2nd-Dispatch DLPD IIT-JEE Class-XII English PC(Maths)

CONTENT

MATHEMATICS CLASS : XII Preface

1.

Functions Exercise

2.

79 - 100

Application of derivatives Exercise

6.

54 - 78

Method of differentiation Exercise

5.

29 - 53

Continuity and derivability Exercise

4.

01 - 28

Limits Exercise

3.

Page No.

101 - 126

Solution of triangle Exercise

127 - 149

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FUNCTIONS EXERCISE # 1 PART - I Section (A) : A–3.

(i)

f(x) =

f(x) =

x 3  5x  3 x2  1 x 3  5x  3 ( x  1)( x  1)



Division by zero is undefined Domain x  R – {1, –1}

 

x±1 x  (–, –1)  (–1, 1)  (1, )

(ii)

f(x) =

sin1 x x For sin–1x, x  [–1, 1] and division by zero is undefined x  0



Domain x  [–1, 0)  (0, 1]

1 (iii)

(iv)

(v)

f(x) =

x | x |

for function to be defined x + |x| > 0 for x > 0, x + |x| = 2x > 0 for x  0, x + |x| = 0  Domain is x  (0, ) f(x) = ex + sin x Domain x  R as there is no restriction for exponent of e.

1 f(x) = log (1  x ) + 10

x2



1 – x > 0 and x + 2  0 x  (– , 1) – {0} and x  – 2

(vi)

f(x) =

and 

1–x1 x  [–2, 0)  (0, 1)



x

 3x  1 –1 1 2x + 3 sin  2 

1 – 2x  0 and – 1 

3x  1 1 2

1 2

and –

1 x1 3

Taking intersection 

 1 1 Domain x   ,   3 2

(vii)

f(x) = 2 sin

1

x

x

1

1 

x



x  (0, 1)  (1, ) and 0 < x –

  1  f(x) = logx  log 2  x  1/ 2      In case of composite function in log. We start with outer log. x > 0, x  1 and



1/2

–

x2 – 1  x  1 and x > 2 (viii)

–1/3

1

1 >1 x 2 1 3 x  (0, ) – {1} and
0

RESONANCE

1/2

1

3/2

1 <1 2

x

S OLUTIONS (XII) # 1

Taking intersection

A–6.



1   3 x   , 1   1,  2    2

(i)

f(x) = 3 sin x + 4 cos x + 5

 

– 3 2  4 2  3 sin x + 4 cos x  3 2  4 2 – 5  3 sin x + 4 cos x  5  0  3 sin x + 4 cos x + 5  10 Range y  [0, 10]

(ii)

f(x) =

1

>  (iii)

1 x x 0

1

0<

1 x f(x) = n (sin–1x) Domain sin–1 x > 0 Range 0 < sin–1 x 

(iv)

1



>1+

x 1



Range y  (0, 1]



x  (0, 1]

 2

 –  < n (sin–1x)  n   2 Inequality doesn't change as n is increasing function f(x) = 2 – 3x – 5x2 Domain x  R Method 1 y = – 5x2 – 3x + 2 opening downward parabola 0

 D  Range y     , 4a  



49   y    ,  20   Method 2 5x2 + 3x + (y – 2) = 0

D0 (v)

– D/4a 2

9 – 20 (y – 2)  0





20y – 49  0



y

49 20

f(x) = 3 |sin x| – 4|cos x| f(x) is a periodic function with period . So analysis is limited in [0, ]

 , |sin x| = 1, |cos x| = 0 2 fmin = 3.0 – 4.1 = – 1 at x = 0, |sin x| = 0, |cos x| = 1 fmax = 3.1 – 4.0 = + 3 at x =

(vi)

f(x) =

sin x

Range y  [–4, 3]

cos x

1  tan x 1  cot 2 x f(x) = sin x |cos x| + cos x |sin x| periodic period = 2   sin 2x    0 f(x) =   sin 2x    0 



+

2



  x   0,  2    , x   ,  2    , x   ,  2   3  , x   , 2  2   ,

Range y  [–1, 1]

RESONANCE


Section (B) : B–1.

(ii)

2 x 2 and g(x) = ( x ) Domain x  R, Domain x  [0, ) non-identical functions f(x) = sec(sec –1x) and g(x) = cosec (cosec –1x) Domain x  (–, –1]  [1, ), Domain x  (–, –1]  [1, ) f(x) = x g(x) = x Identical functions

(iii)

f(x) =

(i)

(iv)

B–5.

(i)

(ii)

(iii)

(iv)

f(x) =

1 cos x and g(x) = cos x 2 f(x) = |cos x| non-identical function f(x) = x and g(x) = enx, Domain x  R+ Domain x  R non-identical function f(x) = ex and g(x) = n x fog(x) = en x = x, x > 0 gof(x) = n ex = x, x  R f(x) = |x| and g(x) = sin x fog(x) = f(sin x) = |sin x| gof (x) = g(|x|) = sin |x| f(x) = sin–1x and g(x) = x2 fog(x) = sin–1(g(x)) = sin–1x2 gof(x) = (f2(x)) = (sin–1x)2 f(x) = x2 + 2, g(x) =

fog(x) = g2(x) + 2 =

gof(x) =

B–6.

x x 1

x2 ( x  1)2

+2=

3x 2  4x  2 ( x  1)2

f (x) x2  2 = 2 f (x)  1 x 1

1  x 2 , x 1 f(x) =   1  x , 1  x  2 g(x) = 1 – x, – 2  x  1

 1  g2 , g( x )  1  x  [0,1] fog(x) =  1  g( x ) , 1  g( x )  2  x  [–1,0)



1  (1 – x )2 fog (x) =   1  (1 – x )

,

x  [0,1]

, x  [–1, 0)



2 – 2x  x 2 , x  [0,1] fog (x) =   2 – x , x  [–1, 0)

Section (C) : C–1.

(i)

y = |(x + 2) (x + 3)| many - one function

(ii)

y = |nx| many - one function

RESONANCE


(iii)

   f(x) =sin 4x, x   – ,   8 8

 2 one-one function period =

1 1 , x  (0, ) x x many one function

(iv)

f(x) = x +

(v)

f(x) =

1   –1 

1 – e x 1   –1 .

e x f =

2 1– e

1 x2  0

1   –1 x 

increasing function

Hence one - one

C–5.

3x 2 – cos( x ) 4 Hence many - one

(vi)

f(x) =

(vii)

f(x) = sin–1 x – cos–1 x = 2 sin–1 x –

(i) (ii) (iii)

f(x) = sin (x2 +1)  f(x) = x + x2  f(x) = x – x3 

f(– x) = f (x) = even function f(– x) = x2 – x  f (x) or – f(x) Neither even nor odd function f(–x) = – x + x3 = – f(x) odd function

(iv)

 a x – 1   f(x) = x  x   a  1

 ax – 1    f(–x) = – x  – x   a  1



even function

 monotonically increasing. 2

 a x – 1   = f(x) even function f(–x) = x  x   a  1

(v)

f(x) = log (x +

(vi)

2 2   f(x) + f(– x) = log ( x  x  1)(– x  x  1) = log [(x2 + 1) – x2] = 0 hence odd function   f(x) = sin x + cos x  f(– x) = – sin x + cos x  f(x) or – f(x) Neither even nor odd. f(x) = (x2 – 1) |x|  f(–x) = f(x) even function.

(vii)

(viii)

x2  1 )



f(–x) = log (–x +

x2  1 )

 | tan(tan 1 x ) | x    2 x [ 2 x ] 1        x 1 f(x) =   sec(sec 1 x ) x   x   | x |     x  x0   f(x) =  3 0  x 1   x x 

RESONANCE



x    x     x  x0   f(x) =  3 0  x 1   x x 


C–6.

(i)

x 2 – sin x , – 1  x  0 even extension of f(x) = f(–x) =   – x  e x , x  –1

(ii)

– x 2  sin x , – 1  x  0 odd extension of f(x) = – f(– x) =   x – e x , x  –1

Section (D) : D–2.

(i)

f(x) = 2 + 3 cos (x – 2)

(ii)

f(x) = sin 3x + cos2 x + |tan x|



fundamental period = 2 

 2  period of f(x) = L.C.M.  , ,   = 2   3  f(x + ) = – sin x + cos2 x + |tan x|  f(x)

f(x) = sin

(iv)

f(x) = cos



for fundamental period 

fundament period = 2

3x sin 2x – 5 7

period 

10 ,7 3

 10  , 7  = 70  period of f(x) = L.C.M.  3  

Fundament period = 70

f(x) = [sin 3x] – |cos 6x|

period 

 2   2 period of f(x) = L.C.M.  ,  =  3 3 3

Fundamental period =

(vi)

f(x)=

(vii)

f(x) =

2 3

 3 2 3

1 fundamental period = 2 1  cos x sin12x

   period of f(x) = L.C.M.  ,  = 6 3 3

2

1  cos 6 x for fundamental period

  fx   = 6 

(viii)

2 , , 3

x x + sin 4 3 period  8, 6 period of f(x) = L.C.M. (8, 6) = 24 fundamental period = 24

(iii)

(v)

period 

  sin12 x   6    1  cos 2 6 x   6 

= f(x)



f(x) = sec3 x + cosec3 x  period  2 Fundamental period = L.C.M. (2, 2) = 2

Fundament period =

 6

2

Section (E) : E–1.

(i)

f:DR f(x) = 1 – 2–x



f (x) = 2– x n2 > 0 increasing function  one one function

D : [x  R), Range : (–, 1)  codomain 

function is not bijective

RESONANCE



f –1 does not exist S OLUTIONS (XII) # 5

(ii)

f(x) = (4 – (x – 7)3)1/5 f (x) =

1 (4 – (x – 7)3) 5

– 4/5

. (– 3 (x – 7)2)  0 decreasing function  one one function

Lim f ( x )  –  x 

Lim f ( x )  

x – 

D:R

Range : R = codomain

 onto function



function is bijective (invertible)

y = (4 – (x – 7)3)1/5 4 – y5 = (x – 7)3 x = 7 + (4 – y5)1/3 (iii)

f –1 (x) =

E–6.

f –1(x) = 7 + (4 – x 5)1/3

or

x=



f(x) = 1 ± xn

 or

f(x) = 1 – x3 f(1) = – 3

f(x) = n  x  1  x 2    D : x  R, Range : R y = n  x  1  x 2   

E–5.

or

e y  ey 2

e x  e x 2

 1  1 f(x) . f   = f(x) + f   x x f(3) = – 26 f(x) = – 3x2

f(x + y) = f(x) . f(y) and f(1) = 2 10

 f (n) = f(1) + f(2) + ........... + f(10)

n 1

 210  1    = 21 + 22 + 23 + ....... + 210 = 2  2  1  = 2046  

PART - II Section (A) : A–2.

For domain – log0.3(x – 1)  0  log0.3(x – 1)  0  (x – 1)  1  x2 Taking intersection x  [2, )

A–3.

f(x) = cot–1

x( x  3 ) + cos –1

for domain x(x + 3)  0  x  (–, –3]  [0, ) Taking intersection x  {–3, 0}

RESONANCE

and and and

x2 + 2x + 8 > 0  (x + 1)2 + 7 > 0  xR

x 2  3x  1

and and

0  x 2 + 3x + 1  1 x 2 + 3x + 1  0 and

x 2 + 3x  0  x  [–3, 0]


A–5.

f(x) = 4x + 2x + 1 Let 2x = t > 0,  x  R

f(x) = g(t) = t2 + t + 1,



t>0

2

1  3 g(t) =  t   + 2 4  2

1  1 t   > 2  2

1  1 t   > 2 4 



2



1  3 t   + >1 2 4 

Range is (1, )

Section (B) : B–2.*

–1

f(x) = en(sec x ) = sec–1x, x  (–, – 1]  (1, ) –1 g(x) = sec x, x  (–, – 1] [1, ) non-identical functions f(x) = tan (tan–1 x) = x, x  R g(x) = cot (cot–1 x) = x, x  R identical functions

(A)

(B)

 1 x 0  f(x) = sgn (x) =  0 x  0 – 1 x  0 

(C)



 1 x 0  g(x) = sgn(sgn x) =  0 x  0 – 1 x  0 

Identical functions f(x) = cot2 x . cos2 x, x  R – {n }, g(x) = cot2 x – cos2 x = cot2 x (1 – sin2 x) = cot2 x. cos2 x Identical functions

(D)

B–4.

B–6.*

Domain of f(g(x)) Range of g(x)  Domain of f(x)  – 5  |2x + 5|  7  – 12  2x  2 f(x) =

1– x , 1 x fog(x) =

0x1

n I x  R – {n },

n I

 

0  |2x + 5|  7 –6x1



g(x) = 4x (1 – x),0  x  1



–7  2x + 5  7

1 – g( x ) 1 – 4 x(1 – x ) 1 – 4 x  4x 2 = = 1  g( x ) 1  4 x(1 – x ) 1  4x – 4x 2

 1– x   1– x  gof(x) = 4f(x) . (1 – f(x)) = 4  1  x  1 –  1  x      

=

8 x(1 – x ) (1  x )2


One One / Many One f(x) =

f(x) =

f(x) =

2x 2  x  5 7 x 2  2x  10

, Domain x  R

( 4x  1)(7 x 2  2x  10)  (14 x  2)(2x 2  x  5) (7 x 2  2x  10)2 11x 2  30x  20 2

(7 x  2x  10)

2

 30  > 0  x  (– , 0)   ,    11   30   f (x) < 0  x   0,  11 

f(x) = 0

RESONANCE



x = 0,

30 11 S OLUTIONS (XII) # 7

Function is increasing and decreasing in different intervals, so non monotonic  Many one function. Onto / Into f(x) =

2x 2  x  5

7 x 2  2x  10 2x 2 – x + 5 > 0,  x  R and 7x 2 + 2x + 10 > 0  x  R a = 2 > 0 and  a = 7 and D = 4 – 280 < 0 D = 1 – 40 = – 39 < 0  f(x) > 0  x  R Also f(x) never tends to ± as 7x 2 + 2x + 10 has no real roots, Range  Codomain so into function.

C–3.

f(x) = x 3 + x 2 + 3x + sin x, f(x) = 3x 2 + 2x + 3 + cos x 

3x 2 + 2x + 3 

–1  cos x  1

lim f(x) = + 

xR

32 as a = 3 > 0 and D < 0 12 so f(x) > 0  x  R

lim f(x) = – 

x 

x  

Hence f(x) is one-one and onto function (as f(x) is continuous function) C–6.

f(g(x1)) = f(g(x2)) as f is one - one function hence f(g(x1)) = f(g(x2))  x1 = x2

 

g(x1) = g(x2) x1 = x2



f(g(x)) is one - one function

as

g is one - one function

Section (D) : 2 D–2.

f(x) = sin

 [a] x  .

[a] = 4 D–3.

Period =

[ a]

=

a  [4, 5)



f(x) = x + a – [x + b] + sin x + cos 2x + sin (3x) + cos (4x) + ........ + sin (2n – 1) + cos (2px) f(x) = {x + b} + a – b + sin (x) + cos (2x) + sin (3x) + cos (4x) + .... + sin (2n – 1) + cos (2nx) Period of f(x) = L.C.M (1, 2,

2 2 2 2 , , ........., , )=2 3 4 2n  1 2n

 period of f(x) = 2 since f(1 + x)  f(x) , hence fundamental period is 2 D–7.*

(A) (B)

f(x) = cos (cos–1 x) = x, x  [–1, 1] odd function f (x + ) = cos (sin (x +)) + cos (cos (x + )) f (x + ) = cos (sin x) + cos (cos x) = f(x)

          f  x   = cos  sin x    + cos  cos x    2  2 2       = cos (cos x) + cos (sin x) = f(x) fundamental period = (C)

 2

f(x) = cos (3 sin x), x  [–1, 1] – 3 sin1 3 sin x 3 sin 1  cos (3 sin 1) cos (3 sin x) 1



Range is [cos (3 sin1), 1]



 1 x   f–1(x) = n   1 x 


ex  ex y = x 1 e  e x By compnendo and dividendo

1 y 2e x = 1 y 2



RESONANCE

 1 y  x = n  1  y   


E–7.

E–8.

f(1) = 1 = 2 – 1 f(n + 1) = 2f(n) + 1  f(3) = 7 = 23 – 1 f(4) = 15 = 24 – 1 Similarly f(n) = 2n – 1

f(2) = 2f(1) + 1 = 2. 1 + 1 = 3 = 22 – 1

Method 1 : (usual but lengthy) x2 f(x) + f(1 – x) = 2x – x4 .....(1) replace x by (1 – x) in equation (1) (1 – x)2 f(1 – x)+ f(x) = 2 (1– x) – (1 – x)4 .....(2) eliminate f(1 – x) by equation (1) and (2) we get f(x) = 1 – x2 Method 2 : Since R.H.S. is polynomial of 4th degree and also by options consider f(x) = ax2 + bx + c x2 f(x) + f(1 – x) = 2x – x4  x 2 (ax2 + bx + c) + a (1 – x)2 + b (1 – x) + c = 2x – x4 by comparing coefficients a=–1 b=0 c=1  f(x) = – x2 + 1

EXERCISE # 2 PART - I 1.

(i)

f(x) =

or

3  2 x  2 .2  x 3 – 2x – 2 .2–x  0 (2x – 1) (2x – 2)  0

(ii)

f(x) =

(iii)

1 – 1 x 2  0 f(x) = (x2 + x + 1)–3/2 D:xR

(iv)

f(x) =

or  

(2x)2 – 3.2x + 2  0 2x  [1, 2] x [0, 1]

1 1 x2

x2 + x2



1 x 2 1



0  1 – x2  1

or

x 

or

x  2n

or

x  (0, 1]  [4, 5)



x [– 1, 1]

1 x 1 x

x2 1 x 0 and 0 x2 1 x x  (– , –2)  [2, ) and x  (–1, 1] D: (v)

f(x) =

tan x  tan 2 x

tan x – tan2x  0

or

0  tan x  1

 n 

  n , n  4   

1

(vi)

(vii)

f(x) = 2 sin x 2

f(x) =

sin

x 0 2

 5x  x 2   log1 / 4    4  

5x  x 2 1 4

RESONANCE

and

5x – x2 > 0


(viii) or 2.

(i)

f(x) = log10 (1 – log10(x2 – 5x + 16)) 1 – log10 (x2 – 5x + 16) > 0 x  (2, 3) f(x) = 1 – |x – 2| |x – 2|  [0, )

or

x2 – 5x + 6 < 0



f(x)  (– , 1]



1  f(x)   , 1 3 

1 (ii)

(iii)

f(x) =

x5 D : x  (5, ) R : f(x)  (0, ) 1 f(x) = 2  cos 3 x range of cos 3x is [–1, 1] cos 3x [–1, 1]

(iv)

x2

f(x) =

=y x  8x  4 x + 2 = yx2 – 8yx – 4y for x to be real D  0 (8y + 1)2 + 4y (4y + 2)  0 64y2 + 16y + 1 + 16y2 + 8y  0 2

80y2 + 24y + 1  0 (v)

f(x) =

(viii) (ix)

f(x) = 3 sin

(i)

1   1  ,  y     ,     4   20 

or

1  y   , 3 3 

2  x2 16

   D : x   ,   4 4



f(x) = x3 – 2x2 + 5 = (x2 – 1)2 + 4 R : [4, ) f(x) = x3 – 12x , x  [–3, 1] = x (x2 – 12) f(x) = 3x2 – 12 = 0 or f(x) = sin2x + cos4x = sin2x + 1 + sin4x – 2 sin2x = sin 4x – sin2x + 1 1  2 3 =  sin x   + 2  4

7.

or

=y x 2  2x  4 x2 – 2x + 4 = yx2 + 2xy + 4y x2 (1 – y) – 2x(1 + y) + 4(1 – y) = 0 D0

  2  x 2  0 ,   4 16

(vii)

yx2 – x (8y + 1) – (4y + 2) = 0

x 2  2x  4

4(1 + y)2 – 16(1 – y)2  0

(vi)

or

f(–x) =

 3  f(x)  0 ,  2 

x=±2

R : [–11, 16]

3  R :  , 1 . 4 

(2 x  1)7 (2 x )6

neither even non add (ii) (iii)

sec x  x 2  9 = f(x) x sin x f(–x) = – f(x) odd f(–x) =

RESONANCE

even


(iv)

 x  1  x2  2 [ x ] [ x ] 1      x  1 , even by graph of function f(x) =  2  x x 1 

(v)

f(x) =

2x(sin x  tan x ) x 2   1 

if x = n, f(n) = 0 if x  n

8.

 x   x    = – 1        



f(–x) = – f(x) odd function

(i)

f(x) = 1 –



  cos2 x sin2 x – fx   = 1 –  f(x) 2  1 – tan x 1 – cot x fundamental period = 

(ii)

(iii)

(iv)

sin2 x cos2 x – 1  cot x 1  tan x period of f(x) = L.C.M. (, ) =  For fundamental period

  f(x) = tan  [ x ]  : [x]  2n + 1 2  f (x) = 0 By graph fundamental period = 2 f(x) = log (2 + cos 3x) fundamental period of f(x) = fundamental period of (2 + cos 3x) (as log is a monotonic function) f(x) = en sin x + tan3x – cos (3x – 5) f(x) = sin x + tan3x – cos (3x – 5), sin x > 0

period  2 ;  ,

2 3

2    = 2 Period of f(x) = L.C.M.  2 , , 3  

(v)

x x x   x x x x   f(x) =  sin x  sin 2  sin 4  ....  sin n –1    tan  tan 3  tan 5  ...  tan n  2 2 2 2   2 2 2  

Period of f(x) = L.C.M. of 2, 23 , 25 ,......2n , 2, 23 .....2n  = 2 n



(vi)

f(x) =



sin x  sin 3 x cos x  cos 3 x

2 2   period of f(x) = L.C.M.  2. , 2,  = 2 3 3   For fundamental period

f(x + ) = 11.

f(x) =

sin ( x  )  sin(3 x  3 ) – sin x – sin 3 x = cos( x  )  cos(3 x  3) – cos x – cos 3 x

1 x



1 x2



f(x) = 0 at x = 1 ±



Fundamental period = 

2



for x   2  1, 1  2 f is bijective function hence f is invertible.

RESONANCE


1 x 1 x2 or

or

=y x2y + x + (y – 1) = 0 x=

 1  1  4 y( y  1) 2y

=

 1  4y  4 y 2  1

  1  4x  4x 2  1  , f–1(x) =  2x  , 1

2y

x0 x0

as f (1)  0

n

 f (a  k ) = 16 (2

n

13.

k 1

or

or Now or 15.

(i)



 (ii)

– 1)

f(a + 1) + f(a + 2) + ......... + f(a + n) = 16 (2n – 1) f(x + y) = f(x) . f(y) f(0) = 1, f(1) = 2 f(x) = 2x f(a + 1) + f(a + 2) + ........ + f(a + n) = 2a [2 + 4 + ......... + 2n] = 2a . 2(2n – 1) 16 = 2a + 1 or a=3 f(x) = Ax2 + Bx + C x   and f(x)   at x = 0, f(0) = C  at x = 1, f(1) = A + B + C C is integer  at x = –1, f(–1) = A – B + C f(1) + f(–1) = 2A + 2C  2A is also integer f(x) = A x(x – 1) + (A + B) x + C f(x) = 2A

C is integer A + B is also integer C is integer

x( x  1) + (A + B)x + C 2

x( x  1) is also an integer and 2A, (A+ B), C   2 f(x) is also an integer.

If x is an integer then 

PART - II 2.

f(x) = here

1

x

 1 cos 1 (2 x  1) tan 3 x

– 1  2x + 1 < 1



– 2  2x < 0



–1x<0

 x  [–1, 0) But x  – 1 as |x| – 1  0  x  (–1, 0) for x  (–1, 0), (|x| – 1) is –ve  tan 3x < 0 0 > 3x > –

 2

or

   x   , 0  6 

      Domain :   , 0   (–1, 0)    , 0  6    6 

RESONANCE


3.

 1 x3    f(x) = sin  3 / 2  + sin(sin x ) + log(3{x} + 1) (x 2 + 1)  2x  Domain : 3{x} + 1  1 or 0  x   –1

and

6.

1 x 3

1 2x 3 / 2 – 2x 3/2  1 + x 3  2x 3/2 1 + x 3 + 2x 3/2  0 (1 + x 3/2)2  0  xR 1 + x 3 – 2x 3/2  0 or (1 – x 3/2)2  0 3/2 or 1–x =0 or x=1 Hence domain x f(x) = (sin–1x + cos –1x)3 – 3 sin–1x cos –1x (sin–1x + cos –1x) –1

=

 3 3  1   3 2 – 3 sin–1x   cos x  = – sin–1 x + 3 (sin–1x)2 2  2 8 8 2 4

3 3 = + 8 2

2   2   2 1 1 3 3 3 3  1 sin x (sin x )  sin x   –    = + 2 16  4  32 32 2 

maximum value of f(x) at x = – 1 8.



f maximum =

9 3 7 3 3 3 + × = 32 2 16 8

Here (2 – log2 (16 sin2x + 1) > 0 

0 < 16 sin2x + 1 < 4





1  16 sin2x + 1  4



3 16 0  log2 (16 sin2x + 1) < 2



log



2

2  2 – log2 (16 sin x + 1) > 0

0 sin2x <

2

2  log

2

(2 – log2 (16 sin2x + 1)) > – 

 2y>– Hence range is y  (– 2] 11.

(A)

f(x) = e1/2 n x = g(x) =

12.

D:x>0

x,D:x0

D : x  ± (2n +1)

 2

(B)

tan–1 (tan x) = x

(C)

cot–1 (cot x) = x D : x  ± n f(x) = cos 2x + sin4x = cos 2x + (1 – cos 2x)2 = 1 – cos 2x + cos 4x = sin2x + cos 4x g(x) = sin2x + cos 4x

(D)

f(x) =

|x| , D:x0 x g(x) = sgn (x), D : x  R

f(6{x}2 – 5{x} + 1) (3{x} – 1) (2{x} – 1)  0

13.

x,



f((3{x} – 1) (2 {x} – 1)) or

 1 1 {x}  ,  3 2



x



n 

1 1  n  3 , n  2   

  R+   0 ,   2 g(x) = 2x – x2 R  R f(g(x)) = cot–1 (2x – x2), where x  (0, 1]

f(x) = cot–1x

   hence f(g(x))   ,  4 2 

RESONANCE


20.

21.

25.

1 1 2 1     f  x   =  x   +  x   + [x + 1] – 3  x   + 15 3 3 3 3         1 2   =  x   +  x   + [x] – 3x + 15 = f(x) 3 3     f(x) = |x – 1| f : R+  R x g(x) = e , g : [–1, )  R fog(x) = f[g(x)] = |ex – 1| D : [–1, ) R : [0, )

f(x) = (A) (B) (C)

(D)



fundamental period is 1/3

sin( [ x ]) =0,x { x} By graph fundamental period is one f(–x) = 0 = f(x)  even function Range y  {0}  { x }  y = sgn  sgn – 1, x  { x }   y = sgn (1) – 1  y=1–1 y = 0, x   Identical to f(x)

28.

f(x) = sin x + tan x + sgn (x2 – 6x + 10) f(x) = sinx + tan x + sgn ((x – 3)2 +1) f(x) = sin x + tan x + 1 period = L.C.M. (2, ) = 2 fundamental period = 2

29.

f:NI

n – 1 , n  odd  f(n) =  2 n – , n  even  2 For For

n  odd numbers f(n)  0, 1, 2, 3, ...... n  even numbers f(n)  –1, –2, –3, ...... range I

 

f(n) is one -one onto function.

EXERCISE # 3 2.

 x = 2

(A)

sin–1 x + cos–1

(B)

2   sin–1 x + cos–1 1  x  = 0  

 (C)

x  [0, 1]



x  [–1, 0]



x [0, )



 1 x   tan–1 x + tan–1= tan–1   1– x 

 1 – x2    g 2  = 2h(x)  1 x 



(D)

cos–1 1 x 2 = – sin–1(x)



cos

–1

 1 – x2    –1  1  x 2  = 2 tan x  

 1 x   h(x) + h (1) = h   1– x   x (–, 1)

RESONANCE


Comprehension # 2 (6, 7, 8) Period of e

x tan   4

is 4

 (1 2 [ x])   =0 2  

cos 

xR

 [x] 

Period of sin  2  is 4   then y =

8  2[ x ]  [ x ] 2



p=4

   

[x]2 – 2[x] – 8  0 – 2  [x]  4 q=–2 , r=5 r–q–1=5+2–1=6 x  2 , f2 (x) =  2  x ,



Period of f(x) is 4



– [x]2 + 2 [x] + 8  0

i.e., 

([x] – 4) ([x] + 2)  0 –2  x<5

x0 x0

2  f2 (x) , f2 (f2 (x)) = 2  f ( x ), 2  2  x  2  2  ( x  2)  = 2  2  x 2  (2  x)

f2 (x)  0 f2 (x)  0 ,

x  2  0,

x0

,

x  2  0,

x0

,

2 x  0,

x0

,

2  x  0,

x0

x0 4  x , =  x0 4  x , Range of f2 (f2 (x)) is [4,  )  (4 ,  ) = [4 ,  ) = [p ,  )

11.

f : [0, 3]  [0, 13] y = f(x) = x2 + x + 1 

x=

– 1  1 – 4(1 – y ) 2



x=

– 1  4y – 3 2

– 1  4x – 3 2 Hence option ‘B’ is correct



14.

f–1(x) =

as

f–1 [1, 13] [0, 3]

  f(x + 4) = sin  [ x  4]  2       = sin  [ x ]  2  = sin  [ x ]  = f(x) 2  2 

17.

f(x) =

2x 2  x  1 (7 x 2  4 x  4) f(x) = 

 x 2  2x (7 x 2  4 x  4 ) 2



f(x) is not monotonic

f(x) is many one.

RESONANCE


22.

1<2<3 f(1) f(2)  f(3)



f (1) f (2) f (3 ) No. of maps 1 1 1, 2, 3, 4, 5 5

2

3

4

2

2, 3,4,5

4

3

3,4,5

3

3

3,4,5

3

4

4,5

2

4

4,5

2

5 2

5 2, 3, 4, 5

1 4

5 2

5 2, 3, 4, 5

1 4

3

3,4,5

3

3

3,4,5

3

4

4,5

2

4

4,5

2

5

5

1

5

5

1

3,4,5

3

3,4,5

3

4,5

2

5

1

5

24.

5

2, 3,4,5

4

5

Q y uksad h la[;k

2

3

4

f (1) f (2) f (3 ) 1 1 1, 2, 3, 4, 5

2

3

3

4,5

2

5

1

4

4,5

2

5

5

1

4 5

4

4,5

2

5

5

1

5

5

4

1

5

5

5

1

4

f(x) = e cos x  { x }  cos(x ) Since ex is a monotonic function fundamental period = L.C.M. (1, 1, 2) = 2


f : [0, )  [0, ) f(x) =

x 1 x

x1 x2 1  x1 = 1  x 2 for given domain f(x) < 1 2.

y=



x2  x  2 x2  x  1 D0

,xR

(y – 1) (3y – 7)  0



x 1 = x 2 only



function is into



x 2 (y – 1) + x(y – 1) + (y – 2) = 0



(y – 1)2 – 4(y – 1) (y – 2)  0



1y

7 3



1=2

not possible

but for y = 1 quadratic vanishes so put y = 1

1=

RESONANCE

x2  x  2 x2  x  1




3.

y=

No real values of x.

sin 1 2 x 

4.

–

y = 1 is not in the range

 6

For domain sin–1 2x +





 0 6

1  2x  1 2



–

   sin–1 2x  6 2



–

1 1 x 4 2

g(f(x)) = (sin x + cos x)2 – 1 = 1 + sin 2x – 1 = sin 2x 



   2x  2 2

   gof(x) is invertible in   ,   4 4

5.

 x, x  Q y = (f – g) (x) =   x, x  Q

6.

For option A

 Which is one-one and onto function

f –1 (f(A)) = A  A  A

Hence A is wrong

For option B

f(X) = Y 

f is onto but it will not effect on mapping of function

Hence B is wrong For option A & B other explaination can be given else if Y is a singleton set then the function f is constant function and hence is trivially onto (unless X = ). But in such a case, even if A consists of just one point, f(A) is entire set Y and so f –1 (f(A)) is the entire set X, which could be much bigger than A. So A and B are wrong even if f(X) = Y For option C

f(X)  Y (range  co-domain) f(X) is a proper subset of Y (so that f is not onto), then for B = Y option C is wrong because f –1(Y) = X but f(f–1(Y)) = f(X)  Y. For option D

If B = Y, then f(f –1(Y)) is the range of the function f. If this is equal to Y, then function must be onto, thus f(X) = Y is necessary condition Hence D is correct

RESONANCE


7.

A = {x |x2 + 20  9x} = {x |x  [4, 5]} Now, f(x) = 6(x2 – 5x + 6) f(x) = 0  x = 2, 3 f(2) = –20, f(3) = –21, f(4) = –16, f(5) = 7 from graph, maximum of f(x) on set A is f(5) = 7

8.

g(f(x)) = x  g(f(x)) f(x) = 1 ........(i) if f(x) = 1  x = 0, f(0) = 1 substitute x = 0 in (i), we get

1 g(1) =  f (0) 9.

(f(x) = 3x2 +

g(1) = 2



f(x) = x2 ; g (x) = sin x gof (x) = sin x2   (fogogof) (x) = (sin (sin x2 ))2 = sin2 (sin x2) Now sin2 (sin x2) = sin (sin x2)  

sin x2 = n, (4n+1)



x2 = n

 ; 2

 I



x =  n ; n  W

11.

cos4 =

2 2 – sec 2 

=

sin (sin x2) = 0, 1 sin x2 = 0

F : [0, 3]  [1, 29] f(x) = 2x3 – 15x2 + 36 x + 1 f(x) = 6x2 – 30 x + 36 = 6(x2 – 5x + 6) = 6(x – 2) (x – 3) in given domain function has local maxima, it is many-one Now at x = 0 f(0) = 1 x = 2 f(2) = 16 – 60 + 72 + 1 = 29 x = 3 f(3) = 54 – 135 + 108 + 1 = 163 – 135 = 28 Has range = [1, 29] Hence given function is onto

Now f(cos4) =

gogof (x) = sin (sin x2)



10.

1 1 2  2cos22 – 1 =  cos22 = 3 3 3

1 x/2 1 e  f(0) = ) 2 2

 cos2 = 

2 3

1  cos 2 1 =1+ cos 2 cos 2

 1 3  f  = 1 ± 3 2 NOTE : Since a functional mapping can't have two images for pre-image 1/3, so this is ambiguity in this question perhaps the answer can be A or B or AB or marks to all.

PART - II 1.

We have f : N  If x and y are two even natural numbers then

x y =  x=y 2 2 Again if x and y are two odd natural numbers then f(x) = f(y)  

x 1 y 1 = x=y  f is one-one 2 2 Also each –ve integer is an image of even natural number and each +ve integer is an image of odd natural number.  f is onto. f(x) = f(y) 

RESONANCE


2.

f(x + y) = f(x) + f(y) function should be y = mx f(1) = 7  m = 7 n



n

f (r ) = 7

r 1

3.

4.

r

=

r 1

 f(x) = 7x

7n (n  1) . 2

4 – x2  0 , x3 – x > 0 x  ±2 and –1 < x < 0 or 1 < x <   D = (–1, 0)  (1, ) – {2} or D = (–1, 0)  (1, 2)  (2, ). 2   f(x) = log  x  x  1   

  x 2  x 2  1 2     f(–x) = log   x  x  1  = log   2   x  x  1   2   = –log  x  x  1  = –f(x)  

5.

 f(x) is an odd function.

We know that – a 2  b 2  a sin  + b cos  

a2  b2

 –2  sin x –

 –1  sin x –

3 cosx  2

3 cosx + 1  3

 f(x)  [–1, 3]. 6.

Let us consider a graph symmetric w.r.t. line x = 2 as shown in figure

from figure f(x1) = f(x2) where x1 = 2 – x & x2 = 2 + x  f(2 – x) = f(2 + x) 7.

f(x) =

sin 1( x  3 ) 9  x2

is defined if

–1  x – 3  1 & 9 – x2 > 0 Hence from (1) & (2) we get 2  x < 3  Domain = [2, 3). 8.

2x4  –3 < x < 3

...(1) ...(2)

7x

Px 3 is defined if 7 – x  0, x–30  3  x  5 and x   

f(3) =

7 3

P3 3 = 4 P0 = 1

f(4) =

74

P4 3 = 3 P1 = 3

and 7–xx–3  x = 3, 4, 5

P5 3 = 2 P2 = 2 Hence range = {1, 2, 3}. f(5) =

7 5

RESONANCE


9.

 2x   = 2 tan–1 x for x  (–1, 1) f(x) = tan–1   1 x2      tan–1 x    ,   4 4

If x  (–1, 1)     2 tan–1 x    ,   2 2

   Clearly, range of f(x) =   ,   2 2

10.

    Codomain of function = B =   ,  .  2 2 f(2a – x) = f(a – (x – a)) = f(1) f(x – a) – f(a – a) f(a + x – a) = f(1) f(x – a) – f(0) f(x) = – f(x) [ x = 0, y = 0, f(0) = f 2(0) – f 2(1)  f 2(1) = 0  f(1) = 0]  f(2a – x) = –f(x).

11.

f(x) is defined if –1 

for f to be onto, co-domain = range

or 0 

x –1  1 2

x    2 and – < x < 2 2 2

12.

y = 4x + 3

13.

f(x) =

x=

and cos x > 0

or 0  x  4 and –

 
y 3 4

   x  0,  .  2

 f –1 (y) = g(y) =

y 3 . 4

1 | x | x

|x|–x>0 |x|>x 14.

x< 0

 x  (– , 0) Ans.

f(x) = (x – 1)2 + 1, x  1 f : [1, )  [1, ) is a bijective function  x=1±

y – 1  f –1(y) = 1 ±

 y = (x – 1)2 + 1  (x – 1)2 = y – 1

y –1

 f –1(x) = 1 + x – 1 { x  1} so statement-2 is correct Now f(x) = f –1(x)  f(x) = x  x2 – 3x + 2 = 0  x = 1, 2 so statement-1 is correct

 (x – 1)2 + 1 = x

ADVANCE LEVEL PROBLEM PART-I 1.

f(x) =

2x  1   log x  4  log  2  3 x  2

2x  1    0 For domain : log x  4  log2 3x   2

Case-I 0<

x4 <1 2



RESONANCE

–4
..........A


2x  1    then log x  4  log 2 0 3x    2

2x  1 1 3x  on A  B x  (–4, –3)

log2

  Case-II

x4 >1 2

or

x > –2

2x  1 2 3x 



x < –3 ..........B



1<

..........(i) ..........A

2x  1    0 log x  4  log 2 3x   



0 < log2



(i)  (ii)

2





x  (4,  ) ..........(ii)

2.

f(x) = (x 12 – x 9 + x 4 – x + 1)–1/2 Dr : x 12 – x 9 + x 4 – x + 1 > 0 For x  0 it is obvious that for f(x) to be defined Dr > 0. For x  1, (x 12 – x 9) + (x 4 – x) + 1 is positive Since x 12  x 9, x 4  x. For 0 < x < 1, Dr = x 12 + (x 4 – x 9) + (1 – x) > 0 Since x 4 > x 9, x < 1. Hence Dr > 0 for all x  R Domain is x  R

3.

f(x) =

2x  1 1 3x 

2x  1 2 3x 

Domain x  (– 4, – 3)  (4, )

e x  e |x| e x  e | x|

if x  0, f(x) =

e x  e x 2e x

=

1 1 1 – x 2 = 2(e ) 2 2

 1 1   (e x )2 

  ; f(x)  0, 1   2    

........(i)

 1 f(x)  0,   2

if x < 0, f(x) =

4.

f(a) =

ex  ex x

e e

x

=0

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



a(2a – 1)  0

Let g(x)  x 2 + (a + 1)x + (a – 1) = 0 (i) D0 (a + 1)2 – 4(a – 1)  0  aR

B <1  2A (iii) g(– 2) > 0  (iv) g(1) > 0  Now (i) (ii) (iii) (iv) we get 5.

 1 range of f(x) is (i)  (ii) = 0,   2



1  a  (–, 0]   ,   2 

2a 2  a for domain of f(x)

2a2 – a  0

(ii)



–2 < –

 2 1 f(x) = sin–1  x   + cos –1 2 

...(i)

(a  1) <1 2 4 – 2(a + 1) + (a – 1) > 0 1+a+1+a–1>0 –2<–

a   3, 3  ....(ii)

 

a<1 a > –1/2

 2 1 x  2   

Domain :

 2 1 –1   x    1 2 



 5 5   x   2, 2   

and

 2 1 – 1  x    1 2 



 3 3  , x    2 2  

RESONANCE




 3 3    domain is x    2 , 2   

6.

7.

8.

9.

 3 x 2  0,   2

or

if

(i)

 1 x 2  0,  , then  2

f(x) = 

if

(ii)

1  x 2   ,1 , then 2 

f(x) = 

if

(iii)

 3 x 2  1,  , then  2

f(x) = 



range = {}

1  1 2   1  1  1 1999   1 f  =   +   +  +.......      = 1000  2   2 2000   2 2000   2 2000  2

f(x) = ax 2 + bx + c f(0) = c  c   f(1) = a + b + c  (a + b + c)  

6 x  6  x f(x) =   2.6 x

f(x) =



(a + b)  

, x0 , x0

sin 2 x  4 sin x  5

=y 2 sin 2 x  8 sin x  8 sin2 x (1 – 2y) + 4sinx (1 – 2y) + (5 – 8y) = 0 vertex (–2, 1) Let sin x = t, where t  [–1,1] 5  g (–1) · g (1)  0 or 2(1 – y) . 2(5 – 9y)  0 or y   ,1 9 

y

10.

0

–2

2

4

x

y = f(x + 2) is drawn by shifting the graph by 2 units horizontally.

11.

 0  2  x sin x f(–x) =  x x 



x0 x  (1,1)  0 |x||

= – f(x)

odd function

PART - II 1.

(i)

log1/ 3 log 4 ([ x ]2  5) Domain (i)

log1/3 log4 ([x]2 – 5)  0 or or [x]2  9 or

RESONANCE

log4 ([x]2 – 5)  1 x  [–3, 4)

.........(i) S OLUTIONS (XII) # 22

log4 ([x]2 – 5) > 0 or [x]2 – 5 > 1 or x  (–  , –2)  [ 3,  ) [x]2 – 5 > 0 x  (–  , –2)  [ 3,  ) Now (i)  (ii)  (iii)  x  [–3, 2)  [ 3, 4)

(ii)

(iii)

(ii)

.........(ii) .........(iii)

1 f(x) = [ | x  1 | ]  [ | 12  x | ]  11 Case- x > 12 f(x) =

1 [ x ]  1  [ x ]  12  11

Now for f(x) to be defined [x]  12 x  (12, 13) Case- 1  x  12 f(x) =



x  [12, 13)

1 1  [  x]  [  x]  12  11 xI xI

x  (0, 1)



x  (–0.5,  ) & x  0.5



x<1

x2  2x  3

2

or



f(x) = x 0.5 log( 0.5  x ) 4x2  4x  3 x + 0.5 > 0, x + 0.5  1

&

but x > 12

1  if x  I 1  =  [ x ]  ( 1  [ x ]) [ x]  1  12  [  x]  11  not defined if x  I

1   2(1  [ x ]) if f(x) =  1  if   2 [ x ] (iii)

1 f(x) = 2 ([ x ]  12)

x  {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

 Case- x<1 f(x) =



x  2x  3 2

4x  4x  3

> 0 or

.....(A)

( x  3)( x  1) >0 (2x  3)(2x  1)

 1  3  x  (–  , –3)    ,1   ,    2  2 

.....(B)

 1   3   1 (A)  (B)  Domain of f(x) : x    ,1   ,   –   2    2  2

cos x 

(iv)

f(x) =

1 2

6  35 x  6 x 2

cos x –

   1 0 or x  2n  , 2n   , n   3 3 2 

and

6 + 35x – 6x 2 > 0



 1    5  Domain   ,    ,6   6 3  3 

RESONANCE

or

 1  x    ,6   6 


(v)

2.

f(x) =

( 7 x  1) ! 5 sin1 x 2  3 + x  1 x1    2 

 x  1  2   0  



x  [1, 3)

&

x  [–1, 1]

&



x  (–1,  )

&

7x + 1  w

x+1>0



 1 1 2 3 4 5 6 Domain  ,0, , , , , ,   7 7 7 7 7 7 7

(i)

f(x) =

x 1 + 2 3  x D:x–10 & 3–x0

1 Range : f(x) =

(ii)

x  [1, 3]

or

f(x) = 0 at x =

7 5

maxima at x =

7 5

1

2 x 1

–

f  5  > 0  

&

 7  f  5 

For domain

(i)

 7 



3x

=0

  < 0    [x] > 0 and [x]  1

for range if x  [2, ), then

|x| =1 x

so

Range :

 2,

10

[x]  2, so x  [2, )

so f(x) = cos –10 =

 2

 Range of f(x) =   2

log1/ 2 log 2 [ x 2  4 x  5]

(iii)

f(x) =



D : 0 < log2 [x 2 + 4x + 5]  1 [x 2 + 4x + 5] = 2 D : x  (–2 – R : {0}

2 , – 3]  [–1, –2 +

(iv)

  x2  f(x) = sin–1 log 2  2  



1 x2  <4 2 2

and (v)

or or

1 < [x 2 + 4x + 5]  2 2  x 2 + 4x + 5 < 3

2)

   



 x2    –1  log2  2  < 2  



x  (– 8 , –1]  [1,

   R :  , 0,  2 2   f(x) = log[x – 1] sinx sin x > 0  x  (2n,(2n+1)) here [x – 1] > 0 & [x – 1]  1 

8)

x  [3,  ]

  Domain x  [3, )    2n , (2n  1)   . n  1   For range sin x  (0, 1] and [x – 1]  [2, ) so range  (–, 0] (vi)

f(x) = tan–1

[ x ]  [  x ]  2 | x | +

Domain : (i) [x] + [–x]  0 (ii) 2–|x|0 (iii) x  0 For domain (i) (ii) (iii)

 

1

x2 x |x|  2  x  [–2, 2]

Domain : {–2, –1, 1, 2}

1  Range :  , 2 4 

RESONANCE




3.

3 f(x) = sin

x x  sin5 2 5

T 1 = 2, T 2 = 5 4.

(i)



T = 10

y = cos (sin x) sin x  [–1, 1]

1 cos 1



3 2

–



O

 2

(ii)

y = | n | x 2 – x | | (rough sketch)

(iii)

y = min (x – [x], – x – [–x]) 0  =  min ({ x }, {  x })

 2



3 2

5 2

, xI ,x  I

½

–1

(iv)

y = (sin 2x)

RESONANCE

0

1

2

3

1  tan 2 x


(v)

(vi)

5.

y = sin x + | sin x | Ans. y = 2 sin x =0

y = In x –

if if

, ln x  0 , x 1  0 (In x )2 =  2 ln x , ln x  0 , 0  x 1 | In x | 

[x] [y] = x + y (i) if x, y   or (ii) and 

sin x  0 sin x < 0

x x 1 if x, y  I y = I2 + f 2 f1 + f2  I 0 < f1 + f2 < 2 I 1 + I 2 + 1 = I 1I 2 y=

then

xy = x + y



(x, y) is (0, 0), (2, 2)

Let then

x = I1 + f 1 I 1 + I 2 + f 1 + f 2 = I 1I 2



f 1 + f 2 = 1.

I2  1 2 I1 = I  1  1  I  1 . 2 2

 6.

I2 – 1 = ± 1, ± 2, I2 = 2, 0, 3, – 1 \ I1 = 3, –1, 2, 0 I1 I2 = 6, 0 x + y = I 1I 2 x + y = 0 or x + y = 6

(i) | [x] – 2x | = 4 or [x] – 2x = ± 4 let x =  + f then (a)  – 2 – 2f = – 4  = 3, f = 1/2, (b)  – 2 – 2f = 4

RESONANCE

or

f=

 = 4, f = 0

4I 2

 

4I <1 2 x = 4, 7/2 0



= 3, 4


(4  I) 2  = – 4, f = 0 f=–

 

(4  I) <1 2 x = –4 0 –

1 9  x=– 2 2 [x – 1] + [1 – x] + x – {x} > 0 [x] + [–x] + [x] > 0 x  , x > 0  x : {1, 2, 3, 4, .......... } x   [x] – 1 > 0 [x] > 1  x  [2, ) A x  {1}  [2, )  = –5, f =

(ii) If If   7.

9 7  x =  4,  , 4,  2 2 

.......(A)

.......(B)

Case 

y=x x<1 x=y y<1 f –1(x) = x x<1 Case  y = x2 1x4 x2 = y 1  y  16 x=

y

f –1(x) = Case  x=

1  y  16

x

1  x  16

y= 8 x

y2 64

f –1(x) = 8.



x>4

y > 16

x2 64

x > 16

Let x = y = 1 f(x) + f(y) + f(xy) = 2 + f(x) . f(y) 3f (1) = 2 + (f(1))2  f(1) = 1, 2. But given that f(1)  1 so f(1) = 2 1 Now put y = x  1  1  1  1 f(x) + f   + f(1) = 2 + f(x) . f    f(x) + f   = f(x) . f   x x     x x so f(x) = ± x n + 1 Now f(4) = 17  ± (4)n + 1 = 17  n = 2 2 f(x) = +(x) + 1.  f(5) = 52 + 1 = 26

9.

Put x = 1, y = 1 (f(1))2 = f(1) + 6  f(1) = 3, – 2 f(1) = 3 [Since f(x) > 0] Put y = 1 in given relation f(x) f(1) = f(x) + 2(x + 2) 2f(x) = 2x + 4 f(x) = x + 2

10.

| f(2k) – f(2i)| = | f(2k) – f(2k –1) + f(2k–1) – f(2k–2)......... f(2i+1) – f(2i)|  | f(2k) – f(2k–1)| + | f(2k–1) – f(2k–2)| + ...........| f(2i+1) – f(2i)| Consider | f(2k–1 + 2k–1)| – f(2k–1) |  1 So | f(2k) – f(2i)|  1 + 1 + ........(k – i) term

RESONANCE


k



| f ( 2k ) – f (2 i ) | 

i 1

k (k – 1) 2



11.

 (k – i) Hence proved.

 1   = f(x + 1) (x + 1)2 f   x  1

1  f ( x)  1   = f  ( x  1)2  x  1

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

x  x   1     = f 1 –  = 1+ f  –  Also f  x  1  x  1   x  1  x   = 1–f   x  1  x  1  x     =1–f   x   x  1

= 1–

2

x2

f (x)  1 x2   1  1 –    1  2  .........(ii) f = 2 2 x x  ( x  1)  ( x  1) 

from equation (i) and (ii), we can say that (i) = (ii) 

1  f ( x) ( x  1)

2

= 1–

x 2  f (x) ( x  1)2

 1+ f(x) = 1 + 2x – f(x) 12.

f(x) = x

(i)

Put x = y = 1 in given relation, we get f(f(1)) = f(1)

(ii)

Now put x = 1, y = f(1) in given relation, we get f(f(1)) =

from (i) and (ii)  f(f(1)) = 1 Put x = 1, f(f(y)) = f(f(f(x))) = 13.





f (1) =1 f (1)

f(1) = 1

f (1) 1 y  f(f(y)) = y now substitute y = f(x)

1 1  1  f  = f (x) f (x) x

f(x, y) = f(2x + 2y, 2y – 2x) = f(2(2x + 2y) + 2 (2y – 2x), 2(2y – 2x) – 2(2x + 2y))  f(x, y) = f(8y, –8x) = f(8(–8x), –8(8y))  f(x, y) = f(–64x, –64y) = f((–64) (–64 x), (–64) (–64y))  f(x, y) = f(212x, 212y)  f(x, 0) = f(212x, 0) Replace x by 2y  f(2y, 0) = f(212 . 2y, 0)  f(2y, 0) = f(212+y , 0)  f(2x, 0) = f(212+x , 0)  g(x) = g(12 + x) [ given g(x) = f(2x, 0)] Hence g(x) is periodic function with period 12. 

RESONANCE

.......(i) from (i) .......(ii) (using (ii)) .......(iii) (using (iii))


LIMITS EXERCISE # 1 PART - I Section (A) : A-6.

(i) (ii) (iii)

  0 im [ x ] =    Not an indeterminate form x 0 x   ve value  im

im (tan x)tan2x = ()º form  Yes  x

(iv)

x 2  1 – x   +  =  Not an indeterminate form

x  

2

im

x

x 1

im {1  h}

1 nx 1 n(1h )

h 0

im {h}

1 n(1 h )

h 0

= (0 form) = 0  Not an indeterminate form

SECTION (B) : B-2.

(i)

im x 0

1  cos 4 x 1  cos 5 x

0   form  0  2

 sin 2x  22   2 sin 2x 16  2x   im  im = x 0 = x 0 2 = 2 5x 25 5x   2 sin 2 sin  2  5   2     2   5 x   2  2

(ii)

3 sin x  cos x  x 6 using L' Hospital rule im x

=

(iii)

 6

im x

im x 0

im x 0

(iv)

 6

0   form  0 

3 1 3 cos x  sin x =  =2 2 2 1

tan 3 x  2x 3x  sin2 x 3 . tan 3 x –2 1 3(1) – 2 3–2 3x = = = 3 – 0 . 1 3 3 sin x 3 – sin x. x

2 2 0  im (a  x ) sin(a  x )  a sin a  form  x 0 0   x using L' Hospital rule

2(a  x ) sin(a  x )  (a  x )2 cos(a  x ) = x im 0 1 = 2a sina + a2 cos a

RESONANCE


5

B-4.

5

2 2 im ( x  2)  (a  2) x a xa

R.H.L. = im

(x 

5 2) 2

– (a  x–a

x a

= im

(a 

h 0

5 2  h) 2

– (a  ah–a



(a 

5 2) 2 1 

  

im = h 0

5 2) 2

(a 

5 2) 2

5  h 2   – 1 a2 

5  2) 2  1 

 

= im

h

h 0

  5  3      2 5 h h 2 2  .     .  ..... – 1 2 a2 2!  ( a  2 )2   h 5

3

5 ( a  2) 2 = 2

5

(a  2 – h ) 2 – ( a  2 ) 2 L.H.L. = im h 0 –h

   5  3     2  5  h h 2 2     a  2 1 –          .... – 1 2!  2a2  a2     = im h 0 –h 5 2

3

5 2 = ( a  2) 2

L.H. L. = R.H.L. 5

So

 im

x a

5

3

( x  2 ) 2 – (a  2 ) 2 (x – a)

=

5 ( a  2) 2 2

SECTION (C) : C-1.

(i)

2 x   1 im  2  2  ....  2  = im x  x  x x x   im = x  

(ii)

x( x  1) 2x

im cos x 

 

im = x  

2



1  2  .....  x  x2

1 1  1  1  x =  2  2

 

x1  cos x

 x  x 1  x – x 1    im 2sin  = x  . sin      2 2      x  x 1   x – x –1    im 2 sin  = x  . sin      2  2 . ( x  x  1)     x  x 1   –1  im im sin    =2 x  . sin    x   2  2. ( x  x  1)    = 2x (oscillating –1 to 1) × 0 =0

RESONANCE


(iii)

im x 

x

2

 8x  x



= ( + ) =  3 3

(iv)

3

2

n  2n  1  n  1

im

n 4

4

n2 1–

4

= im

n  6n5  2  5 n7  3n3  1 6

n 

3

2 1 1   n3 1 4 n n3 n 7

n2 1

6 2 3 1  – n5 1 4  7 n n6 n n

1

– 2 1 1  3  n 6 1 4 n n n

1– im = n  1

(v)

im x 

6 2  –n n n6

–

1 10

2 2   x  13 – ( x – 1) 3  

1

=

3 1  n 4 n7

1 0 =1 1– 0

   

4 2 2 4 2 2  ( x  1) 3 – x – 13  ( x  1) 3  x  13 ( x – 1) 3  ( x – 1) 3      im  = x  4 2 4  2   ( x  1) 3  x  13 ( x – 1) 3  ( x – 1) 3   

( x  1)2 – ( x – 1)2

im = x   

( x 

4  1) 3

2 3 (x

 x  1

2 – 1) 3

 (x

4 – 1) 3 

(x

 4x

im = x   ( x  1)( x

1   1) 3 1 

 

2 – 1 3

4 – 1 3 

x x      x  1  x  1  

4x

im = x  

=

2 4 x – 1 3  x – 1 3       x  1  x  1  

4   1) 3 1 

 

4 =0 (1  0)  ()  [1  1  1]

SECTION (D) : 1

1

D-1.

(i)

5 2 im ( x  2)  (15 x  2) x 2

(7 x 

Let

im = h 0

1 4 2)

x

x=2+h

(4

1  h) 2

(16

– (32  15h)

1  7h) 4

– (2  h)

1

1

1 5

im = h 0

15 h  5 h2   2 1   – 2 1  32  4    1

7h  4  2 1   – (2  h) 16  

     1  1   1  4   h    –   h2    3h    –  15h 2  2 5 2   5   21       .... – 2 1        ....   2 2  8   32   32   16     im  = h    0  7h  21   ...... – ( 2  h)  64 

im = h 0

1 3 1 9  1 3  2 – – h –   h –   ... 16 4 16 4 64 256     = 7  7  –1 – 1  ..... h 32  32 

RESONANCE

=–


(ii)

im

e x  1  sin x 

tan 2 x 2

x3

x 0

2 3 4 3 5 3        x  x  x  x  ...  –  x – x  x  ...  – 1  x  x  ...        2! 3! 4! 3! 5! 3    2  im  h 0 x3

2

 1 1  1 1  1 1 x 2  –   x 3  –   x 4  –   .... 2 2 6 6  4! 3  im = h 3 0 x

1 1 1 + = 6 6 3

=

SECTION (E) : E-2.

(i)

im (tan x)tan2x  x

=

4

im (tan x –1) tan 2 x  x e 4

= e–1 =

=

–2 tan x im  1 tan x x e 4

1 e x

(ii)

 1  2x  im   x   1  3x 

(iii)

im 1  nx  x 1

x

sec

= (iv)

    im = x     x 2

nx im x 1 cos x 2 e

1  2  x 2    = =0 1 3  3  x 

im ( nx ). sec

= e x 1

x 2

lim

1 x

x 1 –  sin x 0 2 2 ( form) = e 0

im x x 2

=

 –2    e  

((0)0 form)

x 0 

im x x 2 Let y = x0 1

y=



(v)

im

e x0



x 2(nx)

im (tan x) cosx – x







(0 form)

y=



nx im 1 x 0 2 x e

y=

im cos x.( n tan x ) – x 2 e 1 sec 2 x im  tan x sec x tan x – x e 2

 2

y=

n tan x im sec x –  x 2 e



y=

y=

cos x im 2x – sin  x e 2



y = e0 = 1

RESONANCE

y=



x im 2 x  0 – 3 x e

= e0 = 1


(vi)

im ([x])1–x

x 1

im [1 – h]1 – (1 – h)

h0

im

h0

E-3.

(0)h = 0

im [1 . 2x ]  [2 . 3 x ]  .....  [n . (n  1) x ] n  n3 (1.2)x – 1 < [1.2x]  (1. 2)x (2.3)x – 1 < [2.3x]  (2.3)x    n(n + 1) x – 1 < [n (n + 1)x]  n(n + 1)x so (1.2)x + (2.3)x + ... n (n + 1)x – n < [1.2 x] + [2. 3 x] + .... [n(n +1)x]  (1. 2) x + (2. 3) x + .... n(n +1)x x . (n2 + n) – n [1. 2x] + [2. 3x] +......[n(n+1)x] x (n2 + n)



 n(n  1) (2n  1) n (n  1)  x.   –n [1 . 2 x ]  [ 2 . 3 x ]  .....  [ n ( n  1) x ] 6 2  im  <  im n  n  n3 3 n

im  n 

 n(n  1) (2n  1) n (n  1)  x  6 2   n3

  1  1  1 1   1.1  n   2  n   n  2   1 [1.2 x ]  [2.3 x ]  .....  [n(n  1)x ] n    im x   – 2  im n  n 6 2   n n3       1 1  1 1  1 . 1  n   2  n   n  2 n     im x    n  6 2    x [1.2 x ]  [2.3 x ]  .....  [n(n  1)x ] x  im  3 n  3 n3

so

E-5.

im

[1.2 x ]  [2.3 x ]  .....  [n(n  1)x ]

n 

n

3

=

     

x 3

2n im x  1 f(x) = n  x 2n  1

case (i) when x = 1

im 1 – 1 = 0 f(x) = n  1 1

case (ii) when x > 1

 1 1–   1– 0 x im f(x) = n =1  2n = 1 0  1 1   x

case(iii) when x < 1

2n im x  1 = 0 – 1 = – 1 f(x) = n  x 2n  1 0  1

2n

range of f(x) is {–1, 0, 1}

RESONANCE


PART - II Section (A) : A-3*.

f(x) =

x 2  9 x  20 x  [x] x 2  9 x  20 25 – 45  20 = =0 x  [x] 1

im

x 5

2 2 im x  9 x  20 = im (5  h) – 9(5  h)  20 x 5 h0 x  [x] 5  h – [ 5  h]

25  10h  h2 – 45 – 9h  20 h 0 h

= im

h(h  1) h2  h = im =1 h  0 h h0 h

= im im



x5–

im f(x)  x5  f(x)

im f(x) does not exist x 5

so

SECTION (B) : 3

B-2.

 4x – 1  x3   x  x 3 ( 4 1 )    im = x im 2 x 0 0   x  x x  sin   n 1    sin     p 3 p   x x 2          .n 1   3   p  x    p    3

 4x – 1  x2       x   3      = 3p . im .   x  x 0  x2  sin    n 1  p  3      x   p   

B-6.

= 3 p (n 4)3    

im tan2 x  2 sin 2 x  3 sinx  4  sin 2 x  6 sinx  2    

x

2

 2 sin2 x  3 sinx  4  (sin2 x  6 sinx  2)    = im tan2 x   2 2 x  2 sin x  3 sinx  4  sin x  6 sin x  2  2 = im tan2 x . x

2

(sin2 x – 3 sin x  2) 2  3  4  1 6  2

1  sin2 x – 3 sin x  2   = im 6  cos2 x x   2

0   form  (use L'Hospital rule) 0 

=

1 im 2 sin x cos x – 3 cos x 2 cos x(– sin x ) 6 x  2

=

 1  1  1 im 2 sin x – 3 1 =    =  6 2    6 x  2 – 2 sin x 12

RESONANCE


B-7*.

Let f(x) =

cos 2  cos 2x x2  | x |

f (x) (A) xim 1

im

Now

x  1

for x = – 1 cos 2  cos 2x x2  x

cos 2  cos 2x x2  x

(B) xim 1

|x| = – x

f(x) =

(

0 2 sin 2x form) = im = 2sin2 x   1 0 2x  1

(

0 2 sin 2 x form) = im = 2sin2 x  1 0 2x  1

cos 2  cos 2x x2  x

SECTION (C) : C-1.

sinh < h < tanh h 1 sinh

,

  h   0,   2

–h  –1 sin h



     h  2   h  im  2 LHL = h0  = –2    = h im  0  sinh     cos 2  h           h  2   h  im  2 RHL = h0  = –2    = h im  0  sinh     cos 2  h     

C-3.

C-6*.

im

 3n  ( 1)

n 

im

x–

im

(B)

x 

1 n 3 –3  0 =– n 1 = 4  ( 1) . 4–0 4 n

 3  ( 1)n

n

im = n 

4n  ( 1)n

(A)

 LHL = RHL = –2

2

x 2 3 x6

2 x2 1 im x 2 = im = x 6 =–  x  – 3 x6 3 –3– x

x2  2 im = x  3x – 6

1

2

2 x2 6 3– x 1

=

1 3

SECTION (D) : D-4.

im e x 0

–

x2 2

– cosx x sinx 3

 x x4   x x4  1 1   1 1 –   ..... x4  –  – x6   – ...... – 1 –  – ...... 1 – 2 4 . 2! 2! 4!      8 4!   8 . 3! 6 !  = x im = x im 0 0 2 3     x x x 4 1 –  .......  x3 x –  ....... 3 ! 3!     1 1 1 =  –  =  8 24  12

RESONANCE


D-5*.

1 a cos x For x im 0 x2 0 for  0  form   1+a=0

a=–1



b sin x im for x im = x 0 3 0 x b=0

so Now

b x2

1 a cos x b sin x  = x im – x im 0 0 x2 x3

2 sin2 x 1 – cos x  im  im 2 = x 0 = x 0 x2 x2 (a, b) = (–1, 0) and  =



2 = x im 0 4

  sin x  2  x   2

     

2 =

1 1 . (1)2 = 2 2

1 2

SECTION (E) :

E-3.

im (1  [ x ]) x

E-7.

1 n (tan x )

 4

im n 

= im ( exact 1) x

1 n (tan x )

=1

4

1 [13 x]  [23 x] ... [n3 x ] n4





13 x –1 < [13 x]  13 x 23 x –1 < [23 x]  23 x . . . . . . . . . n3 x –1 < [n3 x]  n3 x Adding all these inequilities (13 + 23 + 33 ...........+ n3) x – n < [13x] + [23x] + ...........[n3x]  (13 + 23 + ..........n3) x 2

 n(n  1)  n 2 (n  1)2   x x n 3 3 3 [1 x ]  [2 x ]  ....  [n x ]  2  4 <  n4 n4 n4 2

2

1 x 1 1 x   [13 x ]  [23 x ]  .........  [n 3 x ] im 1   im 1    3 < im  n n   n  n 4 n n 4 n4 x [13 x ]  [2 3 x ]  ........[ n3 x ]  im  4 x n4

E-9*.

im f(x)= im | 0 – h| h0

x 0 

sin(0 –h)

x 4



im x 

im hsin(–h) = him0(– sin h ) (  nh ) = h e 0

[13 x ]  [2 3 x ]  ........  [n3 x ] x  4 n4

= e0 = 1

im f(x) = im | 0  h |sin ( 0 h ) = im hsin h = im (sin h ) ( nh ) = e0 = 1 h e h  h 

x 0 



im f(x) = im f(x) = 1 x 0

x 0 

RESONANCE



im f(x) = 1 x 0



(i)

Pn =

3 23  1 3 3  1 4  1 n3  1 . . .......... 3 23  1 3 3  1 4  1 n3  1

(2  1)(2 2  2  1)

 Pn =

(2  1)(2 2  2  1) 2 . 13

1.7

(3  1)(3 2  3  1)

.

(3  1)(3 2  3  1)

...........

(n  1)(n 2  n  1) (n  1)(n 2  n  1)

(n  1)(n 2  n  1)

3 . 21

1 . 2 . 3 ...........(n  1)

7 . 13 . 21...........(n 2  n  1)

 Pn = 3 . 3 . 4 . 7 . 5 . 13 ........ Pn = 3 . 4 . 5..........(n  1) . 3 . 7 . 13..........(n2  n  1) (n  1)(n 2  n  1)  Pn =

(ii)

im

n

im = n 





im

n

k 1 



n



r 1







n

im = n 

( 2k  1)(n  1  k )(n  2  k ) 2n

k 1

4

 (2k  1) (n  1)(n  2)

  k 1

2n 4





( 2k  1)(n  k  1)(n  k  2) 2n 4

k 1

n

im

=

n



( 2k  1) (n  1)(n  2)  k( 2n  3)  k 2 2n

k 1

(2n  3) (2k 2  k ) 2n 4





2.

 (2k  1) (n  1)(n  2) ( 4n  7) k 2 2k 3 (2n  3) k    4    = n 2n 4 2n 4 2n 2n 4   k 1  im

(i)



im

x 1

(n(1  x ) – n2) (3.4 x –1 – 3 x ) [(7  x )1/ 3 – (1  3 x )1/ 2 ] sin( x – 1)



4

2k 3  k 2   2n 4 

n

=

2 (n2  n  1) 2  3 n (n  1) 3

 n k 1 

n4

n

im



 (2k  1)  r  n

=

im P = im n n  n 

n    n 1  n2  1     3 k   5   ......  (2n  1) . 1 1 k k      n 4   k  1   k 1   k 1   

n

=

2 (n 2  n  1)  Pn = 3 n(n  1)

1.2 n2  n  1 . n (n  1) 3

im = h 0

1 12

[n (2  h) – n2] [3.4 h – 3 – 3h] [(8  h)1/ 3 – ( 4  3h)1/ 2 ] sin h

 h n  1    h 2   3( 4h – 1)   n 1   3( 4 h – 1) – 3h · – 3 2 (h / 2)  h   im im = h = h 1/ 3 1/ 2  0 0   h   3h   h 3h   2 1   – 1  1   – 1     sin h  sin h   24   8   8 4      4 h  h     



=

(ii) 



9 4 1.(3n 4 – 3) 9 – (n 4 – 1) = – n   4 e 4  1 4–   3

1 x 11 2   (1  x ) x = e 1 – 2  24 x  .....  

im e  (1  x ) Now  = x 0 tan x

RESONANCE

1 x

 x 11 2  e – e 1 –  x – ..... 2 24   e im = x = 3 5 0 x 2x 2 x   ..... 3 15


8.

AT AQ

tan  = sin 2 =

AP AT  2r 2r

2 tan  2

1  tan  2 tan  2

1  tan 

AQ =

11.



AT 2r



AQ tan  2r

4r

im AQ = im 0

2

P A

1  tan 

14.

1  tan2 

= 4r = 2 (Diameter)

im (f(x) – f(x)) L1 = x  

– x im e f  ( x ) –  f ( x ) L1 = x  e – x





d f ( x )e – x dx im L = L1 = x  d  – x 1  – e .  dx  





L = im – f(x)



L = – im f(x)



L = – L2



L2 = –



12.

4r

– x – x im e f  ( x ) –  f ( x )e L1 = x  e – x



x 

– x im f ( x )e L = x   – x  – e      

x 

L 

1<2


n 

im n! < 0 0 < n  nn



n 

im

1 im n! < im 1 n –1 < n  nn n  n n

im n! = 0 nn

AN = tan CN CN = AN cot

Area of ABC =

cos  1 (AB) (CN) = AN.CN = AN. AN sin  2

 r cos 2   r 2 cos3  = r cos.  sin   = sin   

RESONANCE


1 (DE) (CM) = (DM) (CM) = (CM)2 tan = (OC – r)2 tan 2

Area of DEC =

2

r 2 (1  sin )2  r  – r  tan =  sin  cos   sin   now

im r 2 cos 3  . cos  . sin  ABC =   sin  . r 2 (1 – sin )2 DEC 2

im

AB  0

=

im 

2

cos 4  (1 – sin )2

= im 

(1 – sin2 )2

2

(1 – sin )2

= im (1 + sin)2 = (2)2 = 4 

2

PART - II 1.

n im x = 0 x  ex case(i) when n = 0

(n  integer)

im xn = im x  ex

then

x 

1 =0 ex

case(ii) when n is +ve integer

im xn

x 

  

ex



form  

n!

im = x  ex = 0 case(iii) when n is – ve n = –m where m  z+ –m 1 im xn im x im = x x   ex = x    xm.e x = 0 ex so

3.

n im x = 0 x  ex

im x0 f(g (h (x))) L.H.L. x  0– im + x 0– h (x) = 0 im x0 f(g(x))

then

im g(x) = 1+ x0

 im f(x) = 1 – 1 = 0 x 1 R.H.L. x  0+ im + x0 h (x) = 0

5.

so

im f(g(x)) = 0 x0

L.H.L. = R.H.L. = 0

     im  x sin 1   sin 1     2  x   x  x 

 sin  1    x  1    im   sin   = 1 + 0 = 1 x   1  x 2    x  

RESONANCE


6.

 3 3   im  |x| –  x   (a > 0) x a–  a  a  

x=a–h



   3  |a –h|3  a –h 3   |a|3  h       im –  im – 1 –    = h0  a = h0  a  a   = a2 – 0  a        

12.

= a2

 x  im sec–1    x  1

x 

replace x 

1 y

im sec –1 = y 0

 1     y  im sec –1 1  = y 0  y  1  

when

im   im y0



x 

 1     y  1

1 <1 y 1

y  0+ ;

 1  im sec –1  so y  y  1  does not exist. 0  

14.

(i)



im sin x  1  sin x  = x 



 x 1 x    . sin x  1  x  im 2 cos  = x      2 2    



 x 1 x    x  1 x . sin   im 2 cos  = x      2 2 x  1  x    



 x  x 1  1  sin  im 2 cos  = x     2    2 x  x 1  = (oscillating value –1 to 1) × 0 = 0





 (ii)





sin x  1  sin x m = xim 





   



when x   then x undefined m is undefined 16.

im f(x) To find x  0 L.H.L. = xim f(x) 0– = xim 0–

im { x } cot { x } = h 0

im f(x) = im R.H.L. = x0 x 0

(1 – h) cot (1 – h) =

cot 1

2 tan2 [ x] tan2 [0  h] im im tan 0 = = h0 (0  h)2 – [0  h]2 h0 x 2 – [ x ]2 h2

=0



im f(x) does not exist. f(x) is not continuous at x = 0. L.H.L.  R.H.L. so x 0

Now

cot

2 –1

   im f ( x )  –  x 0 

= cot–1 ( cot 1)2 = cot–1 (cot 1) =1

RESONANCE


18.

sin <  < tan

   0,    2

,

sin  tan   1   n sin  n tan  n  

 n sin    n tan   im            

0

im

L.H.L. =

  0

n N

  n sin    n tan             

im   n sin     n tan              

R.H.L. =

20.

  ; 

  0

  = n – 1 + n = 2n – 1    = n – 1 + n = 2n – 1 



L.H.L. = R.H.L = 2n – 1

 1  1 1 1  im     .......... ...   2 2 2 n  n 2 n 1 n 2 n  2n 

using sandwitch theorem

1 

2

n

1 n

1 2

n 1

1 n







1 2

n  2n



1 n

adding all these inequilities 1 n

2



1 2

n 1



1 2

n 2

 .......... ... 

1 2

n  2n



2n n

im Taking both side n   1  1 1 1  im     .......... ...  = 2 n  n 2 n2  1 n2  2 n 2  2n 

21.

im  2x cot 1 x  f(x) = t 0   t2  Case-I : when x = 0 f(x) = 0 Case-II : when x > 0 x  2x 2x 2x im  cot 1 2  =  cot 1(  ) = f(x) = t  .0=0 0     t  Case-III when x < 0



x  2x 2x im  cot 1 2  = 2 x × cot–1 (– ) = f(x) = t  .  = 2x 0    t   f(x) = 2x

RESONANCE


25.

  ay  by     exp x n 1  x    exp x n 1  x           im  im im  = y  y 0 0 x  y      

x x  1  ay  1  by   x   x    im  x  y   

      

 2 2 2 2      1ay  x(x – 1) . a y  ..... 1 by x(x – 1) . b y  ....       2! 2! x2 x2     im   im   x y = y 0      

by expansion

  y2 2 (a – b 2 )  .....   y( a – b )  2  im  = y  y  =a–b 0    29.

im

(ax  1)n

xn  A (A) If n  N x

n

1  a   x  (a  0)n im = an A = x  1 0 1 n x –

(B) If n  Z & a = A = 0 n im (ax  1) = im 1 =  x  x  xn xn  A (C) If n = 0

then

nZ

–

n 1 1 im (ax  1) = im = x  x  1  A 1 A xn  A

then –

(D) If n  Z , A = 0 & a  0 n

n n im (ax  1) = im (ax  1) = im  a  1  = (a + 0)n = an x  x  x  x  xn  A xn

then

EXERCISE # 3 1.

(A)

2 2 2 2 im tan[ e ]x  tan[ e ]x x 0 sin 2 x

[e2 ]x2

= xim 0

2 2 tan[e2 ]x2 2 2 tan[e ]x  [– e ] x [e2 ]x2 [e2 ]x2 = [e2] – [– e2] = 15 2 2 sin x x x2

im  min(t 2  4t  6) sin x  = im  2 sin x   x 0 x   x  



(B)

x 0



sin x < x

RESONANCE





2 sin x <2 x

 2 sin x     =1  x 

So

im  2 sin x  = 1  x   

x 0


  1 1     – 1 2    4 x 2  ....  1  x  ....  – 1 – x  4  4     2 3 2!     1   = xim = 0 2 2 xx

1

(C)

(D)

2 1/ 3 im (1  x ) – (1 – 2x ) 4 x 0 x  x2

im =

2  1  cos x

= xim 0

2

x 0

sin x

2 2 sin2 2

sin x

x 4 = xim 0

2 2 sin2

x 4

x 2 sin2 x 16. . 2 16 x

2 8

=

Comprehension # 1 (For Q.No. 3 to 5) n



im  cos  x  im  cos  = e n   f(x) = n   n 

Substituting, n =

1 t2

4b g(x) = – x



g(x) = – x

4.

1 2

x n

 1 . n 

 cos tx 1  im   2 e t 0  t 

= e

 1 cos tx  im  x 2  2 2  t 0  t x 

= e

1  x2 2



f(x) =



 x2  x  1 x2  1  1  im  x 2  x  1  x 2  1  = im  b = x =   2 2   x    2  x  x 1 x 1

= – x2

By observation, graphs of f(x) and – g(x) intersect each other at two points  Number of solutions is 2. 9.

Statement -1 is true as  sin x  im  and  =0 x 0  x  Statement - 2 is true as

im h( g( x )) = h ( im( g( x )) xa x a 10.

sin x   im  x   =1  0 x 

if h(x) is continuous at x = g(a).

Statement -1

im

x 0

1– cos2x 2 | sin x | = xim  0 x x



L.H.L. = – 1

&

R.H.L. = 1

Statement -2 is true 14.

True indeterminate form of type  – 

18.

im

x 1

(5  x )2 (2  x)1

(2 – x )  1 1 1– x 2 im (5 – x ) – 4   = = xim x 1 ( 2 – x ) – 1 1 (5 – x )  2 1– x 4 2

RESONANCE


21.

im

 ( x  a)(x  b)  x 

x 

im = x  im = x 

( x  a)( x  b) – x 2 ( x  a )( x  b )  x

(a  b) x  ab

ab ab = 1 1 2

=

  (a  b) ab x  1  2  1 x x  


im

x tan 2x  2x tan x

x0

(1  cos 2x )

2

im = x 0

x 2

(2 sin x )

2

 2 tan x   2 tan x   2 1  tan x  3

 tan x    2 1 x  2 . x tan x . tan x 1 1 1   im  im . = x0 = x0 = .  4 4 2  sin x  2 1 2 4. sin x    x 

2.

 x3  for x  R , im  x   x  2 

3.

im

=

 x–3  im  –1 · x x  2  e x   

= e

 –5  im  · x x 2 

x 

= e–5

sin(  cos 2 x ) x2

x 0

= im

sin( (1  sin2 x ) x2

x 0

4.

x

sin(  sin2 x )  sin2 x  = 1..1 =  x 0  sin2 x x2

= im

x im (cosx 1) (cos x  e ) x 0 xn

 x 2 x 4    x 2 x 4    x2 x3  – .........  1 1   – .........  1  x   ......... 1       im  2! 4 ! 2! 3 !    2! 4!    x 0  n x

 1 x2    1 1  1 x2     1 1  – .........   1  x    – .........      – .........   1 x   – ......... x   2! 4!  2! 2!   2! 4!    im im   2! 2!   = x   x 0 0 n–3 x xn is finite and non-zero if n = 3 3

5.

im ((a  n)nx  tan x ) sin nx = 0 x0 x2

 

tan x   sin nx   im  0  (n) (a  n) n  x  0  nx  x   (1) (n) [(a – n) (n) – 1] = 0 n(a – n) –1 = 0



a–n=



1 n

RESONANCE



a=n+

[ n  0]

1 n


6.

1  1 im x x 0 sin x     x

for x > 0 1 x

 1 im(sin x )  im   x 0 x0 x = e

7*.

1 x im x  0 (cos ecx cot x )

=

sin x

sin x

 1 = 0 + im   x 0  x 

2 im sin x x cos x e x0

sin x

im (sin x ). n  1  x

= e x 0

im

e x 0

=

 n( x ) – cos ec x

   form   

= eº = 1

  1   1 x 2 1  2  1 x 4  x2 .  a  a . 1  . 2   ....  2 2  2 a  4 a4 x2 2 2 a a x    4 = im   L = im x  0 4 4 x 0 x x

( a > 0)

x2 1 x4 x2  . 3  ......  4 im 2a 8 a = x  0 x4

Since L is finite

8.

lim

xn(1 b2 ) e x



2a = 4 

= 1 + b2 = 2b sin2 



sin2 =

1 b  1    b 2 

but

sin2  1

x 0

We know b + 

9.

1 2 b

sin2 =1



sin2 1



= ±

1 8.a

3

=

1 64

 2

 x2  x  1  im  – ax – b  = 4   x  x 1   x 2 (1 – a)  x(1 – a – b)  (1 – b)   =4 im   x   x 1   Limit is finite it exists when 1–a=0

then 10.

im  L = x  0

a=2

a=1

1– b    1– a – b   x  im  1 x   =4 1   x  



1–a–b=4



b=–4

((1 + a)1/3 – 1)x2 + ((a+1)1/2 – 1)x + ((a+1)1/6 – 1) = 0 let a + 1 = t6  (t2 – 1)x2 + (t3 – 1)x + (t – 1) = 0 (t + 1)x2 + (t2 + t + 1)x +1 = 0 2x2 + 3x + 1 = 0  x = – 1 and x = –

As a  0 , t  1

1 2

PART - II 1.

im

x 0

im

x 0

1 – cos 2x 2x

2 sin2 x 2x

=

im | sin x | x

x 0

im sin h  – 1 , LHL = h 0 –h LHL  RHL So limit does not exist

RESONANCE

im sin h  1 RHL = h 0 h


2.

im  x – 3  x x2

= e 3.

x

 x–3  im  –1 x x    x 2 

(1 form)

im –

= ex

im x f (2) – 2f ( x ) x–2

f (x)  1

x 1

x 1

 2  im  x  5 x  3  2 x     x x3 

im f (2) – 2f ( x ) = f(2) – 2f(2) = 4 – 8 = – 4 = x 2 1

0   form  0 

f( x ) 2 x = xim 1 2 f (x)

=

f (1) f (1)



2 2 1

n [ x] im nx – im 1 = x  x [ x] [ x]

=0–1=–1

x

im

(1 form)

x ( 4 x 1) 2  x3

x = e x

= e4

x   x 1 – tan  (1 – sin x ) tan  –  (1 – sin x ) 2  im  im 4 2 = x x  x 2  2 3 (  – 2 x )3 1  tan  (  – 2x ) 2   put x =

im = y 0

 y 2

y 0+

 y y  tan  –  (1 – cos y )  tan  (1 – cos y ) 2  2  im  = y 0 (– 2y )3 8y3

y   tan  2  im  = y y  0     2  8.

0   form  0 

x

4x  1  im 1   = x   x2  x  3 

7.

= e– 5

n im n x – [ x ] x  [ x]

n im nx – im = x  x [ x]

6.

= e

–5 2 x

1

f(1) = 1, f(1) = 2

im

5.

im

x 

f(2) = 4, f(2) = 4 x 2

4.

5x x 2

   sin2 y  2 · 2   y 2  64    4 

= (1) (1) ·

1 1  32 32

im n (3  x ) – n (3 – x )  k x 0 x





x x   n  1   – n  1 –  3 3     k im x 0 x 1 1  k 3 3

RESONANCE





   x x   n 1  3  1 n 1 – 3   · –    – 1      im  x 3  x   3  = k x 0 –    3  3  

k=


Alt.

im n(3  x ) – n(3 – x )  k x 0 x Apply L' Hospital rule

9.

im  1 



x 0



k=

1 1  k 3 3



2 3

 f (a) g( x ) – g(a) f ( x )  im   = 4 x a g( x ) – f ( x )  

im k  g( x ) – f ( x )   4 x a  g( x ) – f ( x )    k=4

 

im 1  a  b  x   x x2  

e

2x

 e2

a b  im   2  2 x x x 

x 

 e2

im 2a  2b  2 x 2 ax + bx + c = 0 a(x – ) (x – ) = 0 

11.

1  k 3– x

im f (a) g( x ) – f (a) – g(a) f ( x )  g(a)  4 x a g( x ) – f ( x ) Apply L' Hospital rule 

10.

3 x

x

im

x

im (ax  b)2  2 x



x



b  R, a = 1

1  cos(ax 2  bx  c ) ( x   )2

 ax 2  bx  c  a  2 sin2  ( x –  ) ( x – )   2 sin2   2   2 im   = x = im  2 2 ( x – ) x ( x  ) 2

  a( x –  ) ( x – )    sin   2 2 2     2a ( x – )  im = x    a( x –  ) ( x – )  · 4   2   12.

lim

x 0

= (1)

2a2 ( – )2 a 2 ( – )2  4 2

f (3 x ) =1 f ( x)

f(x) < f(2x) < f(3x) Divide by f(x)

1

f ( 2 x ) f (3 x )  f(x) f ( x)

using sandwitch theorem 

lim

x 

f (2x ) =1 f ( x)

RESONANCE


13.

lim 2

x2

sin ( x  2) ( x  2)

 does not exist

14.

im

( f ( x )2 ) – 9

x 5

|x–5|

0

im[( f ( x ))2 – 9]  0 x 5

im f(x) = 3 x 5

ADVANCE LEVEL PROBLEM 1

1.

x  x im  a1  a 2 x x x 0   b1  b 2

1

= e2

2.

 a x a

(log e a1  loge a 2  log e b1  loge b 2 )

p q im p  q  qx  px x1 1  x p  x q  x pq

= xim 1

x

im  1x 2 x x  = e x  0 b1 b 2  

1  1 x 

1

= e2

=

 a x  1 a 2 x  1 (b1x  1) (b 2 x  1)   1  im  1      b x b x x x x x 1 2    e x  0

aa log e  1 2  b1b 2

  

=

   

a1a 2 b1 b 2

pqx p 1  pqx q1 0  0   form  = im  form  x1  px p 1  qx q1  (p  q)x p  q1  0 0  

pq(p  1)x p 2  pq(q  1)x q 2  p(p  1)x p 2  q(q  1)x q 2  (p  q)(p  q  1)x p  q2

=

pq 2

1

3.

t 1  t  t 1 im  b  a  t 0    ba 

 b t 1  a t 1  1  b(bt 1) a(a t 1)   1   bnb  ana 1  im    bb  t   ba  b  a t 0  t t   = im e    e ba = t  = = e  aa 0 

4.

1

 b a   

im f(0 – h) = im f(– h) = im h2  =1 LHL = h0 h 0 h 0 – 1 im f(0 + h) = im f(h) = im  im 1  = 1 RHL = h0 h 0 h 0  n  1  hn 

5.

im

x0

nx  k x2 2 (sin x  tan x)

e  n x  e n x  2 cos

 n2 x 2 n 4 x 4   n2 x 2 n 4 x 4    .....  2 1    ........  k x 2 2 1  2! 4! 4.2! 16.4!     im  3 3     = x 0  x  x  ........    x  x  2 x 5  .......      3! 3 15    

   2n 4 2n 4  n2  x 2 n2   k   x4      4! 16.4!  4     im = x 0 1 1 3 x      ..........  3! 3 

RESONANCE

limit exists, if coff. of x 2 is zero.


n2 –k=0  4 so the possible value match that is n2 +



6.

L=

n = 2, k = 5

1 3 im cos (3 x  4 x )   cos 1( 4 x 3  3 x ) = im1 1 1 x 1 x 2 x 2 x 2 2

Let

 = cos–1 x

As x 



1 2





 3

1 L = im   cos (cos 3)  1 3 cos   2



1 im   cos (cos 3) im =    1   3 cos   3 2 3 = 2 3 = im    sin  3

LHL =

Also,

RHL =

= im

3 0  im =– 2 3  form  =  1 0    sin   3 cos   2 LHL  RHL  Limit does not exist



im

n

1 cos   2

1 1   ( 2  3) im   cos (cos 3) = im   cos (cos(2  3)) = im   1   1 1    cos   3 cos   cos   3 3 2 2 2

3  

n

n

r

0   form  0  

  3

Now

  3

7.

4k = 5n2

2



(n  r  1) = im

r 1

r 1

r

3

r 1

n

n

n

r

n

(n  1) r 2 

r

3

3

r 1

r 1

   (n  1) (n) (n  1) (2n  1)  4 1 6 1/ 3    im  1 = n   = 1/ 4 – 1 = 3  1  3 n 2 (n  1)2   4   8.

9.

n1999

1 = x x   2000 C C 3 2 n x 1 x C1    ...  2   2 n 6 n   the limit obviously exists if 2000 – x = 0 

im

n 

Let  = sin–1 x

 as

x

1 2



x = 2000



1 1 im cos (2 sin  cos ) im im cos (sin 2)  = =     1  1 4 4 4 sin   sin   2 2

 4    cos 1 cos  2   2    1 sin   2

    cos 1 cos  2    2 2 2   = im  Left hand limit = im    1   1  sin   4 sin   4 2 2

RESONANCE

0   form  0  


= im–   4

2 = 2 2 cos 

Right hand limit =

= im   4

10.

im 

 4

        cos 1 cos  2   cos cos 2    2   im  im 2 2     2      =   =   1 1 1 4 4 sin   sin   sin   2 2 2

2 = 2 2 cos 

LHL  RHL



x x x f ( x )  f    f    ...  f   im 2 3 k x 0 x

Limit does not exist



0   form  0 

im f ( x )  1 f  x   1 f  x   ...  1 f  x  = x   0  2 2 3 3 k  k   1 1 1 =1+ + + .... + = 3 2 n n

= n C1 –

C2 + 2

n

1

1 2



(1  x  x  ...  x

n 1

) dx =

 0

0

C3 – ........ + (–1)n–1 . 3

n

1 xn dx = 1 x

1

 0

1  (1  x )n dx x

Cn n

PART - II 1.

im 1 – cos(a1x). cos(a2x).cos(a3 x).......cos(an1x) Ln+1 = x  0 x2

Let

im 1 – cos (an1x )  cos (an1x ) – cos (a1x ). cos (a2 x )..... cos (an1x ) Ln+1 = x 0 x2 im 1 – cos(an1x)  cos (an1x) {1 – cos(a1x).cos(a2 x).....cos (an x)} Ln+1 = x 0 x2 im 1 – cos(an1x ) + im cos (a x) . L Ln+1 = x  n+1 n 2 0 x 0 x

Ln+1 =

an12 + L n 2

a12

a22

a22

a12

a32

a32

a22

a12

n

1 ai2 L1 = , L2 = + L1 = + , L3 = + L2 = + + , Ln = 2 i 1 2 2 2 2 2 2 2 2

2.



im  x 2  x  1  a x  b = 0

x  





im  x 2  x  1  ax  b  = 0 

x

Let x =

1 h

So

1–a=0

im

h0

im

h0

1  h  h 2  a  bh 0 h 



limit exists

a=1

1  h  h 2  1  bh 0 h

RESONANCE


3.

( 2h  1)

im



h  0

So

a = 1, b =



x 

2 1 h  h

1 –b=0 2



–b=0

2



b=

1 2

1 2

im xnf(x) = p

n1 im x f ( x ) = p x  x using L- Hospital rule, we get



n n 1 im (n  1) x f ( x )  x f ( x ) = p x  1

4.

Let f(x) =

im xn+1.f(x) = p   (n + 1) p + x 

im xn+1 f(x) = –np.

x 

ex – x – 1 x2 x3

Since limit exists

im f(0 + h) = im f(0 – h) = L (say) h 0



h 0 

h 2 im e – h – 1  h L = h0 h3

h –h –h 2 im e  h – 1  h  2L = im e – e – 2h Also, L = h0 h 0 h3 – h3 Put h = 3t 3t –3 t im e – e – 6t  54 L = im ( e t – 1)3 – ( e – t – 1)3  3( e 2 t – e –2 t ) – 3( e t – e – t ) – 6t  2L = t0 t 0 27 t 3 t3

3 3  t   e – t – 1 e – 1  (e 2 t – e – 2 t – 4t ) ( e t – e – t – 2t )    im     3 – 3 54 L = t 0    –t   t   t3 t3     3 3  t – 2t   –t   2t  e t – e – t – 2t   – 4t   e – 1    e – 1   24  e – e   3 –  im 54 L = t 0   t   –t      ( 2t )3 t3         

 54 L = 1 + 1 + 48 L – 6L

5.

 12 L = 2

im  n2  ( n  1)  n  1   n  1  .......  n       n  n 2





2 

2 

 1 1     (n  1)  n  ....... n  n –1   2 2    im  = n    nn     im 1  1  = n   n

n

2n 2

 1 1    2n  1

1

=e .e

1/2

.e

1/4

....... e 2

n –1



L=

1   2n 1  

1 6

n

n

n

  1  1  1  im  1   1  .......1  n –1   = n    2n  2 . n     n 

(1 form )

2n – 1. n

 1  ........  1  n –1  2 .n  

......  term

2n – 1

im 1  1  { using n   n

pn

1   e p } { im 1   n n  

pn

 ep }

= e2

RESONANCE


6.

Let x 2= 1 1 =

2 ; x3 =

2 1 =

n  n 1 x   im  n 1  = im   n n x n  n  

=e

7.

  1     im  1 1 1  n    n    

Now,

1 2

n n





<

8.

1

<

n k

2

n n

n



n

=e

 n  n  1  n 

 im  n   

n 1

k

<

2

n k

n

k 2

k

k 1

n

k

<

2

n 1 k 2

k 1

k

=

loge x sinsinx  1 let y = x –1

<

n

k 2

k 1

n

n

n



n

<

n k 1

k 2

k

n

<

n

k 2

k 1

1

nn  1 2 n2  1





n

n

k 2

k 1

k

im < n 

1 2

nn  1 2 n2  1





(By Sandwich theorem)

im  x 1

x = 1 + y

log e 1  y   log e 1  y  y 1 im  im = y  × × sin y 0 sinsin y  = y 0 y sinsin y  sin y

im x 3  x 2   



2 4 2  x  1  x  2x   im im 3 = x  x  2  = x  4

 x  1  x

im = x 

im = x 

 x 2 

x3

(1  x 4  x 4 )

4 2  2    x  1  x 4  x 2   1  x  x     

x3  2   x  1  x 4  x 2   1  x 4  x 2     x3  x3  1  

Let y = x–

1 x4

1 

   1 2   1  4  1   x  

1 2 2

.

1 1  2 4 2

 2

im e  cot y = im 

y 0 

= –1 × 1 × 1 = –1

 1  x4  x 2  

x 



10.

n 1  n

1

 1  . nn  1 im  im  < n n   n 2  n   2 im

=e



im n 

2

 1  . nn  1  2  < 2 n n

 Given limit

9.

n  1  1 n  n 

xn = n

e

2

n



=e

 im  n   

;......;

1

= e2 =

k



n

3 ; x4 = 4 ; x5= 5

y 0

1 e

cot y

RESONANCE

=0


11.

12.

log e log e x 

Let

im L = x 



im L = x 

Since

x log e x im x  0 (using L.H. Rule) = 0 and x  x  e x

e

x

2 log e loge x  loge x x . . loge x x e x

= 2×0×0×0 = 0

im

im log e sin 4m  1x   x log e sin4n  1x  2

0   form  0  

1 . cos4m  1x.4m  1 im sin 4m  1x im tan 4n  1x. . 4m  1  = x = x  tan4m  1x 4n  1 1 2 2 . cos4n  1x.4n  1 sin 4n  1x   =y x= +y 2 2  tan (4m+1)x = – cot(4m+1)y similarly tan (4n+1)x = – cot(4n+1)y Put x–



im cot 4n  1y . 4m  1  Given limit = y 0 cot 4m  1y ( 4n  1)

13.

Let

tan 4m  1y . 1 . 4m  1 ( 4m  1) y tan4n  1y 4n  12 ( 4n  1) y x–1 = y x=1+y loge y  cot

given limit = im



14.

cot y

=

4m  12 4n  12

    cot y      2  im loge y  tan y   = y 0   cot y    

   y    tan y  tan y     2. . 2 y loge y   = 0. – 2.1.1 = – 2 = yim  y 0     y  tan y     2   n1 nx  1 x 1/ x 1/ x x e n 1  nx    1  nx  1  nx  . 1  nx  x e im  im  im  = im1  = x ( e 1 )  x  e 1 x  e  x  e 1 x  e 1 x  e –1  x  e –1  x  e 1

=

15.

y 0

 y 2

 +(4m+1)y 2

tan 4m  1y . 4m  1 im = y  0 tan 4n  1y ( 4n  1)

2

im = y  0

(4m+1)x = (4m+1)

im

 

x e

(i)

–1 

1 x en ex .1  nx  x = e.1. (0)e–1 ex  1

nx    1 = 0  xim 1 x 1  

Since x 2>0 and limit equals 2, f(x) must be a positive quantity. Also since im

f x 

= 2, the x2 denominator  zero and limit is finite therefore f(x) must be heading towards zero. If f(x) tends to some non-zero number then limit will cease to be a finite quantity.  f(x) > 0 in small neighbourhood of x = 0 thus im f x  = 0 x0

x 0

(ii)

f x  xf x  f x  f x  f x   2  im = im x. 2 = im x. im 2 = 0 ×2 = 0 x0 x0 x 0 x x x0 x x x

 f x  im  im if we find RHL of x   then it is zero but if we find LHL of x  0  x  0 Hence limit does not exist.

RESONANCE

 f x  im  x  = x  0  

 x.f x   2  = –1  x 


CONTINUITY AND DERIVABILITY EXERCISE # 1 PART - I Section (A) : A-3.

(i) h(x) = {x} [x] at x = 1 at x = 2 (ii) h(x) = {x} + [x] = x (iii) h(x) = {x} – [x] = x (iv) h(x) =

RHL = 0 RHL = 0

{ x } + [x]

at x = 1 at x = 2 A-4.

h(1) = 0 LHL = 0 h(2) = 0 LHL = 1 at x = 1 continuous – 2[x] discontinuous at x = 1

h(1) = 1 h(2) = 2

LHL = 1 LHL = 2

RHL = 1 RHL = 2

f (x) = (x + 2) (x – 2) (x – 3) ( x  2) ( x  2) , x  3 h (x) =  , x 3 k

for continuity

k = xlim h(x) = 5 3

h(x) = (x + 2) (x – 2) = x2 – 4 which is even  x  R

Section (B) : B-2.

1  { x }, Since {–x} =   0,

f(x) = x + {–x} + [x]  x  1  { x}  [ x ] , =  , x  [ x] 1  2 [ x ] =   2x

x x

x x

,

x

,

x

Curve of y = f(x) discontinuous at all integers in [–2, 2]

B-5._

u=

1 is discontinuous at x = – 2 x2

f(u) =

3 2

2u  5u – 3

=

discontinuous at u =





1 1 = x2 2 x=0

3 2

2u  6u – u – 3

=

1 & –3 2 and

and

1 =–3 x2 x=–

7 3

Hence y = f(u) is discontinous at x = –

RESONANCE

3 is (2u – 1) (u  3)

7 , – 2, 0 3 S OLUTIONS (XII) # 54

Section (C) :

C-5.

 x m sin 1  ; f (x) =  x  0 ;

x 0 x0

for continitity f(0) = 0 = RHL (x = 0)  1 lim hm sin   0 h 0 n lim hm [a finite quantity between [–1, 1]] = 0

h 0

It hold only when m > 0 if m  0 neither continuous  nor derivable for derivability

lim

h 0

f (h)  f (0) = finite h



 1 lim h (m 1) sin  h 0 h

it is finite and unique

and equal to zero if m > 1  when m >1 continuous and derivable if 0 < m  1 continuous but not derivable

Section (D) :

D-2.

y = f(x)

Section (E) : E-2.

f : R  R and f(x + y) = f(x) f(y)  x, y  R Put x = y = 0 f(0) = f2(0) since f(0)  0  f(0) = 1 f '(x) = lim

h 0

f (h)  1 f ( x  h)  f ( x ) = lim f ( x ) = f(x) f '(0) h 0 h h

dy = y.f '(0) dx On solving ny = x f '(0) + c y = f(x) = ec. ex f '(0)  xR Thus f(x) = ex.f '(0)

Let f(x) = y 

E-3._

f(x) is continuous and

 f(0) = 1  c = 0

7  [f(–2), f(2)], by intermediate value theorem (IVT), there exists a point 3

c (–2, 2) such that f(c) =

7 3

PART - II Section (A) : A-2.*

(A) f(x) is continuous no where (B) g(x) is continuous at x = 1/2 (C) h(x) is continuous at x = 0 (D) k(x) is continuous at x = 0

RESONANCE


Section (B) : B-4.

1 1 , where t = , y = f(x) is discontinuous at x = 1, where t is discontinuous and y = t t2 x 1

y =

2

1 at t = – 2 and t = 1 ( t  2)( t  1) 1 x 1

 1–

1 x 1



–2x + 2 = 1,



x=2

x=

1 2 f(g(x)) is discontinuous at x =

1 , 2,1 2

Section (C) : C-9.

3 Lim = f(2)  x  2 f(x) = 1 5 f(x) is not continous at x = 2 Lim

x  2

Lim x  3

f(x) =

f(x) =

Lim x  3

f(x) = f(3) =

9 2

1 9 ((3  h)3  (3  h)2 )  Lim  4 2 h0 h

Now LHD (x =3) is

2 Lim  h  8 h  21  21 4 4 9 9 Lim  4 (| h  1 |  | 1  h |)  2  0  f(x) is not differentiable at x =2 and x = 3 h0 h



and RHD (x = 3) is

h0

Section (D) : D-3.*

(sin1 x )2 cos1/ x y  f (x)    0

x0 x0

f(x) can be discontinuous only at x = 0 in [–1,1] So we check only at x = 0  1 (sin1  h) 2 cos    0  n LHD (x = 0) = lim h 0 h 2

 sin 1  h   . h cos 1  = –1. 0. [finite quantity between [–1,1]] = 0 lim    h 0  n n  

RHD (x = 0) is lim h h 0

sin

1

h

n2

2

 .cos  1   0 n

Hence f(x) is differentiable as well as continuous in [–1,1]

D-4.

RESONANCE


D-6.

x  x 2  y  g( x )   1/ 4  sin x 

0  x  1/ 2 1/ 2  x  1 x 1

Section (E) : E-1.

f(x + 2y) = f(x) + f(2y) + 4xy  x, y  R Replace 2y with y we have f(x + y) = f(x) + f(y) + 2xy  x, y  R diff. w.r.t. x f '(x+y) = f '(x) + 2y Put x = 1 y = –1 f '(0) = f '(1) –2


f(0) = 0 f(0+) = hlim 0

h h h   + ............... h  1 (h  1)(2h  1) (2h  1)(3h  1)

  1 1 1 1 1      .......... ..  1  = hlim 0  h  1 (h  1) 2h  1 2h  1 3h  1  f(x) is not continous at x = 0 since f(0)  f(0+)

4.

f(1) = lim

n

log 3  12n sin1 2n

1

lim f(1+) = hlim 0 n 

1



log 3  sin1 2

log(3  h)  (1  h)2n  sin(1  h)

lim f (1–) = hlim 0 n 

(1  h)2n  1 log(3  h)  (1  h)2n  sin(1  h) (1  h)2n  1

  sin1

 log 3

discontinous at x = 1 6.

y = f(x)

RESONANCE


y = |f(x)|

y = f(|x|)

9.

y = f(x) = xsin 1/x. sin y = 0, x = 0,

1 r

Let t = x sin1/x

1 x sin 1/ x

when x  0,

where r = 1,2,3,............... as x  0  , t  0 and as x 

y = t sin1/t

also

1 , r = 1,2,3 r

1 , t 0 r

lim y  lim t sin t  0 = f(0)

x 0

t 0

1 lim y  lim t sin t  0 = f  r  1   t 0 x r

1  f(x) is continous  x  [0, 1] r We know that t = xsin1/x is not differentiable at x = 0

 f(x) is continous at x = 0 and

therefore y = tsin1/t = xsin1/x. sin

1 x sin

1 x

is not differentiable at x = 0

Section (D) : 10.

y = |sinx|

y = sin|x|

RESONANCE


y = f(x) = |sinx| + sin|x|

f(x) is continous every where f(x) is not differentiable at x = n f(x) is not periodic 16.

Differentiability at x = 1 sin[(1  h)2 ]  (1  h)2  3 (1  h)  8 f(1–) = Lim h0

 a (1  h)3  b  (a  b ) h

2 a (1  h)3  a  0 form Lim 3a (1  h)   = Lim = h0 h0 0  h 1 – f(1 ) = 3a

2 cos(1  h)   tan 1(1  h)  a  b f(1+) = Lim h0 h Function is differentiable 

–2+

 =a+b 4

.....(1)

 2 cos h  tan 1(1  h)  2   / 2 = Lim h0 h Now f(1–) = f(1+) a=

1 6

3a =

( 2 cos h  tan 1(1  h)  a  b) = Lim h0 h

= Lim 2 sin h + h0

1 1  (1  h)2

=

1 2

1 2

....(2)

by (1) and (2) b =

13  – 4 6

PART - II

3.

1  cosh n (cosh)  f   = hlim 0  4h2 n [1  4h2 ] 2 2  sin2 h / 2  4h2 n(1  2 sin 2 h / 2) 2 sin 2 h / 2 lim . = h0 16  16  2 . 2 .  2 sin 2 h / 2 h2 / 2  h / 2  n (1  4h ) =

1 1 . 1. 1.(–1) .1 =  64 64

RESONANCE


5.

[sin 0]  0  [sin 1]  0 [sin 2]  0  f(x) = [sin[x]] = [sin 3]  0 [sin 4]  1  [sin 5]  1  [sin 6]  1

0  x 1 1 x  2 2x3 3x4 4x5 5x6 6  x  2

f(x) is discontinuous at (4, –1)

9.

10.

x , x 1  f(x) =  2 ax  bx  c , otherwise  f(x) should be continous at x =1 it gives a+b+c =1 f(x) should be differentiable at x= 1 it gives 2a+b=1  b =1–2a c= 1–a–b= a lim f ( x )  lim x 2 e 2( x 1)  1

x 1

x 1

f(1) = 1

lim f ( x )  lim a sgn (x +1) cos2(x–1) + bx2 = a.1.1+ b x 1

x 1

for continuity a + b = 1

 e 2h  1   2h  2h (1  h)2 e 2h  1   lim 2 e  he  LHD (x = 1) is lim = h 0   h h 0 h   = 2+0+2=4 a sgn(2  h) cos 2h  b(1  h)2  1 h 0 h

RHD (x = 1) is lim

a cos 2h  b  bh 2  2bh  (a  b) = h 0 h f(x) is differentiable at x = 1 if 2b = 4 b = 2 a = –1 

= lim

11.

f(x) = [x] [sin x],

13.

, x  ( 1, 0) 1 =  , x  [0, 1) 0 f(x) is continuous in (–1, 0) Given f(0) = 4

x  (–1, 1)

lim 2f ( x )  3f ( 2x )  f ( 4 x ) x2 using L Hospital rule

0   0  form  

lim 2f ( x )  6f (2x )  4f ( 4 x )

0   0  form  

x 0

x 0

 cos 2h  1  lim a   bh  2b = 2b h  

h 0

2x

using L Hospital rule

lim 2f ( x )  12f (2x )  16 f ( 4 x ) = 2.4  12 .4  16.4 = 12 2 2

x 0

17.

f(x) = [n + psinx] , x  (0, ) graph of y = n + p sinx

RESONANCE


obviously f(x) = [n +psinx] is discontinous at points mark in above curve  number of such points  (p –1) + 1 + p –1 = 2p –1 21.

f(x) = [x] +

{x}

 1  x  1 ,  1  x  0  x , 0  x 1 Curve of y = f(x) =  1 x  1 , 1  x  2 

Method : II y = f(x) can be discontinuous only at x  so we check continuity only at x = n  f(n) = [n] + {n}  n  0  n LHL (x = n) is

h0

Lim [n – h] +

{n  h} = (n–1) + 1 = n

RHL (x = n) is

h0

Lim [n + h] +

{n  h} = (n+0) = n

f(x) is continous for x R n

22.

f(x) =

a

k

| x |k

k 0

= a0 + a1 |x| + a2|x|2 + a3|x|3 + ............+ an|x|n f(0) = a0

| x | 0 we know that xlim 0

lim f ( x )  a0

x 0

f(x) is continous for x = 0 |x|n is differentiable if n  1, nN f(x) is not differentiable at x = 0, due to presence of |x| If all a2k+1 = 0, f(x) does not contains |x|  f(x) is differentiable at x = 0

RESONANCE


EXERCISE # 3 2.

(A)

f(x) = |x3| is continuous and differentiable

(B)

f(x) =

| x | is continuous

1

f (x) = (C)

{does not exist at x = 0}

f(x) = |sin–1 x| is continuous f (x) =

(D)

x |x|

.

2 |x|

sin 1 x

1

.

1

1 x 2 f(x) = cos–1 |x| is continuous | sin x |

1

f (x) =

1 x

2

.

x |x|



Comprehension # 1_ 3.

(A)

1 f (x) = n | x |

(B)

f (x) = x sin

(C)

f (x) =

LHL = 0 = RHL

 x

LHL = 0 = RHL

1

f (0) = not define

1  2cot x

LHL = 1 RHL = 0 (D)

 | sin x |   f (x) = cos  x 

 LHL  RHL

LHL (at x = 0) = cos (– 1) = cos 1 RHL (at x = 0) = cos 1 LHL = RHL

4.

(A) (B) (C) (D)

5.

Lim x  0

Lim x  0

Lim x  0

Lim x  0

Lim

f(x) = 1 f(x) = –

x  0

 2

f(x) = –1 f(x) =

 1 tan tan x x  f(x)   n [ x]  1 x  

Lim x  0

   

Lim x  0

Lim x  0

f(x) = 0 f(x) =

 2

f(x) = 1

f(x) = 0

   f  = 1 4   

   f  = x =  4  4   Jump P - 1 –

 4

RESONANCE


Comprehension # 3 (9 to 11)  0  1 x Given function f(x) can be rewritten as, f(x) =  1  x  0



also,

Now,

0   1  ( x  1) f(x – 1) =  1  ( x  1)  0

,

0    1 ( x  1)   f(x + 1) = 1  ( x  1)  0

,

0  x 1 1

,

x 1 1

,

,

x  1  1

or

, 1 x 1 0 , 0  x 1 1

or

x 1 1

 0  2  x   x g(x) = f(x – 1) + f(x + 1) =   x 2  x   0

x  1

, 1 x  0 , 0  x 1

x  1  1

, 1 x 1 0

,

,

x 1  0   x f(x – 1) =  2  x  0

,

 0  2  x f(x + 1) =   x  0

,

x0

, 0  x 1 , 1 x  2 ,

x2

x  2

,  2  x  1 ,

1 x  0

,

x0

, x  2 ,  2  x  1 ,

1 x  0

, ,

0  x 1 1 x  2

,

x2

It is easy to check that g(x) is continuous for all x  R and non-differentiable at x = – 2, –1, 0, 1, 2. 13.

Statement - 1 f (x) = {tanx} – [tan x]

 ,  tan x  f(x) = tanx – 2 [tan x] = tan x  2 ,  obviously at x =

0x

 4

  x  tan 1 2 4

 f(x) is continuous. (True) 3

Statement-2 y = f (x) & y = g(x) both are continuous at x = a then y = f(x) ± (g(x) will also be continuous at x = a (True) Statement-1 can be explained with the help of statement-2.

21.

 x  f(x) = |x sgn (1 – x 2)| =  0  x 

x  ( ,1)  ( 1,  ) x  1, 0, 1 x  (0, 1)  (1,  )

function is discontinous at x = –1, 1 and non differentiable at x = –1, 0,1

RESONANCE


24.

1  3 3  (x )  x , x0 1 1 f(x) = (x2 |x|)1/3 =  (  x 5 ) 3  ( 1) 3 x   x , x  0 

=–x f(x) is differentiable every where except at x = 0

EXERCISE # 4 PART - I 1. 2.

f (2h  2  h 2 )  f (2) (h  h 2  1)  (1) (2h  2  h2 )  (2) f ( 2) h(2  h) lim 6 = lim . = hlim =3 h0 h0 ( 2h  2  h 2 )  ( 2) f (h  h 2  1)  f (1) (h  h 2  1)  (1) 0 f (1) h(1  h) 4 We have given L.H.D. at x = a is zero  f(a–) = 0

f (a  h )  f ( a ) =0 h 0 h Now, L.H.D. at x = – a lim



 f(–a–) = lim h 0

= lim h 0

...........(1)

 f (a  h )  f (a ) f ( a  h )  f ( a ) = lim h 0 h h

f (a  h)  f (a) = h

lim h 0 

[  f is an odd function]

f ( a  h )  f (a ) f [ 2a  ( a – h )]  f ( a ) = lim h h0 h

[  f(2a – x) = f(x)  x  [a, 2a]] f (a  h)  f (a ) _ = – f(a ) h = – (0) = 0 Thus, f(–a–) = 0

= – lim h 0

3.

[using (1)]

Since f(x) is differentiable at x = 0 i.e.,



continuous at x = 0

lim f ( x )  lim f ( x )  f (0) x  0

x  0–

e ah / 2 – 1 e ah / 2 – 1 a a  lim ·  h h0 h0 2 2 h a 2

lim  f ( x )  lim

x 0

also,

lim x 0

–

c –h c  = b sin–1 f(x) = lim b sin–1  2 2 h0  

c a 1   2 2 2



b sin–1

  

a=1 ... (i) Also, f(x) is differentiable at x = 0 f(0+) = f(0–)

eh / 2 – 1 1 h/2 – –2–h 1 h 2 = lim 2e f (0+) = hlim = 0 h0 8 h 2h 2

and

f(0–) =

lim h0

c –h 1 b sin –1 –  2  2 = h

b/2 1–

RESONANCE

c2 4


b  

4.

2



1 8

4–c a = 1 and

or

64b2 = (4 – c 2)

64b2 = (4 – c 2)

2 lim f ( x )  f ( x ) . Put x = 0 and we get 0 form. Also because ‘f’ is strictly increasing and differentiable. x 0 f ( x )  f (0 ) 0

Apply L-Hospital rule, weget  2  lim 2xf ( x )  f ( x ) = – 1. Since ‘f’ is strictly increasing f(x)  0 in an internal x 0 f ( x )

5.

Graph of f(x) is

From graph f(x) is not differentiable at x = 0,  1. 6.

As f is continuous

lim f ( x ) = lim f (1/ n) = lim 0  0 f(0) = x 0 n  n  Also f(0) = lim

x 0

= 7. put put

f ( x )  f (0 ) f (x) = lim x 0 x x

f (1/ n) 0(exact zero) lim = =0 n  (1/ n) n  (1/ n) lim

f(x – y) = f(x).g(y) – f(y) g(x) x = y in (1) we get f(0) = 0 y = 0 in (1) we get g(0) = 1

given that f(0+) = lim

h 0

So f(0) = 0 = f(0)

...................(1)

f ( 0  h)  f ( 0 ) f (0 ).g( h)  g(0)f ( h)  f (0 ) = lim h 0 h h

f ( h) lim f (0  h)  f (0) = h 0 = f(0–) h h0  h Hence f(x) is differentiable at x = 0 Let g(x – y) = g(x) . g(y) + f(x). f(y) ..................(2) Put y = x in (2)  1 = g2(x) + f 2(x) g2(x) = 1 – f 2(x)  Put x = 0  g(0) = 0 = lim



2g(x). g(x) = –2f(x). f(x)

8*.

From graph broken at x = 2, 3 & does not have definite tangent at x = 2, 3 9*.

Here, f(x) = min. {1, x2, x3} which could be graphically shown as

RESONANCE


 1, x  1 f(x) =  3 x , x  1  f(x) is continuous for x  R and not differentiable at x = 1 due to sharp edge. Hence (A) and (C) are the correct answer.



10.

11.

(A)

y = x|x|

(A)  p, q, r

(B)

y=

(B)  p, s

(C)

y = x + [x]

(C)  r, s

(D)

x 1  2x ;  ; 1 x  1 y = |x – 1| + |x + 1| =  2  2 x ; x  1 

(D)  p, q

|x|

f(x) = g(x) cos x + g(x) sin x f(0) = g(0) = 0

f ( x )  f (0) g( x ) cos x  g( x ) sin x  g(0) f(0) = Lim = Lim x 0 x 0 x x g( x ) cos x  g(0) g( x ) sin x = Lim + Lim x 0 x 0 x x g( x ) cos x  g(0) g( x ) cos x  g(0) = Lim = Lim = Lim (g(x) cot x – g(0) cosec x) x 0 x 0 x 0 x sin x 12.

f (1  h)  f (1) |h|  0 Clearly ‘p’ = hLim = =–1 0  h h Now

Lim x 1

Lim x 1

( x  1)n =–1 log cos m ( x  1) ( x  1)n =–1 log [1  cos( x  1)  1] m  [cos( x  1)  1] cos( x  1)  1

RESONANCE


or,

n Lim ( x  1) x 1 m

1

=–1 ( x  1) 2 (C) is correct answer.

 2 sin

Hence n = 2, m = 2.

2

Aliter : LHD of |x – 1| at 1 is – 1  p

lim g(x) = lim x 1 t 0 =–



–

2 m

2 m

tn log cos m t

=

lim tn–2 t 0 

lim tn–2 = – 1

t 0 

1 m

cos t  1 t2 lim tn–2 . . t 0 log cos t cos t  1

( tlim 0 

1 cos t

=

t2

1 ) 2

n = 2, m = 2

13.

f(x) = kx Hence f(x) is continuous & differentiable at x  R & f ’(x) = k (constant)

14.

(A)

at x = –

 2

  Lf  –  = 0  2

  = f–   2

  Rf  –  = 0  2

(B) (C)

(D)

15.

 continuous at x = 0 Rf(0) = 1 Lf(0) = 0 at x = 1 Rf(1) = 1 Lf(1) = 1



not differentiable



differentiable at x = 1

3  > – 2 2



f(x) = – cos x

at x = –



differentiable at x = –

1    b x b 1 b   f(x) = = + bx  1 b (bx  1)

1     b  b, f(x) < 0 x (0, 1) f(x) =  b (bx  1)2

Range of f(x) is (–1, b) so range  co-domain so f is not invertible f–1 doesnot exist No comparison with f–1 16.

(I) for derivability at x = 0

  h2 . cos –  – 0  f ( 0 – h ) – f ( 0 ) im im im – h . cos  h L.H.D. = f '(0–) = h0 = h0 = h0 =0 h –h –h  h2 . cos  – 0 h im f (0  h) – f (0) = im RHD f '(0+) = h0 =0 h0 h h So f(x) is derivable at x = 0 (ii) check for derivability at x = 2

RESONANCE


3 2

im f (2  h) – f (2) = h0 h

RHD = f '(2+)

      (2  h)2 . cos (2  h)2 . cos  –0  im im 2h 2h = h0 = h0 h h  h      ( 2  h)2 . sin (2  h)2 . sin –   2( 2  h)     im  im 2 h 2  .   = = h0 = (2)2 . = h0 2 ( 2) 2( 2  h)    h  h  2( 2  h) 

   (2 – h)2 . cos  –0 f ( 2 – h ) – f ( 2 ) im im 2–h = h0 = h0 –h –h

LHD

       (2 – h)2 . – cos (2 – h)2 cos   – 0   im  2 – h  2–h  = h0 h –h

im = h0

   (2 – h)2 . sin –   im h 2 2 –  = h0 h

 h   ( 2 – h)2 . sin –  2( 2 – h)  . –   im = h0 =– 2( 2 – h)  h   –   2(2 – h) 

So f(x) is not derivable at x = 2 f (2n)  a n

an  b n  1

  f (2n )  a n  f (2n  )  b n  1 

17.

an  bn  1 So B is correct

f (2n  1)  a n

an  b n 1  1

  f ((2n  1) )  a n  f ((2n  1) )  b n1  1 

a n  b n 1  1 an 1  b n  1

PART - II 1.

Continuity at x = 0  1 1   –   |– h| (– h )   e

(0 – h) LHL = lim– f ( x ) = hlim 0

= lim (–h)e

x0

f(x) RHL = xlim 0  

 1 1 – –  h h

h 0

 1 1   –  | | e  h h

= lim (0  h) h0

= lim

 1 1 –    he  h h 

h 0

lim (–h) = 0 = h 0

= lim

h0

h 2 h e

=0

lim f ( x ) = lim f ( x ) = f(0) x 0

x0–

Therefore, f(x)is continuous for all x. Differentiability at x = 0  1 1 – – 

(–h)e  h h  – 0 e0 = 1 Lf  (0) = lim = hlim  0 h 0 (–h) – 0  1 1 –   he  h h 

– 0 = hlim 0

1

2 =0 eh h–0 Therefore, f (x) is not differentiable at x = 0.

Rf  (0) = lim



Rf  (0) Lf(0)

h 0

RESONANCE


2.

Since, f(x) =



1 – tan x 4x – 

 1 – tan x  lim f ( x )  lim      4x –   x x 4

(Using L'Hospital's rule)

4

  – sec 2 x  – sec 2    lim  4  – 2  = =  4 x   4 4 4





lim f ( x )  –

 x 4

1 2

   Also, f(x) is continuous in 0,  , so f(x) will be continuous at . 2   4

1  f   = lim f ( x )  – 2  4  x 4

3.

Since |f(x) – f(y)|  (x – y)2

| f ( x) – f ( y ) |  lim | x – y | xy |x– y|  | f (y)|  0  f (y) = constant  f (1) = 0 lim

xy

4.

f (1) = lim

h 0

Since,

f (y) = 0 f (y) = 0

[ f(0) = 0 given]

f (1  h) f (1) f (1  h) – f (1) – lim = lim h 0 h 0 h h h

lim

h0

f (1  h) f (1) = 5, so lim must be h0 h h

finite as f (1) exist and lim

h0

only, if f(1) = 0 and lim

h0

5.

 

f (1) can be finite h

f (1) =0 h

Since, f(x) =

x 1 | x |

Let

g( x ) x = h( x ) 1 | x |

f(x) =



f (1) = lim

h0

f (1  h) = 5. h

It is clear that g(x) = x and h(x) = 1 + |x| are differentiable on (–, ) and (–, 0)  (0, ) respectively. Thus, f(x) is differentiable on (–, 0)  (0, ). Now we have to check the differentiability at x = 0



x –0 f ( x ) – f (0 ) 1  |x| lim = lim x0 x–0 x 0 x = xlim 0

1 =1 1 | x |

Hence, f (x) is differentiable on (–, ).

6.

Now,

= xlim 0

2  e 2 x – 1 – 2x 1 lim  – 2 x  = lim x0  x x 0 x( e 2 x – 1) e – 1

2e2 x – 2 (e 2 x – 1)  2xe2 x

RESONANCE

(using L'Hospital's rule) S OLUTIONS (XII) # 69

= lim

x 0

7.

4e 2 x 4e 2 x  4 xe2 x

=1

(using L'Hospital's rule)



f(x) is continuous at x = 0, then xlim f(x) = f(0) 0



1 = f(0)

f(x) = min {x + 1, |x| + 1} f(x) = x + 1,  x  R. It is clear from the figure that f (x) 1,  x  R.

8.

1   f (1 – h – 1) sin –0  1  1 – h – 1 f (1 – h) – f (1) lim sin  –  = – lim sin 1 Now, f (1) = lim = h 0 = hlim  0 – h h0 h 0  h –h h

and f (1+)= lim

h 0

f (1  h) – f (1) h

1   (1  h – 1) sin –0 1 1 h – 1  = lim = lim sin h 0 h h 0 h f is not differentiable at x = 1. Again, now



f(1–)  f(1+)

1   (0  h  1) sin  – sin1 0  h  1   f (0) = hlim 0 –h    1   1   1  sin   – (h  1) cos     2  h  1   (h  1)   h  1    = lim h 0 –1

(using L' hospital's rule)

= cos 1 – sin 1

and

1   (0  h – 1) sin   – sin1 0  h – 1   f (0+) = hlim 0 h

 1   – 1   1  (h – 1) cos  sin   2  h – 1   (h – 1)   h – 1 = lim h 0 1 = cos 1 – sin 1  f (0–) = f (0+)  f is differentiable at x = 0.

9.

(using L'Hospital's rule)

gof (x) = sin (x |x|) x  ) , x0 cos( x | x |)(| x |  x . |x| gof (x) =   | x | cos( x | x |) , x0

 gof is differentiable at x = 0 and the derivatives is continuous  statement-1 is true Statement-2

lim h0 

 h  2  cos(h | h |) | h | h (gof )(0  h)  (gof )(0) | h |  = cos(h )(h  h) = 2  = hlim  0 h h h

RESONANCE


10.

 h 2  cos(h | h |) | h |  2  | h |  lim (gof )(0  h)  (gof )(0) = lim lim cos( h )( h  h) = – 2  = h0 h0 h0 h h h  not differentiable  statement-2 is false f(0) = q 1 1/ 2 1  x  ..... – 1 1 ( 1  x ) – 1 2 f(0+) = im = im = x 2  x0 x x0

f(0–) = xim 0–

sin (p  1)x  sin x x



f(0–) = im– x 0

(cos(p  1)x )(p  1)  (cos x ) 1

= (p + 1) + 1 = p + 2



11.

p=–



p+2=q=

1 2

3 1 ,q= 2 2

x sin(1/ x ) , x  0 f(x) =  0 , x0  at x = 0

  1  – h sin  –  = 0 × a finite quantity between – 1 and 1 LHL = hlim 0   h  1 RHL = hlim h sin =0  f(0) = 0 0 h  f(x) is continuous on R. f2(x) is not continuous at x = 0 12.

x 2 f (a) – a 2 f ( x ) 2xf (a) – a2 f ( x ) = im = 2af(a) – a2f (a) xa x a x–a 1

im

x 2 f (a ) – a 2 f ( x ) x 2 f (a ) – a 2 f ( a )  a 2 f (a ) – a 2 f ( x ) ( x 2 – a 2 )f (a) – a 2 ( f ( x ) – f (a)) = im = im xa x a xa x–a x–a x–a

im

Alter

 f ( x ) – f (a )  im (x + a) f(a) – a2   = 2af(a) – a2f (a) = x  a  ( x – a) 

13.

Doubtful points are x = n, n  I  2x – 1   2n – 1    = (n – 1) cos   = 0 L.H.L = lim– [x] cos   2   2  x n

R.H.L. =

[x] cos

 = n cos

=0

f(n) = 0 14.

f(x) = 3 f(x) = 0

2x5 2
RESONANCE

f(4) = 0


ADVANCE LEVEL PROBLEMS PART - I 1.

·

=

2.

R.H.L =



L.H.L =



=–

=

–



=

=

=

3.



L.H.L  R.H.L



L.H.L. =

and R.H.L =

=1×–1=–1

f(x) =

=

f(x) =

=



L.H.L R.H.L

 (D)

4.

f(x) = discontinuity may arise at the points where ,

x=

; x=

5.

,

and

;x=

five points

=

=1



6.

,

=0

=

similarly

f(x) = 1

=2

now, L.H.D. =

=

=



 L.H.D does not exist.

RESONANCE


7.

As 0 < {ex} < 1 

8.

f(x) = – 1  x  R

 f(x) =



(1 + 0) = 1



f(x)



(0 + 1) = 1

f(1) = 1



f(x) = f(1)

and

 continuous at x = 1 similarly we check for another integers 9.

As f(x) is continuous for all x R Thus

f(x) = f

since f(x) =

f(x) =

 Thus

10.

,x

=2

=

Let f(x) =



f() = 2

=



Now

=

= f()

f(x) =

f() = 0 11.

Function f(x) = (x 2 – 1) |x 2 – 3x + 2| + cos (|x|) ....... (i) Imp Note : In differentiable of |f(x)| we have to consider critical points for which f(x) = 0 |x| is not differentiablity at x = 0 but

cos | x | =



cos | x | =

Therefore it is differentiable at x = 0 Next, |x 2 – 3x + 2| = |(x – 1) (x – 2)| = Therefore,

RESONANCE


f(x) =

Now, x = 1, 2 are critical point for differentiability Because f(x) is differentiable on other points in its domain Differentiability at x = 1 L f (1) =

=

= 0 – sin 1 = – sin 1

(

=

(cos x)

at x = 1 – 0 = – sin x at x = 1 – 0 = – sin x at x = 1 = – sin1 and

R  (1) =

=

= 0 – sin 1 = – sin 1 (same approach)  Lf (1) = Rf (1). Therefore, function is differentiable at x = 1. Again Lf (2) =

=

and

= – (4 – 1) (2 – 1) – sin 2 = – 3 – sin 2

R f (2) =

= (22 – 1) – sin 2 = 3 – sin 2

=

So L f (2)  R f(2), f is not differentiable at x = 2 Therefore, (d) is the answer. 12.

f(x) = [x sin x] f(0) = 0 [–h sin(–h)] = 0 ; f(0+) =

f(0–) =

[h sin h] = 0

+

Here f(0 ) = f(0 ) = f(0) = 0 So continuous at x = 0 Since graph of f(x) is as shown in the figure Now all options can be checked from graph. –

PART - II 1.

f(x) = (i)

for 0 < sin x < 1,

f(x) =

(ii)

=1

for sin x = 0, f(x) =

RESONANCE


(iii)

for – 1  sinx < 0,

f(x) =



f(x) =



f(x) is discontinuous at integral multiples of 

2.

=

=

=

=

for g(x) to be continuous (na)2 = (n2a)2

3.



a=



f(x) =



 g(0) =

(na + n2a) = 0

(n 2)2

by definition of g(x)

g(x) =

g(x) =

clearly g(x) is discontinuous at x = 0 and not differentiable at x = 0, 2

4.

f(x) =

graph of f(x) is as shown is figure  f(x) is continuous for all x but non-differentiable for integral points.

5.



f(x) =

RESONANCE


f(x) =

 and

= 0 and

f(x) =

=0

f (0) = 0  f(x) is continuous at x = 0 f(x) =



=

and

f(x) =



 f(x) is discontinuous for x = 1 Similarly we can check that f(x) is discontinuous at x = – 1



L.H.D. at (x = 0) is =

=



R.H.D. at (x = 0) is =

=

=1

=1

 L.H.D. = R.H.D.  at x = 0, f(x) is derivable. 6.

is 1 or 0 according to x is a rational number or an irrational number. As m!x will become integral multiple of  when x is rational, then cos (m!x) = ± 1. And when m!x is not an integral multiple of  i.e. when x is irrational then –1 < cos (m!x) < 1  f(x) =  f(x) is discontinuous and non-differentiable at every real number.

7.

f(x) + f(y) =

f ’(x) = f ’

f ’(x) = f’ put x = 0 f ’(0) = f ’(y) (1+y2) = f ’(y) Integrating both sides f ’(0) tan–1y = f(y) + c ..... (1) Now put x = 0 & y = 0 in f(x) + f(y) = f we get 2f(0) = f(0)  f(0) = 0 ............. (2)

and =2



=2

 

f(x) = 2



f(0) = 2

........... (3)

from (2) & (3) & (1) 2tan-1y = f(y) + c now put y = 1, we get 2tan-11 = f(1) + c =

+c



c=0

 f(x) = 2tan-1x

RESONANCE


8.

R.H.L. =



f(0 + h) =

=

.

(putting 1–h2 = cos2) =

=

L.H.L =



=

f(0 – h) =

=

=

=

=

since R.H.L.  L.H.L, therefore no value of f(0) can make f continuous at x = 0 9.

As f is continuous on R, so f(0) =

Thus f(0) =

10.

=

f(1 + h) =

L.H.L. =

cos –1

f(1 – h) =

=

cos –1

f(–1) = 0 f(x) is discontinuous hence non-derivable at x = 1



f(–1+) =

=0

and

f(–1–) =

=0

 

f(–1+) = f(–1–) = 0 f(x) is derivable at x = – 1



f(x) =

=0

=

=

so

cos –1

= cos –1 (0) =

 

= f(x)

12.

=0+1=1

f(1) = 0



R.H.L. =

11.

f

f(x)

dx =

 f(x) =

=

we have f(x) =

(sin(–h) + cos(–h))cosec(-h)

RESONANCE

=

(cosh – sinh)–cosech


=

=

=e

Now we have f(x) =

=

=

If ‘f’ is continuous at x = 0 , then e=a= 13.

gives a = e and b = 1

Given that f(xy) = exy–x–y (eyf(x) + ex f(y)) putting x = y = 1 , we get f(1) = e–1 (ef(1) + ef(1))  f(1) = 0

now



x,y  R+

f(x) =

=

=



= f(x) +

f(x) = f(x) +



=



=

Integrating both sides w.r.t. ‘x’ , we get n |x| + c = or f(x) = ex (n|x| + c) since f(1) = 0  c = 0  f(x) = exn|x| 14.

Given f(x + y3) = f(x) + [f(y)]3 and f ’(0) > 0 putting x = y = 0, we get f(0) = f(0) + (f(0))3  f(0) = 0 also f(0) =

=

Let L = f ’(0) =

= L3

=

or L = L3 or L = 0 , 1, –1 as f(0) > 0  f(0) = 0, 1 Thus f ’(x) =

=



f ’(x) =

 f ’(x) = 0 , 1

Integrating both sides, we get f(x) = 0 or f(x) = x + c As f(0) = 0 , we have f(x) = 0 or f(x) = x Now f(x) = 0 is imposible as f(x) is not identically zero  f(x) = x and f(10) = 10

RESONANCE


METHOD OF DIFFERENTIATION EXERCISE # 1 PART - I Section (A) : A-1.

(i)

f(x) = sinx 2 f(x) =

=

=

=

=

(2x + h) .

·

.

cos



· cos

(f(x)) = (1) . (2x) . cosx 2

f(x) = 2x cosx 2 (ii) f(x) = e2x + 3 f(x + h) = e2(x + h) + 3 (f(x)) =

= 2e2x + 3

A–2.

(i)



y = x2/3 + 7e – =

(ii)

= e2x + 3

=

f(x) = 2e2x + 3 . 1

.2

 f(x) = 2e2x + 3

+ 7 tan x + 7 sec2 x

x–1/3 +

y = x2.n x.ex = 2x.n x .ex + x.ex + x2.nx.ex

(iii)

y = n

=

(iv)

.

sec2

=

.

=

= sec x

y=

=

=

RESONANCE


(v)

y = tan

y=

y = tan

=



sec2


(i)

ax 2 + 2hxy + by2 + 2gx + 2fy + c = 0 =–

 (ii)



=–



=

xy + xe–y + y. ex – x 2 = 0 =–


(i)

y = tan–1

+ tan–1

y = tan–1

+ tan–1

y = tan–1 5x – tan–1x + tan-1

+ tan–1 x

=

(ii)

y = sin–1

y=

,0
– cos–1

Let

tan–1 x = ,  

y=

– cos–1 (cos 2)

0 < 2 < y=

– 2 tan–1 x =

RESONANCE


(iii)

y = sin–1

y=

,–1
– cos –1

let

2 = cos –1x

y=

– cos –1

y=

–

y=

cos –1x

– cos –1 (cos) =

.


(i)

y = tan–1

,z=

y = tan–1 (1) + tan–1 (2x) =

=

(ii)

.8x

y = tan–1

,

.

=

z = tan–1x, z 

x = tanz



=





y = tan–1

y = tan–1 (tan z/2) y=

 =


(i)

ey (x + 1) = 1 x + 1 = e–y 1 = –e–y = e–y

= –ey

 (ii)

...(1)

y = sin (2sin–1x) =



= –ey (–ey) =

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

=

RESONANCE


=

(1 – x 2)



E–5.

(i)

= – 4y + x

y=

(1 – x 2)



=

=x

– 4y

= e1/3

=

(cosecx)1/nx

(ii) n y =

E–7.

ny =

· (– cosec x cotx) ·

ny = –

·

By L.H.Rule

cosx

y = e–1



H(1) = 1, g(1) = 2, H(1) = 1, g(1) = 2

=

By L.H. Rule

=

=0


f(x) = logx(nx)

f(x) =



f(x) =



f(e) =



= 1/e


B–3.

y = x 3 – 8x + 7 and x = f(t) 

t=0,

= 2 and x = 3



=



t = 0, x = 3



t = 0,





=

=

sin(xy) + cos (xy) = 0



=

=–

RESONANCE

 cos(xy)



– sin(xy)

=0

=–



C–4.

f(x) = |x||sinx| at x = /4, |x| = x and |sinx| = sinx  f(x) = x sinx 

n(f(x)) = sinx . nx



f(/4) =

y = sin–1

+ cos–1

f(x) = cosx nx +

, |x| > 1

y=





=0

Section (D) : D–2.*

y = sin–1

,

z = cos –1

<1



–1<

let

sin–1

=

sin 

cos =

(i)

if  

and

0<

 1 t  R



 (– /2, /2)

, then

y = , z =  

= 1

(ii)

if 

y = , z = – 

, then



= – 1

Section (E) : = f(ex).ex

E–2.

= ex f(ex) + e2x f(ex) E–5.*

u = ex sin x, v = ex cos x v

–u

= v(ex cos x + ex sin x) – u(ex cos x – ex sin x) = ex sin x(v + u) + ex cos x( v – u) = u(v + u) + v(v – u) = v2 + u2

RESONANCE


= ex sin x + ex cos x

again

= ex sin x + ex cos x + ex cos x – ex sin x = 2v similarly other options can be checked.


3

x = at

2

and y = bt

= 3at2 , =

= 2bt

.

=

.

=– 4.

.

y = 1+

=–

.

.

=

+

=

.

=

+

+

+

=

=

=

·

+

·

 ny = nx – n(x – c 1) + nx – n(x – c 2) + nx – n(x – c 3) Differentiating both sides w.r.t. x , we get =



=



10.

=–

y = xn







=

= n

–

...(1)

–

=–

 (1 + x)

 (1 + x)

=–

 (1 + x)

= –x

RESONANCE

+



=–

=

–

from equation (1)

–

+y–1

 (1 + x)

+x

=y–1


12.

let g(x) =

 g’ (x) =

+0+0

 g() = g() = 0 i.e.  is the repeated root of g(x) = 0 and h(x) is a polynomial of degree 3  f(x) = 0 has repeated root  hence g(x), is divisible by f(x)

PART - II : OBJECTIVE QUESTIONS 3.

x

+y 2

=0 2

x (1 + y) = y (1 + x) x 2 – y2 + x 2y – y2x = 0 (x + y) (x – y) + xy (x – y) = 0 (x – y) (x + y + xy) = 0 =



5.

y = sin–1

y=–

=–

+

f(x) =



+ sin–1 x

=

6.

xy



=

+

P=

, g(2) = a & g(2) = ?

g is inverse of f f(g(x)) = x Differentiating w.r.t. x f(g(x)) . g(x) = 1 g(x) = 10.



g(2) =

=



g(2) =

u = ax + b Let y = f(ax + b) = a f(ax + b)

=a2 f(ax + b)

= a3 f(ax + b)

= an f n (ax + b)

RESONANCE



f(ax + b) = an f n (u) = an

f(u)


12.

y=f

&

= f



14.

17.

20.

f(x) = sin x

.

= sin

.

y2 = P(x) 

2y

= P (x)





2

=



2y3



2y3



y2 = P(x)



2

= P(x) . P(x) + P(x) . P(x) – 2



2

= P(x) . P(x)

= P(x) P(x) –

=

= y2 P(x) – P(x) . y &

y

=

f(x) = – f(x) .... (i) f(x) = g(x) .... (ii) h(x) = (f(x))2 + (g(x))2 .... (iii) h(0) = 2, h(1) = 4 Differentiating equation (ii) w.r.t. x f(x) = g(x) = – f(x) Differentiating equation (iii) w.r.t. x h(x) = 2f(x) . f(x) + 2 g(x) . g(x) = 2f(x) . f(x) – 2f(x) . f(x) = 0 { g(x) = – f(x)}  h(x) is constant  h(x) is linear function  h(0) = 2  h(x) not passing through (0, 0) Let y = h(x) = ax + b at x=0 y=2=b  y = ax + 2 at x=1 a+2=4 a=2  curve is y = 2x + 2

y = cos –1

=

·



=



=

RESONANCE

·

·

when

x<0



=



=

·


when

x>0

=



Alternate : put x = tan tan–1x = 



y = cos –1

y = cos –1 (cos/2)



y=

= cos –1

=



EXERCISE # 3 2.

y = f(x3)

(A)

= f(x3) . 3x2



= f (1) . 3 = 9



(B)

f(xy) = f(x) + f(y) f(1) = f(1) + f(1)



f(1) = 0

(C) 



f (x) = – f(x), f(x) = g (x) g(x) = f  (x) = – f (x) h (x) = (f (x))2 + (g (x))2 h (x) = 2 f (x) . f (x) + 2g(x) . g (x) = 2 f (x) . g (x) + 2g(x) (– f (x)) = 0 h (x) = c, x  R  h (10) = h (5) = 9

(D)

y = tan–1 (cot x) + cot–1(tan x),



f(1) = f(e) + f



=

f(e) + f



=0

+

=–1–1=–2 Comprehension # 2 (6, 7, 8)

=–



=–

=

=

at (1, 1)

=

=



For question 8 Slope of normal at (1, 1) = –

=

Equation of normal y–1=

(x – 1)

RESONANCE



5y – 5 = 8x – 8



8x – 5y – 3 = 0


11_.

y = x2 = 2x

=2

again 2x

2

=1

+ 2x

=0 x

=–

=–





1

Statement-2 : =



15.



x = 1/2

then

tan–1 x =



u = sin–1



=1

19.



=–



0
,0<

=–

<

= sin– 1 (sin y) = y

Statement is true.



= =

21.



=



= 4y

=

a=±

F(x) = f(x).g(x).h(x) F(x0) = 21F(x0), f(x0) = 4f(x0), g(x0) = –7g(x0) h(x0) = kh(x0) F(x) = f(x).g(x).h(x) + f(x).g(x).h(x) + f(x).h(x).g(x) at x = x0 21F(x0) = 4f(x0)g(x0)h(x0) – 7f(x0).g(x0).h(x0) + kf(x0)g(x0).h(x0) 21(f(x0).g(x0).h(x0)) = (4 – 7 + k) f(x0).g(x0).h(x0)  k = 24


We have x 2 + y2 = 1 Differentiating w.r.t. x. 2x + 2yy = 0 Again diff. w.r.t. x. 1 + yy + yy = 0 yy + (y)2 + 1 = 0

RESONANCE


2.

p(x) = a0 + a1x + a2x2 + ........ + anxn  p(x) = a1 + 2a2x + ......... + nanxn – 1  p(1) = a1 + 2a2 + ........ + nan .......(i) Now | p(x) |  |ex – 1 – 1|  x  0 (given)  |p(1)|  |e0 – 1| = |1 – 1| = 0 But | p(1) |  0  p(1) = 0 Now for – 1 < h <  , h  0, 1 + h > 0 and |p(x)|  |ex – 1 – 1|  x  0  |p(1 + h)|  |eh – 1|  h > –1, h  0  |p(1 + h) – p(1)|  |eh – 1|  p(1) = 0 



Taking limit as h  0, we get 



 |a1 + 2a2 + 3a3 + ........ nan|  1 Hence proved 3.

| p(1)|  1

{from (i)}

f : R  R, f(1) = 3, f(1) = 6 y=

4.



ny =



ny =

n

=

At x = 0, ny = 0

=2





ny =



y = e2

y=1

Also on differentiating w.r.t x on both sides, we get

Putting x = 0, we get, 

(1 + y) = 2y + 2xy

= 2y..

y = 2y2 – 1 = 1.

5.

Let P(x) = bx2 + ax + c b0 P(0) = 0 c=0 P(1) = 1 a+b=1 P(x) = (1 – a) x2 + ax P(x) = 2(1 – a)x + a Now P(x) > 0 for x  [0, 1] P(0) > 0 & P(1) > 0 a>0 & 2(1 – a) + a > 0 0
6.

x sin y + y cos x =   x = 0, y =  x cos y

By L.H. Rule

×

+ sin y + y (– sin x) + cosx

RESONANCE

=0


=

y(0) = 0



=

=

 7.

g(x) = f(x) g(x) = f(x) = – f(x)

 

F(x) =

+



F(x) = 2f(x/2) . f(x/2) .



F(x) = f



F(x) = 0

8.

. f

+ 2 g(x/2) . g(x/2) .

f

– f



F(x) =

+g

g

{Using (i) & (ii)} F(x) = constant 



F(5) = F(10) = 5

=

=– 9.

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

.

=–

g(x + 1) = log [f(x + 1)] = log [xf(x)] = log x + log [f(x)] = log x + g(x)  g(x + 1) – g(x) = log x 

g(x + 1) – g(x) = –



g

=–

– g

Adding all, g

– g



g



g

– g

=–4

=–

– g

=–4

PART - I

1.

=n

.

=

=

=



(1 + x2)

RESONANCE

= n2y – x



(1 + x2)

+x

= n2y


2.

x=

= 3.

xy = ex–y x – y = y n x 1–

= n x .

5.

+

(1 + n y) =

 4.

=



Let f(a) = a1x2 + a2x + a3  f(1) = f(–1)  a1 + a2 + a3 = a1 – a2 + a3  f(x) = a1x2 + a3 f(x) = 2a1x f(a) = 2a1a f(b) = 2a1b f(c) = 2a1c  a, b, c are in AP



=



a2 = 0



f(a), f(b), f(c) are also in AP

=



f(x) = nxn–1 f(x) = n(n – 1) xn–2 f(x) = n(n – 1) (n – 2) xn–3 fn(x) = n(n – 1) ..... 3 . 2 . 1 Now

6.

f(1) –

+

= 1 – nC1 + nC2 – nC3 + ..... + (–1)n nCn = 0

+ ....... +

–

x = ey+x x + y = n x 1+

= =

7.

xmyn = (x + y)m+n taking log both sides m n x + n n y = (m + n) . n (x + y) +

=



=

–



=

8.

x2x–1 . 2x + 2x2x n x – 2 (xx n x + x . xx–1 ) cot y + 2xx cosec2y . y = 0  at x = 1, y = /2  2 + 0 – 0 + 2y = 0 y = – 1

9.

g(x) = 2f(2f(x) + 2) . f (2f(x) + 2) . 2f(x) g(0) = 2f (2f(0) + 2) f (2f(0) + 2) . 2f(0) = 2f(0) f(0) 2f(0) = (2) (–1) (1) (2) (1) = – 4 Hence correct option is (1)

RESONANCE


10.

=

=–

=

ADVANCE LEVEL PROBLEM PART - I 1.

y= ... (i) taking loge both sides, we get ny = (n x)n (n x) . nx Again taking loge , we get n (n y) = n (n x) . n (n x) + n (n x) = {n (n x)} [n (n x) + 1] Diff. w.r.t. x, =

2.



= y ny



= y ny

At x = –

[n (n x) + 1] + {n (n x)}

,

cos(4x) = cos

= – cos



|cos 4x| = – cos 4x

and

sin x = sin

= – sin



|sin x| = – sin x.



y = logu | cos 4x| + |sin x|



y=

... (1)

 (y + sinx) n sec (2x) = n (– cos 4x) Diff. w.r.t. x, (y1 + cosx) n sec(2x) + (y + sin x)

=



(y1 + cosx) n sec(2x) + (y + sin x) 2 tan (2x) = – 4 tan 4x

Put

x=–

in (1), y =

Put this in (2) and x = –



y1 =

... (2)

, we get

n 2 +

=

RESONANCE


3.

Here

=

.....(i)



but

y=

.....(ii)



From (i) and (ii)

4.

  and

x = cosec – sin x2 + 4 = (cosec + sin)2 y2 + 4 = (cosecn + sinn)2

Now

=

=

=

Squaring both sides, we get

or 5.

=0



(x2 + 4)

=

= n2 (y2 + 4)

(y1/5 + y –1/5)2 = (y1/5 – y –1/5)2 + 4 Then

y1/5 – y –1/5 = 1/5

Given y + y From (1) & (2)

–1/5

... (1)

= 2x

... (2)

y= Differentiating both sides w.r.t. x, we get or

Again differentiating w.r.t. x, 2x

Dividing by 2

6.

(x2 – 1)

= 25y2

+ 2(x2 – 1)

·

= 50 y

on both sides

x

+ (x2 – 1)



1<



If x =



[x] = 1

so,

f(x) = sin x3



f(x) = 3x2 cos x3



f

= 25y

<2

RESONANCE


7.

y = (tan–1 (x + 1) – tan–1x) + (tan–1 (x + 2) – tan–1(x + 1)) + .........+ (tan–1 (x + n) – tan–1 (x + (n – 1)))  y = tan–1 (x + n) – tan–1 (x) Hence,

8.

–

Given that g–1(x) = f(x)  x = g(f(x)) 

9.

=

or

g(f(x)) =

g(f(x)) f(x) = 1 g(f(x)) · f(x) =



y = sin(m sin–1 x)



g(f(x)) = –

= cos(m sin–1 x) .



Squaring and cross multiplying, (1 – x2) y12 = m2 (1 – y2), Again differentiating w.r.t.x (1 – x2) 2y1y2 + (– 2x) y12 = m2 (– 2yy1) Dividing by 2y1, (1 – x2) y2 – xy1 = – m2y 10.

11.

Take  = tan–1x, then

y = cos–1

or

y=

so |sec| = sec



xy . yx = 1  Diff. w.r.t. x, we get . nx +

y=

tan–1 x



=

y nx + x ny = 0

+ ny +

=–



12.

so (–/2, /2)

y = x(nx – n (a + bx)) = nx – n (a + bx) +



... (1) ... (2)





x3

... (3)

By (1) and (2), x

x3

–y=

=

RESONANCE


PART - II 1.

x = a(t – sint), y = a(1 – cost) = a(1 – cost)



... (1)

= a sint

=



=



= cot t/2



=

=

=–

from (1) ;

=–

2.

y=



+ k n

Differentiating both sides w. r. t. x, we get =2

+

·

=

n

= 2 n



– 2 n a + k

Differentiating both sides w.r.t. x, we get +

or

or

=

·

+

(x2 – a2)

+x

=

=2

hence value of (x2 – a2)

RESONANCE

is 2


3.

Let f(x) = xx From definition f(x) =

=

=

=

=

=

= xx (1 + n x)

=

4.

(a + bx) ey/x = x or

ey/x =

... (1)

Taking loge both sides, we have

= nx – n (a + bx)

Differentiating both sides w.r.t.x, we have or

(xy – y) =

= aey/x

=

=

from (1)

Again taking loge on both sides, we have n(xy – y) = n a + Again differentiating both sides w.r.t.x, we have (xy+y – y) = 0 +  5.

x3y = (xy – y)2 = a tan–1 (a n y)





a n y = tan

.... (1)

Differentiating both sides w.r.t.x, we get = sec2



= sec2

= 1 + tan2



= 1 + a2 (n y)2

RESONANCE

(from (1))


= 2a2 (ny)

Again differentiating both sides w.r.t.x, we get

. y

y y– (y)2 = yy (ny)  y y – yy(ny) = (y)2 6.

Let

y1 = tan–1

and

y2 = tan–1

Let

a = 2 n x,

then

y1 = tan–1

= tan–1

= tan–1

=

Similarly y2 = tan–1



7.

=  + , where tan  = 3

y=

+ ( + ) =

=0



y=

– , where a = tan 

+ tan–1 3 = constant

=0



tan–1



=



=

= 8.

xy = ex – y  Diff. w.r.t. x, we get nx = 1 – y

y nx = x – y



... (1)

y + y . xnx = x – xy 

y =

From (1), y=



x–y=

RESONANCE



y =


9.

y3 – y = 2x

... (1)

Differentiating both sides w.r.t.x, we get (3y2 – 1) or

=2

... (2)

Again differentiating w.r.t.x,

=

......... from (2)

=

... (3)

Now L.H.S. =

=

+

=

= 3y2 – 1 = 

Let =

10.

=

y=1+

 =

=

+

= R.H.S.

+ ..... (n + 1) terms

+ ....... (n + 1) terms.

=

+ ...... (n + 1) terms,

finally we get y= Taking loge of both sides, we have ny = n nx – n (x – a1 ) – n (x – a2) – n (x – a3) – .......– n (x – an) Differentiating both sides w.r.t. x, we get =



RESONANCE


=

=



=

11.

Let

z = cos x + i sinx



2 i sin x =

z–1 = cos x – i sin x



(2 i sinx)9 =



29 i sin9x =

y=



Binomial coefficients of

.... (1)

(with sign) are

1, – 9, 36, – 84, 126, – 126, 84, – 36, 9, – 1 = z9 – 9z7 + 36z5 – 84z3 + 126z –



=

–9

+ 36

+

+

–

+ 126

– 84

= 2 i sin 9x – 9.2 i sin 7x + 36.2 i sin 5x – 84.2 i sin 3x + 126. 2 i sinx From (1) & (2), we get y=

(sin 9x – 9 sin 7x + 36 sin 5x – 84 sin 3x + 126 sinx)



yn =

– 9.7n sin

+ 36.5n sin

+ 126 sin

12.

Given

..... (2)

– 84.3n sin

, where n  w..

y=

or (x2 + c)y = ax + b Differentiating both sides w.r.t. x, we get (x2 + c)y + y . 2x = a Again differentiating both sides w.r.t. x, we get (x2 + c) y + y . 2x + 2 (xy + y.1) = 0 or

(x2 + c) y=– 4xy – 2y

or

– x2 – c =

Now differentiating both sides w.r.t. x, we get

RESONANCE


– 2x = or or 13.

– x(y)2 = 2x(y)2 + 3y y – (2xy + y) y (2xy + y) y = 3 (xy+ y) y

We have sec–1 (2x) + 2x tan (n (x + 2)) = 0

sec–1 (2x) + 2x tan (n (x + 2)) = 0





.... (1)

esin(x/2)n sin y +

sec–1 (2x) + 2x tan (n (x + 2)) = 0

Differentiating both sides w.r.t. x, we get .

+ 2x . sec2 (n (x + 2)) .

.2

+ tan (n (x + 2)) . 2x n 2 = 0

... (2)

Putting x = – 1 in (1), we get (sin y)–1 = –



sin y = –

.... (3)

and putting x = – 1 in (2), we get e–n sin y

.2+

=0





Hence

{from (3)}

=

RESONANCE


APPLICATION OF DERIVATIVES EXERCISE # 1 PART - I Section (A) : A-3.

Let AC be pole, DE be man and B be farther end of shadow as shown in figure From triangles ABC and DBE

3y = 1.5 x = 2,

(x + y) = =4+2=6

A-4_.

Let r be the radius of the sphere and r be the error in measuring the radius. Then r = 8 cm, and r = 0.03 cm, Now the voume V of the sphere is given by Let r be the radius of the sphere and r be the error in measuring the radius. Then r = 8 cm, and r = 0.03 cm, Now the voume V of the sphere is given by

r3

V=

= 4r2

V =

r = (4r2) r = 48)2 × 0.03 = 7.68 cm 3

Section (B) : B-3.

Here x 2/3 + y2/3 = a2/3 Differentiating w.r.t. to x.

Equation (y – k) = –

P(0, k 1/3 (h2/3 + k 2/3)) or P(0, a2/3 k 1/3) Q (h1/3 a2/3, 0)

And, PQ = B-7.

(x – h)

= |a| = constant.

f(x 1) = g(x 1) = 0 m 1m 2 = – 1 and |m 1| = |m 2|  m1 = 1 ; m2 = – 1 or [(0 + h) (0 – h)] =

RESONANCE

m1 = – 1 ; m2 = 1

[0 – h2] = – 1


B-9.

Minimum distance = 10

Section (C) : C-3.

f'(x) = 3x 2 – 3 = 3(x – 1) (x + 1)

(i)

at at at at at at

(ii)

C-6.

x = 1point of minima  Neither increasing nor decreasing x = 2increasing x=–2 decreasing x = 0 decreasing x = 3 neither increasing nor decreasing x = 5 increasing

g(x) is monotonically increasing  g(x)  0 & f(x) is M.D. (fog) (x) =

C-7.

=

as  Also  

f(x)  0 & g(x)  0 (fog) (x) is monotonically decreasing x+1>x–1 f(x + 1) < f(x – 1) as f(x) is M.D. g(f(x + 1) < g(f(x – 1)) as g(x) is M..

Let

f(x) =

, x

f(x) = Let



f(x)  0



.

g(x) = x sec2x – tan x g(x) = 2x sec2x tanx > 0 x>0 g(x) > g(0), g(x) > 0 x1 < x2 f(x1) < f(x2) <

RESONANCE





f(x) > 0



f(x) is M.I.

< S OLUTIONS (XII) # 102

Section (D) : D-2.

f(x) = 3x 3 f(x) = 0  x=0 x = – 2, f(–2) = – 8 x = 0, f(0) = 0 x = 2, f(2) = 8 Minimum = – 8, maximum = 8 (ii) f(x) = cos x – sin x (i)

f(x) = 0

 x=

x = 0, x=

f(0) = 1 ,

x = ,

f

=

f() = – 1

Minimum = –1, Maximum = (iii) f(x) = 4 – x f(x) = 0  x=4 x = – 2,f(–2) = – 10 x = 4, f(4) = 8 x=

,f

=

Minimum = –10, (iv)

Maximum = 8

f(x) = cos x – sin 2x f(x) = 0

x = 0, f(0) = Minimum = D-6_.



cos x = 0, sin x =



x=

, x=



x=

f

=

 x=

, f

=

, Maximum =

Let No. of children of john & anglina = y  x + (x + 1) + y = 24  y = 23 – 2x Number of fights F = x(x + 1) + x(23 – 2x) + (x + 1) (23 – 2x) F = – 3x2 + 45x + 23  But 'x' wil be integral.

D-8.

,

2

2

= 0  – 6x + 45 = 0

 x = 7.5

check x = 6 or x = 7, F = 191

2

R +r =h R2 = h2 – r2 volume of cylinder , V = R2 (2h) =  (2h) (

)2

= 2 (r2 – h2) + 2h(–2h) = 0 

r2 = 3h2



h=

< 0 at h =



Vmax = 2

RESONANCE

= S OLUTIONS (XII) # 103

D-10.

2 + 2r = 440 A =  2r = – 2r2 + 440r = – 4r + 440 = 0

D-12.

at

r=

Let  Let

x, y be dimensions of rectangle. 2x + 2y = 36. V be volume sweeped V = x 2y V = x 2(18 – x) = x.3.(12 – x)

At

x = 12, V has maximum value 

y=6

Section (E) : E-2.

Let (h, k) be point of inflection h sin h = k y = sin x + x cos x y = cos x + cos x – x sin x y = 0  2 cos h – h sin h = 0  2 cos h = k sin2 h + cos 2 h = 1 = 1  4k 2 + h2k 2 = 4h2

+

...(1)

...(2)



locus y2 (4 + x 2) = 4x 2

Section (F) : F-2.

F-4.

Let

f(x) = 3x2 + px –1



f(x) = x3 +

 

f(x) satisfies conditions in Rolle's theorem 3x2 + px –1 = 0 has atleast one root in (–1,1).

–x+c



f(–1) =

+ c = f(1) 

f(c) = 0 for atleast one c (–1,1)

Let h(x) = f(x) g(x) h(a) = 0 = h(b) By Rolle’s theorem on [a, b] h(x) = 0, for at least one c  (a, b).  f (c) g(c) + f(c) g(c) = 0

PART - II Section (A) : A-1.

V=

V=

=

77 × 103 =

× 70 × 70 ×

( 1 litre = 103 c.c.)

= 20 cm/min.

RESONANCE


A-3_.

= Let x = 25 and x = 0.2 such that f(x) =  f (x) =

 f(x + x) = f(x) + f(x). x

=

+

.x

= A-4_.

+

× 0.2 =

=5+

= 5 + 0.02 = 5.02

V = x3 x = (3x 2) x = (3x 2) (0.04x) = 0.12x 3m 3

V =

Section (B) : B-5*.

2y3 = ax 2 + x 3 6y2

= 2ax + 3x 2

=

=

Tangent at (a, a) is 5x – 6y = – a =

,

2 +  2 = 61

= = 61



a2 = 25.36 a = ± 30 B-8.

P1 : y2 = 8x C1 : x 2 + (y + 6)2 = 1

Equation of normal of parabola y = mx – 2am – am 3 if passes through (0,–6) –6 = – 2am – am 3  a=2  3 = 2m + m 3 3 m + 2m – 3 = 0 m = 1. Point on parabola (am 2 , – 2am)  (2, –4).

Section (C) : C-2.

5x 4 – 3

f(x) =

It is sufficient to solve for p, the condition f(x)  0

xR

5x 4 – 3  0  x  R

RESONANCE


Case - 

1–p<0 p>1 Inequality holds true. 1–p>0 p<1

Case - 

Inequality holds if  

p  – 4, p + 4  (1 – p)2 p  – 4, p2 – 3p – 3 0



– 4  p 

Hence p 

C-5*.

–10

 (1, )

g(x) =

and

g(x) = f(x/2) – f(1 – x) Now g(x) is increasing if g(x)  0 f

f(1 – x)

[ f(x) < 0 i.e. f(x) is decreasing] 

1–x



x  2 – 2x



x  2/3



3x  2

 and

g(x) increases in 0 x  2/3 g(x)  0 for decreasing

  

 f(1 – x)





0 x 

1–x

x  2/3 2/3  x  1

Section (D) : D-3.

f(x) = 2 – |x + 1|

From figure it is clear that greatest, least values are respectively 2, 0 D-6*.

f(x) = The only factor in f(x) which changes sign is cosx – x. Let us consider graph of y = cos x and y = x It is clear from figure that for x  (0, x 0), cos x – x > 0 and for x  cos x – x < 0, 

RESONANCE

f(x) has maxima at x 0


D-8*.

f(x) =

,x>0

=

x > 0.

f(x) is decreasing On

x > 0.

, greatest value is

f(

)=

–

n

f

=

n

and least value is

.

D-10. S = 2rh = 2H

= 2H

Maximum at r =

D-13.

Let d be distance between (k, 0) and any point (x, y) on curve. d= ( y2 = 2x – 2x 2).

d=

Maximum d =

Maximum d =

Section (E) : E-3.

x=1

a=–

E-4.



3=a+b

=0



,

b=

3a + b = 0

f(x) = n(x – 2) –

f(x) =

=

=

=

.

As n(x – 2) is defined when x > 2  f(x) is M.. for x  (2, )



f –1(x) is M.. wherever defined

Also



f(x) is always concave downward

f(x) =

RESONANCE

<0



y = 50 – 16t2

So,

tan =



y=

.x

=

=

= – 16



= – 1500 ft/sec. 2.

at t = 0

;

x = 0, y = 1

= 2cm/sec

A=

(At, t = 7/2 sec, change in y -co-ordinate = 7 hence, pt. C has y-co-ordinate = 8 and x- co-ordinate = 6 at t = 7/2 sec.) = (4 + 7) ×

7.

×2 =

cm 2/sec

f(x) =

=

=



0

f(x) is increasing

RESONANCE

x  [0, )



f(x) is injective.


12.

Let (x) = x + (x) = 1 + If x < 0 , –|x| = x (x) =

=

>0

If x > 0, (x) > 0 Hence (x) is increasing As we know ex  x + 1  (ex)  (x + 1) ex + 13.

x+1+

Let x > –1 Consider f(x) = (1 + x)n(1 + x) – tan–1 x f(x) = n(1 + x) + 1 – f(x) =

>0

 f(x) is increasing  f(x) < 0

For x < 0, f(x) < f(0)  f(x) decreasing  f(x) > f(0)

 (1 + x)n(1 + x) – tan–1 x > 0 For x > 0,

f(x) > f(0)  f(x) > 0

 f(x) > 0

n(1 + x) >  f(x) is increasing

 f(x) > f(0)

 f(x) > 0

 n(1 + x) > Hence larger of these is n(1 + x). 15.

f(x) = a0 + a1x + a2x 2 + a3x 3 + a4x 4 + a5x 5 + a6x 6 a0 = a1 = a2 = a3 = 0 = e2

a4 = 2

f(x) = 2x 4 + a5x 5 + a6x 6 f(x) = x 3 ( 8 + 5a5x + 6a6x 2) f(1) = 0, f(2) = 0 , 17.



f(x) = 2x 4

xy = 18 Area of printed space

=

=

Maximum when x=

RESONANCE

y= S OLUTIONS (XII) # 109

21.

f(x) = 0  sin x=

=0

= n

,nN

x = .......,

,

, .......

,

Consider interval

, 1.

=0=



By Rolle’s theorem f(x) vanishes at least once in Infinite number of such intervals are there. Hence f(x) vanishes at infinite number of points in (0, 1) 24.

Let g(x) =

, x  [a, b]

By Rolle’s theorem, g’(x0) = 0 =0

 25.

f(x0) =



Let f(x) = (x + a) – (x)  f(x) = (x + a) – (x) + k f(0) = (a) – (0) + k f(2a) = (3a) – (2a) + k  f(0) = f(2a) By Rolle's theorem on [0, 2a], f(c) = 0 for at least one c  (0, 2a)  (x + a) = (x) has at least one root in (0, 2a)

PART - II 3.

f(x) =

>0

 f(x) increasing hence g(x) is also increasing function 7.

Let

 be quantity y=

y2 = y

=

=

=a 9.



= = a

f(x) = ; g(x) = for a > 1, a  1 and x  R n a h(x) = n f(x) + ng(x) 

(n a) h(x) =



h(x) =

na +

n a

+ |x|

 h(x) = a sgn x Now h(–x) = a|–x| sgn (–x) = –h(x)  h(x) is an odd function Also graph of h(x) is It is clear from the graph that h(x) is an increasing function

RESONANCE


12.

f(x) =

f(x) = For 13.

16.

0 < x < 1,

tan

f(1–) f(1) and f(1+)  f(1) – 2 + log2 (b2 – 2)  5 



f (x) > 0

0 < b2 – 2  128



f(x) is increasing.

2 < b2  130

2 = h2 + x 2 Area of base (triangle) is

3x =

a

Volume V = h

. 4 . 3.x 2 = 3

=h

=3 18.

a2

(2 – 3h2)



h (2 – h2)

V is maximum when h =

Maximum of f(x) is Given expression is f(x 1) + f(x 2)

20.

.



f(x 1) 



f(x 2) 

f(x 1) + f(x 2) 



f (x) = 1 +1– 2– At x = 0, f(x) is least. Least value = f(0) = 1

23.

f(x) = m – n is odd. f (x) < 0

25.

>0

x  (–, 0) 

f (x) > 0

x  (0, )

(x) = (3(f(x))2 – 6(f(x)) + 4)f(x) + 5 + 3 cos x – 4 sin x 5–

5 + 3 cosx – 4 sin x  5 +

adding (3(f(x))2 – 6(f(x)) + 4)f(x) (3(f(x))2 – 6(f(x)) + 4)f(x)  (x)  (3(f(x))2 – 6(f(x)) + 4)f(x) + 10  3(f(x))2 – 6f(x) + 4 = 3 (f(x) – 1)2 + 1 > 0 2 (3(f(x)) – 6(f(x)) + 4)f(x)  0 when ever f(x) is increasing.  (x) 0  (x) is increasing, when ever f(x) is increasing. If f(x) = – 11 then (3(f(x))2 – 6f(x) + 4) f(x) + 10 = – 33 (f(x) – 1)2 – 1 < 0  (x) < 0  (x) is decreasing.

RESONANCE


28.

f(x) = f(x) = 0 at x = 0, x = – 3, x = 1 so at x = 0, f(x) has local minima. and at x = –3, x = 1 ; f(x) has local maxima f(1) = 

,

f(– 3) =

. f(–3) < 0, f(1) > 0 and f(x)  0

f(x) is undefined at point(s) in (–3, 1). Hence f(x) has no absolute maxima.


(A) (B)

f(x) is continuous and differentiable f(0) = f() Hence condition in Rolle’s theorem and LMVT are satisfied. f(1–) = –1, f(1) = 0, f(1+) = 1 f(x) is not continuous at x = 1, belonging to Hence, atleast one condition in LMVT and Rolle’s theorem is not satisfied

(C)

f’(x) =

(x – 1)–3/5, x  1

(D)

At x = 1, f(x) is not differentiable. Hence at least one condition in LMVT and Rolle’s theorem is not satisfied. At x = 0

L.H.D. =

3.

=

= –1

R.H.D. = 1 At x = 0, f(x) is not differentiable Hence at least one condition in LMVT and Rolle’s theorem is not satisfied. (A) Let PQ = x Then BP =



PS =

tan60º =



area A of rectangle =

=

=–

(4 – x) x

(4 – 2x) = 0



x=2

<0



A is maximum, when x = 2.



Maximum area =

2.2 =

.

Square of maximum area = 12

RESONANCE


(B)

Dimensions be x, 2x, h 72 = x . 2x . h 36 = x2h ....(1) S = 4x2 + 6xh S = 4x2 + 6

= 8x –

=

For least S, x = 3 and least S is 108. (C)

Let y = x2 + y2 – 4x + 3 = 0 (x – 2)2 + y2 = 1, center C = (2, 0) Consider point P(5, – 4) CP =

=5 is (5 + 1)2 = 36.

Maximum value of

(D)

x2 + y2 = 5 = Let

cos, b =

sin

f() be perimeter f() = 2a + 2b =2

(2cos + sin)

f()= 2

(– 2sin + cos)

f()= 2

(–2cos – sin)

f() = 0  tan = 

and f () < 0

f() is greatest a = 4, b = 1 a3 + b3 = 65

Comprehension # 2 (7 to 9) Let g(x) =

, x [0, ]. g(x) is increasing function of x.



range of g(x) is



f(x) =

, x  [0, ]

Now let  t  2, then f(t) + f(2 – t) = 

RESONANCE


i.e

f(t) +

=

i.e

f(t) +  –

i.e

f(t) =



f(x) =

–

=

for  x  2

Thus f(x) =

for 0  x  2

Also f(x) = f(4 – x) for all x  [2, 4]  f(x) is symmetric about x = 2  from graph of f(x)   = 2 – 0 = 2  = Maximum value is f(2) = = Comprehension # 3_ (10 to 12) 7. Sin x is concave downward in (0, ) 8. ex is concave upward 9. f(x) concave downward (f(x) < 0) f(x) increasing (f(x) 0) Let g(x)= f –1(x) = x f(g(x)) = x f(g(x)). gx = 1 g(x) =

>0

g(x) = –

× f(x).g(x)

g(x) > 0 g(x) = f –1 (x) concave upward

15.

f(x) =

 17.

(1 – nx) f(x)  0, when x  e  f(x) is decreasing function, when x  e >e  f() < f(e) 1/ < e1/e  e > e Statement-1 is True, Statement-2 is False

f ’ (x) = 50x49 – 20x19 = 10x19(5x30 – 2) x = 0 is stationary point. Statement-2 is ture. f(0) = 0 =

–

<0

f(1) = 0  Global maximum is 0. Statement-1 is true.

RESONANCE


19.

y=

is even funciton.

Even function is nonmonotonic. 27.

From figure z2 = x2 + y2 z

=x

If z = 500 then x = 400 500



= 400(5) =4




In case of function given in (A), f is continuous on [0, 1] but not differentiable at x = 1/2  (0, 1). Note that Lf (1/2) = – 1 and Rf (1/2) = 0 Thus, the Lagrange's mean value theorem is not applicable. The function in (B) is continuous on [0, 1] and differentiable on ]0, 1[ and hence the Lagrange's mean value theorem is applicable. The function in (C) is f(x) = x|x| = x . x = x2 and in (D) is f(x) = |x| = x on [0, 1]. As both are polynomial function, the Lagrange's mean value theorem is applicable.

2.

Given that 2(1 – cosx) < x 2 , x  0 To prove sin(tanx)  x,  x  Let f(x) = sin(tanx) – x f(x) = cos(tanx) sec 2 x – 1 f(x) =

2(1 – cosx) < x 2 , x  0

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

cosx > 1 –

Similarly cos(tanx) > 1 –

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

By equation (i) & (ii)

f(x) >

f(x) >

f(x) >

f(x) > 0

x

f(x) is an increasing function.

For x  x0 f(x)  f(0) sin(tanx) – x  0 sin(tanx)  x Hence proved

RESONANCE


3.

f is differentiable function on [0, 4]  f is continuous on [0, 4] also. By using LMVT, there exists a  (0, 4) such that f(a) = Now

........ (i) [(f(4))2 – (f(0))2 ] =

[(f(4))2 – (f(0))2 ] = 4 f(a) [f(4) + f(0)] ....... (ii) f is a continuous function on [0, 4], so By intermediate mean value theorem, there exists b  (0, 4) such that f(b) =

..... (iii)

now by equation (ii) & (iii) [(f(4))2 – (f(0))2 ] = 8f(a) f(b) hence proved 4.

In OSP tan  =

and

=

sin  =



cos  =

Area of  PSM =

=

SM × PN

. 2SN × PN = SN × PN

= SP sin  × SP cos  =

× sin cos 

= (100 – r2)

=



=

.

(100 – r2)1/2 (–2r) r +

=0

(100 – r2)1/2 (– 3r2 + 100 – r2) = 0, r 10 as P is outside the circle 4r2 = 100  r2 = 25  r = 5. Also

= 0,

=0

Thus for r = 5, areawould be maximum. 5.

. Put x = 0 and we get

form. Also because ‘f’ is strictly increasing and differentiable.

Apply L-Hospital rule, weget = – 1. Since ‘f’ is strictly increasing f(x)  0 in an internal

6.

f(x) = For Rolles theorem to be applicable in [0, 1], f(x) should be continuous in [0, 1], differentiable in (0, 1) and f(0) = f(1). Last two conditions hold and

x  n x = 0

Put x = e–t, then as x  0, t  =0

which is true for all  > 0

RESONANCE


7.

=

– 23 x101 – 45

+ 1035x + c.

Now consider a function f(x) =

(x100 – 45) – 23x (x100 – 45) =

(x – 46) (x100 – 45)

It satisfies all conditions of Rolle’s theorem in . Hence there exist at least one c  which f(c) = 0  P(c) = 0 which proves that there exist a root of P(x) = 0 in 8.

f(x) = 3x2 + 2bx + c whose discriminant is 4(b2 – 3c) which is negative as 0 < b2 < c. Thus f(x) is always positive and f(x) is strictly increasing.

9.

f(x) = sinx + 2x – f(x) = cosx + 2 – f(x) = – sinx – Also f(

< 0 x 

f(0) =



>0

)= 2 –

... (i)

... (ii) =2–3–

Equation (1), (2) & (3) shows that value of x 

f(x) is decreasing function

=–1–

<0

... (iii)

a certain

for which

f(x) = 0 & this point must be a point of maxima for f(x) since sign of f(x) changes from + ve to – ve 10.

As |f(x 1) – f(x 2)|  (x 1 – x 2)2 |f(x 1) – f(x 2)|  (x 1 – x 2)2

x 1, x 2  R {As x 2 = |x|2} =

|x 1 – x 2|

Taking limit both sides as x 1  x 2 So

or 11.

12.



|f(x 2)|  0, x1  R  |f(x)| 0 |f(x)| = 0 f(x) = 0 or f(x) is constant function Equation of tangent at (1, 2) is y – 2 = f(1)(x – 1) y = 2 is required equation of tangent.

Let g(x) = f(x) – x2 g(x) has atleast 3 real roots which are x = 1,2,3 By langrange mean value theorem (LMVT) g(x) has atleast 2 real roots in x  (1, 3) f(x) – 2 = 0 for atleast 1 real roots in x  (1, 3)

|x 1 – x 2|

xR

g(x) has atleast 1 real roots in x  (1, 3) f(x) = 2 for atleast one x  (1, 3)

Let P(x) = ax 3 + bx 2 + cx + d P(x) = 3ax 2 + 2bx + c P(x) = 6ax + 2b given that P(–1) = –a + b – c + d = 10, P(1) = a + b + c + d = –6 P(–1) = 3a –2b + c = 0 P(1) = 6a + 2b = 0  a = 1, b = –3, c = –9, d = 5 P(x) = x 3 – 3x 2 – 9x + 5 P(x) = 3x 2 – 6x – 9 = 3(x + 1)(x –3) x = – 1 is point of maxima & x = 3 is point of minima Hence distances between (–1, 10) and (3, –22) is 4

RESONANCE

units. S OLUTIONS (XII) # 117

for

13.

g(x) =

(f(x). f(x))

to get the zero of g(x) we take function h(x) = f(x). f(x) between any two roots of h(x) there lies at least one root of h(x) = 0 i.e. g(x) = 0 Now h(x) = 0 f(x) = 0 or f(x) = 0 As f(x) = 0 has 4 minimum solutions. and f(x) = 0 minimum 3 solutions. h(x) = 0 minimum 7 solutions. h(x) = g(x) = 0 has minimum 6 solutions. 14*.

Since f(x) has local maxima at x = –1 and f(x) has local minima at x = 0. f(x) = x f(x) =

{f(–1) = 0}

+c=0

 = –2c

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

again, Integrating both sides we get f(x) =

+ cx + d

f(2) =

and

f(1) =

+ 2c + d = 18

+c+d=–1

using (i),(ii) and (iii) we get =

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

..............(iii)

 f(x) =

(19x3 – 57x + 34)

(x –1)(x + 1), using number line rule

and

f(x) =

f(x) is increasing for [1, 2

(57x2 –57)

]

f(x) has local maximum at x = –1 and f(x) has local minimum at x = 1

also, f(0) = 15.

Slope of the line joining the points (c – 1, ec–1)

and (c + 1, ec+1) is equal to

tangent to the curve y = ex will intersect line to the left of the line x = c. Comprehension # 1 (16, 17, 18)

16.

RESONANCE


17. Consider y = kex and y = x Let (, ke) be a point on y = kex if it lies on y = x also then  = ke = kex

{ y = x is tangent to y = kex at one point}

= ke =  = 1



 1 = ke i.e. k = 1/e 18. Consider y = kex and y = x from above question ex = slope of y =



if we decrease the value of k from

, then

increases

y = ex and y =

intersect at two distinct points 

k

19.

f(x) = 2 + cos x f(x) = – sin x f(x) = 0  x = n , n   Now, we can easily see that, the interval [t, t + ] for all values of t, contain atleast one integral multiple of   Statement - 1 is true f(t) = 2 + cos t f(t + 2) = 2 + cos (t + 2) = 2 + cos t = f(t)  Statement - 2 is true but we can see that Statement - 2 is not a correct explanation of Statement - 1

20.

g(u) = 2 tan–1 (eu) –

21.

g(u) = tan–1 (eu) – cot–1 (eu) g(u) = tan–1 (eu) – tan–1(e–u)

odd function.

g(u) =

Strictly increasing function

+

>0

f(x) =

one point of local maxima at x = – 1 one point of local minima at x = 0  (C) is the correct option

22*.

f(x) = x cos



f(x) =

,x1 sin

RESONANCE

+ cos S OLUTIONS (XII) # 119



f(x) = –

Now

cos

f(x) = 0 + 1 = 1

x  [1, )

option ‘B’ is correct

  (0, 1]



 f(x) < 0 As f(1) = sin 1 + cos 1 > 1

option ‘D’ is correct



f(x) is strictly decreasing and

f(x) = 1

so graph of f(x) is as below Now in [x, x + 2], x  [1, ), f(x) is continuous and differentiable so by LMVT, f(x) = as

f(x) > 1 for all x  [1, ) >1



f(x + 2) – f(x) > 2



for all x  [1, ) 23.

p(x) = ax4 + bx3 + cx2 + dx + e p (x) = 4ax3 + 3bx2 + 2cx + d p (1) = 4a + 3b + 2c + d = 0 p (2) = 32 a + 12 b + 4c + d = 0

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

=2

=2 c + 1 = 2, d = 0, e = 0 c=1 Now equation (i) and (ii) are 4a + 3b = – 2 and 32 a + 12 b = – 4 

a=

and b = – 1

24.

f(x) = 2010 (x – 2009) (x – 2010)2 (x – 2011)3 (x – 2012)4 f(x) = n (g(x))  g(x) = ef(x)  g(x) = ef(x) . f(x) only point of maxima [Applying first derivative test]

25.

Clearly f(x) = f(x) = 2x ( g(x) = x 

–

)  0 increasing

+

 g'(x) =

fmax = f(1) = e +

 + 2x2

– 2x

> 0 increasing

gmax = g(1) = e +

h(x) = x2 

+

+

hmax = h(1) = e +

RESONANCE

h(x) = 2x ,

+ 2x3 so

– 2x

= 2x

>0

a=b=c S OLUTIONS (XII) # 120

26.

(A)

Re

= Re

= –1  y  1 =

1 or

= Re

= Re (–1/y) =

– 1

Alternate

Re

= Re

= Re

= Re

= Re

= Re as –1  sin  1 (– , 0 )  (0, ) (B)

(C) (D)

–1 

1



–1 



0

1





–1 



0

0 

0

1



– 1 0



 t  (– , –9]  [–1 , 1]  [9, )  x  (– , 0)  [2 , ) f() = 2 sec2  f()  2  f(x) = x3/2 (3x – 10) 

f ’(x) = x3/2 3 +

0

f()  [2, )

x1/2 (3x –10)

asf ’(x)  0 0



 27.

– 15  0



x 2



3x +

– 15  0



x  [2, )

f(x) = x4 – 4x3 + 12x2 + x – 1 f(x) = 4x3 – 12x2 + 24x + 1 f(x) = 12x2 – 24x + 24 = 12 (x2 – 2x + 2) > 0 xR  f(x) is S.I. function Let is a real root of the eqution f(x) = 0  f(x) is MD for x (– , ) and M.I. for x (, ) where < 0  f(0) = – 1 and < 0  f() is also negative  f(x) = 0 has two real & distinct roots.

RESONANCE


28.

p = (x – 1) (x – 3) = (x2 – 4x + 3) p(x) = (x3/3 –2x2 + 3x) +  p(1) = 6 6 = (1/3 – 2 + 3) +  6 = (1/3 + 1) +  18 = 4 + 3 ...(i) p(3) = 2 2 = (27/3 – 2 × 9 + 9) +  2= =2=3 p(x) = 3(x – 1) (x – 3) p(0) = 3(–1)(–3) =9

29.

f(x) = |x| + |x2 – 1|

f(x) =

f(x) =

PART - II 1.

Let

f(x) = x +

f(x) = Minimum occures at x = 1. 2.

3.

f(x) = 6x2 – 18ax + 12a2 f(x) = 0  x = a, 2a f(x) = 12 x – 18a f(a) = – 6a < 0 and f (2a) = 6a > 0  p = a, q = 2a  p2 = q a = 2 (a > 0) Let

f (x) = ax2 + bx + c



f(x) =



a2 – 2a = 0



a = 0, 2

f(0) = d = f(1) (2a + 3b + 6c = 0) By Rolle's theorem, at least one root of f(x) = 0 lies in (0, 1) 4.

y2 = 18x Differentiating w.r.t. t

...... (1)



From equation (1), x =

RESONANCE

2y.2 = 18



y=

Required point is S OLUTIONS (XII) # 122

5.

x = a (1 + cos ), y = a sin 



Equation of normal at point (a(1+ cos ), a sin ) is y – a sin =

(x – a (1 + cos ))

y cos = (x – a)sin It is clear that normal passes through fixed point (a, 0) 6.

y = x2 – 5x + 6 = 2x – 5

= – 1,

product of slopes = –1

=1

angle between



tangents is 7.

Let f1(x) = x3 + 6x2 + 6 Increasing in (–, – 4] Let

f2 (x) = 3x2 – 2x + 1



f1(x) = 3x (x + 4)



f2(x) = 6x – 2

Not increasing in Let

f3(x) = 2x3 – 3x2 – 12x + 6



f3(x) = 6x2 – 6x – 12 = 6(x +1) (x – 1)

Increasing in [2, ) Let f4(x) = x3 – 3x2 + 3x + 3 Increasing in (–, ) 8.

V=

 4 (10 + r)2



, 0  r  15 = – 50. = – 50 

=

9.

Any point on ellipse is P(x, y) = (a cos , b sin ) Area of rectangle = 4ab cos sin  = 2ab sin 2   Maximum area = 2ab.

10.

By LMVT and f(x) 2

f(x) = 3(x –1)2 0

= f(x),

x (1, 6)



11*.

(where r = 5)

2



2

f(6) 8.



 Equation of normal is y – a(sin – cos ) =

(x – a (cos  + sin ))

 x cos + y sin = a. Distance from origin = |a| (constant) Slope of normal = –

RESONANCE

= tan



angle made with x-axis is

+ . S OLUTIONS (XII) # 123

12.

f (x) = f (x) changes sign as x crosses 2. f(x) has minima at x = 2.

13.

= f(c)  1 < c < 3 

c =

14.

f(x) =

15.

Graph of y = x3 – px + q cuts x-axis at three distinct pints

(cos x – sin x)

=0 16.

= 2 log3e



x = 

Maxima at x = –

, minima at x =

Condition for shortest distance is slope of tangent to x = y2 must be same as slope of line y = x +1. 

=1



y=

,

x=

,x–y+1=0

Shortest distance =

17.

P (x) = 4x 3 + 3ax 2 + 2bx + c and P(0) = 0  c=0  P (x) = x(4x 2 + 3ax + 2b) D = 9a2 – 32b < 0



b>

>0

(P(x) = 0 has only one root x = 0)

P (– 1) < P (1)  a>0 P(x) has only one change of sign.  x = 0 is a point of minima. P(–1) = 1 – a + b + d, P(0) = d P(1) = 1 + a + b + d  P(–1) < P(1), P (0) < P(1), P (–1) > P(0)  P(–1) is not minimum but P(1) is maximum.

18.

y=x+ y = 1 –

=0

y=2+

=3



x3 = 8



x=2

(2, 3) is point of contact Thus y = 3 is tangent Hence correct option is (3)

RESONANCE


19.

f(x) = 1 f(–1) = k + 2 f(x) = k + 2 f has a local minimum at x = – 1 1k+2k+2 possible value of k is – 1 Hence correct option is (3)

 



20.

ex + 2e–x  2

f(–1+)  f(–1)  f(–1–) k–1

(AM  GM)



 f(x) > 0

so statement- 2 is correct

As f(x) is continuous and

f(c) =

 21.

belongs to range

for some C.

of f(x),

Hence correction option is (4).

y–x=1 y2 = x 2y

=1

=

  y =

= 1 , x=

tangent at



y=

 y=x+

y–x=

distance =

22.

=

=

f(x) =

In right neighbourhood of ‘0’ tan x > x

RESONANCE


In left neighbourhood of ‘0’ tan x < x as(x < 0) at x = 0 , f(x) = 1  x = 0 is point of minima so statement 1 is true. statement 2 obvious

23.

r3

V=

4500  =

= 4r2

45 × 25 × 3 = r3

r = 15 m after 49 min = (4500 – 49.72) =972 m3 972 =

r3

r3 = 3×243 = 3× 35 r=9 72  = 4 × 9 × 9

=

24.

f '(x) = at x = – 1

+ 2bx + a –1 – 2b + a = 0 a – 2b = 1

at x = 2

+ 4b + a = 0

a + 4b =

On solving (i) and (ii)

f '(x) =

...(i)

=

a=

...(ii)

, b=

=

So maxima at x = – 1, 2

RESONANCE


SOLUTION OF TRIANGLE EXERCISE # 1 PART - I Section (A) : A-1.

(i)

L.H.S. = a sin (B – C) + b sin (C – A) + c sin (A – B) = k sin A sin (B – C) + k sin B sin (C – A) + k sin C sin (A – B) = k (sin2 B – sin2 C) + k (sin2C – sin2 A) + k (sin2 A – sin2 B) = 0 = R.H.S.

(ii)

L.H.S. =

first term =

= = k 2 sin (B + C) sin (B – C) = k 2 (sin2 B – sin2 C) = k 2 (sin2 C – sin2 A)

Similarly

= k 2 (sin2 A – sin2 B)

and

L.H.S. = k 2 (sin2 B – sin2C + sin2C – sin2A + sin2 A – sin2 B) = 0 = R.H.S. (iii) L.H.S. = 2bc cos A + 2ca cos B + 2ab cos C = b2 + c 2 – a2 + a2 + c 2 – b2 + a2 + b2 – c 2 = a2 + b2 + c 2 = R.H.S 

(iv) L.H.S. = a2

– 2ab

= a2 + b2 – 2ab cos C = a2 + b2 – (a2 + b2 – c 2) = c 2 = R.H.S. (v)  L.H.S. = b2 sin 2C + c2 sin 2B = 2b2 sin C cos C + 2c 2 sin B cos B = 2k 2 sin2 B cos C sin C + 2k 2 sin2 C sin B cos B = 2k 2 sin B sin C [sin B cos C + cos B sin C] = 2(k sin B) (k sin C) sin (B + C) = 2bc sin A (vi)  R.H.S =



(b = ksin B, c = ksin C)

c = a cos B + b cos A, b = c cos A + a cos C

=

=

=

A–4.

    

= L.H.S.

= sin(B + C) sin(B – C) = sin(A + B) sin(A – B) sin2 B – sin2 C = sin2 A – sin2 B 2 sin2 B = sin2 A + sin2 C 2b2 = a2 + c 2  a2, b2, c 2 are in A.P.

RESONANCE


A–7.



x 3 – Px 2 + Qx – R = 0



a2 + b2 + c 2 = P a 2b 2 + b 2c 2 + c 2a 2 = Q a2b2c 2 = R  abc =

+



+

=

[a2 + b2 + c 2] =


(i)

L.H.S. = 2a sin2 = = = =

(ii)

+ 2 c sin2

a(1 – cos c) + c(1 – cos A) a + c – (a cos C + c cos A) a+c–b R.H.S.

 L.H.S. =

+

=

.

+ +

= (iii)

.

+

=

.

.

L.H.S. = 2bc(1 + cos A) + 2ca(1 + cos B) + 2ab(1 + cos C) = 2bc + 2ca + 2ab + 2bc cos A + 2ca cos B + 2 ab cos C + a2 + b2 + c 2 = (a + b + c)2

=2 = R.H.S. (iv)

 L.H.S. = (b – c)

 (b – c) cot

+ (c – a)

+ (a – b)

= k(sin B – sin C)

= 2k cos

sin

= 2k sin

sin

= k [cos C – cos B] similarly (c – a) cot and (a – b) cot

= k[cos A – cos C]

= k[cos B – cos A]

 L.H.S. = k[cos C – cos B + cos A – cos C + cos B – cos A] =0 = R.H.S.

RESONANCE


(v) L.H.S. = 4 (cot A + cot B + cot C) = 4 = 2bc cos A + 2 ca cos B + 2ab cos C = a2 + b2 + c 2 = R.H.S. (vi) L.H.S. =

cos

. cos

. cos

= =

=  = R.H.S.

B–3.

Let ADB =   we have to prove that tan = if we aply m – n rule, then (1 + 1) cot= 1.cot C – 1.cotA. =

–

=

–

= =

[2(a2 – c 2)]

2cot =





tan =


(i)

r. r1 .r2 .r3 =

(ii)

r1 + r2 – r3 + r = 4R cosC

= 2

L.H.S. = =

=

=

RESONANCE


=

=

=c

=



L.H.S. =

(iii)



=



   

= R.H.S.

=

(s – a + s – b + s – c) =

·

=

=



=

=

=

=

=

 = 24 sq. cm 2s = 24  r1, r2, r3 are in H.P.

=

=

=r

=r

.... (i) s = 12 .... (ii)

are in A.P..





[4s 2 – 2s(a + b + c) +a2] =

R.H.S. =

similarly we can show that 

= 4RcosC

L.H.S. =

=

C–4.

=

(s + s – a + s – b + s – c)2 = 4



(v)

=

L.H.S. =



=

cos C =



[s 2 + (s – a)2 + (s – b)2 + (s – c)2] =

(iv)

=

=





a, b, c are in A.P. 2s = 24 a + b + c = 24 3b = 24 b=8

are in A.P.. 

2b = a + c



a + c = 16

But  =   = 24 × 24 = 12 × (12 – a) × 4 × (12 – c)  2 × 6 = 144 – 12 (a + c) + ac 12 = 144 – 192 + ac  ac = 60 and a + c = 16  a= 10, c = 6 or a = 6, c = 10 and b = 8

RESONANCE



(i) 

,=

=

, =

=



R.H.S. =

=

=

L.H.S. = R.H.S.



(ii)

=

=

=

R.H.S. =

=

= L.H.S.= R.H.S.


(b + c)2 – a2 = kbc

 (a + b + c) (b + c – a) = kbc



b2 + c2 – a2 = (k – 2) bc



 In a ABC –1 < cos A < 1



–1 <



b

=

= cos A

<1

0 < k < 4.


b cos 2





+ a cos 2

[ s – a + s – b] =

=

c.

c

=



+a



×c=



a + b = 2c

=

c.

c

 a, c, b are in A.P. B–5.

B–6*.

 = (a + b – c) (a – b + c)  = 4(s – c) (s – b)





tan

=



(A)



tan

 tan2

=

RESONANCE

=

cot

=

=

tan A =

 tan A =

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

= S OLUTIONS (XII) # 131

 tan

=

 a = 5 and b = 4

 from equation (i), we get =

cot

=



 cos C =

=

=

 Area =

=

=

ab sinC  cosC =

Area =

Area =

cot



 c 2 = a2 + b2 – 2ab cos C

 cos C =

(B), (C)

cot

 sinC =



c=6

=

×5×4×

sq. unit.  From Sine rule

=

=

 sinA =

=

 sinA =


=

=

=4 = C–5*.

(A)

. +



=

+

+

+

=

(B)

(C)

+



=

+

=

=



+

+

=

cot A = cot B = cot C



A=B=C

true for equilateral triangle only

RESONANCE


(D)

=

=



=

=



true for equilateral triangle only

cot A = cot B = cot C



 A = B = C

Section (D) :

D–1.

D–4*.

=2

 a =

cos

(A) correct

(C)

(B) incorrect

=

(D) 

cosec

=

=

=

.

=

cos

.

=

cos


If we apply Sine-Rule in  ABD , we get =

sin  =



AB =

and

cos  =



AB =

=

...(i)

 from equation (i), we get

AB =

7.

RESONANCE


required distance = inradius of  ABC  2s

=a+b+b+c+c+a = 2 (a + b + c) s=a+b+c

 = =  required distance =

=

=

= 8.

(i)

L.H.S. = (r3 + r1) (r3 + r2) sin C =

sin C

=

sin C

=

sin C

=

=

= 2 sr3

R.H.S. = 2r3



= 2r3 = 2sr3 L.H.S. = R.H.S.

(ii)

L.H.S. = –

=– (iii)

=

= R.H.S.

First term = (r + r1) tan =

=

cot

.

.

=b–c similarly second term = c – a & third term = a – b  L.H.S. = b – c + c – a + a – b = 0 = R.H.S.

RESONANCE


(iv)   and  

11.

(i)

 r1 + r2 + r3 – r = 4R (r1 + r2 + r3 – r)2 = r12 + r22 + r32 + r2 – 2r (r1 + r2 + r3) + 2(r1r2 + r2r3 + r3 r1) r(r1 + r2 + r3) = ab + bc + ca – s2 r1r2 + r2r3 + r3r1 = s 2 from equation (i) 16R2 = r2 + r12 + r22 + r32 – 2 (ab + bc + ca – s2) + 2s2 r2 + r12 + r22 + r32 = 16 R2 – 4 s2 + 2 (ab + bc + ca) = 16R2 – (a + b + c)2 + 2 (ab + bc + ca) = 16R2 – a2 – b2 – c2

........(i)

EFA is a cyclic quadrilateral =A



A = r cosec A/2 EF = r cosec A/2.sin A = 2 r cos A/2 similarly DF = 2 r cos B/2 and DE = 2r cos C/2. (ii) ECD is a cyclic quadrilateral  



CE = DE =

similarly DF = BF =

(iii)

area of DEF =

FDE =



=

=

–

FD . DE sin FDE

=

= 2r2

= 2r2

=

=

=

=

=

=

=

PART - II 3.

 ED =

– c cos B

=

–c

=

–

=

=

RESONANCE


5.

f = RcosA , g = R cos B, h = R cosC. +

+

=

+

+

=2 =8

+ = .8

 9.

+

=

=



MNA is a cyclic quadrilatral = A



 MN = r cosec

sin A = 2r cos

M = N = r

x=

=

similarly y =

=

r2 R

=

 r1 + r2 = 

(r1 + r2) =

 

=

=

=

= 4Rs 2

=4

 14.

=

and z =

 xyz =

12.

,

A, C1 , G and B1 are cyclic BC1 . BA = BG . BB1 .c=

= 

16.

18.

(2c 2 + 2a2 – b2)

c 2 + b2 = 2a2

a=1

 2s = 6

2s = 2

R=1

 sin C =

RESONANCE

= 2R  sin A =

A=

 1  cos (A – B)  1

 cos (A – B) = 1  sin C =



A–B=0 =1

A=B

 C = 90º S OLUTIONS (XII) # 136

20.

if we apply m-n Rule in  ABE, we get (1+1) cot = 1.cot B – 1.cot  2 cot = cot B – cot 3 cot = cot B tan  = 3 tan B

..........(1)

Similarly, if we apply m-n Rule in  ACD, we get (1+1) cot (–) = 1.cot – 1.cotC. cotC = 3 cot

 tan  = 3 tanC .......(2)

form (1) and (2) we can say that tan B = tan C



B=C

A+B+C=  A =  – (B + C) =  – 2B

B=C

 tan A = – tan2B

=–

=–

 tan A =

22.

r1 – r =

–

 (r1 – r)

=

= a tan

= abc tan

tan

tan

= abc  tan

= abc

RESONANCE

=

=

=

=

= 4Rr2


EXERCISE # 3 2.

Match the column (A) AA1 and BB1 are perpendicular  a2 + b2 = 5c2 

c2 =

=5

cos C =

=

2 = 11

(B)

 G.M.  H.M.

(r1 r2 r3)1/3 



=

=

=

a = 5, b = 4

=

cos B =

=



2a2 + 4b2 + c2 = 4ab + 2ac. 

(D)

(

ab sin C =

 (r1 r2 r3)1/3  3r

tan2

(C)

c =

=

sin C = 



 27



2s = 9 + c

 c2 = 36

(a – 2b)2 + (a – c)2 = 0

 c=6 a = 2b = c

=

 8 cos B = 7 COMPREHENSION # 2 (Q. No. 7 to 10) 7. Clearly 8.

Let  3 1 2 =  Then angle of pedal trinagle =  – 2 = A =

9.

Side of pedal triangle

= I2I3cos = BC

I 2I 3 =

I2I3 = 4Rcos

10.

1 = 4 R sin I2I3 = 4 R cos

RESONANCE



12 + 232 = 16R2


12.

1 2 = 4R cos

if we apply Sine-Rule in  1 2 3 , then

2 Rex =

=

=

14.

2Rex 

= 4R Rex = 2R  ABC is pedal triangle of  I1 I2 I3



statement - 1 and statement - 2 both are correct and statement -2 also explains Statement - 1

 sin

=

=

similarly sin

 3 sin

 19.

–

=

– 4 sin3

=

=

ax 2 + bx + c = 0

 r2 =

 r = a.

cm.

...(1)

x2

+x +1=0 ...(2)  roots of (2) are imaginary and a, b, c are real 

b a c = = =k 2 1 1

 cos C =

2  1 1 b2  a2  c 2 = = 2  1 2 2ab

1 2

C=

 4


We have a2  a2 – (b – c)2 = (a + b – c) (a – b + c)  a2  (2s – 2c) (2s – 2b) = 4(s – b) (s – c) similarly b2  4 (s – c) (s – a) and c 2  4 (s – a) (s – b). Multiplying the above inequalities, we get a2b2c 2  64 (s – a)2 (s – b)2 (s – c)2  (a + b + c) abc  16 s (s – a) (s – b) (s – c) = 162

1 (a  b  c ) abc 4 Equality occurs if and only if (b – c)2 = 0 (c – a)2 = 0 and (a – b)2 = 0 i.e if and only if a = b = c. 2.





(A)

a, sin A, sin B are given one can determine

a a sin B c = sin C So the three sides are unique. So option (a) is incorrect option sin A sin A S OLUTIONS (XII) # 139 RESONANCE

b=

(B) (C)

The three sides can uniquely determine a triangle. So option (b) is incorrect option. a , sin B, R are given one can determine b = 2R sin B, sin A =

(D)

. So sin C can be determined. Hence side c can also be uniquely determined

for a, sin A, R = 2R

But this could not determine the exact values of b and c 3.

n = 2n × area of OA11 

n = 2n ×

×A A11 × O1



n = n × sin

× cos



n =

.

sin

.........(1)

On = 2n × area of  OB1O1 

On = 2n ×

× B1O1 × O 1O = n × tan



On = n tan

......(2)

Now

R.H.S. =

= 4.

× 2 cos 2

× 1 = n tan

=

= On. cos 2

Let angle of the triangle be 4x, x and x . Then 4x + x + x = 180° Longest side is opposite to the largest angle. Using the law of sines

= n tan

.cos 2



x = 30°



2S =

=

sin

= n

= L.H.S

= 2R



a = R, b = R, c =

5.

Clearly the triangle is right angled. Hence angles are 30º, 60º and 90º are in ratio 1 : 2 : 3

6.

Consider



=

=

=

RESONANCE

=

=


7.

Clearly P is the incentre of triangle ABC. r=

8.

=

Here

2s = 7 + 8 + 9 

Here

r=

=

=

. b . b . sin 120º =

Also and  =

and s =

=

s = 12

b2

.........(1)

a=

.........(2)

(a + 2b)

(a + 2b)

..........(3)

From (1), (2) and (3), we get  = 9.*

We have ABC = ABD + ACD



bc sin A =

c AD sin

+

b × AD sin

AD =



Again AE = AD sec =

AE is HM of b and c.

EF = ED + DF = 2DE = 2 × AD tan =

=

× cos

sin

As and DE = DF and AD is bisector Hence A, B, C and D are correct answers. 10.

× tan an

AEF is isosceles.

In  ABC , by sine rule =

=

 C = 45º, C = 135º

When C = 45º  A = 180º – (45º + 30º) = 105º When C = 135º  A = 180º – (135º + 30º) = 15º Area of  ABC =

AB . AC.sin BAC =

Area of  ABC =

AB . AC .sinA =

×4×

Absolute difference of areas of triangles = | 2

RESONANCE

sin (15º) =

×4×

×

=2

sin (105º) = 2 –2

|=4


Aliter

AD = 2 , DC = 2 Difference of Areas of triangle ABC and ABC = Area of triangle ACC = 12.

cos B + cos C = 4 sin2



2 sin



cos



– cos



tan

= 4 sin2

cos

=0

=0

– 2 cos cos tan

+ 3 sin

sin

as sin

0

=0

=

=

=



2s = 3a





b + c = 2a

Locus of A is an ellipse



12.

×2×4=4

2 cos





11.

AD × CC =

sin 2C +

cos

sin 2A =

(a cos C + c cos A) =

= 2 sin B = 2 sin 60º =

=

=

=

=

 

(



– 2) x2 + (

– 2) x + (

(x2 + x + 1) = 2x2 + 2x – 1

+ 1) = 0

on solving x2 + x – x=

= 0 we get + 1, –

RESONANCE



At x = –

, Side c becomes negative.



x=


13.

Area of triangle =

. 6 . 10 sin C = 15



14.

ab sin C = 15



C=

Now

cos C =



–



r=



sin C =



c = 14



r2 = 3

(C is obtuse angle )

=

=

=

a = 2 = QR b=

= PR

c=

= PQ

s=

=

=4

=

=

=

=

= tan2

=

=

=

PART - II 1.

Let a = 3x + 4y, b = 4x + 3y and c = 5x + 5y as x, y > 0, c = 5x + 5y is the largest side   C is the largest angle. Now cos C =

=

<0

 C is obtuse angle  ABC is obtuse angled. 2.

r1 > r2 > r3



>

>

 s – a < s – b < s – c  –a < –b < –c;

3.

tan

=

; sin

r+R=

RESONANCE

a>b>c

=



r+R=

.cot


4.

a

=

=

  a + c = 2b 5.



a + b + c = 3b.



a, b, c are in A.P.

AD = 4 

AG =



Area of ABG =



=



Area of  ABC = 3(Area of  ABG)

6.

cos =

7.

C = /2

×4=

×

×

× AB × AG sin 30º

×

=



Sin 60º =



AB =

=



 = 120º





=

=–

r = (s – c) tan



C = 90º

r = s – 2R  2r + 2R = 2 (s – 2R) + 2R. = 2s – 2R = (a + b + c) –



C = 90º

=a+b+c–c =a+b

8.

are in H.P.

are in A.P.

9.



a,b,c are in A.P.

= cos

Let cos

=

As





for some n  3, n  N



cos

 cos

 cos



 3  n < 4, which is not possible so option (2) is the false statement so it will be the right choice Hence correct option is (2)

RESONANCE


ADVANCE LEVEL PROBLEMS PART - I 1.

From figure, AD = c sin B Hence number of triangle is 0 if b < c sin B one triangle for b = c sin B two triangles for b > c sin B

2.

C = 60° Hence c2 = a2 + b2 – ab

3.

=



= 2 cos

Using properties of pedal triangle, we have

 MLN = 180° – 2A  LMN = 180° – 2B  MNL = 180° – 2C

Hence the required sum =

sin2A + sin2B + sin2C

=

4.

5.

4sinA sinB sinC

From figure, we can observe that

OGD is directly similar to PGA

BD = s – b, CE = s – c and AF = s – a Hence BD + CE + AF = s

6.



, as cos



= cos

A = B, in either case

 7.

, Using cosine rule in

ABO, we get

h=

8.

In

ABD,



RESONANCE


Comprehension # 1

9.

10.

+

+

= b sin B + c sin C + a sin A =



k = 2R



cot A + cot B + cot C =

=

(b2 + c2 + a2) =

(b2 + c2 – a2 + c2 + a2 – b2 + a2 + b2 – c2)

=

=

.

=

11.

=

k=



=6

Comprehension # 2 (12 to 14) 12.

 PG =

AD

= =

=  PG = =

.ab sin C

or

b sin C

(  =

ac sin B)

ac sin B c sin B

13.

 Area of GPL =

and

Area of ALD =

(PL) (PG) (DL) (AD)

=



 PL =

=

DL and PG =

AD

=

14.

 Area of PQR = Area of PGQ + Area of QGR + Area of RGP ...(1)



Area of PGQ = = =

RESONANCE

×

×

PG.GQ.sin(PGQ) AD ×

×

BE sin ( – C) sin C S OLUTIONS (XII) # 146

=

×

=

bc sin A ×

ac sin B × sin C

sin A.sin B.sin C

Similarly Area of QGR =

sin A.sin B .sin C and Area of RGP =

sin A.sin B.sin C

 From equation (1), we get (a2 + b2 + c 2) sin A.sin B.sin C

Area of PQR = 15.

In

CDB ,

=

 Also from same triangle 16.

17.

 BD =

cosAcosB + sinAsinBsinC = 1



(cosA – cosB)2 + (sinA – sinB)2 + 2sinAsinB(1 – sinC) = 0



a:b:c=1:1:



A = B & C = 90°

We have

 18.

=

a:b:c=5:4:3

from figure, OO = ON – ON = R –

ZO = ZM + = from

RcosA +

OZO, using Pythagorous theorem,

we get (R –

)2 = (RcosA +

)2 +



=

PART - II 1.

from

 from

ABC ,

=

AB = 2Rsin(A + )

ACB,

=



AC’ = 2Rsin( – A)

=

4RcossinA = 2acos

similarly CA = 2bcos =

RESONANCE



 area

BC = 2R(sin (A + ) – sin( – A))

ABC = =

4cos2. S OLUTIONS (XII) # 147

2.

c2 – 2bc cosA + (b2 – a2) = 0 c1 & c2 are roots of this quadratic equation Hence (c1 – c2)2 + (c1 + c2)2tan2A = 4a2

3.

Area =

=

= =

2Rs

= 4.

We know that OA = R, HA = 2RcosA and applying Appoloneous theorem to

AOH, we get

2.(AQ)2 + 2(OQ)2 = OA2 + (HA)2



2.(AQ)2 = R2 + 4R2cos2A –



5.

=



+

using sine rule, diameter of required circle =

 6.

=

= 20

radius = 10

L.H.S. =

(a2 (b + c – a) + b2 (c + a – b) + c2 (a + b – c))

=

=

=

abc

=

4R

RESONANCE


7.

from the parellelogram ABAC, AA = 21 , from



AAC, AA < b + c 21 < b + c

similarly 22 < c + a

...(1) ...(2)

and 23 < a + b ...(3) (1) + (2) + (3) gives 1 + 2 + 3 < 2s 8.

ZXY =

and

 Area of

=

 Area of

 Area of XYZ = 2R2 cos

9.

cos

cos

=



Drop a perpendicular from the apex P to the base

ABC.

The foot of perpendicular is at circum centre O of

ABC

Using given data, we get from, right angle

POB, we get

= = 8.83 m 10.

from cyclic quadrileteral CQFP, we get

from cyclic quadriletral AQMF, we get  FQM =  FAM = 90º – B



 AQM = 90º + 90º – B = 180º – B

 

P, Q, M are collinear

similarly P, Q, N are collinear hence, P, Q, M, N are collinear

RESONANCE


2nd-Dispatch DLPD IIT-JEE Class-XII English PC(Maths)

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