CXC CSEC ADD MATHS 2013 P2 SOLUTIONS

1.(a ) f ( x) = x 3 − x 2 − 14 x + 24 Factor (x + 4)

(i ) f ( x ) =

(i) ( x + 4) x 3 − x 2 − 14 x + 24 -1

-14

+24

↓ -4

20

-24

6

0

1

-5

 x + 2

 fg ( x ) =

Through Synthetic Division -4 1

2 x − 1

 fg ( x ) =  fg ( x ) =

Since the Remainder is Zero we can

( x + 1) + 2 2 x + 2 − 1  x + 1 + 2 2 x + 1  x + 3

(ii)The other other linear linear factor factorss can be found from from the the quadratic expression.

5 3 x − 2 = 7 x + 2

( x + 4)( x 2 − 5 x + 6)

( c ) G i v e n 5 3 x

 

−2

(3 ( 3 x − 2 ) lo g 5 = ( x + 2 ) lo g 7 3 x lo g 5 − 2 lo g 5 = x lo g 7 + 2 lo g 7

Factorize : ( x2 − 5 x + 6)

 

( x 2 − 2 x)( −3 x + 6)

3 x lo g 5 − x lo g 7 = 2 lo g 7 + 2 lo g 5

 

 x( x − 2) − 3( x − 2)

   x (3log 5 − lo g 7) 7) = 2 ( lo g 7 + lo g 5 )

 

( x − 2)( x − 3)

Therefore there are 3 factors. ( x + 4)( x − 2)( x − 3)

  (b )G iv e n

2  x − 1

f ( x ) =

(i) In v e r s e

f

− 1

 x + 2

( x )

 f ( x ) =  y

=

 x =

2  x − 1  x + 2 2  x − 1  x + 2 2  y − 1  y + 2

 x ( y + 2 ) =

2 y − 1

 x y + 2 x =

2 y − 1

 x y − 2 y =

 

2( x + 1) − 1

= 7 x + 2 S h o w th a t   2 ( lo g 5 + lo g 7 )  x = ( lo g 12 12 5 − l o g 7 )

concl nclude ude that ( x + 4) is a factor of f (x ).

 

; g (x) = x + 1

− 1 − 2 x  y ( x − 2 ) = − 1 − 2 x − 1 − 2  x  y =  x − 2 − 1 − 2  x 1 =  f  −  x − 2

 x =  x =

2 ( lo g 7 + lo g 5) 5) ( 3 lo lo g 5 − lo g 7 ) 2 ( lo g 5 + lo g 7 ) ( lo g 12 12 5 − lo g 7 )

3.( a ) ( x − h) 2 + ( y − k ) 2 = r 2  

2.(a) f (x) =3x +6x −1 2

(i) Express f (x) in the formof a ( x+ h)2 + k  

( x − 2) 2 + ( y − 1) 2 = r 2

 

Find the value of the radius (r),

3 x2 +6x −1

 Note that tha t the radius radiu s r is the distance di stance AB.

 

3( 3( x2 +2) −1

 

r=

( x2 − x1 ) 2 + ( y2 − y1 ) 2

 

3( 3( x2 +2x +12) −1−3

 

r  =

(1 (10 − 2) 2 + (7 − 1) 2

 

3( 3( x +1)2 −4

 

r  =

64 64 + 36

(ii) Therefore Therefore the minimum inimumis k  = - 4.

r  = 100 r  = 10

(iii) The The minimumvalue value of x = -1. -1.

 

2 (b) Find the values of x for which:

  ( x − 2)( x − 2) + ( y − 1)( y − 1) = 100

2 x2 +3x −5≥ 0

 

(2 (2 x +5x)(−2x −5) ≥0 2

   

 x ≥

2

(x −1) ≥0 Zero Product Theorem

2

 x 2 − 4 x + y 2 − 2 y = 100 − 4 − 1  x 2 − 4 x + y 2 − 2 y = 95 95

   

 x 2 + y 2 − 4 x − 2 y − 95 = 0  x 2 + y 2 + hx + gy + k  = 0

(2 (2 x +5)(x −1) ≥0

−5

2

 

 x(2 (2x +5) −1(2x +5) ≥0

(2 x +5) ≥ 0

 x − 4 x + 4 + y − 2 y + 1 = 100

 

2 x2 +5x −2x −5 ≥ 0 Factorize AC Method    

( x − 2) 2 + ( y − 1) 2 = 10 2

(ii )Equa )Equati tion on of of line line AB. AB. Use points given, A and B.

x ≥1

7 −1

m=

10 − 2 3 m = gradient of line 4 3  y = x + c 4 3   1 = (2) + c 4 3 1= +c 2 1 c=− 2 3 1  AB → y = x − 4 2 Equation of line l is perpendicular therefore;  y = −

 

7 =− c=

3 4 4

3 61

x+c

(10) + c

3 3

61

4

3

∴  y = − x +

5.( a) y = x − 3x + 2 3

2

∫

6 .( a ) 5 x 2 + 4 d x =

(i ) S tati tation onar ary y Poi Point ntss of of y. y.  y = x − 3x + 2 3

2

=

 y ' = 3x − 6 x = 0 2

∴ Factorizing   3 x( x − 2) = 0 3 x = 0; ( x − 2) = 0  x = 0; x=2 to find the y coordinates of these stationary points.  y = (0) − 3(0) + 2 3

2

 y = 2

∴ (0,2)  y = ( 2) − 3( 2) + 2 3

2

 y = ( 2, −2)

To determine the nature of the points. we find the second derivative and test: y '' = 6 x − 6  

= 6( 6( 0 ) − 6 =-6

 since -6 -6 ≤ 0 we we can can conc conclu lude de its its a maxi maximu mum m. y '' = 6 x − 6  

= 6( 6(2) − 6 = 6;

sinc sincee 6 ≥ 0 we can can con concl clud udee its its a mini minimu mum. m.

5 x 2 +1 2 +1

5 x 3 3

+

4 x 0 +1 0 +1

+C

+ 4 x + C 

π 

2

∫

(b ) 3 s i n x − 5 c o s x d x 0 π

π 

2

2

0

0

= 3 ∫ si  xd x − 5 ∫ co s xdx s in  xdx = 3(co s  x ) − 5( − sin x) = 3 cos  x + 5 sin x π π  ⎞ ⎛ = ⎜ 3 co s + 5 si n ⎟ − ( 3 co s 0 + 5 si n 0 ) 2 2⎠ ⎝ = ( 3(0) + 5(1) ) − ( 3(1) + 5(0) )

= (5) − (3) =2

 

3. (b)

(c)

Given that they are prependicular  then we know that the dot product is Zero. Dot Product

×

a · b= a

×

b

× cos ( 90°)

a · b = a

×

b

×0

b

× cos(θ )

0

∴a · b = ax × bx + ay × by  

0 = 10× λ + −8×10

 

0 = 10λ − 80

  −10 10λ  = −80 λ  = 8

 

OA = −2i + 5 j 

OB = 3i − 7 j  

a · b= a

a · b =

Given vectors



 

 AB = AO + OB  note the change in direction 

of vector OA .

= 2i − 5 j + 3i − 7 j = 5i − 12 j  Now we find the magnitude of the resultant resultant vector. 2 2  AB = 5 + (12)

 AB = 25 + 144  AB = 169  AB = 13 square square root of 169. 169.

∴ the unit vector in the direction of AB. 1 13

( 5i − 12 j )

4(a ) Ar ea of sec to = or  =

1

θ 

360

× π r 

2

r  θ 

2 we nee  to find the radius "r".

we are given the perimeter, tha is, the arc lenght plus the rad us twice. Recall hat: Arc lenght = r θ , w ere θ  is in adians. 5

∴ r θ  θ  + r + r  = (12 + π ) r θ  r (θ 

dy

= 2 - 2 x is the deriv tive dx th n integrati g would gi e us the

6( ) Given that

fu ction of th  curve.

∫2−

 x dx

= 2 x  x 2 + C  = 2 x  x 2 + 8 C  8; since C is the y int rcept,(0,8).

w  integrate t  find the ar  a under cur  e.

6 5 2r  = (12 π ) 6 5 2) = (12 π ) Factori e, r. 6 5 (12 + π ) r  = 6 θ  2 5 (12 + π ) π  r  = 6 , note = π  6 2 6 r=5

 Area o  sector =

4

∫

2 x −  x 2 + 8 dx

 Area =

0

=

2

2

−

 x

3

2 3 E aluate from 0 to 4.

+ 8 x

⎛ 2(4)2 (4)3 ⎛ 2(0) 2 (0)3  ⎞ =⎜ − + 8(4) ⎟ − ⎜ − + 8(0) ⎟ 3 3 ⎝  2 ⎝  2  ⎠ = 48 − = 26

2 3

64 3

+ 32 − 0

uni s

2

would be the Area u der the cur  e.

 Area =  Area =

1

1 2

2

r  θ 

52 *

2 1

π 

6

using

× π  25

12 25π  12

units 2

=

π 

6

.

4(c) Compound Angle Formula for  ta tan( A ± B ) =

 

tan(θ − α ) = Given  ta ta tan(θ − α ) =

 

tan θ − tan α  1 + tan θ tan α  3 − tan α     

=

tan  A ± tan B 1 ∓ tan  A tan B 1 2 tan θ − tan α  1 + tan θ tan α  1 2 1

= ; note that ta tan θ  = 3

1 + 3 tan α  2 2 ( 3 − tan α ) = 1(1 + 3 tan α ) 6 − 2 tan α = 1 + 3 tan α 

−2 tan α − 3 tan α  = 1 − 6 − 5 tan α  = −5   −5 tan α  = −5   ta tan α  = 1  

α  =

ta tan −1 (1)

α  =

45