (d) : lim

h→0

⎛ sin 4 x ⎞ ⎛ x ⎞ cos 2 x = lim 4 ⎜ =4 ⎟⎜ ⎟ x →0 ⎝ 4 x ⎠ ⎝ sin x ⎠ cos 4 x

y = f(x)

and lim f (x ) = lim 2 − (1 + h) = 1 h→0

1

2

6. (d) log cos x 7. (a) : lim = lim x x →0 x →0 36

x⎤ ⎡ log ⎢1 − 2 sin2 ⎥ 2⎦ ⎣ x

MATHEMATICS TODAY | OCTOBER ‘17

x2

x [5 C1 + 5C2 x + 5C3 x2 + 5C4 x3 + 5C5 x 4 ] 5 = . 3 x →0 x [3 C1 + 3C2 x + 3C3 x2 ]

y

O

cos ax − cos bx

13. (c) : lim

lim f (x ) = lim (1 − h) = 1

Hence limit of function is 1.

2

⎛b −a ⎞ ⎛a +b⎞ 2 sin ⎜ x ⋅ sin ⎜ x ⎝ 2 ⎟⎠ ⎝ 2 ⎟⎠ b2 − a2 = = lim 2 2 ⎛b −a ⎞ 2 x →0 ⎛ a + b ⎞ ⎜⎝ ⎟⎠ x ⋅ + ⋅ − ⋅ ⎜⎝ 2 ⎟⎠ x 2 a b b a

n3 + 5n2 + 5n − 2

5. (a) : lim f (x ) = lim (1 − h) = 1

x →1+

log sin x π cot x x→

π 2

2

⎛ 4 1⎞ n3 ⎜ 4 + + ⎟ ⎝ n n2 ⎠ =4 = lim n→∞ 3 ⎛ 5 5 2⎞ n ⎜1 + + − ⎟ ⎝ n n2 n3 ⎠ 4. (b)

x →1−

esin x − 1 esin x − 1 sin x = lim × x x x →0 x →0 sin x

9. (a) : lim

1 cos x = lim sin x = 0 (Applying L-Hospital’s rule) 2 π x → − cosec x

2. (c)

x →1−

2 x

x→

x →0

n→∞

=

4 f (9) 3 = =4 1 1 3 9

11. (a) : lim tan x log sin x = lim

In this way lim f (x ) = 0.

= lim

2 f (x ) lim 1 x →9

f ′(9)

⋅ f ′(x )

10. (a)

x →0

n→∞ (n + 2) (n2

1

sin x esin x − 1 × lim = 1×1 = 1 . x →0 sin x x →0 x

We know that lim | x | = 0 and lim − | x | = 0

3. (c) : lim

8. (b) : Applying L-Hospital’s rule,

= lim

1. (b) : Here f(0) = 0 1 1 Since − 1 ≤ sin ≤ 1 ⇒ − | x | ≤ x sin ≤ | x | x x x →0

2 ⎤ ⎡ x⎞ ⎛ ⎥ ⎢ 2 sin2 x ⎜ 2 ⎟ + ......⎥ − ⎢2 sin2 + ⎜ 2 2 ⎟ ⎥ ⎢ ⎟ ⎜ ⎥ ⎢ ⎠ ⎝ ⎦= ⎣ = lim 0 x x →0

x

15. (b) : lim x log sin x = lim log(sin x )x x →0

x →0

x (sin x −1) ⎤ ⎡ = log ⎢ lim (1 + sin x − 1) sin x −1 ⎥ ⎢ x →0 ⎥ ⎣ ⎦ lim x (sin x −1)

= loge [e x →0

] = loge 1.

16. (a) 2

26. (c) : Apply L-Hospital’s rule, we get 1 1 − sin x − cos x − 1 (1 − x )2 1 − x = lim =− lim 2x 2 2 x →0 x →0

2

17. (c) : lim (a + h) sin(a + h) − a sin a h h→0 2 (a + h)sin(a + h) + (a + h)2 cos(a + h)

= lim

h→0

1 27. (a) : Putting x = , the given limit t sin t −1 1−1 = lim t = = 0, 0 −1 t →0 t − 1

1

= 2asina + a2cosa. 18. (b)

19. (a)

20. (a)

21. (c) : Put cos–1x = y and x o 1 y o 0 ∴

1− x

lim

−1

x →1 (cos

2

x)

= lim

28. (b) ⎡ ⎤ 29. (b) : lim ⎢ x + x + x − x ⎥ x →∞ ⎣ ⎦ x+ x+ x −x = lim x →∞ x+ x+ x + x

1 − cos y

y →0

y

2

(1 − cos y ) Now rationalizing it, we get lim 2 y →0 y (1 + cos y ) = lim

1 − cos y y2

y →0

1 1 1 . lim = × = . 2 2 4 y →0 1 + cos y

x →∞

cot x − 1

x → π/ 4

Apply L'Hospitals's rule, we get

∴ f (1) = 2, f (1+ ) = lim

2

1 lim 2 sin3 x = = . x → π/ 4 2 2 2

x →1

f (1− ) = lim

x →0

x

= 1 and lim

−

x →0

|sin x | x

1 ⎞ ⎛ = lim ⎜1 + ⎝ x + 1 ⎟⎠ x →∞

x +3

( x − 3) = −1 x →1 ( x + 1)

= lim

+

x −1

2

x − 4x + 3

= −1 x2 − 1 Hence the function is discontinuous at x = 1. 31. (c) 32. (b) : When 0 d x < 1 f(x) doesn't exist as [x] = 0 here. Also, lim f ( x ) and lim f ( x ) does not exist. x →1+

x →1−

Hence f(x) is discontinuous at all integers and also in (0, 1). 33. (b) : lim f ( x ) = 0, f (0) = 0, lim f ( x ) = −4 x → 0+

x →0−

= −1

? f(x) is discontinuous at x = 0 and lim f ( x ) = 1 and lim f ( x ) = 1, f (1) = 1

Hence limit does not exist. ⎛x +2⎞ 25. (b) : Let A = lim ⎜ ⎟ x →∞ ⎝ x + 1 ⎠

2

x →1

1 (1 − cos 2 x ) |sin x | 24. (d) : lim 2 = lim x x →0 x →0 x +

1 2

−

1 23. (b) : Let tan 2 x = θ ⇒ x = tan θ 2 and as x o 0, T o 0 1 tan θ x 1 ⇒ lim = lim 2 = . − 1 2 θ x →0 tan 2 x θ→0

So, lim

x2 − 4 x + 3

+

−1

|sin x |

1 + x −1 + x −3/2 + 1

=

⎧⎪ x2 − 4 x + 3 ⎪⎫ 30. (c) : f (x ) = ⎨ ⎬ , for x ≠ 1 and f(x)= 2, 2 x − 1 ⎪ ⎪ ⎩ ⎭ for x = 1

2 cos x − 1

22. (b) : lim

1 + x −1/2

= lim

1

x →1−

x +3

x →1+

Hence f(x) is continuous at x = 1 ( x +3) x +1 ⎤ ( x +1) ⎞

⎡⎛ 1 = lim ⎢ ⎜1 + ⎝ x + 1 ⎟⎠ x →∞ ⎢ ⎣

⎥ ⎥⎦

Also, lim f ( x ) = 4(2)2 − 3 ⋅ 2 = 10 x → 2−

=e

f (2) = 10 and lim f ( x ) = 3(2) + 4 = 10 x → 2+

Hence f(x) is continuous at x = 2. MATHEMATICS TODAY | OCTOBER‘17

37

x → 4−

x →3+

h→0

4−h−4 + a = lim − h + a = a − 1 = lim h→0 | 4 − h − 4 | h→0 h = lim f (x ) = lim f (4 + h) = lim x→4

+

4+h−4

h→0 | 4 + h − 4 |

h→0

h→0

= lim | 3 + h − 3 | = 0 h→0

x →3−

+ b = b +1

|3 − h − 3| − 0

a – 1 = a + b = b + 1 b = –1 and a = 1.

R f ′(3) = lim

37. (c) : For continuity at all x R, we must have ⎛ π⎞ f ⎜− ⎟ = lim (−2 sin x ) = lim ( A sin x + B) ⎝ 2 ⎠ x →(− π/2)− x →(− π/2)+ 2 = –A + B ....(i) ⎛π⎞ and f ⎜ ⎟ = lim ( A sin x + B) = lim (cos x ) ⎝ 2 ⎠ x →( π/2)− x →( π/2)+ ....(ii)

38. (b) 1 39. (a) : f (x ) = x p sin , x ≠ 0 and f (x ) = 0, x = 0 x Since at x = 0, f(x) is a continuous function 1 ∴ lim f (x ) = f (0) = 0 ⇒ lim x p sin = 0 ⇒ p > 0 x x →0 x →0 f (x ) − f (0) exists x −0 x →0 lim

1 lim x p −1 sin exists x x →0

p – 1 > 0 or p > 1 ⎛1⎞ If p d 1,then lim x p −1 sin ⎜ ⎟ does not exist and at ⎝x⎠ x →0 x = 0 f(x) is not differentiable. ? for 0 < p d 1 f(x) is a continuous function at x = 0 but not differentiable. 40. (b) 41. (d) : lim f (x ) = lim f (3 − h) = lim | 3 − h − 3 | = 0 38

h→0

h→0

MATHEMATICS TODAY | OCTOBER ‘17

h = −1 − h→0 h

= lim

|3 + h − 3| − 0 f (3 + h) − f (3) = lim =1 h h h→0 h→0

36. (c)

x →3−

−h

h→0

x →4+

f(x) is differentiable at x = 0, if 1 x p sin − 0 x ⇒ lim exists ⇒ x −0 x →0

x

x=3

Hence f is continuous at x = 3 f (3 − h) − f (3) Now L f ′(3) = lim −h h→0 = lim

0=A+B From (i) and (ii) we get, A = –1 and B = 1.

O

x →3+

Therefore lim f (x ) = f (4) = lim f (x )

35. (c)

|x – 3|

' lim f (x ) = lim f (x ) = f (3)

and f(4) = a + b Since f(x) is continuous at x = 4 x →4−

y

lim f (x ) = lim f (3 + h)

34. (d) : lim f (x ) = lim f (4 − h)

'

Lf c(3) z Rf c(3). Hence f is not differentiable at x=3 .

42. (b)

43. (c) [k − h]sin π(k − h) − [k]sin πk −h h→0

44. (a) : f ′(k − ) = lim

(−1)k −1 (k − 1)sin πh − k × 0 = (−1)k ⋅ (k − 1)π −h h→0 f (1 − h) − f (1) 45. (b) : L f ′ (1) = lim −h h→0 = lim

= lim

m (1 − h)2 − m

h→0

−h

= lim

m[1 + h2 − 2h − 1]

h→0

−h

= lim m (2 − h) = 2m h→0

and R f ′ (1) = lim

f (1 + h) − f (1)

h→0

h

= lim

h→0

2 (1 + h) − m h

=2

For differentiability, Lf c(1) = Rf c(1). Hence m = 1 ⎧| x | −1, | x | −1 ≥ 0 46. (b) : = ⎨ ⎩− | x | +1, | x | −1 < 0 ⎧| x | −1, x ≤ −1 or x ≥ 1 =⎨ ⎩− | x | +1, − 1 < x < 1 ⎧− x − 1, ⎪ x + 1, ⎪ =⎨ ⎪− x + 1, ⎪⎩ x − 1,

x ≤ −1 −1 < x < 0 0 ≤ x <1 x ≥1

From the graph. It is clear that f(x) is not differentiable at x = –1, 0 and 1.

47. (a, b) : If f(x) is continuous at x = a, lt f ( x ) = f (a) x →a

Let a ∈ R then lt f ( x ) = lt f ( a + h ) x →a

h→0

= lt ( f (a) + f (h)) = f (a) + lt f (h) h→0

x →0

= f(a) + f(0) = f(a + 0) = f(a) 'f c is continuous ∀x ∈ R, as ‘a’ is arbitrary ' f (x + y) = f (x) + f (y) f (0) = 0 f (1) For any +ve inteteger ‘n’ f (1) = f (1 + 1 + ..... + n times) = n f (1) For any –ve integer 'm' we have 0 = f (0) = f [m + (–m)] = f (m) + f (–m) f (m) = –f (–m) = –(–m) f (1) = m f (1) Let p/q be any rational number where ‘q’ is a +ve integer and p is any integer, +ve, –ve or zero. ⎛ p⎞ ⎛p p ⎞ Then f ⎜ q. ⎟ = f ⎜ + + ........... q times ⎟ q q q ⎝ ⎠ ⎝ ⎠ ⎛ p⎞ = f ⎜ ⎟+ ⎝q⎠

⎛ p⎞ f ⎜ ⎟ + ........q times = q. f ⎝q⎠

⎛ p⎞ ⎝⎜ q ⎟⎠

p · f(1) = q · f(p/q)

⇒ f ( p / q) =

48. (a, b, c)

p . f (1) q

⎛ B −C ⎞ sin ⎜ ⎝ 2 ⎟⎠ A B −C =r = 4 R sin sin B C 2 2 sin sin 2 2 B +C B −C sin = |2R(sinB – sinC)| = |b – c| 2 2 f (x ) − f (0) 1 = lim x x →0 x →0 1 + e1/ x

51. (b, c) : f ′(0) = lim f ′(0 ) = lim

1

f ′(0− ) = lim

1

+

1/ x x →0+ 1 + e − 1 + e1/ x

x →0

=0 =1

∴ f ′(a+ ) = lim ( x − a)φ ′( x ) + φ( x ) = φ(a) x →a

f ′(a− ) = lim (a − x )φ ′( x ) − φ( x ) = − φ(a) x →a

1 54. (a, b, c) : If | x |< 1 and | x |≠ , n ⎡ 1 ⎤ 1 1 −1 < ⎢ ⎥ < then |x| ⎣| x |⎦ | x | ⎡ 1 ⎤ ⇒ 1− | x |<| x | ⎢ ⎥ < 1 ⇒ f ( x ) = 0 ⎣| x |⎦

Then f(x) = 0

⎛1⎞ ⎛1⎞ 49. (a, b) : Clearly φ′( x ) = 2 x ⋅ sin ⎜ ⎟ − cos ⎜ ⎟ ⎝x⎠ ⎝x⎠ if x z 0 and 0 if x = 0. ⎛1⎞ Ic(x) is discontinuous at x = 0, as cos ⎜ ⎟ is ⎝x⎠ oscillating in the neighbour hood of '0'. B C⎞ ⎛ 50. (a, c) : | BD − CD |= r ⎜⎝ cot 2 − cot 2 ⎟⎠

= 4 R cos

⎧(x − a)φ(x ) if x ≥ a 53. (a, b, d) : f (x ) = ⎨ ⎩(a − x )φ(x ) if x < a

If |x| > 1, then 0 <

f(p) = q · f(p/q) But f(p) = p · f(1) from previous cases. ?

52. (a, d) : f is continuous in its domain [–1, 1] x , x ≠ 0, x ≠ ±1 f ′( x ) = 2 2 2 1− 1− x 1− x

1 ⎛ 1 ⎞ < 1 and hence ⎜ = 0. |x| ⎝ | λ | ⎟⎠

Hence f(x) = 0 for all x R 55. (b, d) : f(x) = 2 – sin–1x if –1 d x d 0 2 + sin–1x if 0 < x d 1 Hence f is continuous everywhere on the domain 1 −1 if − 1 < x < 0, f ′( x ) = if 0 < x < 1 1 − x2 1 − x2 ? f is not differentiable at x = 0 ⎧ − cos πx , if − 1 < x < 0 ⎪ f ′ x = ( ) 1 , if x = 0 56. (a, c, d) : ⎨ ⎪ cos πx , if 0 < x < 1 ⎩ ? f is not continuous at x = 0 57. (d) : Degree of p(x) is 2 58. (a) : '

p′( x ) 2ax + b is discontinuous at x = 0 = x x

2ax is discontinuous. We know if f is continuous x

and 'g' is discontinuous then f + g is discontinuous. 59. (b) : p(1) = 3, p(–1) = 5 a = 4, b = –1, ? p(x) = 4x2 – x pc(x) = 8x – 1, ' gc is inverse of pc(x) x +1 1 g (x ) = ; g ′(0) = 8 8 MATHEMATICS TODAY | OCTOBER‘17

39

60. (c) : Let 0 d x < 1, then n ⎞ ⎛ f ( x ) = log(2 + x ) ⎜' lt x 2 = 0 ⎟ ⎠ ⎝ n→∞ 1 61. (b) : Let x = 1, f (x ) = (log 3 + sin 1) 2 62 (a) : If x > 1, then ⎤ ⎡ log(2 + x ) + sin x ⎥ ⎢ n n 2 ⎥⎦ log(2 + x ) + x 2 sin x = lt ⎢⎣ x lt 1 n n→∞ n→∞ +1 1 + x2 n x2 ⎧ log(2 + x ) , if 0 < x < 1 ⎪1 ⎪ ∴ f ( x ) = ⎨ (log 3 + sin 1) , if x = 1 ⎪2 sin x , if x > 1 ⎪⎩ ⎞ ⎛ 1 = sin x ⎜' lt = 0⎟ n n →∞ x 2 ⎠ ⎝ π 63. (d) : f (x ) = lim (sin2 x )m = 1, x = (2n + 1) , n ∈ I 2 m→∞ π = 0, x ≠ (2n + 1) , n ∈ I 2 π ? f(x) is discontinuous at x = (2n + 1) , n ∈ I 2 sin(x − 1) ⎞ 64. (c) : L.H.L. zR.H.L., as lim − 2 ⎜⎛ ⎟=− 2 − ⎝ x −1 ⎠ x →1 and lim

x →1+

⎛ sin(x − 1) ⎞ 2⎜ = 2 ⎝ x − 1 ⎟⎠ lim cot x (1+ x −1)

65. (b) : f ( x ) = lim (1 + x )cot x = e x →0 =

x tan x x → 0 e lim

x →0

=e

66. (A)o(p, q), (B)o(p, q), (C)o(r, s, p), (D)o(r, s, p) ⎧ π − 2 tan−1 x , x >1 ⎪⎪ (A) y = ⎨ 2 tan−1 x −1 ≤ x ≤ 1 ⎪ −1 x < −1 ⎪⎩ − π − 2 tan x −2 ⎧ , | x |> 1 ⎪ 2 ⎪ 1+ x dy ⎪ 2 ∴ =⎨ , | x |< 1 dx ⎪ 1 + x2 ⎪does not exist, at | x |= 1 ⎪ ⎩ Hence, f(x) is not differentiable at x = –1, 1 40

MATHEMATICS TODAY | OCTOBER ‘17

⎧ π + 2 tan−1 x , π < −1 ⎪⎪ −1 (B) y = ⎨ 2 tan x , −1 < x < 1 ⎪ −1 x >1 ⎪⎩− π + 2 tan x , ∴

2 ⎧ , x ∈ R − {−1,1} dy ⎪ =⎨ 1 + x2 dx ⎪ ⎩does not exist, at x = {−1,1}

Hence, f(x) is not differentiable at x = –1, 1

1 ⎧ −1 ⎪2 π + 3 cos x , −1 ≤ x ≤ − 2 ⎪ 1 1 ⎪ −1 (C) f ( x ) = ⎨2 π − 3 cos x , − ≤ x ≤ 2 2 ⎪ 1 ⎪ 3 cos−1 x , ≤ x ≤1 ⎪⎩ 2 −3 1 ⎧ , < | x |< 1 ⎪ 2 2 x − 1 ⎪ ⎪ dy 1 ∴ = f ′( x ) = ⎨does not exist, | x | = dx 2 ⎪ 3 1 ⎪ , |x|< ⎪ 2 2 x − 1 ⎩ −1 1 Hence, f(x) is not differentiable at x = , 2 2 1 ⎧ −1 ≤ x ≤1 ⎪ π − 3 sin x , 2 ⎪ 1 1 ⎪ − ≤x≤ (D) f (x ) = ⎨ 3 sin−1 x , 2 2 ⎪ ⎪− π − 3 sin−1 x , −1 ≤ x ≤ − 1 ⎪⎩ 2 1 −3 ⎧ , <| x |< 1 ⎪ 2 2 1− x ⎪ ⎪ 1 f ′(x ) = ⎨does not exist, | x |= 2 ⎪ 3 1 ⎪ , | x |< ⎪ 2 2 1− x ⎩ Hence, f(x) is not differentiable at x =

−1 1 , . 2 2

67. (A)o(p, s), (B)o(p, r), (C)o(p, s), (D)o(q, s) ⎧− x , x < 0 (A) f (x ) = ⎨ ⎩ x, x ≥ 0 continuous but not differentiable at x = 0

(B) f(x) = xn|x| L.H.D = R.H.D = 0 at x = 0 (C) L.H.L = R.H.L = f(0) = 0 but L.H.D and R.H.D are not finite. e1/ x x →0 1 / x

(D) L.H.L. = 0, R.H.L. = lim = lim

x →0

e1/ x (−1 / x2 ) 2

(−1 / x )

= lim e1/ x = ∞ x →0

68. (A) o (p), (B) o (q, r, s), (C) o (s), (D) o (q, r) (A) We know that [x] ∈ I , ∀ x ∈ R ∴ sin(π[x]) = sin πx = 0 ∀ x ∈ R Since every constant function is differentiable in its domain. ? sin(S[x]) is differentiable everywhere. (B) f(x) = sin[(x – [x])S] Since x – [x] is not differentiable at integral points ? f(x) = sin(S(x – [x])] is not differentiable at x I ? It is not differentiable at –1, 1 (C) lim f (x ) = 0 (a finite quantity between –1 and 1) x →0 = 0 = f(0) ? f(x) is continuous at x = 0 and f ( x ) − f (0) ⎛1⎞ lim = lim sin ⎜ ⎟ ⎝x⎠ x−0 x →0 x →0 Which does not exists. ? f(x) is not differentiable at x = 0. (D) |2 – x| is continuous everywhere and [2 + x] is discontinuous at all integral values of x. ? f(x) is discontinuous at x = 2. ? f(x) is not differentiable at x = 2. 1⎤ ⎡ 2⎤ ⎡ 3⎤ ⎡ 69. (4) : [ x ] + ⎢ x + ⎥ + ⎢ x + ⎥ + ⎢ x + ⎥ = [4 x ] 4 4 4 ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ? f(x) = [4x] which will become discontinuous at 1 2 3 4 x= , , , 4 4 4 4 70. (5) : Given f (x ) = x 2 + (α − 1)| x | −α f(x) = |(|x| – 1)(|x| + D)| ? f(x) is not differentiable at ‘5’ points. 71. (1) : In the vicinity of x = 0,we have ⎡1⎤ ⎢ ⎥ ⎣ |x| ⎦

⎛ ⎡ 1 ⎤⎞ x 2 ∑ r = x 2 ⎜1 + 2 + 3 + ... + ⎢ ⎥ ⎟ ⎝ ⎣| x |⎦ ⎠ r =0 Use sandwich theorem

⎛ ⎡ 1 ⎤⎞ x 2 ⎜1 + ⎢ ⎥ ⎟ ⎛ ⎡ 1 ⎤⎞ ⎝ ⎣| x |⎦ ⎠ ⎡ 1 ⎤ P = x 2 ⎜1 + 2 + 3 + ... + ⎢ ⎥ ⎟ = ⎢ ⎥ 2 ⎝ ⎣| x |⎦ ⎠ ⎣| x |⎦ 1 1 (1− | x |) < P ≤ (1+ | x |) 2 2 Then the limit is 1/2. So,

0 < x <1 ⎧ 0 ⎪ 1≤ x < 2 ⎪ sin πx ⎪ 2≤x< 3 72. (6) : f ( x ) = ⎨ 2 sin πx ⎪ 3≤x<2 ⎪ 3 sin πx ⎪ 4 sin πx 2 ≤ x < 5 etc. ⎩ The function is discontinuous at x = 2 , 3 , 5 ,.... K where K is not a perfect square. ? Points of discontinuity are 6. 73. (3) : Since f(x) is continuous at x = 0 ?

f (0) = lim

x2

x →0

= lim

(1 + x )log(1 + x ) − x

x →0

?

log (1 + x )1+ x − x

x

2

1 + log(1 + x ) − 1 1 = 2x 2 x →0

= lim

6f(0) = 3

⎧⎪ ax 3 + b , 0 ≤ x ≤1 74. (8) : f (x ) = ⎨ −1 ⎩⎪2 cos πx + tan x , 1 < x ≤ 2 ⎧ 3ax 2 ⎪ f ′(x ) = ⎨ 1 ⎪−2 π sin πx + ⎩ 1 + x2

, 0 < x <1 , 1< x ≤ 2

As the function is differentiable in [0, 2] function is differentiable at x = 1 1 1 ⇒ a= 2 6 Function will also be continuous at x = 1

?

f c(1–) = f c(1+) ⇒ 3a =

∴ lim f (x ) = lim f (x ) −

x →1

+

x →1

⇒ a + b = −2 +

π 4

1 π π 13 ∴ b = −2 − + = − ⇒ k1 = 4 & k2 = 12 6 4 4 6 k2 – k1 = 8 MATHEMATICS TODAY | OCTOBER‘17

41

*3(::?00

:LYPLZ

CBSE Differential Equations IMPORTANT FORMULAE z

z

z

z

z

An equation involving derivatives of the dependent variable with respect to independent variable (variables) is known as a differential equation. Order of a differential equation is the order of the highest order derivative occurring in the differential equation. Degree (when defined) of a differential equation is the highest power (positive integer only) of the highest order derivative in it.

y1 = y ′ =

dy d2 y d3 y , y2 = y ′′ = , y3 = y ′′′ = , .... dx dx 2 dx 3

dn y dx n

A differential equation is linear if it can be expressed as a0

dn y n

+ a1

d n −1 y n −1

+ .... + an −1

dy + an y = Pn dx

dx dx Where a0, a1, a2, ......., an and pn are constants or functions of independent variable x. z

42

Variable Separable : If the equation is of the form

z

dy dy = f (x ) ⋅ g ( y ), then ∫ = f (x )dx + C. dx g ( y) ∫ Reducible to Variable separable : If the equation dy i s of the for m = f (ax + by + c), the n put dx ax + by + c = z.

z

If the equation is

dy = f (x ), then y = ∫ f (x ) dx + C. dx

MATHEMATICS TODAY | OCTOBER ‘17

Homogeneous Equation : If the equation is of dy f (x , y ) = , where f(x, y), g(x, y) are dx g (x , y ) homogeneous functions of the same degree in x and y, then put y = vx. the form

The solution which contains as many arbitrary constants as the order of the differential equation is called a general solution and the solution free from arbitrary constants is called particular solution.

.., yn = z

z

z

Linear Equation : If the equation is of the dy + Py = Q, where P, Q are functions dx of x, then solution of differential equation is form

Pdx ∫ Pdx = ∫ Qe ye ∫ dx + C w h e r e e ∫ Pdx i s t h e integrating factor (I.F.).

Or dx + Px = Q, where P, dy Pdy Pdy = ∫ Qe ∫ dy + C Q are functions of y, then xe ∫

If the equation is of the form

Pdy is the integrating factor. where e ∫

WORK IT OUT VERY SHORT ANSWER TYPE

1. Find the differential equation corresponding to y = Aex + Be–x. dy = (ex + 1)y 2. Solve: dx dy + xy = ax. 3. Find the I.F of (1 – x2) dx 4. Find the order and degree of y = px + a2 p2 + b2 , dy where p = . dx 5. Find the differential equation of all non-vertical lines in a plane. SHORT ANSWER TYPE

B is a solution of x d2 y dy –y=0 the differential equation, x 2 +x 2 dx dx ⎛ π⎞ 7. Find the particular solution of yc = y cot2x, y ⎜ ⎟ = 2 ⎝ 4⎠ 8. The slope of the tangent at a point P(x, y) on a curve is (–x/y). If the curve passes through the point (3, – 4), find the equation of the curve.

LONG ANSWER TYPE - II

16. Form the differential equation of the family of circles in the first quadrant which touch the coordinates axes. 17. If the tangent at any point P of a curve meets the axis of X in T. Find the curve for which OP = PT, O being the origin. 18. Find the particular solution of (x + y)dy + (x – y)dx = 0, given that y = 1 when x = 1. dy 19. Solve: + y = cos x – sin x, y (0) = 1. dx 20. Experiments show that the rate of inversion of cane sugar in dilute solution is proportional to the concentration y(t) of the unaltered sugar. Suppose that the concentration is 1/100 at t = 0 and is 1/300 at t = 10 hours. Find y(t).

6. Show that the function y = Ax +

dy x + y cos x =− 1 + sin x dx 2 10. Show that y = x + 2x + 1 is the solution of the initial

9. Solve :

value problem

d3 y dx 3

= 0, y(0) = 1, yc(0) = 2, ycc(0) = 2.

LONG ANSWER TYPE - I

11. Form the differential equation corresponding to (x – a)2 + (y – b)2 = r2 by eliminating a and b. 12. 13.

14.

15.

⎛ −2 x y ⎞ dx = 1. Solve : ⎜ e − ⎟ ⎝ x x ⎠ dy Solve the following differential equations : dy = x2 + 4x – 9, x z –2 (i) (x + 2) dx dy (ii) = sin3x cos2x + xex dx Find the equation of the curve passing through origin if the slope of the tangant to the curve at any point (x, y) is equal to the square of the difference of the abscissa and ordinate of the point. dy Solve (x + y)2 = a2 dx

SOLUTIONS

1. We have, y =

Aex

+ Be–x

...(i) dy Differentiating (i) w.r.t. x, we get = Aex – Be–x dx Differentiating again w.r.t. x, we get d2 y

= Aex + Be–x = y is the required differential dx 2 equation. dy = (ex + 1)y dx

dy = (ex + 1)dx y dy On Integrating both sides, we get ∫ = ∫ (e x + 1)dx y log y = ex + x + c.

2. We have,

dy + xy = ax dx dy ax x + y= dx 1 − x 2 1 − x2

3. We have, (1 – x2)

∫

x

? I.F. = e 1− x =

2

−

dx

1 − log (1− x 2 ) e 2

= (1 – x2)–1/2 =

=e

1 −2 x dx ∫ 2 1− x 2 2 –1/2

= elog(1 – x ) 1 1 − x2

4. Given, y − px = a2 p2 + b2 (y – px)2 = a2p2 + b2 (x2 – a2)p2 – 2xyp + (y2 – b2) = 0 2 ⎛ dy ⎞ ⎛ dy ⎞ 2 2 (x – a ) ⎜ ⎟ – 2xy ⎜ ⎟ + (y2 – b2) = 0 ⎝ dx ⎠ ⎝ dx ⎠ Hence, order 1 and degree 2. MATHEMATICS TODAY | OCTOBER‘17

43

5. The equation of the family of non-vertical lines in plane is ax + by = 1, where b z 0 and a R. Differentiating w.r.t. x, we get dy a+b =0 dx On differentiating again w.r.t. x, we get b

d2 y

=0

dx 2 equation.

d2 y dx 2

= 0 is the required differential

Integrating both sides, we get y2 x2 = cc y2 + x2 = 2cc + 2 2 x2 + y2 = c, where c = 2cc. Since it passes through (3, –4) ? (3)2 + (4)2 = c c = 25. Hence, the required equation of the curve is x2 + y2 = 25. dy x + y cos x =− 1 + sin x dx dy x cos x + y=− dx 1 + sin x 1 + sin x

9. We have,

B x B dy Differentiating w.r.t. x, we get =A– dx x2 Again differentiating w.r.t. x, we get

6. The given function is : y = Ax +

d2 y

2B

...(i) ...(ii)

cos x

dx ∫ ? I.F. = e 1+ sin x = elog(1 + sin x) = (1 + sin x)

y(1 + sin x) =

d2 y

2B = –B(–2x–3) = ...(iii) ⇒x = 2 3 2 x x dx dx 2

B⎞ ⎛ B⎞ ⎛ dy – y = x ⎜ A − 2 ⎟ − ⎝⎜ Ax + ⎟⎠ ⎝ x dx x ⎠ [Using (i) and (ii)] dy 2B B B x – y = Ax – – Ax – = − ...(iv) x dx x x Adding (iii) and (iv), we get d y

dy 2B 2B –y= =0 − dx x x dx B Hence, y = Ax + is a solution of above differential x equation. dy 7. We have, yc = y cot 2x = y cot 2x dx dy = cot 2x dx y Integrating both sides, we get dy ∫ y = ∫ cot 2x dx 1 log |y| = log |sin 2x| + log c 2 log |y| = log sin 2x + log c x2

2

+x

log |y| = log c sin 2 x y = c sin 2 x ⎛ π⎞ Given y ⎜ ⎟ = 2 ⎝ 4⎠ 2 = c sin 8. Given : 44

π c = 2 ? y = 2 sin 2 x 2

x dy = − y dy + x dx = 0 y dx

MATHEMATICS TODAY | OCTOBER ‘17

∫ − x dx + c

[Using : y(I.F.) = ∫ Q(I.F.) dx + c ] y(1 + sin x) = −

Now, x

2

...(i)

x2 +c 2

2c − x 2 , which gives the required 2(1 + sin x )

y =

solution. 10. We have, y = x2 + 2x + 1

d2 y d3 y dy = 2x + 2, = 2 and = 0, dx dx 2 dx 3 ⎛ dy ⎞ Also, y(0) = 0 + 0 + 1 = 1, ⎜ ⎟ =2 ⎝ dx ⎠ x =0 2 ⎛ d y⎞ =2 and ⎜ ⎟ ⎝ dx 2 ⎠ x = 0

y(0) = 1, yc(0) = 2 and ycc(0) = 2. Hence, y = x2 + 2x + 1 is the solution of the initial value problem. 11. We have, (x – a)2 + (y – b)2 = r2 ...(i) Differentiating both sides w.r.t. to x, we have dy 2(x – a) + 2(y – b) =0 dx dy (x – a) + (y – b) =0 ...(ii) dx Again differentiating both sides w.r.t to x, we get 2

d2 y

⎛ dy ⎞ + ⎜ ⎟ =0 2 ⎝ dx ⎠ dx Form (ii) and (iii), we get 1 + (y – b)

y–b= −

⎛ dy ⎞ 1+ ⎜ ⎟ ⎝ dx ⎠ d2 y dx 2

2

...(iii)

⎡ ⎛ dy ⎞ 2 ⎤ dy ⎢1 + ⎜ ⎟ ⎥ ⎣ ⎝ dx ⎠ ⎦ dx and x – a = d2 y dx 2

Now, susbstituting these in the given equation, we obtain 2

2 2⎫ ⎧ ⎡ ⎛ dy ⎞ 2 ⎤ dy ⎫ ⎧ dy ⎛ ⎞ ⎪ ⎢1 + ⎜ ⎟ ⎥ ⎪ ⎪1 + ⎪ ⎪ ⎣ ⎝ dx ⎠ ⎦ dx ⎪ ⎪ ⎜⎝ dx ⎟⎠ ⎪ 2 ⎬ +⎨ ⎨ ⎬ =r 2 2 d y d y ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 2 2 dx ⎭ ⎩ dx ⎭ ⎩ 3 2 2 ⎞ ⎧⎪ ⎛ dy ⎞ 2 ⎪⎫ ⎛ d y ⎨1 + ⎜ ⎟ ⎬ = r 2 ⎜ ⎟ is the required ⎝ ⎠ ⎝ dx 2 ⎠ dx ⎭ ⎩ differential equation. ⎛ e −2 x y ⎞ dx − 12. We have, ⎜ =1 ⎟ ⎝ x x ⎠ dy

dy e −2 x y = − dx x x

dy y e −2 x + = ...(i) dx x x This is a linear differential equation of the form

−2 x

1 dy e and Q = . + Py = Q where P = dx x x P dx =e ? I.F. = e ∫

∫

1 dx x

e ye 2 x = ∫ e 2 x ⋅ ye 2 x =

∫

1 x

−2 x

x

2 x =e

dx + c

dx + c ye 2

x

= 2 x +c

y = (2 x + c)e −2 x , which gives the required solution. dy 13. (i) We have, (x + 2) = x2 + 4x – 9 dx dy x2 + 4x − 9 = [∵ x z –2] x+2 dx ⎛ x2 + 4x − 9 ⎞ dy = ⎜ ⎟ dx ⎝ x+2 ⎠ Integrating both sides, we get 13 ⎞ ⎛ x2 + 4x − 9 dx ³dy = ∫ ⎜⎝ x + 2 − ⎟ dx ³dy = ∫ x + 2⎠ x+2 2 y = x + 2 x − 13 log | x + 2 | + c 2 dy 3 (ii) We have, = sin x cos2 x + xe x dx dy = (sin3 x cos2 x + x ex)dx Integrating both sides, we get ³dy = ³(sin3 x cos2 x + x ex) dx

³dy = ³ sin3 x cos2 x dx + ³x ex dx ³dy = ³cos2 x (1 – cos2 x) sin x dx + ³xexdx y = – ³t2 (1– t2) dt + {x ex –³ ex dx}, where t = cos x 3 t 5 ⎪⎫ ⎪⎧ t y = − ⎨ − ⎬ + (xe x − e x ) + c 5 ⎭⎪ ⎩⎪ 3 1 1 y = − cos3 x + cos5 x + x e x − e x + c, ∀x ∈R 3 5 14. The slope of the tangent at any point (x, y) on the dy . curve y = f(x) is given by dx dy Also, ...(i) = (x − y)2 (Given) dx dy dv ? Let x – y = v 1− = . Substituting these dx dx values in (i), we get 1 dv dv 1− = v2 ⇒ = 1 − v2 ⇒ dv = dx dx dx 1 − v2 Integrating, we obtain 1 1+ v 1+ v = x + c or log log = 2x + 2c 2 1− v 1− v 1+ x − y = 2x + 2c or, log ...(ii) 1− x + y It is given that the curve passes through the origin. Therefore, y = 0 when x = 0. Putting x = 0, y = 0 in (ii), we obtain c = 0. Putting c = 0 in (ii), we obtain 1+ x − y 1+ x − y log = e 2x = 2x or log 1− x + y 1− x + y or, (1 + x – y) = ±(1 – x + y) e2x, which is the equation of the curve. 15. Let x + y = v. Then, dy dv dy dv 1+ = ⇒ = −1 dx dx dx dx On substituting in given equation, we get dv ⎛ dv ⎞ = a2 + v 2 v 2 ⎜ − 1⎟ = a 2 v 2 ⎝ dx ⎠ dx v2 dv = (a2 + v2)dx v2 2 2 dv = dx [By separating the variables] v +a ⎛ a2 ⎞ ⎜1 − 2 2 ⎟ dv = dx ⎝ v +a ⎠ 1 2 ∫ 1 ⋅ dv − a ∫ 2 2 dv = ∫ dx + c v +a [On integration] MATHEMATICS TODAY | OCTOBER‘17

45

−1 ⎛ v ⎞ v − a tan ⎜ ⎟ = x + c ⎝a⎠ ⎛x+ y⎞ = x + c, is the Hence, (x + y) – a tan–1 ⎜ ⎝ a ⎟⎠ required solution. 16. Family of circles which touch the coordinates axes in first quadrant is: (x – a)2 + (y – a)2 = a2 …(i) where a is the radius of the circles. Differentiating (i) w.r.t. x, dy 2(x − a) + 2( y − a) = 0 dx dy Putting = p, we get dx (x – a) + (y – a)p = 0 or x – a + py – pa = 0 x + py – a(1 + p) = 0 x + py ∴ a= 1+ p Putting value of a in (i), we get 2

2

⎛ x + py ⎞ ⎛ x + py ⎞ ⎛ x + py ⎞ ⎜x − = +⎜y− ⎟ 1+ p ⎠ ⎝ 1 + p ⎟⎠ ⎜⎝ 1 + p ⎟⎠ ⎝ 2

⎡⎢ x(1 + p) − x − py ⎥⎤ + ⎡⎢ y(1 + p) − x − py ⎥⎤ 1+ p 1+ p ⎣ ⎦ ⎣ ⎦

2

2

2

2

2

⎡ (x − y) p ⎤ ⎡ ( y − x) ⎤ ⎡ x + py ⎤ or, ⎢ ⎥ +⎢ ⎥ =⎢ ⎥ ⎣ 1+ p ⎦ ⎣ 1+ p ⎦ ⎣ 1+ p ⎦ Multiplying both sides by (1 + p)2, we get (x – y)2 p2 + (x – y)2 = (x + py)2 or, (x – y)2(1 + p 2) = (x + py)2 2 ⎡ ⎛ dy ⎞ 2 ⎤ ⎡ dy ⎤ 2 = + ( x − y ) 1 + x y ⎢ ⎥ ⎜ ⎟ ⎢ dx ⎥⎦ ⎢⎣ ⎝ dx ⎠ ⎥⎦ ⎣ This is the required differential equation. 17. Let y = f(x) be given curve and let P (x, y) be a point on it. The equation of the tangent at P is dy Y − y = ( X − x) ...(i) dx This meets X -axis at T. So, the x-coordinate of T is obtained by putting Y = 0 in (i). Putting Y = 0 in (i), we get dy 0 − y = ( X − x) dx 48

MATHEMATICS TODAY | OCTOBER ‘17

2

⎧ ⎛ dx ⎞ ⎫ ? PT = ⎨ x − ⎜ x − y ⎟ ⎬ + ( y − 0)2 dy ⎠ ⎭ ⎩ ⎝ 2 ⎛ dx ⎞

2

= y ⎜ ⎟ + y 2 [Given] ⎝ dy ⎠ Now, PT = OP (Given)

2

2 ⎛ dx ⎞

2

y + y ⎜ ⎟ = x2 + y2 ⎝ dy ⎠ 2

2

2 ⎛ dx ⎞ 2 y +y ⎜ ⎟ =x +y y ⎜ ⎟ =x dy ⎠ ⎝ ⎝ dy ⎠ dx dy dx =± y =± x x y dy 2

2 ⎛ dx ⎞

2

2

log x = ± log y + log c

2

⎛ x + py ⎞ =⎜ ⎝ 1 + p ⎟⎠

dx dy ⎛ dx ⎞ Thus, the coordinates of T are ⎜ x − y ,0 ⎟ dy ⎠ ⎝ ⇒X=x−y

⎛c⎞ log x = log (c y) or, log x = log ⎜ ⎟ ⎝y⎠ c x = c y or, x = , are the equations of curve. y 18. We have, (x + y)dy + (x – y)dx = 0 (x + y)dy = –(x – y)dx dy y − x ⇒ = ...(i) dx x + y dy dv =v+x Put y = vx ⇒ ...(ii) dx dx From (i) and (ii), we get dv (v − 1)x dv v − 1 v+x ⇒x = −v = dx (v + 1)x dx v + 1 dx dv v − 1 − v 2 − v v +1 dv = − ⇒x = ⇒ 2 x dx v +1 1+ v v dx 1 dv +∫ dv = −∫ ⇒∫ 2 2 x 1+ v 1+ v 1 2v dv = − log | x | + c ⇒ tan −1 v + ∫ 2 1 + v2 1 ⇒ tan −1 v + log | 1 + v 2 |= − log | x | + c 2 y 1 y2 ...(iii) ⇒ tan −1 + log 1 + 2 = − log | x | + c x 2 x Given y = 1 when x = 1. Putting in equation (iii), we get

MATHEMATICS TODAY | OCTOBER‘17

49

1 1 tan −1 1 + log 1 + = − log | 1 | + c 2 1 π 1 c = + log 2 4 2 π Putting c = + log 2 in equation (iii), we get 4 ⎡ y2 ⎤ π −1 y 1 tan + log ⎢1 + 2 ⎥ + log | x |= + log 2 x 2 4 ⎢⎣ x ⎥⎦ 19. The given differential equation is dy ...(i) + y = cos x – sin x dx This is a linear differential equation of the form dy + Py = Q, where P = 1 and Q = cos x – sin x. dx Pdx 1dx = e∫ = ex ? I.F. = e ∫ Solution of linear differential y ⋅ (I.F.) = ∫ Q(I.F)dx + c

equation

is

dy + ye x = e x (cos x − sin x) dx ye x = ∫ e x (cos x − sin x) dx + c

ex

x x x ye = ∫ e cos xdx − ∫ e sin x dx + c

ye x = e x cos x − ∫ − sin x e x dx − ∫ e x sin x + c ye x = e x cos x + ∫ e x sin x dx − ∫ e x sin x dx + c

yex = ex cos x + c, Now, y(0) = 1 y = cos 0 + c c = 0 ? yex = ex cos x is the required equation.

dy dy ∝y⇒ = ky dt dt where k is constant of proportionality. dy ⇒ = k dt y Integrating both sides, we get dy ...(i) ∫ y = k ∫ dt ⇒ log y = kt + c 1 Initially, when t = 0, y = 100 1 = 0 + c c = log(100)–1 = –log 100 ∴ log 100 ? From (i), we have, log y = kt – log 100 ...(ii) 1 y = Now it is given when t = 10 hours, 300 1 ∴ log = 10 k – log 100 300 –log 300 = 10 k – log 100 10 k = log 100 – log 300 100 1 log 3 10k = log = log = − log 3 ? k = − 300 3 10 Putting these values in (ii), we get log 3 log 3 log y = − t − log 100 ⇒ log 100 + log y = − t 10 10 log 3 − t log 3 ⇒ log 100 y = − t ⇒ 100 y = e 10 10

20. Given:

log 3

log 3

− t 1 − 10 t ⇒y= e ⇒ y = 0.01e 10 100 Hence, concentration at time t is given by y = 0.01 e–[log 3/10]t

50

MATHEMATICS TODAY | OCTOBER ‘17

Class XII

T

his specially designed column enables students to self analyse their extent of understanding of speciﬁed chapters. Give yourself four marks for correct answer and deduct one mark for wrong answer. Self check table given at the end will help you to check your readiness.

,QGHÀQLWH,QWHJUDWLRQ Total Marks : 80

Time Taken : 60 Min.

Only One Option Correct Type

1 1 (a) A = , B = 8 4 2 1 1 (c) A = − , B = 8 4 2

1. If I = ∫ (7x + 4) 5 − 4x − 2x 2 dx, then I equals (a)

21 2 −1 ⎛ 2(x + 1) ⎞ 7 x 5 − 4 x − 2x 2 + sin ⎜ ⎟+c 4 2 7 ⎠ ⎝

⎛ 2(x + 1) ⎞ 2 21 2 sin −1 ⎜ (b) (7x + 3) 5 − 4x − 2x + ⎟+c 4 7 ⎠ ⎝

21 2 1 (14x 2 − 3x − 4) 5 − 4x − 2x 2 + 6 4 −1 sin 2(x + 1) + c (d) None of these (c)

(

2. If I = ∫ (x + 1)

)

x +2 dx, then I equals x −2

1 (x + 6) x 2 − 4 + 2 log x + x 2 − 4 + c 2 1 2 2 (b) (x + 6) x − 4 + 4 log x + x − 4 + c 2 1 (c) (x + 6) x 2 − 4 + 6 log x + x 2 − 4 + c 2 (d) None of these

5. Let x2 + 1 ƾ nS, n N, then

∫x

(2x12 + 5x 9 )

∫ (x 5 + x 3 + 1)3 (a)

where u = y + 1/y and z = y – 1/y, y = x 4 / 2 then A is

(x 5 + x 3 + 1)2

+c

(c) ln |x5 + x3 + 1| + (d) None of these 4. If ∫

(a) 128 (b) 32 (c) 64 (d) 512 One or More Than One Option(s) Correct Type

dx is equal to

x 2 + 2x

(b)

x10 2(x 5 + x 3 + 1)2

2 sin(x 2 + 1) − sin{2(x 2 + 1)}

dx is 2 sin(x 2 + 1) + sin{2(x 2 + 1)} 1 ⎧1 2 ⎫ 2 (a) ln sec(x + 1) + c (b) ln sec ⎨ (x + 1)⎬ + c 2 ⎩2 ⎭ 1 2 (c) ln |sec(x 2 + 1)| +c (d) ln|sec (x + 1)| + c 2 3 1 x A z u− 2 − log + c, tan −1 dx = 6. If ∫ 16 4096 2 64 u+ 2 4+x

(a)

3.

1 1 (b) A = − , B = − 8 4 2 1 1 (d) A = , B = − 8 4 2

+c

7. A primitive of sin 6x is 1 (a) (sin6 x − sin3 x) + c 3 1 1 (b) − cos2 3x + c (c) sin2 3x + c 3 3 1 ⎛ π⎞ ⎛ π⎞ (d) sin ⎜ 3x + ⎟ sin ⎜ 3x − ⎟ + c 3 ⎝ 7⎠ ⎝ 7⎠

(2x 7 + 5x 4 ) + c

sin x 1 + sin x dx = A log + B log sin 4x 1 + sin x 1 + 2 sin x + c then 1 − 2 sin x

8.

x8 + 4

∫ x 4 − 2x 2 + 2 dx =

5 3 x5 x3 + + 2x 2 + 2x + c (a) x + 2x + 2x + c (b) 5 3 5 3 5 3 3 (c) 3x + 10x + 45x + c (d) 4x + 5x2 + 2x + c MATHEMATICS TODAY | OCTOBER‘17

51

9. If ∫ tan4 x dx = K tan3 x + L tan x + f(x), then (a) K = 1/3 (b) L = – 1 (c) f(x) = x + C (d) K = 2/3 1 10. If ∫ f (x)sin x cos x dx = 2 2 log (f(x)) + c, 2(b − a ) f x dx ( ) then ∫ equals

1 ⎛ a tan x ⎞ ⎛ a tan x ⎞ tan −1 ⎜ (b) ab tan −1 ⎜ ⎟ ⎝ b ⎠ ⎝ b ⎟⎠ ab 1 ⎛ bx ⎞ ⎞ 1 ⎛ b tan x ⎞ −1 ⎛ (c) tan −1 ⎜ ⎟⎠ (d) ab tan ⎜⎝ tan ⎜⎝ a ⎟⎠ ⎟⎠ ⎝ ab a 11. If the primitive of sin(ln x) is f(x){sin g(x) – cos h(x)} + c (c being the constant of integration), then g (x) f (x) = 1 (a) xlim (b) lim =1 →2 x →1 h(x) (c) g(e3) = 3 (d) h(e5) = 5 dx = 12. ∫ x e 2e x − 1 1 −1 +c (a) 2 sec −1 2e x + c (b) −4 tan 2e x + 1 (a)

(c) 2 sec −1( 2e x ) + c

x m+1(a + bx n ) p (m + 1) –f(m, n, p, b) Im+n, p–1, then the value of f(1, 2, 3, 4) is equal to (a) 8 (b) 10 (c) 12 (d) None of these Matrix Match Type 16. Match the following :

15. If Im, p = ∫xm. (a + bxn)pdx and Im, p =

Column I

m−2

x cosn+2 x dx and

sinm−1 x cosn+1 + f (m, n)Im−2, n+2 then (n + 1) f(2, 3) is equal to Im,n = −

(a)

1 2

(b)

1 3

(c)

1 4

(d)

1 5

e (2 − x )

∫

Q.

∫ (x + 2)2 dx

(1 − x) 1 − x 2 e x (x + 1)

dx 1.

R.

∫ x(1 + xe x )2 dx

ex +c x +2

2. e x

x +1

3.

1+ x +c 1− x

(x – 2)u – log

u −1 u +1 e x +1

+ c, u = x 4. log xe

1 + xe

(a) (b) (c) (d)

13. If

14. If Im–2, n+2 = ∫ sin

Column II

2

P.

(d) 2 tan −1 2e x − 1 + c

2 x4 +1 –1 –1 ∫ x 6 + 1dx = tan f(x) – 3 tan g(x) + c, then (a) both f(x) and g(x) are odd functions (b) f(x) is monotonic function (c) f(x) = g(x) has no real roots f (x) 1 3 dx = − + 3 + c (d) ∫ g (x) x x Comprehension Type Repeated application of integration by parts gives us the reduction formula if the integrand is dependent of n, n N.

x

P 2 1 2 2

x

+

1 1 + xe x

+c

Q R 3 4 2 3 1 4 1 3 Integer Answer Type

17. Let f(x) = ∫xsin x (1 + x cos x. ln x + sin x) dx and 2 ⎛π⎞ π f ⎜ ⎟ = . Then the value of |cos f(S)| is ⎝2⎠ 4 xdx λ ⋅ (7x − 20) 18. If ∫ = + c, (7x − 10 − x 2 )3 9 7x − 10 − x 2 then the value of O is 19. If ∫x2 · e–2x dx = e–2x (ax2 + bx + c) + d, then the value of |a/bc| is x x ⎡⎛ x ⎞x ⎛ e ⎞x ⎤ ⎛x⎞ ⎛e⎞ 20. If ∫ ⎢ ⎜⎝ ⎟⎠ + ⎜⎝ ⎟⎠ ⎥ ln x dx = A ⎜⎝ ⎟⎠ + B ⎜⎝ ⎟⎠ + C, e x x ⎥⎦ ⎢⎣ e then the value of A + B is

Keys are published in this issue. Search now! -

Check your score! If your score is No. of questions attempted …… No. of questions correct …… Marks scored in percentage ……

52

> 90%

EXCELLENT WORK ! You are well prepared to take the challenge of ﬁnal exam.

90-75%

GOOD WORK !

You can score good in the ﬁnal exam.

74-60%

SATISFACTORY !

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< 60%

MATHEMATICS TODAY | OCTOBER ‘17

NOT SATISFACTORY! Revise thoroughly and strengthen your concepts.

M

aths Musing was started in January 2003 issue of Mathematics Today. The aim of Maths Musing is to augment the chances of bright students seeking admission into IITs with additional study material. During the last 10 years there have been several changes in JEE pattern. To suit these changes Maths Musing also adopted the new SDWWHUQ E\ FKDQJLQJ WKH VW\OH RI SUREOHPV 6RPH RI WKH 0DWKV 0XVLQJ SUREOHPV KDYH EHHQ DGDSWHG LQ -(( EHQH¿WWLQJ WKRXVDQG RI RXU readers. It is heartening that we receive solutions of Maths Musing problems from all over India. Maths Musing has been receiving tremendous response from candidates preparing for JEE and teachers coaching them. We do hope that students will continue to use Maths Musing to boost up their ranks in JEE Main and Advanced.

JEE MAIN

1. The equation (cos p – 1)x2 + (cos p)x + sin p = 0 in the variable x has real roots. Then p can take any value in the interval (a) (0, 2S) (b) (–S, 0) ⎛ π π⎞ (c) ⎜ − , ⎟ (d) (0, S) ⎝ 2 2⎠ 2. The bisectors of 'ABC meets its circumcircle at the Area of ΔDEF points D, E, F respectively, then = Area of ΔABC 2r R r R (a) (b) (c) (d) R 2r r 2R π 1 2π 1 3π 3. sin + sin + sin + ........∞ = 3 2 3 3 3 (a) 0 (b) S/2 (c) S/4 (d) S/3 4. The curve described parametrically by x = t2 + t + 1, and y = t2 – t + 1 represents (a) a pair of straight lines (b) an ellipse (c) a parabola (d) a hyperbola 5. The number of integers greater than 6,000 that can be formed, using the digits 3, 5, 6, 7 and 8, without repetition, is (a) 216 (b) 192 (c) 58 (d) 46

COMPREHENSION The graph of a certain function ‘f ’ contains the point (0, 2) and has the property that for each number t the tangent line to y = f(x) at (t, f(t)) intersects the x-axis at t + 2. 7. The value of lim f (x) is x →∞

(a) 1

x2

Let F(x) =

∫

f ( t )dt for x [0, 2]. If F c(x) = f c(x)

0

for all x (0, 2) then F(2) equals (a) e2 – 1 (b) e4 – 1 (c) e – 1 (d) e4

(d) f

(c) 0

8. Equation of normal at (0, 2) to the curve is (a) x – 2y + 4 = 0 (b) x + 3y – 6 = 0 (c) x – y + 2 = 0 (d) None of these INTEGER TYPE 9. A curve passing through the point (1, 1) has the property that the perpendicular distance of the origin from normal at any point ‘P’ of the curve is equal to the distance of P from the x-axis is a circle with radius MATRIX MATCH 10. Match the following ([.] denotes the greatest integer function) List-I List-II

(cos x − 1) (cos x − e x )

1.

does not exist

Q. lim =

2.

1/2

R.

3.

1/6

4.

3

P.

lim

x →0

JEE ADVANCED

6. Let f : [0, 2] o R be a function which is continuous on [0, 2] and is differentiable on (0, 2) with f(0) = 1.

(b) –1

S.

(a) (b) (c) (d)

x

n

1+ x − 3 1+ x = x →0 x 4 lim (x − [x]) = x→

3 2

⎛ e1/x − 1 ⎞ 4 lim[x] ⎜ 1/x ⎟ x →0 ⎝ e +1⎠ P 4 2 4 1

Q 3 1 3 4

R 2 4 1 2

=

S 1 3 2 3

See Solution Set of Maths Musing 177 on page no 84

MATHEMATICS TODAY | OCTOBER‘17

53

54

MATHEMATICS TODAY | OCTOBER ‘17

Trigonometric Identities Trigonometric Identities 1. If A + B + C = S, then z sin2A + sin2B + sin2C = 4sinA sinB sinC z cos2A + cos2B + cos2C = –1 – 4cosA cosB cosC z tan2A + tan2B + tan2C = tan2A tan2B tan2C z cos2A + cos2B + cos2C = 1 – 2cosA cosB cosC z sin2A + sin2B + sin2C = 2 + 2cosA cosB cosC z

sinA + sinB + sinC = 4 cos

A B C cos cos 2 2 2

2. If A + B + C = S/2, then z

sin2A + sin2B + sin2C = 1 – 2sinA sinB sinC

z

cos2A + cos2B + cos2C = 2 + 2sinA sinB sinC

Values of Trigonometric Ratios of Some Angles 1. sin15° = sin(45° – 30°) = 2. cos15° = cos(45° – 30°) =

3 −1 2 2

= cos 75°

3 +1 2 2

= sin 75°

3. tan15° = tan(45° – 30°) = 2 − 3 = cot 75° 4. cot15° = cot(45° – 30°) = 2 + 3 = tan 75° 5. 6.

1 1 ⎛ 45 ⎞ ° tan 22 ° = tan ⎜ ⎟ = 2 − 1 = cot 67 ° ⎠ ⎝ 2 2 2 ° 1 1 ⎛ 45 ⎞ cot 22 ° = cot ⎜ ⎟ = 2 + 1 = tan 67 ° ⎠ ⎝ 2 2 2

7. sin18° =

5 −1 = cos 72° 4

8. cos18° =

10 + 2 5 = sin 72° 4

9.

tan18° =

25 − 10 5 5

10. cos36° = sin54° =

5 +1 4

11. sin36° = cos54° =

10 − 2 5 4

12. tan 36 ° =

sin 36° = 5−2 5 cos 36°

5 4 1 14. cos 36° cos 72° = sin 54° sin 18° = 4

13. sin 72° cos 54° = sin 36° cos 18° =

15. cos 36° − cos 72° = sin 54° − sin 18° =

1 2

Important Series 1. tan3x tan2x tanx = tan3x – tan2x – tanx 2. tan2x tan(x + D) tan(x – D) = tan2x – tan(x + D) – tan(x – D) 3. 4sinx sin(60° – x) sin(60° + x) = sin3x 4. 4cosx cos(60° – x) cos(60° + x) = cos3x 5. tanx tan(60° – x) tan(60° + x) = tan3x 1 6. cotx cot(60° + x) cot(120° + x) = − tan 3x 7. cosD cos2D cos22D .... cos2n – 1D ⎧ sin 2n α ⎪ n ⎪ 2 sin α ⎨ =⎪ 1 ⎪⎩ −1

if if if

⎫ ⎪ ⎪ ⎬ α = 2k π ⎪ α = (2k + 1)π⎪⎭ α ≠ nπ

By : R. K. Tyagi, Retd. Principal, HOD Maths, Samarth Shiksha Samiti, New Delhi

MATHEMATICS TODAY | OCTOBER‘17

55

8. cosD cos2D cos22D ..... cos2n – 1D ⎫ ⎧ sin 2n α α ≠ nπ , if ⎪ ⎪ n ⎪ 2 sin α ⎪ ⎪ 1 ⎪ π if α = n , n ∈ I ⎬ , =⎨ n 2 +1 ⎪ 2 ⎪ ⎪ −1 ⎪ π α if , , = n I ∈ ⎪ ⎪ ⎩ 2n ⎭ 2n − 1

Summation of Series 1.

n−1

∑

r =0

sin(α + rβ) = sin α + (sin α + β) + sin(α + 2β) +

.... + sin(D + (n – 1)E) ⎛ nβ ⎞ ⎛ ⎛ n −1⎞ ⎞ β sin ⎜ ⎟ sin ⎜ α + ⎝ 2 ⎠ ⎟⎠ ⎝ 2 ⎠ ⎝ = ... (*) sin(β / 2) If E = D then (*) reduces to sinD + sin2D + sin3D + ... + sin(nD) ⎛ n +1⎞ ⎛ nα ⎞ sin α sin ⎝ 2 ⎠ ⎝ 2 ⎠ = sin(α / 2) 2.

n−1

∑ cos(α + rβ) = cos α + cos(α + β) + cos(α + 2β) +

r =0

.... + cos(D + (n – 1)E) n 1 ( − ) β ⎛ ⎞ ⎛ nβ ⎞ = cos ⎜ α + ⎟ sin ⎝ ⎠ ⎝ 2 ⎠ 2 ... (**) sin(β / 2) If E =D then (**) reduces cos D + cos 2D + cos 3D + ... + cos(nD) ⎛ n +1⎞ ⎛ nα ⎞ cos α ⋅ sin ⎝ 2 ⎠ ⎝ 2 ⎠ = sin(α / 2)

Properties of Inverse Trigonometric Functions 1. • sin–1(–x) = –sin–1(x) x ∈ [–1, 1] • tan–1(–x) = –tan–1(x) x ∈ R • cosec–1(–x) = –cosec–1(x) x ∈ R – (–1, 1) • cos–1(–x) = S – cos–1(x) x ∈ [–1, 1] • cot–1(–x) = S – cot–1(x) ∀ x ∈(− ∞, ∞) • sec–1(–x) = S – sec–1(x) x ∈ R – (–1, 1) 1 2. • sin −1 ⎜⎛ ⎟⎞ = cosec −1x ∀ x ∈ R − (−1, 1) ⎝x⎠ 1 • cos −1 ⎜⎛ ⎟⎞ = sec −1x ∀ x ∈ R − (−1, 1) ⎝x⎠ −1 ⎛ 1 ⎞ ⎪⎧ cot x, x > 0 • tan −1 ⎜ ⎟ = ⎨ ⎝ x ⎠ −π + cot −1 x, x < 0 ⎪⎩

56

MATHEMATICS TODAY | OCTOBER ‘17

π ∀ x ∈[−1, 1] 2 π • sec −1 x + cosec −1x = ∀ x ∈ R − (−1, 1) 2 π • tan −1 x + cot −1x = ∀ x ∈(−∞, ∞) 2 –1 –1 4. tan x + tan y x+y ⎧ tan −1 if xy < 1 x > 0, y > 0 ⎪ 1 − xy ⎪ ⎛ x+y ⎞ ⎪ if x > 0, y > 0, xy > 1 = ⎨ π + tan −1 ⎜ ⎝ 1 − xy ⎟⎠ ⎪ ⎪ −1 ⎛ x + y ⎞ ⎪−π + tan ⎜⎝ 1 − xy ⎟⎠ if x < 0, y < 0, xy > 1 ⎩

3. • sin −1 x + cos −1 x =

5. tan–1x – tan–1y x−y ⎧ if tan −1 ⎪ 1 + xy ⎪ ⎛ x−y ⎞ ⎪ if = ⎨−π + tan −1 ⎜ ⎝ 1 + xy ⎟⎠ ⎪ ⎪ −1 ⎛ x − y ⎞ ⎪ π + tan ⎜⎝ 1 + xy ⎟⎠ if ⎩

xy > −1 x < 0, y > 0, xy < −1 x > 0, y < 0, xy < −1

6. sin–1x ± sin–1y ⎧ sin −1 ⎡ x 1 − y 2 ± y 1 − x 2 ⎤ ∀ x ≥ 0, y ≥ 0, x 2 + y 2 ≤ 1 ⎪ ⎣ ⎦ =⎨ − 1 2 2 2 2 ⎩⎪ π − sin x 1 − y ± y 1 − x ∀ x ≥ 0, y ≥ 0, x + y > 1

7. • cos–1x ± cos–1y ⎧ cos −1[xy B 1 − x 2 1 − y 2 ] ∀ x, y ≥ 0, x 2 + y 2 ≤ 1 ⎪ =⎨ ⎪⎩ π − cos −1[xy B 1 − x 2 1 − y 2 ] ∀ x, y ≥ 0, x 2 + y 2 > 1 • cos–1x – cos–1y −1 2 2 ⎧ ⎪ cos (xy + 1 − x 1 − y ) ∀ x, y ≥ 0, y > x =⎨ ⎪⎩− cos −1(xy + 1 − x 2 1 − y 2 ] ∀ x, y ≥ 0, y < x

8.

⎧ −1 ⎛ 2x ⎞ ⎟ , if | x | ≤ 1 ⎪sin ⎜⎝ 1 + x2 ⎠ ⎪ ⎪tan −1 ⎛ 2x ⎞ , if | x | < 1 ⎜⎝ ⎟ ⎪ 1 − x2 ⎠ ⎪ ⎪ −1 ⎛ 2x ⎞ ,∀ x >1 2 tan −1 x = ⎨ π + tan ⎜ ⎝ 1 − x 2 ⎟⎠ ⎪ ⎪ −1 ⎛ 1 − x 2 ⎞ ∀ x ∈[0, 1] ⎪cos ⎜ ⎝ 1 + x 2 ⎟⎠ ⎪ ⎪ −1 ⎛ 2x ⎞ ⎟ if x ∈(−∞, − 1) ⎪−π + tan ⎜⎝ 1 − x2 ⎠ ⎩

12. • tan–1x + tan–1y + tan–1z

−1 2 9. • sin (2x 1 − x )

⎧ −1 if ⎪ 2 sin x, ⎪ ⎪ = ⎨ π − 2 sin −1 x, if ⎪ ⎪ −1 ⎪⎩−π − 2 sin x, if •

cos–1(2x2

|x |≤ 1 2

⎛ x + y + z − xyz ⎞ = tan −1 ⎜ , x, y, z ≥ 0 ⎝ 1 − xy − yz − zx ⎠⎟

1

• tan–1x1 + tan–1x2 + tan–1x3 + ... + tan–1xn ⎡ S − S + S − S + .... ⎤ = tan −1 ⎢ 1 3 5 7 ⎥ ⎣ 1 − S2 + S4 − S6 + .... ⎦

2

≤ x ≤1

−1 ≤ x ≤ −

1

n

where S1 = ∑ xi , (i = 1,...., n) ∈ R

2

i =1

−1 if x ∈[0, 1] ⎪⎧ 2 cos x, – 1) = ⎨ −1 ⎪⎩2π − 2 cos x, if x ∈[−1, 0]

⎧ 2 tan −1 x, | x | ≤ 1 ⎪ ⎛ 2x ⎞ ⎪ = ⎨ π − 2 tan −1 x, x > 1 • sin −1 ⎜ ⎝ 1 + x 2 ⎟⎠ ⎪ −1 ⎪⎩−π − 2 tan x, x < −1 ⎛ 1 − x 2 ⎞ ⎪⎧ 2 tan −1 x, ∀ x ∈[0, ∞) • cos −1 ⎜ =⎨ ⎝ 1 + x 2 ⎟⎠ ⎪−2 tan −1 x, ∀ x ∈(−∞, 0] ⎩ ⎧ 2 tan −1 x, ∀ | x | ≤ 1 2x ⎞ ⎪⎪ • tan −1 ⎛⎜ = ⎨ π − 2 tan −1 x, ∀ x ∈(−∞, − 1) ⎝ 1 − x 2 ⎟⎠ ⎪ −1 ⎪⎩ −π + 2 tan x, ∀ x ∈(1, ∞) ⎧ ⎡ π⎞ −1 ⎛ 1 ⎞ −1 ⎪⎪ tan ⎜⎝ x ⎠⎟ , if x > 0 i.e. cot x ∈ ⎢0, 2 ⎟⎠ ⎣ −1 10. cot x = ⎨ 1 ⎪ π + tan −1 ⎜⎛ ⎞⎟ , if x < 0 i.e. cot −1 x ∈ ⎛⎜ π , π ⎞⎟ ⎝2 ⎠ ⎝x⎠ ⎪⎩

1 ⎧ −1 ⎪ 3 sin x, ∀ | x | ≤ 2 ⎪ ⎛1 ⎤ ⎪ 11. • sin −1(3x − 4x 3) = ⎨ π − 3 sin −1 x, ∀ x ∈ ⎜ , 1⎥ ⎝2 ⎦ ⎪ 1⎞ ⎡ ⎪ −1 ⎪⎩−π − 3 sin x, ∀ x ∈ ⎢⎣ −1, − 2 ⎟⎠ 1⎤ ⎧ ⎡ −1 ⎪−2π + 3 cos x, ∀ x ∈ ⎢⎣ −1, − 2 ⎥⎦ ⎪ ⎪ 1 • cos −1(4x 3 − 3x) = ⎨ 2π − 3 cos −1 x, ∀ | x | ≤ 2 ⎪ ⎡1 ⎤ ⎪ −1 ⎪⎩ 3 cos x, ∀ x ∈ ⎢⎣ 2 , 1⎥⎦

−1 ⎧ −1 ⎪ π + 3 tan x , ∀x < 3 3⎞ ⎪ ⎛ 3x − x 1 • tan −1 ⎜ =⎪ 3 tan −1 x , ∀ x < ⎝ 1 − 3x 2 ⎟⎠ ⎨ 3 ⎪ 1 ⎪ ⎛ ⎞ −1 , ∞⎟ ⎪ − π + 3 tan x , ∀x ∈⎜⎝ 3 ⎠ ⎩

S2 = x1x2 + x2 x3 + .... + xn−1 xn = ∑ x1x2 S3 = ∑ x1 x2 x3 and so on.

Conversion of Inverse Trigonometric Function in Terms of Others 1. • If asin–1x + bcos–1x = c, then c(a + b) − πab a sin −1 x − b cos −1 x = a −b • If atan–1x + bcot–1x = c, then π(a + b) − 2c b tan −1 x + a cot −1 x = 2 • If asec–1x + bcosec–1x = c, then b sec −1 x − acosec −1x = 2.

2c(a + b) − π(a2 + b2 ) 2(a − b)

sin −1 x = cos −1 1 − x 2 tan −1

x 1 − x2

= sec −1

1 1 − x2

1 − x2 ⎛1⎞ cosec −1 ⎜ ⎟ = cot −1 ⎝x⎠ x Similarly for cos–1 x and tan–1 x.

Medians and Centroid of Triangle 1. The distances of centroid from the vertices A, B, C of the triangle ABC are GA, GB and GC respectively given by z

GA =

2b2 + 2c 2 − a2 3

z

GB =

2a2 + 2c 2 − b2 3

2a2 + 2b2 − c 2 3 2. Distances of centroid (G) from the sides BC, CA and AB respectively given by 2 × area of triangle 2Δ z Ga = = 3a 3a z

GC =

MATHEMATICS TODAY | OCTOBER‘17

57

z

z

z

z

2Δ 3b 1 AD = 2 1 BE = 2 1 CF = 2

•

Gb =

Gc =

4. Half angle formula

2Δ 3c

A

b2 + c 2 + 2bc cos A c 2 + a2 + 2ac cos B

F B

C

D

2

Bisectors of Angles If AD is bisector of angle A and divide the opposite sides BC in two parts l and m then by geometry we have A l AB c = = m AC b c

λ l m l +m a ∴ = = = c b b+c b+c B l D ac ab a ⇒ l= and m = b+c a+c Let AD = Othen we have ar('ABD) + ar('ACD) = ar('ABC) c A b A 1 ⇒ λ sin + λ sin = bc sin A 2 2 2 2 2

⇒ λ = length of AD =

=

(s − b)(s − c) A = 2 bc

x

cos

A s(s − a) = 2 bc

z

sin

(s − c)(s − a) B = 2 ac

x

cos

B s(s − b) = 2 ac

C

C C s(s − c) (s − a)(s − b) x cos = = 2 ab 2 ab 5. Area of 'ABC in different forms If 'denotes the area of triangle ABC z

sin

z

Δ=

c 2 sin A sin B b2 sin A sin C a2 sin B sin C = = 2 sin C 2 sin B 2 sin A

1 1 1 Δ = ab sin C = ac sin B = bc sin A 2 2 2 6. Tangent rule (Napier's analogy) A ⎛ B −C ⎞ b−c z = tan ⎜ cot ⎝ 2 ⎟⎠ b + c 2 C − A⎞ c−a B ⎛ z = tan ⎜ cot ⎝ 2 ⎟⎠ c + a 2 − A B a b − C ⎞ ⎛ z = tan ⎜ cot ⎝ 2 ⎟⎠ a + b 2 z

Circumcircle and Incircle z Circumcircle of a triangle and circumradius (R) OA = OB = OC = R, where O is point of intersection of perpendicular bisector of sides of the triangle ABC

2bc cos(A / 2) ⎛ A A⎞ ⎜⎝ using sin A = 2 sin cos ⎟⎠ b+c 2 2

z

cos B =

c 2 + a 2 − b2 2ac

z

cos A =

b2 + c 2 − a 2 . 2bc

a 2 + b2 − c 2 2ab 3. Projection formula z a = ccosA + bcosC z b = acosC + ccosA z c = acosB + bcosA 58

m

A 2

Properties of Triangles and Circles 1. Sine formula a b c = = sin A sin B sin C 2. Cosine formula

z

b

bc sin A (b + c) sin

sin

E

1 a + b + 2ab cos C = 2a2 + 2b2 − c 2 2 2

z

cosC =

MATHEMATICS TODAY | OCTOBER ‘17

A c B

b D a

C

1.

a b c = = = 2R sin A sin B sin C

2.

R=

abc abc = 4Δ 4 s(s − a)(s − b)(s − c)

Inscribed Circle and Inradius O is point of intersection of internal bisector known as incentre (r). D OF = OE = OD = length of perpendicular from incentre O to any one B of the side of the triangle. Δ 1. r = i.e. Δ = rs s

z

A

O F

E

C

A B C , r = (s − b)tan , r = (s − c)tan 2 2 2 A B C ∴ r = (s − a)tan = (s − b)tan = (s − c)tan 2 2 2 A B C r = 4R sin sin sin 2 2 2 a sin(B / 2) sin(C / 2) b sin(A / 2) sin(C / 2) , r= 3. r = cos(A / 2) cos(B / 2) c sin(A / 2) sin(B / 2) r= cos(A / 2) 4. Let r1, r2, r3 are exradius and a, b, c are sides of triangle ABC then 2.

z

r = (s − a)tan

s(s − b)(s − c) s(s − a)(s − c) Δ Δ r1 = = , r2 = = , s −a (s − a) s −b s −b r3 =

s(s − a)(s − b) Δ = s−c s−c

Thus, r1 =

A a cos(B / 2) cos(C / 2) Δ = s tan = 2 cos(A / 2) s−a

B C A = 4R cos cos sin 2 2 2 A B C A C B r2 = 4R cos cos sin , r3 = 4R cos cos sin 2 2 2 2 2 2 z

1 s 1 1 1 = = + + r Δ r1 r2 r3

z

r1 + r2 + r3 = 4R + r

z

r · r1 r2 r3 ='2

z

1 1 1 a 2 + b2 + c 2 + + + = r 2 r12 r22 r32 Δ2 1

5. If r1, r2, r3 are given then we can find z

s = r1r2 + r2r3 + r1r3

z

r r r rr r Δ = 1 2 3 ,r = 1 2 3 ∑ r1 r2 ∑ r1 r2 (r + r )(r + r )(r + r ) R= 1 2 2 3 3 1 4(r1r2 + r2r3 + r3r1)

z

r (r + r ) r (r + r ) r (r + r ) a= 1 2 3 , b= 2 3 1 , c= 3 2 1 ∑ r1 r2 ∑ r1 r2 ∑ r1 r2 sin B =

2r2

∑ r1r2

(r2 + r3)(r2 + r1)

, sin C =

2r1 ∑ r1r2 (r1 + r2 )(r1 + r3)

Area of Cyclic Quadrilateral Let ABCD is a cyclic quadrilateral with sides AB = a, BC = b, CD = c, DA = d then ∠A + ∠C = ∠B + ∠D = 180° . 1. Area of cyclic quadrilateral 1 1 = ab sin B + cd sin D 2 2 1 = (ab + cd) sin B 2 (' D = 180° – B) ...(i) = (s − a)(s − b)(s − c)(s − d) a +b+c +d 2 2. Again, area of cyclic quadrilateral 1 1 = (ab + cd)sin B = (ab + cd)sin D 2 2 2 ⋅ area of cyclic quadrilateral ∴ sin B = ab + cd where s =

(a2 + b2 ) − (c 2 + d 2 ) 2(ab + cd) (Use this result, if three sides and included angle of one is known) 4. (AC)2 = a2 + b2 – 2abcosB ⎡ (a2 + b2 ) − (c 2 + d 2 ) ⎤ = a2 + b2 − 2ab ⎢ ⎥ 2(ab + cd) ⎣ ⎦ (ac + bd)(ad + bc) ... (ii) (AC)2 = ab + cd and (AC)2 = d2 + c2 – 2cdcosD = d2 + c2 + 2cdcosB ⎡ (a2 + b2 ) − (c 2 + d 2 ) ⎤ = d 2 + c 2 + 2cd ⎢ ⎥ ... (iii) 2(ab + cd) ⎣ ⎦ ? On equating (ii) and (iii) one get (i) area of quadrilateral. 5. Circumradius of a cyclic quadrilateral 3.

cos B =

(ac + bd)(ad + bc)(ab + cd) 4Δ 6. If A, B, C, D taken in order then in a cyclic quadrilateral ABCD we have cosA + cosB + cosC + cosD = 0 R=

z

z

and sin A =

2r3

∑ r1r2

(r3 + r1)(r3 + r2 )

Concept of Pedal Triangle Let ABC be any triangle and AD, BE, CF are perpendicular's drawn from A, B, C to the opposite sides BC, CA, AB respectively. The point of concurrency MATHEMATICS TODAY | OCTOBER‘17

59

of these perpendiculars (altitudes) of the triangle ABC is called orthocentre of the triangle. A The triangle formed by joining the feet of these perpendiculars is called pedal triangle. In E F the figure, 'DEF is the pedal triangle of 'ABC. B C D ? Sides of pedal triangle are • DE = ccosC = 2RsinC cosC = R sin 2C • EF = acosA = 2RsinA cosA = R sin 2A • DF = bcosB = 2RsinB cosB = R sin 2B a b c ⎞ ⎛ = = = 2R ⎟ ⎜⎝' ⎠ sin A sin B sin C • The inradius of pedal triangle= 2RcosA cosB cosC • Circumradius of pedal triangle = R/2 • Perimeter of pedal triangle = acosA + bcosB + ccosC 1 • Area of pedal triangle = 2' cosA cosB cosC = R2 2 sin 2A sin2B sin2C

Ptolemy's Theorem In a cyclic quadrilateral, the product of the diagonals is equal to the sum of the products of the lengths of opposite sides. i.e. AC · BD = ac + bd *) or AC · BD = AB · CD + BC · BD

5.

1.

2.

2sin22x

3.

+ If the sum of the roots of the equation 7cos2x – 7 = 0 over the interval [0, 2017] is mS then the value of [ m ] (where [] represents greatest integer function) is (a) 453 (b) 454 (c) 205761 (d) 642

4.

The number of solutions of ∑ cos r x = 2017 in the interval [0, 2S] is r =1 (a) 2018 (b) 2017 (c) 2015 (d) None of these

60

MATHEMATICS TODAY | OCTOBER ‘17

the

equation

is

(a)

(4 p + 1) π , p ∈I n(n + 1) 2

(b)

(c)

4p +1 π n(n + 1)

(d) None of these

4p −1 π , p ∈I n(n + 1) 2

6.

If logcos x sin x + logsin x cos x = 2, then value of x equals pπ (b) (a) (2 p + 1) π 4 4 π π (c) (4 p + 1) (d) ( p +1) 4 4

7.

If ex-radii r1, r2, r3 of a triangle are in H.P then sides a, b, c are in (a) A.G.P. (b) A.P. (c) G.P. (d) None of these

8.

A triangle is inscribed in a circle. The vertices of the triangle divided the circumference of the circle into three area of length 6, 8, 10 units then the area of triangle is equal to (in sq. units)

9.

36 3 ( 3 − 1)

(b)

72 3 π2

( 3 + 1)

36 3

( 3 + 1) π π2 The number of real roots of the equation sec θ + cos ecθ = 15 lying between 0 and 2S is/are (a) 0 (b) 2 (c) 4 (d) 8 (c)

2

If (cot–1x)2 – 9cot–1x + 20 > 0, then x lies is (a) (cot5, cot4) (b) (−∞, cot 5) ∪ (cot 4, ∞) (c) (cot 4, ∞) (d) None of these

2017

1

of

∑ cos(λ2 x )sin(λx) = 2

(a) 64 3 ( 3 + 1) π2

π cos x + 2 x + 1 + cosec x + x +1 = If 2 x x x =/ 0 , then value of sec−1 − sin−1 ⎛⎜ ⎞⎟ is equal to ⎝2⎠ 2 3π π (a) − (c) 3S/2 (d) S/2 (b) − 2 2 −1

solution

λ =1

PROBLEMS Single Correct Answer Type 2

general

n

}

−1

The

2

(d)

π 10. The number of roots of the equation x + 2 tan x = 2 in the interval [0, 2S] is (a) 3 (b) 2 (c) 1 (d) 0 More than One Correct Answer Type α 11. If fn (α) = tan (1 + secD)(1 + sec2D) (1 + sec4D) 2 .....(1 + sec2nD), n ∈ N then ⎛π⎞ ⎛ π ⎞ ⎛π⎞ ⎛π⎞ (b) f3 ⎜ ⎟ = f4 ⎜ ⎟ (a) f2 ⎜ ⎟ = f5 ⎜ ⎟ ⎝ 16 ⎠ ⎝ 128 ⎠ ⎝ 32 ⎠ ⎝ 64 ⎠

⎛π⎞ ⎛π⎞ (c) f2 ⎜ ⎟ = f4 ⎜ ⎟ ⎝ 16 ⎠ ⎝ 64 ⎠ ⎛π⎞ ⎛π⎞ ⎛π⎞ ⎛ π ⎞ (d) f2 ⎜ ⎟ = f3 ⎜ ⎟ = f4 ⎜ ⎟ = f5 ⎜ =1 ⎝ 16 ⎠ ⎝ 32 ⎠ ⎝ 64 ⎠ ⎝ 128 ⎟⎠

Comprehension Type

12. If y = (sin–1x)3 + (cos–1x)3 then 3

(a) y has least value −π 8

Paragraph for Q.No. 18 to 20 Let cosA cos2A cos4A cos8A ........cos2n – 1A

7 π3 8 5π3 (c) y has greatest value 16 3 π (d) y has least value 32

⎧ sin 2n A , if A =\ nπ ⎪ 2n sin A ⎪ ⎪ π 1 , if A = ⎨ 2n 2n + 1 ⎪ ⎪ 1 π ⎪ − n , if A = n ∀n ∈ N ⎩ 2 2 −1

(b) y has greatest value

13. If x = asin1008D cos1009D, y = asin1009D cos1008D and

(x2 + y2 )l m

(xy ) (a) l = 2017 (c) l = 2016

,(l , m ∈ N ) is independent of D, then (b) m = 2017 (d) m = 2016

(a)

14. If (1 + ax)n = 1 + 16x + 112x2 + ..... and m = number of positive integral solutions of the equation tan–1x + cot–1y = tan–15, then which of the following is true? (a) m = 2 (b) n = 8 n (c) a = 2 (d) =2 ma 15. Let f (α) = e

⎛ ⎛ π ⎞⎞ cos−1 ⎜ sin ⎜ α + ⎟ ⎟ ⎝ ⎝ 3 ⎠⎠

13 π ⎛ 8π ⎞ = e 18

(a) f ⎜ ⎟ ⎝ 9 ⎠

4 π⎞ ⎛ 18. The value of ∏ cos 2r A ⎜ if A = ⎟ is ⎝ 15 ⎠ r =1 1 4

(a) −

π

(b)

⎛ 7π ⎞ f ⎜ − ⎟ = e12 ⎝ 4 ⎠

(a) −

5π

11π

⎛ 8π ⎞ ⎛ 7π ⎞ (c) f ⎜ − ⎟ = e 12 (d) f ⎜ ⎟ = e 18 ⎝ 9 ⎠ ⎝ 4 ⎠ 16. If tan 3 A = λ tan A, A≠ 1 then 1 (a) λ < 3

(b)

(c) O > 3

(d)

17. The solution of the system of equations x + y = and sinx + siny = 2 satisfies by π (a) x = 2k π + , k ∈ I 4 π (b) x = 2k π − , k ∈ I 2 π (c) y = − 2k π, k ∈ I 4 π (d) x = 2k π + , k ∈ I 3

1 64

(c)

1 16

(d) −

(b)

1 64

(c)

1 32

(d) −

1 1 1 (b) (c) (d) 2 2 4 Paragraph for Q.No. 21 to 23 ⎛ x − xr −1 r −1xr

∑ tan−1 ⎜⎝ 1 +r x

r =1

1 32

1 8

⎞ ⎟⎠ =

∑ tan−1 xr − tan−1 xr −1 = tan−1 xn − tan−1 x0 π 2

1 16

3 π then value of ∏ cos 2rA is 7 r =1

n

If

sin 3 A 2 λ = sin A λ − 1 cos A λ − 1 = cos 3 A 2

1 8

7 π⎞ ⎛ 19. The value of ∏ sin(2r − 1)A ⎜ if A = ⎟ is ⎝ ⎠ 14 r =1

20. If A =

then

(b)

∀n ∈ N

21. The sum of infinite terms of the series cot–13 + cot–17 + cot–113 + ...... f is equal to π π (b) tan–12 (c) cot–12 (d) (a) 4 2 22. The value of the infinite terms of the series ⎛ ⎞ ⎛ ⎞ 2 4 tan−1 ⎜ + tan−1 ⎜ ⎟ ⎟ ⎝ 2 + 12 + 14 ⎠ ⎝ 2 + 22 + 24 ⎠ ⎞ ⎛ 6 + .....is + tan−1 ⎜ 2 4 ⎝ 2 + 3 + 3 ⎟⎠ (a)

3π 2

(b)

3π 4

(c)

π 2

(d)

MATHEMATICS TODAY | OCTOBER‘17

π 4 61

23. The value of the infinite forms of the series 1 ⎛2⎞ ⎛4⎞ ⎛ 8 ⎞ tan−1 + tan−1 ⎜ ⎟ + tan−1 ⎜ ⎟ + ... + tan−1 ⎜ ⎝9⎠ ⎝ 33 ⎠ ⎝ 129 ⎟⎠ 3 ⎛ 16 ⎞ ⎛ 32 ⎞ + tan−1 ⎜ ... ∞ + tan−1 ⎜ ⎟ ⎝ 513 ⎠ ⎝ 2049 ⎟⎠ π π (b) (c) 2S (d) S (a) 4 2 Paragraph for Q.No. 24 to 26 Let AD, BE, CF are perpendicular from angular points of a triangle ABC on the opposite sides let ', r, R respectively represents the area, in radius and circumradius of the triangle. 24. If perimeters of 'ABC and 'DEF are P1 and P2 P respectively then the value of 1 equals P2 (a) Δ rR

r R

rR (d) R r Δ 25. If area of ''s AEF, BFD and CDE are '1, '2 and '3 respectively then '1 + '2 + '3 equals (a) 2'(1 + cosA cosB cosC) (b) 2'(1 + sinA sinB sinC) (c) '(1 – 2cosA cosB cosC) (d) '(1 – 2 sinA sinB sinC) Δ′ 26. If area of 'DEF is '' then value of equals Δ (a) 4cosA cosB cosC (b) 2cosA cosB cosC (c) 4sinA sinB sinC (d) 2sinA sinB sin C (b)

(c)

Matrix-Match Type

27. Match the trigonometric ratio with the equations whose one of the roots is given Column-I A. cos6° 1. B. 2cos10° 2. C. tan15° 3.

Column-II – 6x – 1 = 0 3 x – 3x2 – 3x + 1 = 0 x6 – 6x4 + 9x2 – 3 = 0

D.

32x5 – 40x3 + 10x –

cos20°

4.

8x3

3 =0

28. Match the following. Column-I

62

A.

1 3 cos −1 + 2 tan −1 5 3

B.

cos −1

5 3 63 + sin −1 + tan −1 13 5 16

MATHEMATICS TODAY | OCTOBER ‘17

Column-II 1.

S

2.

S/6

C.

sin −1

1

+ 2 tan −1

1 3

3.

S/2

x 3 4. and 2λ − x 2x − λ then A – B is B = tan −1 λ 3

S/4

5 2

D. If A = tan −1

Integer Type

29. If in a 'ABC, BC = 5, CA = 4, AB = 3 and D, E are the points on the side BC such that BD = DE = EC then the value of 8 tan(∠CAE ) is 30. The two adjacent sides of a cyclic quadrilateral are 2 and 5 and angle between them is 60°. If area of the quadrilateral is 4 3 then find the length of larger of the rest two sides. 31. If D + E + J = S and ⎛ γ +α −β⎞ ⎛ γ +β−α⎞ ⎛α +β− γ ⎞ tan ⎜ ⎟⎠ tan ⎜⎝ ⎟⎠ tan ⎜⎝ ⎟⎠ = 1then ⎝ 4 4 4 the value of 1 + cosD + cosE + cosJ is (m –1) thus the value of m equals SOLUTIONS 1. (c) : ' cos−1 x2 + 2 x + 1 + cos ec−1 x2 + 2 x + 1 =

π 2

⎛ ⎞ π 1 ⇒ cos−1 x2 + 2 x + 1 + sin−1 ⎜ ⎟= ⎝ x2 + 2 x + 1 ⎠ 2 ⎛ ⎞ 1 ⇒ cos−1 x2 + 2 x + 1 − cos−1 ⎜ ⎟=0 ⎝ x2 + 2 x + 1 ⎠ 1 ⇒ cos−1 | x + 1 | − cos−1 = 0 (x ≠ −1) | x + 1| 1 ⇒ | x + 1 |= ⇒ | x + 1 |2 = 1 | x + 1| or, x + 1 = ±1 ⇒ x = −2, 0 but x z 0 x = –2 ⎛x⎞ ⎛x⎞ Then, sec −1 ⎜ ⎟ − sin −1 ⎜ ⎟ ⎝2⎠ ⎝2⎠ π 3π −1 = sec (−1) − sin−1 (−1) = π + = 2 2 2. (b) : ' (cot–1x)2 – 9cot–1x + 20 > 0 ⇒ (cot−1 x − 5)(cot−1 x − 4) > 0

⇒ cot−1 x < 4 and cot−1 x > 5 x

x 4

–

5

As cot–1x is decreasing function then x ∈ (−∞, cot 5) ∪ (cot 4, ∞) 3. (a) : ' 2sin22x + 7cos2x – 7 = 0 ⇒ 2(1 − cos2 2 x ) + 7 cos 2 x − 7 = 0 ⇒ 2 cos2 2 x − 7 cos 2 x + 5 = 0 ⇒ 2t 2 − 7t + 5 = 0, where t = cos 2 x 5 ⇒ (t − 1)(2t − 5) = 0 ⇒ t = 1, t = (rejected) 2 ∴ t = 1 ⇒ cos 2 x = 1 = cos 0° ⇒ 2 x = 2nπ ⇒ x = nπ, n ∈ I ? The roots over the interval [0, 2017] are 0, 2S, 3S, .... 641S, ('642 S >2017) ⇒ Sum of roots = S + 2S + 3S + .... + 641S = 205761S = mS (given) ? m = 205761 ⇒ [ m ] = 453 2017

4. (d) : '

∑ cos rx = 2017

r =1

⇒ cosx + cos2x + cos3x + cos4x + cos5x + ...... + cos2017x = 2017, which is possible only if cosx = cos2x = .... = cos2017D = 1 and is satisfied for x = 0 and x = 2S. ? Number of solutions is 2. n

5. (a) : ' Given, ∑ 2 cos( λ2 x ) ⋅ sin(λx ) = 1 ⇒

n

λ =1

∑ [sin{λ(λ + 1)}x − sin{λ(λ − 1)}x] = 1

λ =1

Putting O = 1, 2, ...., n – 1, n (sin2x – sin0) + (sin6x – sin2x) + (sin12x – sin 6x) .... + {sin ((n – 1)n)x – sin (n – 2) (n – 1)x} + {sin (n(n + 1))x – sin(n(n – 1))x} ⇒ sin[n(n + 1)x] − sin 0 = 1 ⇒ sin n(n + 1)x = 1 π (4 p + 1) π ⇒ n(n + 1)x = 2 pπ + , p ∈ I ⇒ x = ⋅ , p ∈I 2 n(n + 1) 2 6. (c) : logcosx (sinx) + logsinx (cosx) = 2 ⇒

⎛ log sin x log cos x log b ⎞ + = 2 ⎜ Using loga b = log cos x log sin x log a ⎟⎠ ⎝

⇒ m+

⎛ log(sin x ) ⎞ 1 = 2 ⎜ where m = m log(cos x ) ⎟⎠ ⎝

⇒ m2 − 2m + 1 = 0 ∴ m = 1 log sin x ∴ = 1 log(sin x) – log(cos x) = 0 log cos x π ⇒ log(tan x ) = 0 ∴ tan x = e0 = 1 = tan 4 π π ⇒ x = pπ + = (4 p + 1) 4 4

7. (b) 8. (d) : From problem, we have 10 assumed that p = 10, CA p = 8, BC p =6 AB Let 'O' be the centre of circumcircle of the 'ABC and OA = OB = OC = r be circumradii of the circle. ? rD = 6, rE = 8, rJ = 10 ? rD + rE + rJ = 6 + 8 + 10

A 8 g

O

b

a C

B 6

arc ⎞ ⎛ ⎜⎝' Angle= radius ⎟⎠ 12 ⇒ r (α + β + γ ) = 24 ⇒ r (2 π) = 24 ⇒ r = π ? Area of triangle ABC = ar('OBC) + ar('OCA) + ar('OAB) 1 2 1 2 1 2 = r sin α + r sin β + r sin γ 2 2 2 1 2 = r (sin α + sin β + sin γ ) 2 1 2 ⎛ ⎛6⎞ ⎛8⎞ ⎛ 10 ⎞ ⎞ = r ⎜ sin ⎜ ⎟ + sin ⎜ ⎟ + sin ⎜ ⎟ ⎟ ⎝r ⎠ ⎝ r ⎠⎠ 2 ⎝ ⎝r ⎠ 2

1 ⎛ 12 ⎞ ⎡ 6 ⎛8 ⎞ ⎛ 10 ⎞ ⎤ = ⎜ ⎟ ⎢sin π + sin ⎜ π ⎟ + sin ⎜ π ⎟ ⎥ ⎝ 12 ⎠ ⎝ 12 ⎠ ⎦ 2 ⎝ π ⎠ ⎣ 12 72 ⎡ π 2π 5π ⎤ = sin + sin + sin ⎥ 2 ⎢⎣ 2 3 6⎦ π 72 = [sin 90° + sin 120° + sin 150°] π2 72 ⎛ 3 1 ⎞ 36 3 = + = ( 3 + 1) 1+ 2 ⎜⎝ 2 2 ⎟⎠ π π2 9. (c) : ' sec θ + cos ecθ = 15 ⇒ sin θ + cos θ = 15 sin θ cos θ 15 2 sin 2θ (after squaring) 4 ⇒ 15 sin2 2θ − 4 sin 2θ − 4 = 0 ⇒ (3 sin 2θ − 2)(5 sin 2θ + 2) = 0 ⇒ 1 + sin 2θ =

2 2 and sin 2θ = − 3 5 As θ ∈[0, 2 π], two solutions from I and II quadrants and two solutions from III and IV quadrants. ? Total number of solutions = 4 ⇒ sin 2θ =

10. (a) : ' x + 2 tan x = If y = tan x = ⇒

π x − 4 2

π x π ⇒ tan x = − 4 2 2

y = tan x ......(i) and y =

π x − 4 2

MATHEMATICS TODAY | OCTOBER‘17

…(ii) 63

y

y=

tan

x

⇒

O

p 2

p

3p 2

2p y=

p x – 4 4

Again − 1 < sin−1 x < 1 ⇒ −

π π ⎛π⎞ ⎛ π ⎞ = tan f4 ⎜ ⎟ = tan , f5 ⎜ ⎝ 64 ⎠ ⎝ 128 ⎟⎠ 4 4 ⎛π⎞ ⎛π⎞ ⎛π⎞ ⎛ π ⎞ =1 ∴ f2 ⎜ ⎟ = f3 ⎜ ⎟ = f4 ⎜ ⎟ = f5 ⎜ ⎝ 16 ⎠ ⎝ 32 ⎠ ⎝ 64 ⎠ ⎝ 128 ⎟⎠ (as tanS/4 = 1) 12. (b, d) : Given, y = (sin–1x)3 + (cos–1x)3 = (sin–1x + cos–1x) {(sin–1x)2 – sin–1xcos–1x + (cos–1x)2} π ∴ y = {sin−1 x + cos−1 x )2 − 3 sin−1 x cos−1 x} 2 2 ⎧ π π ⎞⎫ ⎛π = ⎨ − 3 sin −1 x ⎜ − sin −1 x ⎟ ⎬ ⎠⎭ ⎝2 2⎩ 4 π ⎪⎧ 3π π2 ⎪⎫ = ⎨3(sin−1 x )2 − sin−1 x + ⎬ 2 4 ⎪⎭ 2 ⎩⎪ ⎛ π2 2 y ⎞ 3π −1 sin x + ⎜ − ⎟ = 0, 2 π⎠ ⎝ 4 ⎛ π2 2 y ⎞ 9 π2 − 4(3) ⎜ − ⎟ > 0 For real value D > 0 ⇒ 4 π⎠ ⎝ 4 MATHEMATICS TODAY | OCTOBER ‘17

...(i)

⇒ y >

(1 + sec2D)(1+sec4D).....(1+sec2nD) ⎛ 1 + cos 2α ⎞ = tan α ⎜ (1 + sec 4α).....(1 + sec 2n α) ⎝ cos 2α ⎟⎠ = tan2D(1 + sec4D) ....(1 + sec2nD) = tan4D(1 + sec8D) .... (1 + sec2nD) = tan8D(1 + sec16D) ....(1 + sec2nD) = tan2nD thus fn(D) = tan2nD π ⎛π⎞ ⎛π⎞ ⎛π⎞ ∴ f2 ⎜ ⎟ = tan ⎜ ⎟ , f3 ⎜ ⎟ = tan , ⎝ 16 ⎠ ⎝4⎠ ⎝ 32 ⎠ 4

64

π3 32

x

From the graph, (i) and (ii) intersect at 3 points ? Number of solutions = 3 11. (a, b, c, d) : ' fn(D) = tan(D/2) (1 + secD) (1 + sec2D) (1 + sec4D) ..... (1 + sec2nD) ⎛ α ⎞ ⎛ 1 + cos α ⎞ (1 + sec 2α)(1 + sec 4α)....(1 + sec 2n α) = tan ⎜ ⎟ ⎜ ⎝ 2 ⎠ ⎝ cos α ⎟⎠ ⎛ sin(α / 2) ⎞ ⎛ 2 cos2 (α / 2) ⎞ =⎜ ⎟ ⎜ ⎝ cos(α / 2) ⎟⎠ ⎝ cos α ⎠

⇒ 3(sin−1 x )2 −

24 y 24 y 3π2 9 π2 − 3π2 + >0 ⇒ > 4 π π 4

π π

∴ y (−1) = {(sin−1 x )3 + (cos−1 x )3 }x =−1 = (sin−1 (−1))3 + (cos−1 (−1))3 = −

π3 7 π3 + π3 = 8 8

...(ii)

Similarly, y (1) = (sin−1 (1))3 + (cos−1 1)3 =

π3 π3 +0= 8 8

Form (i), (ii) and (iii) we have ?

Least value of y =

...(iii) π3 π3 7 π3 < < 32 8 8

π3 7 π3 and greatest value of y = 32 8

13. (a, d) : ' x = asin1008D cos1009D and y = asin1009D cos1008D ? x2 + y2 = a2sin2016D cos2018D + a2sin2018D cos2016D = a2sin2016D cos2016D 2 and xy = a sin2017Dcos2017D = a2(sinD cosD)2017 ∴

(x2 + y2 )l

=

a2l (sin α cos α)2016l

(xy )m a2m (sin α cos α)2017m which is independent of D and it will be possible only if 2016l = 2017 m ⇒ l = 2017, m = 2016 n(n − 1) 2 2 a x + ..... 2! = 1 + 16x + 112x2 + ..... On equating the coefficients of x and x2 we get n(n − 1) 2 na = 16 and a = 112 2 ⇒ na(na − a) = 112 × 2 16(16 – a) = 112 × 2 ⇒ 16 − a = 14 ⇒ a = 2 ? n = 8 Now tan–1x + cot–1y = tan–15 ⎛1⎞ 5− x ⇒ tan−1 ⎜ ⎟ = tan−1 5 − tan−1 x = tan−1 1 + 5x ⎝y⎠ 14. (a, b, c, d) : (1 + ax )n = 1 + nax +

1 5− x 1 + 5x = ⇒ y= y 1 + 5x 5− x As x, y are positive integers ⇒

....(*)

MATHEMATICS TODAY | OCTOBER‘17

65

? From equation (*) the possible value of x are 3, 4 when x = 3, y = 8 and when x = 4, y = 21 ? m=2 8 n Also, = =2 ma 2 × 2 15. (a, b) : ' f (α) = e

cos

−1

⎛ π⎞ sin ⎜ α + ⎟ ⎝ 3⎠

=e

⎛ 13 π ⎞ cos−1 ⎜ cos ⎟ ⎝ 18 ⎠ =e

=e

13 π

= e 18

=e

⎛ π 13 π ⎞ cos−1 sin ⎜ + ⎝ 2 18 ⎠⎟

7π π

⎛ ⎞ −1 ⎛ 7 π ⎞ cos sin ⎜⎝ − 4 + 3 ⎟⎠ ∴ f ⎜− ⎟ = e ⎝ 4 ⎠

=e

⎛ 17 π ⎞ cos−1 sin ⎜ − ⎝ 12 ⎠⎟

π⎞ cos ⎜ cos ⎟ ⎝ 12 ⎠ =e −1 ⎛

=

⎧ ⎛ 3π π ⎞ cos−1 ⎨− sin ⎜ − ⎟ ⎝ 2 12 ⎠ ⎩ e

= eS/12

tan 3A =λ tan A tan 3 A − tan A λ − 1 2 cos A = ⇒ = λ −1 tan A 1 cos 3 A cos A λ − 1 …(i) = cos 3 A 2

16. (a, b, c, d) : Given, ⇒ ⇒

tan 3A =λ tan A sin 3 A cos A sin 3 A cos A ⇒ ⋅ =λ ⇒ ⋅ = λ, cos 3 A sin A sin A cos 3 A on using (i) we get sin 3 A 2 λ = sin A λ − 1

1

66

–

&

1 3

3

MATHEMATICS TODAY | OCTOBER ‘17

1 –

...(ii)

18. (c) 19. (b) 20. (d) 21. (a) : cot–13 + cot–17 + cot–113 + cot–121 + cot–131 + ..... We can write the general form of the series 3, 7, 13, 21, 31, .... by successive order of difference, Sn = 3 + 7 + 13 + 21 + 31 + .... + tn Sn = 3 + 7 + 13 + .... + tn – 1 + tn On subtracting we get tn = 3 + (4 + 6 + .... + tn – 1) = 3 + (n – 1) (n + 2) tn = 1 + n + n 2 Now, cot–13 + cot–17 + cot–113 + ..... up to infinite terms = Lt

n

∑ cot−1(1 + r + r2 ) = Lt n

n

⎛

1

⎞

∑ tan−1 ⎜⎝ 1 + r(r + 1) ⎟⎠

n→∞ r =1

⎛ (r + 1) − r ⎞

∑ tan−1 ⎜⎝ 1 + r(r + 1) ⎟⎠ n→∞ r =1 n

[tan−1 (r + 1) − tan−1 r ] ∑ n→∞

= Lt

r =1

2λ 3 sin A − 4 sin3 A 2 λ = ⇒ 3 − 4 sin2 A = sin A λ −1 λ −1 λ−3 or, 4 sin2 A = as 0 < sin2 θ < 1 λ −1 λ−3 < 4 (as sin A ≠ 0 or 1) ⇒ 0< λ −1 λ−3 λ−3 3λ − 1 ⇒ > 0 and < 4 ⇒ >0 λ −1 λ −1 λ −1 x

(using (i))

⇒ x − y = 2 ⋅ 2 k π ⇒ x − y = 4k π , k ∈ I Solving (i) and (ii), we get π π x = 2k π + and y = − 2k π, k ∈ I 4 4

= Lt

⇒

x

...(i)

⎛x− y⎞ ⇒ cos ⎜ =1 ⎝ 2 ⎟⎠

n→∞ r =1

Also

x

π 2

⎛x+ y⎞ ⎛x− y⎞ ⇒ 2 sin ⎜ = 2 cos ⎜ ⎝ 2 ⎟⎠ ⎝ 2 ⎟⎠

8π π

⎛ 9 π 13 π ⎞ cos−1 sin ⎜ + ⎝ 18 18 ⎠⎟

17. (a, c) : Given, x + y = and sin x + sin y = 2

⎛ ⎞ −1 ⎛ 8 π ⎞ cos sin ⎜⎝ 9 + 3 ⎟⎠ ∴ f ⎜ ⎟=e ⎝ 9 ⎠ ⎛ 11π ⎞ cos−1 sin ⎜ ⎝ 9 ⎠⎟

⇒ λ ∈(−∞, 1) ∪ (3, ∞) …(ii)& λ ∈ ⎛⎜ −4, 1 ⎞⎟ ∪ (1, ∞) …(iii) ⎝ 3⎠ From (ii) and (iii) we get, O < 1/3 and O > 3.

x

π π π = Lt [tan−1(n + 1) − tan−1 1] = tan−1 ∞ − tan−1 1 = − = 2 4 4 n→∞ 22. (d) : Given series is ⎛ ⎞ ⎛ ⎞ 2 4 tan−1 ⎜ + tan−1 ⎜ ⎟ ⎟ ⎝ 2 + 12 + 14 ⎠ ⎝ 2 + 22 + 24 ⎠ ⎞ ⎛ 6 + tan−1 ⎜ + .....∞ 2 4 ⎝ 2 + 3 + 3 ⎟⎠ n ⎛ ⎞ 2r = Lt ∑ tan−1 ⎜ ⎟ 2 4 ⎝ ⎠ n→∞ r =1 2+r +r 2 ⎤ ⎡ 2 −1 (r + r + 1) − (r − r + 1) tan ⎥ ⎢ ∑ 2 2 n→∞ r =1 ⎢⎣1 + (r + r + 1)(r − r + 1) ⎦⎥

= Lt

n

n

[tan−1 (r 2 + r + 1) − tan−1 (r 2 − r + 1)] ∑ n→∞

= Lt

r =1

= Lt (tan–13 – tan–11) + (tan–17 – tan–13) + (tan–113 – tan–17) n→∞ + .....+ (tan–1(n2 + n + 1) – tan–1(n2 – n + 1)) = Lt [tan−1 (n2 + n + 1) − tan−1 1] n→∞ −1

= tan

∞ − tan−1 1 =

π π π − = 2 4 4

⎛ ⎞ ⎛ n2 + n ⎞ −1 −1 ⎜ 1 + 1 / n ⎟ or, Lt tan ⎜ ⎟ = Lt tan ⎜ 2 1⎟ n→∞ ⎝ 2 + n2 + n ⎠ n→∞ ⎜ +1+ ⎟ ⎝n n⎠ π −1 = tan 1 = 4 23. (b) : Given infinite series is ⎛1⎞ ⎛2⎞ ⎛4⎞ ⎛ 8 ⎞ tan−1 ⎜ ⎟ + tan−1 ⎜ ⎟ + tan−1 ⎜ ⎟ + tan−1 ⎜ + ... ∞ ⎝3⎠ ⎝9⎠ ⎝ 33 ⎠ ⎝ 129 ⎟⎠ ...(*) r − 1 ⎞ ⎛ 2 ? General term of (*) is tan−1 ⎜ ⎟ ⎝ 1 + 22r −1 ⎠ ⎛1⎞ ⎛2⎞ ⎛4⎞ ∴ Sum of tan−1 ⎜ ⎟ + tan−1 ⎜ ⎟ + tan−1 ⎜ ⎟ ⎝3⎠ ⎝9⎠ ⎝ 33 ⎠ n ⎛ 2r −1 ⎞ ⎛ 8 ⎞ + tan−1 ⎜ + .....∞ = Lt ∑ tan−1 ⎜ ⎟ ⎟ ⎝ 129 ⎠ n→∞ r =1 ⎝ 1 + 22r −1 ⎠ n ⎛ 2r −1 ⎞ = Lt ∑ tan−1 ⎜ ⎟ n→∞ r =1 ⎝ 1 + 2r ⋅ 2r −1 ⎠ n ⎛ 2r − 2r −1 ⎞ = Lt ∑ tan−1 ⎜ ⎟ n→∞ r =1 ⎝ 1 + 2r ⋅ 2r −1 ⎠ n

∑[tan−1(2r ) − tan−1(2r −1 )] n→∞

= Lt

r =1

n→∞

(by putting r = 1, 2, .... n and simplifying) π π π = tan 2 − tan 1 = − = 2 4 4 24. (d) 25. (c) 26.

3

⎛ x2 ⎞ ⎛ x2 ⎞ ⇒ 8 ⎜ 2 ⋅ − 1⎟ − 6 ⎜ 2 ⋅ − 1⎟ = 1 ⎠ ⎠ ⎝ 4 ⎝ 4 (By putting 2cosD = x) ⇒ (x2 − 2)3 − 3(x2 − 2) = 1 ⇒ x6 − 6 x 4 + 9 x2 − 3 = 0 where x = 2 cos 10° C. Let D = 15° ? 3D = 45° ? tan3D = tan45° = 1 ⇒ 3 tan α − tan3 α = 1 − 3 tan2 α ⇒ tan3 α − 3 tan2 α − 3 tan α + 1 = 0 ⇒ x3 − 3x2 − 3x + 1 = 0 where x = tan α = tan 15° 1 D. Let α = 20° ∴ 3α = 60° ⇒ cos 3α = cos 60° = 2 1 ⇒ cos 3α = 2 1 ⇒ 4 cos3 α − 3 cos α = or 8 cos3 α − 6 cos α − 1 = 0 2 or 8x3 – 6x – 1 = 0 where x = cos20° 28. A o 3, B o 1, C o 4, D o 2 ⎛ 2/3 ⎞ ⎛4⎞ ⎛1⎞ ⎛3⎞ A. cos −1 ⎜ ⎟ + 2 tan −1 ⎜ ⎟ = tan −1 ⎜ ⎟ + tan −1 ⎝3⎠ ⎝3⎠ ⎝5⎠ ⎜ 1⎟ 1− ⎝⎜ 9 ⎟⎠ ⎛2 9⎞ ⎛4⎞ = tan −1 ⎜ ⎟ + tan −1 ⎜ × ⎟ ⎝3 8⎠ ⎝3⎠

⎛3⎞ ⎛4⎞ ⎛1⎞ ⎛3⎞ ∴ cos −1 ⎜ ⎟ + 2 tan −1 ⎜ ⎟ = tan −1 ⎜ ⎟ + tan −1 ⎜ ⎟ ⎝4⎠ ⎝3⎠ ⎝3⎠ ⎝5⎠

= Lt (tan−1 2n − tan−1 2D )

−1 ∞

B. Let α = 10° ⇒ 6α = 60° 1 1 ∴ cos 6α = or cos 3(2α) = 2 2 1 3 ⇒ 4 cos 2α − 3 cos 2α = 2 1 2 3 ⇒ 4(2 cos α − 1) − 3(2 cos2 α − 1) = 2 ⇒ 8(2 cos2 α − 1)3 − 6(2 cos2 α − 1) − 1 = 0

−1

(b)

27. A o 4, B o 3, C o 2, D o 1 3 2 ? cos5D = cos(2D + 3D) = cos2D cos3D – sin2D sin3D

A. D= 6° ∴ 5α = 30D ⇒ cos 5α = cos 30D = ⇒ cos 30° = 16 cos5 α − 20 cos3 α + 5 cos α

3 ⇒ = 16 x5 − 20 x3 + 5x 2 ⇒ 32 x5 − 40 x3 + 10 x − 3 = 0 where x = cos α = cos 6°

⎛4⎞ π ⎛4⎞ = tan −1 ⎜ ⎟ + cot −1 ⎜ ⎟ = ⎝3⎠ 2 ⎝3⎠ π⎞ ⎛ −1 −1 ⎜⎝ Using tan α + cot α = ⎟⎠ 2 5 3 63 ⎞ −1 ⎛ ⎞ −1 ⎛ ⎞ −1 ⎛ B. cos ⎜⎝ ⎟⎠ + sin ⎜⎝ ⎟⎠ + tan ⎜⎝ ⎟⎠ 13 5 16 ⎛ 63 ⎞ ⎛3⎞ ⎛ 12 ⎞ = tan −1 ⎜ ⎟ + tan −1 ⎜ ⎟ + tan −1 ⎜ ⎟ ⎝4⎠ ⎝ 16 ⎠ ⎝5⎠ ⎡ ⎛ 12 3 ⎞ ⎤ + ⎥ ⎢ −1 ⎜ 5 4 ⎟ + tan −1 ⎛⎜ 63 ⎞⎟ = ⎢ π + tan ⎜ ⎝ 16 ⎠ 36 ⎟ ⎥ ⎢ ⎟⎠ ⎥ ⎜⎝ 1 − 20 ⎦ ⎣ MATHEMATICS TODAY | OCTOBER‘17

67

⎛ ⎛ x + y ⎞⎞ −1 tan −1 y = π + tan −1 ⎜ ⎜ using tan x + ⎝ 1 − xy ⎟⎠ ⎟ ⎜ ⎟ ⎝ if x, y > 0 and xy > 1 ⎠ ⎛ 63 ⎞ ⎛ 63 ⎞ + tan −1 ⎜ ⎟ = π + tan −1 ⎜ ⎟ ⎝ 16 ⎠ ⎝ −16 ⎠ ⎛ 63 ⎞ ⎛ 63 ⎞ = π − tan −1 ⎜ ⎟ + tan −1 ⎜ ⎟ = π ⎝ 16 ⎠ ⎝ 16 ⎠

73 9 −1 = 64 64 3 3 ∴ tan α = i.e., tan ∠CAE = 8 8 8 tanCAE = 3 30. (3) : Let ABCD be the cyclic quadrilateral in which AB = 2 and BC = 5 and ABC = 60° ? ADC = 180° – 60° = 120° ∠ABC = 60° ∴ ∠ADC = 180° − 60° = 120° Area of quadrilateral ABCD = ar('ABC) + ar('ACD) 1 1 ⇒ AB × BC sin 60° + CD × DA sin 120° = 4 3 2 2 1 3 1 ⎛ 3⎞ ⇒ ×2×5× + xy =4 3 2 2 2 ⎜⎝ 2 ⎟⎠ D Now tan2 α = sec2 α − 1 =

⎛π⎞ π = tan−1 (1) = tan−1 tan ⎜ ⎟ = ⎝4⎠ 4 ⎛ x 3 ⎞ ⎛ 2x − λ ⎞ D. ∴ A = tan−1 ⎜ , B = tan−1 ⎜ ⎟ ⎝ λ 3 ⎟⎠ ⎝ 2λ − x ⎠ x 3 (2 x − λ) − 2λ − x λ 3 x 3 2x − λ ⋅ 2λ − x λ 3

3λx + (2 x − λ)(x − 2 λ)

where CD = x, AD = y ⇒ xy = 6 From 'ABC we get

3[λ(2 λ − x ) + x(2 x − λ)] 1 ⎛ 2 x2 − (5λx − 3λx ) + 2 λ2 ⎞ 1 = ⎟= ⎜ 3⎝ 3 2 x2 − 2 λx + 2 λ2 ⎠ ∴ tan( A − B) = tan

π 6

⇒ A−B =

cos 60° = π 6

29. (3) : As the side of 'ABC are AB = 3, BC = 5, CA = 4 ? Triangle ABC is right angle ' at A, i.e., BAC = 90° 5 In ΔCAE , let ∠CAE = α and CE = 3 2 2 ⎛5⎞ 2 4 + ⎜ ⎟ − ( AE ) 16 + 25 − ( AE )2 ⎝3⎠ 9 we have cos C = = 40 ⎛5⎞ 2⋅4⋅⎜ ⎟ 3 ⎝3⎠ ?

68

169 − ( AE )2 4 9 = 40 5 3

A 4

3 B

MATHEMATICS TODAY | OCTOBER ‘17

169 32 169 − 96 73 73 − = = ∴ AE = 9 3 9 9 3 Again in 'CAE we have ⇒ ( AE )2 =

2

⎛ 1 3 ⎞ + 1 3 ⎟ ⎜ ⎞ ⎞ ⎛ ⎛ = tan −1 ⎜ ⎟ + tan −1 ⎜ ⎟ = tan −1 ⎜ 7 4 ⎟ ⎝4⎠ ⎝7⎠ 1 3 ⎜⎝ 1 − ⋅ ⎟⎠ 7 4 ⎛ 25 ⎞ = tan−1 ⎜ ⎟ ⎝ 25 ⎠

1+

169 4 × 40 32 − ( AE )2 = = 9 5×3 3

⎛5⎞ 73 25 AC2 + ( AE )2 − ⎜ ⎟ 16 + − ⎝3⎠ 9 9 cos α = = 2 ⋅ 4 AC 73 8× 3 144 + 73 − 25 3 8 73 ⇒ cos α = × = ⇒ sec α = 9 8 8 73 73

⎛1⎞ −1 ⎛ 1 ⎞ + 2 tan −1 ⎜ ⎟ C. sin ⎜ ⎝3⎠ ⎝ 5 2 ⎟⎠

∴ tan( A − B) =

⇒

D 5 E

C

....(i)

( AB)2 + (BC )2 − ( AC )2 2 AB ⋅ BC

y 120° x A

C

60° B

⎛1⎞ ⇒ ( AC )2 = ( AB)2 + (BC )2 − 2( AB)(BC ) ⎜ ⎟ ⎝2⎠ 2 = 4 + 25 – 2 × 5 ? (AC) = 19 …(ii) Again from 'ACD, we get (AC)2 = (CD)2 + (DA)2 – 2CD × DA cos120° ⎛ 1⎞ ( AC )2 = x2 + y2 − 2 xy ⎜ − ⎟ = x2 + y2 + xy ⎝ 2⎠ = x2 + y2 + 6 (using (i)) …(iii) From (ii) and (iii) we have x2 + y2 + 6 = 19 ⇒ x2 + y2 = 13 ⇒ x = 2 then y = 3 and x = 3 then y = 2 ? x = 2, 3 and y = 3, 2 ? Length of larger side of the quadrilateral from the rest sides is 3 31. (1)

hallen ing PROBLEMS ON

Vectors and 3D Geometry 1. A plane S is given and on one side of the plane three non-collinear points A, B, C such that the plane determined by them is not parallel to S. Three arbitrary points Ac, Bc, Cc in S are selected. Let L, M, N be the midpoints of AAc, BBc and CCc and let G be the centroid of 'LMN. The locus of G as Ac, Bc, Cc varies across S is (a) a plane parallel to S (b) a plane perpendicular to S (c) a line parallel to S (d) a line perpendicular to S 2. Consider the two statements regarding a tetrahedron ABCS S1 : ABCS has 5 different spheres that touch all 6 lines determined by its edges. S2 : ABCS is a regular tetrahedron. Then (a) S1 S2 (b) S2 S1 (c) S1 S2

/ S1 (d) S2 ⇒

3. If all the angles of a convex n-gon are equal and the lengths of consecutive edges a1, a2, ....., an satisfy a1 t a2 t ..... t an then (a) a1 = an (b) a2 = 2an – 1 (c) a3 = an – 2 + an – 1 (d) a3 = an – 2 – an – 1 4. Consider a tetrahedron ABCD, with O be the centre of the circumsphere of regular tetrahedron ABCD and let P be any point. Let T = OA + OB + OC + OD and Y = PA + PB + PC + PD, then (a) T > Y (b) T < Y (c) T = Y (d) T = 2Y 5. Exactly one side of a tetrahedron is of length greater than 1, then max. volume of the tetrahedron is 1 1 1 (a) 1 (b) (c) (d) 8 4 2 6. In any tetrahedron there is a vertex such that the lengths of its sides through that vertex are sides of a triangle. This statement is

(a) (b) (c) (d)

always true always false true only in regular tetrahedrons not true in general

7. The condition on 'a' such that there exists a tetrahedron of whose one edge has length 'a' and the other 5 edges have length 1 is (a) a < 2 (b) a < 3 (c) a < 2 2 (d) a < 2 3 8. Vertex A of the acute triangle ABC is equidistant from the circumcentre O and orthocentre H. The possible value of A is (a) 30° (b) 60° (c) 75° (d) 90° 9. The lengths of two opposite edges of a tetrahedron are 12 and 15 units, their shortest distance is 10 units. If the volume of the tetrahedron is 200 cubic units then the angle between the two edge is −1 ⎛ 2 ⎞ ⎛1⎞ (b) sin ⎜ ⎟ (a) sin−1 ⎜ ⎟ ⎝3⎠ ⎝2⎠ ⎛3⎞ (c) sin−1 ⎜ ⎟ ⎝4⎠

⎛4⎞ (d) sin−1 ⎜ ⎟ ⎝5⎠

10. The plane lx + my + nz = p intersects co-ordinate axes in A, B and C respectively. If area of 'ABC is ' then 2Δ 2Δ 2 2 (b) p ≥ (a) p ≤ 3 3 3 3 2Δ 2Δ 2 (c) p > (d) p2 < 3 3 3 3 G G G 11. a , b , c are three non-zero non-coplanar vectors with fixed magnitudes. Angles between any of the vector with the normal of the plane containing other two is D. Volume is T and surface area is Y, ⎡ 1 2 1 1 ⎤ then ⋅⎢ G + G + G ⎥ = | cos α | ⎣ | a | | b | | c | ⎦

By : Tapas Kr. Yogi, Visakhapatnam, Mob : 09533632105 MATHEMATICS TODAY | OCTOBER‘17

69

2T 2Y (d) Y T 12. The shortest distance between the lines 2x + y + z – 1 = 0 = 3x + y + 2z – 2 and x = y = z is 1 1 1 1 (b) (c) (d) (a) 2 3 5 4 (a)

Y T

(b)

T Y

(c)

13. Consider the four coplanar points G ^ ^ ^ ^ ^ ^ A(O), B(3 i + f (t ) j + k ), C(2 i + f ′(t ) j + 2 k ) and ^ ^

D( i + j) , then f(x) is of the form (a) 2 + Oe2x (b) 2 + Olog(2x) (c) Oex (d) Ologx G 14. Let a be a vector or rectangular co-ordinate system G with sloping angle 60°. Suppose that | a − ^i | is G G G ^ geometric mean of | a | and | a − 2 i | , then | a | = (a) (c)

2 2 −1

(b) 2 + 1 (d) 2

15. The volume of the solid enclosed by the planes |x| + |y| + |z| = 1 is 1 2 (a) cubic units (b) cubic units 3 3 4 3 cubic units (d) cubic units (c) 3 3 SOLUTIONS

1. (a) : Let d(P) denote the distance of a point P from S, with P being on the same side of Sas A, B, C. Let G1 be the fixed centroid of 'ABC and G1′ be the centroid of ΔA ′B ′C ′ It is easy to prove that G is the midpoint of G1G'1. Hence, varying G1′ across S, we have the locus of G as the plane π ′ parallel to S such that d(G1 ) d( A) + d(B) + d(C ) P ∈π ′ ⇒ d(P ) = = 2 6 2. (c) : The spheres are arranged in similar way, where we have one in-circle and 3 ex-circles. Here, we have one in-sphere and 4 ex-spheres. For the in-sphere, we have SA + BC = SB + CA = SC + AB and for the ex-sphere opposite S, we have SA – BC = SB – CA = SC – AB. Hence SA = SB = SC and AB = BC = CA. By symmetry, AB = AC = SA, i.e. regular tetrahedron. 3. (a) : L e t OA1 , OA2 , ...., OAn b e t h e v e c t o r s corresponding to edges a1, a2, ..., an of the polygon. By the conditions in the question, OA1 + OA2 + ... + OAn = O , 70

MATHEMATICS TODAY | OCTOBER ‘17

∠A1OA2 = ∠A2OA3 = ... =

2π n

and OA1 ≥ OA2 ≥ OA3 ≥ .... ≥ OAn . Let l be a line through O, perpendicular to OAn and B1, B2, ..., Bn – 1 be the projections of A1, A2, ..., An – 1 on l. By the assumptions, ∑ OBi = O. n But, OBi ≤ OBn−i for all i ≤ , 2 because all the sums OBi + OBn−i lies on the same side G of O. Hence, all these sums must be O. i.e. OAi = OAn – i . 4. (b) : Let O(0, 0, 0), A(–a, –a, –a), B(–a, a, a), C(a, –a, a) and D(a, a, –a). For a point P(x, y, z) by using RMS ! AM. We have, PA + PB + PC + PD does not exceed 2 PA2 + PB2 + PC 2 + PD 2 . Now, PA2 + PB2 + PC2 + PD2 = 4(x2 + y2 + z2) + 12a2 > 12a2 = OA2 + OB2 + OC2 + OD2 Hence, PA + PB + PC + PD ≥ 2 OA2 + OB2 + OC 2 + OD 2 = OA + OB + OC + OD 5. (d) : Suppose CD is the longest edge of the tetrahedron ABCD. Let AB = a, CK and DL be the altitudes of 'ABC and 'ABD respectively. Let DM be the altitude of tetrahedron. Then in right triangle AKC, CK 2 ≤ 1 − Similarly, DL2 ≤ 1 −

a2 . 4

a2 . 4

Since DM < DL, so DM 2 ≤ 1 −

a2 4

⎛ a2 ⎞ 1 ⎡1 ⎤ 1 Hence, volume V = ⋅[DM ] ⎢ ⋅ a ⋅ CK ⎥ ≤ ⋅ a ⋅ ⎜1 − ⎟ 3 4 ⎠ ⎣2 ⎦ 6 ⎝ 1 1 1 (2 + a)[1 − (a − 1)2 ] ≤ ⋅ 3 ⋅1 = 24 24 8 6. (a) : Let AB be the largest edge of the tetrahedron ABCD. Then AB ! AC + AD and AB > BC + BD. ⇒ 2AB ≥ AC + AD + BC + BD ⇒ either AB > AC + BC or AB > AD + BD Contradicting the triangle inequality. Hence, the three edges coming out of atleast one of the vertices A and B form a triangle. ≤

MATHEMATICS TODAY | OCTOBER‘17

71

7. (b) : Let AB = a and remaining edges have length 1. Let M be the midpoint of CD. Then since, CDA and BCD are equilateral triangles, we have AM = BM =

3 / 2 and so AB < AM + BM gives a <

3 3 + 2 2

i.e., a < 3 8. (b) : Let origin be the circumcentre, then according to question, G G G G G | A − 0 | = | A | = | A − H | and G G G G G G G circumradius, R = | A | = | B | = | C | and H = A + B + C. G G G G G G G Let a = | B − C |, then R2 = | A |2 = | A − H |2 = | B + C |2 So, R2 = 4 R2 − a2 i.e.

a 3 = 3 ⇒ sin A = R 2

by using sine rule. Hence, ∠A = 60°

^

2

2

p 1 l +m +n ≥ (lmn)2/3 i.e. ≥ 3 3 3 2Δ 2Δ 2 , with equality occurring at l = m = n. Hence, p ≤ 3 3 G G G G G G G G G 11. (a) : T =| a ⋅ (b × c )| = | b ⋅ (c × a )| =| c ⋅ (a × b )| G G G T =| a || b || c || cos α ||sin θ1 | G G G G G G =| a || b || c || cos α ||sin θ2 |=| a || b || c || cos α ||sin θ3 | G G G G G G and surface area Y = 2(| a × b | + | b × c | + | c × a |) G G G G G G i.e., Y = 2(| a || b ||sin θ1 | + | b || c ||sin θ2 | + | c || a ||sin θ3 |) 12. (a) : Any plane passing through the first line can be taken as 2x + y + z – 1 + O(3x + y + 2z – 2) = 0. If this is parallel to the second line then (2 + 3O) · 1 + (1 + O) · 1 + (1 + 2O) · 1 = 0 −2 ⇒ λ= 3 So, plane is y – z + 1 = 0. 1 . Distance from (0, 0, 0) is 2 72

^

^

4 ^^ ^ 4 Hence, reqd. vol. = [ i j k] = cubic units 3 3

2 2 2⎤ ⎡ p4 1 ⎢ ⎛ p2 ⎞ ⎛ p2 ⎞ ⎛ p2 ⎞ ⎥ = + = ⋅ ⎜ ⎟ +⎜ ⎟ ⎜ ⎟ 4 ⎢ ⎝ lm ⎠ ⎝ mn ⎠ ⎝ ln ⎠ ⎥ 4l 2m2n2 ⎢⎣ ⎥⎦ 2 p i.e. Δ = . 2lmn Now, A.M. > G.M. gives 2

⇒ 2 x ⋅ (x − 2)2 + 3x2 = (x − 1)2 + (3x2 ) 1 Simplifying x = ( 2 − 1) 2 15. (d) : Let A(1, 0, 0), B(0, 1, 0), C (0, 0, 1), and O(0, 0, 0) Reqd. volume = 8 × vol. of tetrahedran OABC 1 = 8 × [OA OB OC] 6 where, OA = i , OB = j, OC = k

1 9. (b) : Use, volume = abd sin θ 6 10. (a) : '2 = 'x2 + 'y2 + '2z

2

13. (a) : As A, B, C, D are coplanar ? [ AB AC AD] = 0 Simplifying this gives, f c(t) – 2f(t) = –4 i.e., f(x) = 2 + Oe2x , O R. G 14. (c) : Let a = x ^i + 3x ^j . G | a |= 2 x , x > 0 G G ^ G ^ Now A.T.Q., | a || a − 2 i |=| a − i |2

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Series-4 Time: 1 hr 15 min. The entire syllabus of Mathematics of JEE MAIN is being divided into eight units, on each unit there will be a Mock Test Paper (MTP) which will be published in the subsequent issues. The syllabus for module break-up is given below: Unit No. 4

Topic Mathematical Induction Binomial Theorem & it’s Simple Applications Sequences and Series

Co-ordinate geometry-2D

Syllabus In Detail Principle of mathematical induction and its simple applications Binomial Theorem of positive integral index, general term and middle term, properties of Binomial coefﬁcients and simple applications Deﬁnition of series and sequence. Arithmetic, geometric and harmonic progressions. Arithmetic, geometric & harmonic means between two given numbers, relation between A.M.,G.M. and H.M. Sum up to n terms of special series: ¦ n,¦ n2 ,¦ n3. ArithmeticGeometric progression. Conic Sections: Sections of cones, equation conic sections( parabola, ellipse and hyperbola in standard forms, condition for y = mx + c to be a tangent and point(s) of tangency

1 1 1 1 1. Let P (n) : 1 + + + ... + < 2 − is true for 2 n 4 9 n (a) n (b) for n = 1 (c) n > 1 n ∈ N (d) none of these

5. If (1 + x + x2)n = b0 + b1x + b2x2 + .... + b2nx2n, then b0 + b2 + b4 + .... + b2n equals

2. Let P(n) be the statement represent the sum of three successive natural numbers ∀n ∈ N , then the smallest value of n to which P(n) is divisible by 9 is (a) 1 (b) 3! (c) 3 (d) 9!

3n + 1 (d) none of these. 2 6. Number of irrational terms in the expansion of

3. Let P(n) = 5n – 2n , P(n) is divisible by 3O where O and n both are odd positive integers then the least value of n and O will be (a) 13 (b) 11 (c) 1 (d) 5 4.

Middle term in the expansion of (1 – 3x + 3x2 – x3)2n is 6n ! 3n x 3n !

(a)

6n ! n x 3n ! 3n !

(b)

(c)

6n ! (− x )3n 3n ! 3n !

(d) none of these

(a)

3n − 1 2

1 (b) 3n + 2

(c)

(5 2 + 10 3 )

60

(a) 54

are (b) 61

(c) 30

(d) 31

7. The range of the value of the term independent 10

⎛ cos −1 α ⎞ of x in the expansion of ⎜ x sin −1 α + ⎟ , x ⎠ ⎝ α ∈[−1, 1] is (a) [1, 2] (b) (1, 2) 10 2 10 ⎛ C π − C π2 ⎞ 5 5 , (c) ⎜ ⎟ 5 ⎜⎝ 220 ⎟⎠ 2

By : Sankar Ghosh, S.G.M.C, Mob : 09831244397. MATHEMATICS TODAY | OCTOBER‘17

73

⎛ − 10C π10 10C π10 ⎞ 5 5 , (d) ⎜ ⎟ ⎜⎝ 25 220 ⎟⎠ 8. If C0, C1, C2, C3, ...., Cn denotes the binomial coefficients in the expansion of (1 + x)n, then C C C1 C + 2 2 + 3 3 + ... + n n = C0 C1 C2 Cn−1 (a)

n 2

(b)

n +1 n(n −1) (c) 2 2

(d)

n(n +1) 2

⎛1 ⎞ 9. If the third term in the expansion of ⎜ + x log10 x ⎟ ⎝x ⎠ is 1,000, then x = (a) 102 (b) 103 (c) 10 (d) none of these

5

10. The greatest term in the expansion of (2x – 3y)28 when x = 9, y = 4 is (a) 14th (b) 15th (c) 14th,15th (d) 12th 11. If the sum of n terms of an A.P. is cn (n – 1) where n 0, then the sum of the squares of these terms is 2 2 2 (a) c2n2(n + 1) (b) c n(n − 1)(2n − 1) 3 2 1 (c) c 2n(n + 1)(2n + 1) (d) c 2n2 3 3 12. The interior angles of a polygon are in A.P. with the smallest angle and common difference being respectively 120° and 5°. Then n = (a) 16 (b) 7 (c) 9 (d) 10 13. Between the numbers 1 and 31, m means are inserted so that the ratio of 7th and the (m – 1)th mean is 5 : 9, then the value of m is (a) 14 (b) 10 (c) 7 (d) 3 14. If the arithmetic mean of two positive numbers a and b (a > b) is twice their G.M., then a : b = (b) 2 + 3 : 2 − 3 (a) 6 + 7 : 6 − 7 (d) none of these (c) 5 + 6 : 5 − 6 3 33 333 + + + ...∞ then P equals 17 172 173 3 7 51 3 (a) (b) (c) (d) 17 17 112 7

15. Let P =

16. Between the numbers 2 and 20, 8 A.M's are inserted, then their sum is (a) 88 (b) 44 (c) 176 (d) none of these 74

MATHEMATICS TODAY | OCTOBER ‘17

17. If the roots of the equation ax3 + 3x2 + cx + d = 0 are in G.P., then (a) a3c = bd3 (b) ab3 = cd3 (c) a3b = c3d (d) c3a = b3d 18. The age of the father of two children is twice of the elder one added to four times that of the younger one. If the geometric mean of the ages of the two children is 4 3 and their harmonic mean is 6, father's age (in years) is (a) 36 (b) 40 (c) 50 (d) 56. 19. Let a1, a2, a3, .... be a sequence such that a1 = 2 and an – an – 1 = 2n n ≥ 2, then a1 + a2 + ... + a20 = (a) 3040 (b) 3020 (c) 2020 (d) 3080 20. If three successive terms of a G.P. with common ratio r(r > 1) form the sides of a 'ABC and [r] denotes greatest integer function, then [r] + [–r] = (a) 0 (b) 1 (c) –1 (d) none of these 21. The set of values of m for which a chord of slope m of the circle x2 + y2 = 16 touches the parabola y2 = 8 x is (a) (−∞, −1) ∪ (1, ∞) (b) (−∞, ∞) ⎛ 2 −1 ⎞ ⎛ 2 −1 ⎞ (c) ⎜ −∞, − , ∞⎟ ⎟∪⎜ ⎜⎝ ⎟⎠ 2 ⎟⎠ ⎜⎝ 2 (d) none of these 22. If the tangent drawn at a point (t2, 2t) on the parabola y2 = 4x is same as normal drawn at x2 y2 + = 1 , then 5 4 which of the following is not true ? 1 −1 (a) t = ± (b) α = − tan 2 5 ( 5 cos α, 2 sin α) on the ellipse

−1 (c) α = tan 2

(d) none of these

23. If two points are taken on minor axis of an ellipse x2

+

y2

= 1 at the same distance from the centre as a 2 b2 the foci, the sum of the squares of the perpendiculars from these point on any tangent to the ellipse if a < b is (b) b2 (c) 2a2 (d) 2b2 (a) a2 24. The equation of parabola whose latus rectum is 2 units, axis of line is x + y – 2 = 0 and tangent at the vertex is x – y + 4 = 0 (a) (x + y − 2)2 = 4 2 (x − y + 4)2

(b) (x − y − 4)2 = 4 2 (x + y − 2) 2 (c) (x + y − 2) = 2 2 (x − y + 4) (d) none of these

25. If ASB is the focal chord of the parabola y2 = 8x such that AS = 4, then the length SB = (a) 6 (b) 3 (c) 4 (d) none of these 26. Let y = 4 x 2 and (a) a < (c) a > −

1 2 1 2

x2 a2

−

y2 = 1 intersect iff 16 1 (b) a < − 2 (d) none of these

27. The shortest distance between the parabola's y2 = 4x and y2 – 2x + 6 = 0 is (b) 5 (a) 2 (c) 3 (d) none of these. 28. If chords of contact of tangents from two points 2

2

y x + = 1 are at 52 13 right angle then ratio of the product of abscissa and ordinate is (a) –16 : 1 (b) 4 : 1 (c) 16 : 1 (d) none of these (x1, y1) and (x2, y2) to the ellipse

4 to the ellipse 3 2 2 y x + = 1 intersects the major and minor axes 18 32 at A and B.If O is the origin, then the area of the 'OAB (in sq. units) is (a) 48 (b) 9 (c) 24 (d) 16

29. A tangent having slope −

30. The ellipse x2 + 4y2 = 4 is inscribed in a rectangle aligned with the coordinate axes, which in turn is inscribed in another ellipse that passes through the point (4, 0). Then the equation of the ellipse is (a) x2 + 12y2 = 16 (b) 4x2 + 48y2 = 48 2 2 (c) 4x + 64y = 48 (d) x2 + 16y2 = 16

1 3 6 R.H.S. of P(2) = 2 − = = 2 2 4 ? L.H.S. of P(2) < R.H.S. of P(2) ? P(2) is true. Hence, the statement is true for n > 1 n N. 2. (a) : As P(n) = (n + 1) + (n + 2) + (n + 3) P(n) = 3n + 6 = 3(n + 2) ? P(1) = 3(3) = 9 which is divisible by 9 ? Least value of n is 1. 3. (c) : P(n) = 5n – 2n Let n = 1 ? P(1) = 3O = 3 ? O=1 Similarly, n = 5 ? P(5) = 55 – 25 ⇒ P(5) = 3125 – 32 = 3093 = 3 × 1031 In this case O= 1031 Similarly, we can check the result for other cases and find that the least value of O and n is 1. 4. (c) : The given expression is (1 – 3x + 3x2 – x3)2n = (1 – x)6n Since the power of the expansion is an even number, therefore there will be only one middle term, which 6n is ⎛⎜ + 1⎟⎞ ⎝2 ⎠

term i.e. (3n + 1)th term.

6n ! ∴ T3n+1 = 6nC3n (− x )3n = (− x )3n 3n ! 3n ! 5. (c) : (1 + x + x2)n = b0 + b1x + b2x2 + b3x3 + ... ...+ b2n x2n Putting x = 1, we get 3n = b0 + b1 + b2 + b3 + ... + b2n ... (i) Putting x = –1, we get 1 = b0 – b1 + b2 – b3 + b4 – .... + b2n .... (ii) Adding (i) and (ii), we get 3n + 1 = 2(b0 + b2 + b4 + ... + b2n) 3n + 1 ∴ = b0 + b2 + b4 + b6 + ... + b2n 2 6. (a) : The given expression is

(5 2 + 10 3 ) = (21/5 + 31/10 )

60

60

Now L.C.M. of 5 and 10 is 10. ? Number of rational terms Tr +1 = 60Cr (21/5 )60−r (31/10 )r

SOLUTIONS

1. (c) : For n = 1, L.H.S of P(1) = 1 and R.H.S. of P(1) = 2 – 1 = 1 ? P(1) is not true, as L.H.S of P(1) = 1 < 1 = RHS. of P (1), False 1 5 Again n = 2, L.H.S. of P(2) = 1 + = 4 4

th

= As ? ? ?

60

Cr 2

r⎞ ⎛ 12− ⎟ ⎝⎜ 5⎠

r 10 ⋅3

0 < r 60 r = 0, 10, 20, 30, 40, 50, 60 Number of rational terms are 7. Number of irrational terms equals to 61 – 7 = 54. MATHEMATICS TODAY | OCTOBER‘17

75

7. (d) : Here, Tr +1 =

10

Cr (x sin

−1

10 −r

α)

⎛ cos −1 α ⎞ ⎜ ⎟ ⎝ x ⎠

r

= 10Cr x10 –2r (sin–1D)10 – r (cos–1D)r For independent term of x, let 10 – 2r = 0 ? r = 5, T6 = 10C5(sin–1D cos–1D)5 Here f(D) = sin–1D cos–1D, let sin–1D = w π π ⎛π ⎞ ⇒ f (w ) = w ⎜ − w ⎟ , − ≤ w ≤ ⎝2 ⎠ 2 2 π ⇒ f ′(w ) = − 2w. 2 Now, f ′(w ) = 0 gives w = S/4 π2 ⎛ π ⎞ π2 ⎛ π⎞ ⎛π⎞ ∴ f ⎜ ⎟= , f ⎜− ⎟ = − , f ⎜ ⎟ = 0 ⎝ 4 ⎠ 16 ⎝ 2⎠ ⎝2⎠ 2 ⎛ π2 π2 ⎞ ∴ w ∈ ⎜− , ⎟ ⎝ 2 16 ⎠

1 n n(n − 1) 3n(n − 1)(n − 2) 2 +2 + × + .... + n ⋅ n(n − 1) n 1 2 ⋅n 6

= n + (n – 1) + (n – 2) + ... + 1 n(n + 1) n(n + 1) ['Sum of first n natural numbers = ] = 2 2 5

3

⎛1⎞ Now, T3 = 5C2 ⎜ ⎟ (x log10 x )2 ⎝x⎠ 3

⎛1⎞ Now, 10 ⎜ ⎟ ⋅ x 2 log10 x = 1000 ⎝x⎠ 2 ⇒ x 2 log10 x −3 = 102 ⇒ 2 log10 x − 3 = log x 10

⇒ 2 log10 x − 3 = ⇒ 2a − 3 = 76

2 log10 x

2 [where a = log10 x] ⇒ 2a2 − 3a − 2 = 0 a

MATHEMATICS TODAY | OCTOBER ‘17

29 − r −2 ⋅ ≥1 r 3

⇒ 2(29 – r) > 3r

⎡ −10 C π10 10 C π10 ⎤ 5 5 ⎥ , =⎢ 5 ⎢⎣ 2 220 ⎥⎦ C C C C 8. (d) : Here, 1 + 2 2 + 3 3 + .... + n n C0 C1 C2 Cn−1

⎛1 log x ⎞ 9. (a) : The given expression is ⎜ + x 10 ⎟ ⎝x ⎠

1 ? log10x = 2 x = 100 2 10. (d) : Let (tr + 1)th term is the greatest term. tr + 1 t t will be greatest when r +1 ≥ 1 and r +2 ≤ 1 tr tr +1 Here the given expansion is (2x – 3y)28 ? tr + 1 = 28Cr (2x)28 – r · (–3y)r tr = 28Cr – 1 (2x)29 – r (–3y)r – 1 tr + 2 = 28Cr + 1 (2x)27 – r (–3y)r + 1 t 29 − r −3 y Now, r +1 ≥ 1 gives ⋅ ≥1 tr r 2x ∴ a = 2 and a ≠ −

⇒

5 5 ⎡ ⎛ π2 ⎞ 10 ⎛ π2 ⎞ ⎤ 10 ⎢ ? Required range C5 ⎜ − ⎟ , C5 ⎜ ⎟ ⎥ ⎢ ⎝ 16 ⎠ ⎥⎥ ⎝ 2 ⎠ ⎢⎣ ⎦

=

⇒ 2a(a − 2) + 1(a − 2) = 0 ⇒ (2a + 1)(a − 2) = 0

[' x = 9 and y = 4] ⇒ 5r < 58 r ≤

58 3 = 11 5 5

t 28 − r −3 y ≤1 and r +2 ≤ 1 gives tr +1 r + 1 2x

28 − r 2 28 − r −3 × 4 × ≤1 ⇒ ⋅ ≤1 r +1 2×9 r +1 3 ⇒ 56 – 2r < 3r + 3 ⇒ 53 < 5r 53 3 ⇒ r ≥ = 10 5 5 ? tr + 1 will be greatest when 3 3 10 < r < 11 i.e. when 11 5 5 ? t12 is greatest. 11. (b) : We are given that sum of n terms of an A.P. is cn(n – 1). Let Sn = cn(n – 1) ? tn = Sn – Sn – 1 = cn(n – 1) – c(n – 1)(n – 2) = c[(n2 – n) – (n2 – 3n + 2)] = c(2n – 2) tn = 2c(n – 1) tn2 = 4c2(n2 – 2n + 1) ⇒

Now,

∑ tn2 = 4c2 ∑ (n2 − 2n + 1)

⎡ 2n2 + 3n + 1 − 6n − 6 + 6 ⎤ = 4c 2 n ⎢ ⎥ 6 ⎥⎦ ⎢⎣ 2 2 4c n 2 = (2n − 3n + 1) = c 2n(2n − 1)(n − 1) 6 3 12. (c) : Let the polygon have n sides. Then sum of all the interior angles = (2n – 4)90° n ∴ (2n − 4)90 = [2(120) + (n − 1)5] 2

5n2 – 125n + 720 = 0 (n – 16)(n – 9) = 0 n = 16, 9 When n = 16, the largest angle is given by 120 + (16 – 1)5 = 195, [' tn = a + (n – 1)d] which is impossible and hence n = 9. 13. (a) : Let the progression be 1, *, *, *, ..., *, 31 Here a = 1, tm + 2 = 31. Let the common difference be d. 30 ∴ 1 + (m + 2 − 1) d = 31 ⇒ (m + 1)d = 30 ⇒ d = m +1 th 7 mean 5 Given that, = th 9 (m − 1) mean 7 × 30 m +1 = 5 ⇒ 30(m − 1) 9 1+ m +1 On solving, we get m = 14 14. (b) : According to the given problem, we have a +b = 2 ab ⇒ a + b = 4 ab 2 a2 + 2ab + b2 = 16ab 1 + 7d 5 ⇒ = 1 + (m − 1)d 9

1+

2

⎛a⎞ ⎛a⎞ a2 – 14ab + b2 = 0 ⇒ ⎜ ⎟ − 14 ⎜ ⎟ + 1 = 0 ⎝b⎠ ⎝b⎠ a 2+ 3 2− 3 , = b 2− 3 2+ 3 ? The required ratio is 2 + 3 : 2 − 3 . 3 33 333 + + + .... ∞ 17 172 173 P 3 33 = + + .... ∞ 17 172 173 (i) – (ii) gives, 1 ⎞ 3 30 300 ⎛ P ⎜1 − ⎟ = + + + .... + ∞ ⎝ 17 ⎠ 17 172 173

15. (d) : P =

⎞ 16 3 ⎛ 10 100 P = ⎜1 + + + .... + ∞ ⎟ 2 ⎠ 17 17 ⎝ 17 17 16 51 3 P= ⇒ ⇒ P= 17 7 112 16. (a) : Let the progression be 2, *, *, *, ....., *, 20 Let the common difference be d. ? t10 = 20 2 + 9d = 20 d = 2 First mean = a + d = 2 + 2 = 4 Second mean = a + 2d = 2 + 4 = 6 Third mean = a + 3d = 2 + 6 = 8 ⇒

... (i) ... (ii)

Eight mean = a + 8d = 2 + 16 = 18 8 ? Sum of all the 8 means = [ 4 + 18] = 4[22] = 88 2 17. (d) : Let the roots be x , x , xy. y

Product of the roots = x3 = −

1/3

d ⎛ −d ⎞ ⇒x=⎜ ⎟ ⎝ a ⎠ a

1/3

Now putting x = ⎛⎜ − d ⎟⎞ ⎝ a⎠ is its root, we get 2

in the given equation as x 1

⎛ d⎞ ⎛ d ⎞3 ⎛ d ⎞3 a ⎜− ⎟ + b ⎜− ⎟ + c ⎜− ⎟ + d = 0 ⎝ a⎠ ⎝ a⎠ ⎝ a⎠ 2

1

2

⎛ d ⎞3 ⎛ d ⎞3 ⎛d ⎞ ⎛d ⎞ ⇒ b ⎜ ⎟ = c ⎜ ⎟ ⇒ b3 ⎜ ⎟ = c3 ⎜ ⎟ ⎝a⎠ ⎝a⎠ ⎝a⎠ ⎝a⎠ ⎛d ⎞ ⇒ b3 ⎜ ⎟ = c3 ⎝a⎠

⎡ ⎛ d ⎞⎤ ⎢dividing both sides by ⎜⎝ a ⎟⎠ ⎥ ⎣ ⎦

b3d = ac3 18. (b) : Let the father's age be F and that of his elder and younger child respectively be E and Y. ? F = 2E + 4Y ...(i) 2 EY =6 ...(ii) and ...(iii) EY = 4 3 E +Y Solving (ii) and (iii) we get E = 12 and Y = 4 ? (i) becomes F = 2 × 12 + 4 × 4 = 24 + 16 = 40 Thus, the father's age is 40 years. 19. (d) : a1 = 2, an – an – 1 = 2n n t 2 a2 = a1 + 2.2 a2 = 6 a3 = a2 + 3.2 a3 = 12, a4 = 20 S = 2 + 6 + 12 + 20 + ..... + an ....(i) S = 2 + 6 + 12 + .... + an – 1 + an ....(ii) (i) – (ii) gives 2 + 4 + 6 + 8 + .... + n terms – an = 0 an = 2 (1 + 2 + 3 + ..... + to n terms) n(n + 1) an = 2 = n(n + 1) ? 6an = 6n2 + 6n 2 n(n + 1)(2n + 1) n(n + 1) ? S20 = 3080 ⇒ Sn = + 6 2 20. (c) : Let the sides of triangle be a, ar, ar2. ? r > 1 ? ar2 is greatest side ? a + ar > ar2 r2 – r – 1 = 0 ⇒

1− 5 1+ 5 1+ 5

77

∴ [r ] = 1. Also −

1+ 5 < −r < −1 (∴ [−r ] = 2) 2

23. (c) : Given

?[r] + [–r] = 1 – 2 = – 1 21. (c) : For parabola y2 = 4ax, the line y = mx + c a will be tangent if c = . m Equation of tangent to parabola y2 = 8x 2 is y = mx + . m Now line to be chord of the circle x 2 + y 2 = 16 if distance from (0, 0) to the line will be less than the radius of circle. y 2 m <4 ∴ ⇒

1 + m2 2

(2, 0)

<4

2

O

x¢

m 1+ m 1 < 4m2(1 + m2) 4m4 + 4m2 – 1 > 0

(4, 0) x

y¢ 2

⇒ m4 + m2 −

1⎞ 1 1 ⎛ > 0 ⇒ ⎜ m2 + ⎟ − > 0 ⎝ 2⎠ 2 4

⎛ 1 1 ⎞⎛ 2 1 1 ⎞ ⇒ ⎜ m2 + + ⎟ ⎜m + 2 − ⎟=0 ⎝ 2 2 ⎠⎝ 2⎠ 2 −1 1 1 or m2 > ⇒ m2 > − + 2 2 2 ⇒ |m| >

2 −1 2

⎛ 2 −1 ⎞ ⎛ 2 −1 ⎞ m ∈ ⎜ −∞, − ,∞⎟ ⎟∪⎜ ⎜⎝ ⎟⎠ 2 ⎟⎠ ⎜⎝ 2 22. (d) : Equation of tangent at (t2 , 2t) is x = ty – t2 ...(i) Equation of normal to ellipse at ( 5 cos α, 2 sin α) ...(ii) is 5x sec α − 2 y cos ecα = 1 Given (i) = (ii) 2 cos ecα 1 ⇒ 5 sec α = =− t t2 ⇒ cos α = − 5t 2 and sin α = −2t cos2D + sin2D = 5t4 + 4t2 = 1 1 2t 2 1 2 sin α ⇒ t 2 = . Now, = × = × (±5) = 2 5 cos α − 5t 5 t 5 ? tanD = ±2 tanD = –2, tanD = 2 78

MATHEMATICS TODAY | OCTOBER ‘17

x2 a2

∴ OS = ae = a 1 −

+

y2 b2

=1

b2 a2

= a2 − b2 So two points on the minor axis are S1 (0, a2 − b2 ), S1′(0, − a2 − b2 )

Let tangent to the ellipse be y = mx + c = mx + a2m2 + b2 where m is parameter. Now sum of the squares of perpendiculars on the tangent from the points S1 and Sc1 is ⎛ a2 − b2 − a2m2 + b2 ⎜ 1 + m2 ⎝⎜

2

⎞ ⎛ a2 − b2 − a2m2 + b2 ⎟ + ⎜− ⎟⎠ ⎜⎝ 1 + m2

⎛ a2 − b2 + a2m2 + b2 ⎞ 2a2 (1 + m2 ) = 2a2 = 2⎜ ⎟= 2 2 1+ m 1+ m ⎠ ⎝ 24. (c) : L e t P ( x , y ) b e any p oint on t he parabola and say PM and PT are perpendicular from P on the axis and tangent at the vertex. ? (PM)2 = latus rectum (PT) 2

⎛x− y+4⎞ ⎛x + y −2⎞ ∴ ⎜ ⎟ ⎟ = 2⎜ ⎝ 12 + 12 ⎠ ⎝ 12 + 12 ⎠ ⇒ (x + y − 2)2 = 2 2 (x − y + 4) 25. (c) : We know that semi latus rectum of a parabola is the H.M. between the segment of any focal chord of a parabola. ? AS, 4, SB are in H.P. ⇒ 4 =2× ⇒ 4=

AS ⋅ SB AS + SB

2 × 4SB ⇒ SB = 4 4 + SB

2

⎞ ⎟ ⎟⎠

MATHEMATICS TODAY | OCTOBER‘17

79

2 26. (a) : y = 4x2 ⇒ x =

⇒

2

2

y y y − = 1 becomes − =1 2 16 2 16 a 4a 4y – a2y2 = 16a2 a2y2 – 4y + 16a2 = 0 D t 0 for intersection of two curves 16 – 4a2 (16a2) t 0 1 – 4a4 t 0 1 1 ( 2a2 )2 ≤ 1 ⇒ | 2a | ≤ 1 ⇒ − ≤ a≤ 2 2 x

∴

2

(3 2 cos θ, 4 2 sin θ) and slope of the tangent

1 y 4

27. (b) : Equation of normal to the curve y2 = 4x at (m2 , 2m) is taken as 1 y − y1 = − (x − x1 ) ⎛ dy ⎞ ⎜⎝ dx ⎟⎠ 2 (m ,2m) (y – 2m) = –m (x – m2) y + mx – 2m – m3 = 0

...(i) 1 ⎛ ⎞ 2 Similarly normal to y2 – 2x + 6 = 0 at ⎜ t + 3, t ⎟ is ⎝2 ⎠ 1 ...(ii) y + t (x − 3) − t − t 3 = 0 2 Shortest distance between two curves exist along the common normal. Let (i) and (ii) are same 1 ∴ − 2m − m3 = −4m − m3 ⇒ m = 0, m = ±2 2 Points on the parabolas (m2 , 2m) = (4, 4) ⎛1 ⎞ and ⎜ m2 + 3, m ⎟ = (5, 2) ⎝2 ⎠

b2 x

=−

Again chord of contact of tangent from (x2, y2) is xx2 yy2 x + = 0 ∴ m2 = − 2 52 13 4 y2 Given tangents are right x1 x2 ⇒ ⋅ = −1 ⇒ 4 y1 4 y2

angle ? m1m2 = –1 x1x 2 16 =− y1 y2 1

29. (c) : Any point on the ellipse y2 + = 1 can be taken as (3 2 )2 (4 2 )2 MATHEMATICS TODAY | OCTOBER ‘17

....(ii) π 4

Hence equation of tangent is 1 1 x⋅ y⋅ 2+ 2 =1 ⇒ x + y =1 6 8 3 2 4 2 Hence A (6, 0), B (0, 8) 1 Area of 'OAB = × 6 × 8 = 24 sq. units 2 2 2 30. (a) : The given ellipse is x + y = 1 4 1 i.e., the point A, the corner of the rectangle in 1st quadrant is (2, 1). Again the ellipse circumscribing the rectangle passes through the point (4, 0), so its

y

2 2 equation is x + y = 1 . A(2, 1) 16 b2 A(2, 1) lies on the above x x O (4, 0) ellipse 4 1 4 ⇒ + = 1 ⇒ b2 = 2 y 16 b 3 Thus the equation to the required ellipse is

x2 3 2 + y = 1 ⇒ x2 + 12 y2 = 16 16 4

1. 2. 3. 4. 5. 6. 7. 8. 9.

x2

80

....(i)

From (i) and (ii), we get cot θ = 1 ⇒ θ =

∴ Shortest distance = (5 − 4)2 + (4 − 2)2 = 5 28. (a) : Equation of chord of contact of tangent from (x1, y1) to the ellipse is xx1 yy1 −x + = 0 ⇒ m1 = slope = 1 52 13 4 y1

32(3 2 cos θ)

4 = − cot θ 3 18(4 2 sin θ) a2 y 4 Given slope of the tangent = − 3 −

1.

Solution Sender of Maths Musing SET-176 Satya Dev (Bangalore) Dohita Banerjee (West Bengal) Riku Pramanik (West Bengal) Santak Panda (Odisha) V. Damodhar Reddy (Telangana) Sannibha Pande (West Bengal) N. Jayanthi (Hyderabad) Khokon Kumar Nandi (West Bengal) Abhinav Kumar (Punjab) SET-177 Prashant Pandey (Nagpur)

10 BEST

PROBLE

MS

Math Archives, as the title itself suggests, is a collection of various challenging problems related to the topics of JEE Main & Advanced Syllabus. This section is basically aimed at providing an extra insight and knowledge to the candidates preparing for JEE Main & Advanced. In every issue of MT, challenging problems are offered with detailed solution. The readers’ & comments and suggestions regarding the problems and solutions offered are always welcome. 1. Let Tr be the rth term of an A.P. whose first term is a and common difference is d. If for some positive 1 1 integers m, n, m ≠ n, Tm = and Tn = , then a – d equals n m 1 1 1 (d) (a) 0 (b) 1 (c) + mn m n G G G G 2. If a = 2i − 2 j + k then GGG G GGG G G G [i a a × i ] + [ j a a × j ] + [k aG aG × k ] = (a) –9 (b) 9 (c) 18 (d) –18 10

3.

10

Let S1 = ∑ j( j − 1)( C j ), S2 = ∑ j ( C j ) 10

10

j =1

10

j =1

and S3 = ∑ j2 (10 C j ) j =1

Statement –I: S3 = 55 × 29 Statement –II: S1 = 90 × 28 and S2 = 10 × 28 (a) Both Statement –I and Statement –II are true and Statement –II is correct explanation of Statement –I. (b) Both Statement –I and Statement –II are true and Statement –II is not correct explanation of Statement –I. (c) Statement –I is true but Statement –II is false. (d) Statement –I is false but Statement –II is true. ⎡ π ⎤ 4. Let f :[0, 3 ] → ⎢0, + loge 2⎥ defined by ⎣ 3 ⎦ 2 −1 f (x ) = loge x + 1 + tan x , then f (x) is (a) one – one and onto (b) one – one but not onto (c) onto but not one – one (d) neither one – one nor onto π P Q 5. In ΔPQR, ∠R = , If tan and tan are roots of 2 2 2 the equation ax2 + bx + c = 0, (a z 0) then

(a) a + b = c (b) b + c = a (c) a + c = b (d) b = c 6. T [0, 2S] and z1, z2, z3 are three complex numbers such that they are collinear and (1+ |sin θ |)z1 + (| cos θ | −1)z2 − 2z3 = 0. If at least one of the complex numbers z1, z2, z3 is non- zero then number of possible values of T is (a) f (b) 4 (c) 2 (d) 8 5

7.

∫ (x − 1)(x − 2)(x − 3)(x − 4)(x − 5)dx = 1

(a) 0 (b) 5 (c) 1 (d) 4 8. The differential equation whose solution is Ax2 + By2 = 1 where A and B are arbitrary constants, is of (a) first order and second degree (b) first order and first degree (c) second order and first degree (d) second order and second degree 9. The number of critical points of the curve f(x) = (x – 2)2/3 (2x + 1) are (a) 1 (b) 2 (c) 3 (d) 0 10. The sides of a triangle are 3x + 4y, 4x + 3y and 5x + 5y where x, y > 0 then the triangle is (a) right angled (b) obtuse angled (c) equilateral (d) isosceles SOLUTIONS 1. (a) : Tm = 1 / n = a + (m − 1)d ...(i) ...(ii) Tn = 1 / m = a + (n − 1)d 1 1 a–d=0 From (i) and (ii) we get, a = ,d = mn G mn G G G G G G 2. (d) : − (a × i )2 + (a × j )2 + (a × k )2 = −2 | a |2

{

}

By : Prof. Shyam Bhushan, Director, Narayana IIT Academy, Jamshedpur. Mob. : 09334870021

MATHEMATICS TODAY | OCTOBER‘17

81

10

3. (c) : S1 = ∑ j( j − 1) 10

j =1

10 ! j( j − 1)( j − 2)!(10 − j)!

8! = 90 ⋅ 28 − − − 2 8 2 ( j )!( ( j ))! j =2

= 90 ∑ 10

10 ! j j ( − 1 )!( 9 − ( j − 1))! j =1 10 9! = 10 ∑ = 10 ⋅ 29 − 1 9 − − 1 ( j )!( ( j ))! j =1

S2 = ∑ j

10

S3 = ∑ [ j( j − 1) + j] j =1

10

j =1

j =1

x +1

⇒ f (x ) is increasing in[0, 3 ] x2 + 1 P Q π 5. (a) : + = ⇒ c = a +b 2 2 4 6. (b) : If z1, z2, z3 are collinear and az1 + bz2 + cz3 = 0 then a + b + c =0 .Hence 1+ |sin θ | + | cos θ | −1 − 2 = 0 ⇒ |sin θ | + | cos θ |= 2 ⇒ θ =

10 ! j !(10 − j)!

10

4. (a) : f ′(x ) =

= ∑ j( j − 1)10 C j + ∑ j 10C j = 90 ⋅ 28 + 10 ⋅ 29

7. (a) : Put x – 3 = t 8. (c) : Differentiate the given equations for 2 times and eliminate A, B 9. (b) : f c(x) = 0 or undefined at x = 1, 2 10. (b) : Let cos C =

= 90 ⋅ 28 + 20 ⋅ 28 = 110 ⋅ 28 = 55 ⋅ 29

π 3π 5π 7 π , , , 4 4 4 4

(3x + 4 y )2 + (4 x + 3 y )2 − (5x + 5 y )2 <0 2(3x + 4 y )(4 x + 3 y )

Computer science losing out, mechanical hot pick in BTech Versatile Job Opening Key To New Trend Engineering is being revisited. Even as seats in this professional course are reducing across the country (from 16.3 lakh in 2013-14 to about 14.7 lakh this year), mechanical engineering seems to be emerging as the hottest pick in times of uncertainty in the IT and software industry. While engineering continues to be a big draw, its 70-odd options undergo a life cycle of their own. Experts say industry growth, which translates into more jobs and higher incomes, is what decides the path that colleges and students take. And many feel the sun is setting on the computer science engineering stream. While 25.44% of all students opted for it in 2013-14, that has dropped to around 24% this year. At the same time, mechanical engineering is racing ahead after pipping electronics and communications, which used to be the second most popular choice for four years. While 20.22% of students chose mechanical engineering in the past four years, that proportion has increased to 21.6% now. Core courses like civil and electrical engineering are also expected to be top on the charts.

COURSE CORRECTION STREAM Computer Science Mechanical Electronics and Comm Civil Electrical

2013/14 4,14,762

2014/15 4,08,559

INTAKE 2015/16 3,88,077

2016/17 3,68,267

2017/18 3,53,825

3,30,331 3,43,180

3,58,052 3,41,916

3,47,480 3,22,292

3,35,603 3,02,800

3,17,444 2,80,653

2,22,510 2,10,803

2,55,139 2,18,071

2,46,903 2,09,653

2,40,168 2,00,570

2,28,841 1,90,780

Although data from the All India Council for Technical Education (AICTE) shows that seats in undergraduate engineering are reducing, experts feel the course will continue to have lakhs of takers. “Engineering has become a broad-based course like BA, BCom and BSc. From here, students go on to do several courses, says IIT-Madras director Bhaskar Ramamurthy. “Mechanical engineering is a great branch. 2QH FDQ ¿W LQWR D ORW RI LQGXVWULHV DIWHU 0(³ adds Ramamurthy. “ But given some amount of uncertainty in the IT sector, there are more takers for mechanical-because mechanical students can join IT companies, but the reverse is not possible.” From information technology to leather technology, each of these courses tests the ability of individuals to engineer a lasting impression, may be an innovation, a new process, a fresh thought. But JHQHUDOL]HGFRXUVHVDUHOLNHO\WREHWKHÀDYRXUIRU

the coming few years as several industries see a departure from the old. G D Yadav, vice-chancellor of the Institute of Chemical Technology, says mechanical is rising in popularity because manufacturing industry needs these engineers. “There is so much new construction, new infrastructure, machinery and mechanical engineers are needed everywhere.” Hence, computer science and mechanical engineers will be an equal force in the coming year,” he adds. If the current trend in the USA in anything to go by, the demand for chemical engineers too is likely to rise. Enrolment in this programme has doubled in American universities this year thanks to shale gas companies that have upped their recruitment numbers, says Yadav. Renewable energy and electric cars, many forecast, will also see the energy engineering space powering up. Courtesy: TOI

82

MATHEMATICS TODAY | OCTOBER ‘17

MATHEMATICS TODAY | OCTOBER‘17

83

4. (d) : We have, SOLUTION SET-177

1. (c) : x2 – (y + 42)2 = 2012 – (42)2 = 248 = 124 u 2 ? x + y + 42 = 124, x – y – 42 = 2 x = 63, y = 19, x – 3y = 6 2. (c) : Given g (x ) =

(x − 1)n log cosm (x − 1)

, 0 < x < 2,

⎧ x − 1; x ≥ 1 m 0, n are integers and | x − 1 |= ⎨ ⎩1 − x ; x < 1 The left hand derivative of |x –1| at x = 1 is p = –1 Also, lim g (x ) = p = −1 x →1+

⇒ lim

n

(1 + h − 1)

h →0 log cosm (1 + h − 1)

⇒ lim

hn

h →0 log cosm h

= −1

= −1

n ⋅ hn−1 = −1 [Using LcHospital’s rule] 1 h→0 m (− sin h) cos h

⇒ lim

hn − 2 ⎛n⎞ n ⇒ ⎜ ⎟ lim = 1 ⇒ n = 2 and = 1 ⎝ m ⎠ h→0 ⎛ tan h ⎞ m ⎜⎝ ⎟ h ⎠ m=n=2 3. (d) : We have, dx cos t sin 2t ⎤ −a sin 3t ⎡ = a ⎢ − sin t cos 2t − ⎥= dt cos 2t ⎦ cos 2t ⎣ and

dy sin t sin 2t ⎤ a cos 3t ⎡ = a ⎢cos t cos 2t − ⎥= dt cos 2t ⎦ cos 2t ⎣

dy dy / dt d2 y dt ∴ = = − cot 3t ⇒ = 3 cosec2 3t ⋅ 2 dx dx / dt dx dx =

−3 co sec2 3t ⋅ cos 2t

∴

⇒

84

⎛3⎞ = − ⎜ ⎟ cosec3 3t cos 2t ⎝a⎠

a sin 3t [1 + (dy / dx )2 ]3/2 d 2 y / dx 2

=

(1 + cot 2 3t )3/2 ⎛ 3⎞ 3 ⎜⎝ − ⎟⎠ cosec 3t cos 2t a

[1 + (dy / dx )2 ]3/2 d 2 y / dx 2

t = π/6

2a = 3

MATHEMATICS TODAY | OCTOBER ‘17

sin3 x / 2

∫

x cos cos3 x + cos2 x + cos x 2 1 2 sin(x / 2)cos( x / 2)(2 sin2 x / 2) dx = ∫ 2 2 x 3 2 2 cos cos x + cos x + cos x 2 1 sin x(1 − cos x ) dx = ∫ 2 (1 + cos x ) cos3 x + cos2 x + cos x =−

dx

1 1− t dt , ∫ 2 (1 + t ) t 3 + t 2 + t where t = cosx and –sinxdx = dt 2

=

1 t −1 dt ∫ 2 (t + 1)2 t 3 + t 2 + t

=

1 t2 −1 dt 2 ∫ (t 2 + 2t + 1) t 3 + t 2 + t

1 1− 2 1 1 t dt . = ∫ 2 du = ∫ 2 ⎛ 1⎞ 1 u +1 ⎜⎝ t + 2 + ⎟⎠ t + + 1 t t = tan–1 u + c

1 1⎞ ⎛ where t + + 1 = u2 ⇒ ⎜1 − 2 ⎟ dt = 2udu ⎝ t ⎠ t

1 = tan −1 t + + 1 + c = tan −1 cos x + sec x + 1 + c t 5. (d) : The equation representing the coaxial system of circles is x2 + y2 + 2gx + c + O(x2 + y2 + 2fy + k) = 0 2g 2fλ c + kλ …(i) ⇒ x2 + y2 + =0 x+ y+ + + λ λ 1 1 1+ λ The coordinates of the centre of this circle are g fλ ⎞ ⎛ ,− …(ii) ⎜⎝ − ⎟ 1+ λ 1+ λ ⎠ and radius =

g 2 + f 2 λ 2 − (c + k λ)(1 + λ)

(1 + λ)2 For the limiting points, we must have radius = 0 g2 + f 2O2 –(c + kO)(1 + O) = 0 O2(f 2 – k) – O(c + k) + (g2 – c) = 0 …(iii)

477*3(::?00 1.

(d)

2. (b)

ANSWER KEY 3. (b)

4.

(c)

5. (b)

6. (a) 7. (b,c,d) 8. (a) 9 . (a,b,c) 10. (a) 11. (a,b,c,d) 12. (a,d) 13. (a,c,d) 14. (c) 15. (c) 16. (c) 17. (1) 18. (2) 19. (4 ) 20. (0)

Let O1 and O2 be the roots of this equation. Then, g2 − c c+k and λ1λ2 = 2 …(iv) λ1 + λ2 = 2 f −k f −k Thus, the coordinates of limiting points L1 and L2 are − f λ1 ⎞ g ⎛ and , L1 ⎜ − ⎝ 1 + λ1 1 + λ1 ⎟⎠

⎛ − g − f λ2 ⎞ , L2 ⎜ ⎝ 1 + λ2 1 + λ2 ⎟⎠ [From eq. (iv)] Now, L1L2 will subtend a right angle at the origin, if Slope of OL1 × Slope of OL2 = –1 f λ1 f λ2 2 2 ⇒ × = −1 f O1O2 = –g g g 2 ⎛ g −c ⎞ 2 2 2 2 2 ⇒ f2⎜ 2 ⎟ = − g f (g – c) + g (f – k) = 0 ⎝ f −k⎠ c k ⇒ 2 + 2 =2 g f 6. (a) : Possibilities are (3, 1, 1), (2, 2, 1) 5

Number of ways =

C2 ⋅ 3C2 ⋅ 1C1

= (10 + 15) = 25

2!

5

+

7. (c) : We have, f (x ) = g (x ) = 9 1995

∑

Now,

r =1

⎛ r ⎞ = f⎜ ⎝ 1996 ⎟⎠

⎛ 1 ⎞ + f⎜ ⎝ 1996 ⎟⎠

1

C3 ⋅2C1 ⋅ C1 2! 9x

9x + 3

⎛ 2 ⎞ + f⎜ ⎝ 1996 ⎟⎠

⎛ 1 ⎞ ⎛ 1995 ⎞ ⎫ ⎧ ⎜⎝ ⎟⎠ + f ⎜⎝ ⎟⎬ + ⎨ f 1996 1996 ⎠ ⎭ ⎩

⎧ +⎨ f ⎩

⎧ ⎛ 3 ⎞ ⎛ 1993 ⎞ ⎫ ⎜⎝ ⎟⎠ + f ⎜⎝ ⎟⎠ ⎬ + ... + ⎨ f 1996 1996 ⎭ ⎩

⎡1995 ⎢∑ f ⎢⎣ r =1

⎛ 3 ⎞ f⎜ ⎝ 1996 ⎟⎠

8. (c) :

∑

r =0

=

1 1+ a

+n =

r =1

1 1+ a

+ 987

y−k = −

∴

y12 (1 + y12 )2 + (1 + y12 )2 = r 2 y22

MADHYA PRADESH at • • •

⎛ 1994 ⎞ ⎫ ⎛ 2 ⎞ ⎟⎬ ⎜⎝ ⎟⎠ + f ⎜⎝ 1996 1996 ⎠ ⎭ ⎛ 999 ⎞ ⎫ ⎛ 997 ⎞ ⎟⎬ ⎜⎝ ⎟⎠ + f ⎜⎝ 1996 ⎠ ⎭ 1996 ⎛ 998 ⎞ +f ⎜ ⎝ 1996 ⎟⎠

2n ⎛ r ⎞ = f (0) + ∑ f f⎜ ⎟ ⎝ 2n + 1 ⎠

(1 + y12 ) y , x − h = 1 (1 + y12 ) y2 y2

∴

…(i)

⎛ r ⎞⎤ ⎜⎝ ⎟ ⎥ = 997 1996 ⎠ ⎥⎦

2n

S. (x – h)2 + (y – k)2 = r2, Differentiating twice, (x – h) + (y – k)y1 = 0, 1 + (y – k)y2 + y12 = 0

1 ⎛1⎞ = 1 + ... + + f ⎜ ⎟ = 997 + [from (i)]= 997.5 1 + 1 1 + 1 ⎝2⎠ 2 997 times ∴

1 in the second integral, u e tan x e e ⎤ tdt udu tdt 2 + ∫ = ∫ = ln 1 + t ⎥ = ln e = 1 ∫ 2 2 2 ⎥⎦1/e tan x 1 + u 1/e 1 + t 1/e 1 + t

R. Setting t =

or r 2 y22 = (1 + y12 )3 ∴Degree is 2.

⎛ 1993 ⎞ ⎛ 1994 ⎞ ⎛ 1995 ⎞ +... + f ⎜ +f⎜ +f⎜ ⎝ 1996 ⎟⎠ ⎝ 1996 ⎟⎠ ⎝ 1996 ⎟⎠ ⎧ =⎨f ⎩

9. (8) : 22003 = 23 · (24)500 = 8(17–1)500 = 8(17 O+ 1), O I ? Remainder = 8 1/ x tan x − x 0 ⎛ tan x ⎞ = exp lim = e =1 10. (d) : P. lim ⎜ ⎟ ⎝ ⎠ x →0 x →0 x x2 Q. f(x) is not differentiable at x = 0, ±1

• • • • • • • • • • • • • •

⎛ r ⎞ ⎜⎝ ⎟ 2n + 1 ⎠

•

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(given)

? n = 987 MATHEMATICS TODAY | OCTOBER‘17

85

86

MATHEMATICS TODAY | OCTOBER ‘17

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