Sec 4E5N Prelim AM Paper 2 2016 - Answer Scheme

1 (i) ∫

1

0

p ( x ) dx+∫1

3

p ( x ) dx

¿−3+3=0 (write ∫1

0

p ( x ) dx=−3)

(ii )∫1

3

[ p (x )−k x3 ]dx=26 .

3−(k )(20)=26 (marks for each integral)

k=−2320

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2 At A, y=0 , e0= x2−8

x2=9

x=3∨x=−3(reject)

Gradient = 2 x

x2−8

At A (3, 0) , gradient of tangent=6

Equation of normal y−0=−16

(x−3)

y=−16

x+ 12

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3 (i ) tan 2 x=3 tan x

2 tan x

1−tan2 x¿3 tan x (write tan2 x= 2 tan x

1−tan2 x )

3 tan 3 x−tan x=0 tan x (3 tan2 x−1 )=0

tan x=0x=180 °

tan2 x=13

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1

tan x=±√ 13

basic angle¿30 °

x=30 ° ,210 ° , 150° , 330°

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3

(ii) 1+cosec A

cot A+cos A=sec A.

LHS : 1+cosec A

cot A+cos A

¿1+ 1

sin Acos Asin A

+cos A

¿sin A+1

sin A÷ cos A+sin A cos A

sin A

¿sin A+1

cos A ¿¿¿

¿1

cos A

¿ sec A (RHS)

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factor)

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3 (iii ) y= cos x5−sin x

dydx

=¿¿

tangent // to x-axis → dydx

=0

¿

1−5sin x=0

sin x=¿ 15¿

x=0.201

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2

4(i) (1− x

3 )6

=1−2 x+ 53

x2−2027

x3+…

(ii)(1−x+k x2)¿)

Compare coefficients of x3,

−2027

−53−2 k=−38

27

k=−12

(iii) (10r )¿

Collecting the x term → x20−4 r For independent of x→ 20−4 r=0 r = 5

The term is (105 ) (−2 )5=−8064

B2

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5 a(i) Area = 12

× ( 4√3−2√5 ) × (√5+√3 )

¿12¿

= (√15+1 ) cm2

(ii) H 2=( 4√3−2√5 )2+(√5+√3 )2

= ¿

= 76 −¿14√15

(b) 8x × 52 x=22 x+2× 5x−1

Using laws of indices –

23x−( 2 x+2 )=5x−1−2 x

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2x . 14=5−x . 1

5

10x=45

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6 (i) Stationary point →dydx

=0 at x=23

9− p

49

=0

p=9( 49 )

p=¿4 (shown)

(ii) dydx

=9− 4x2

d2 y

dx2 = 8x3

When x=23

, d2 y

dx2 >0,

( 23

,−6)is aminimum point

(iii ) y=9 x+ 4x+c

At ( 23

,−6) , c = −18

the equation of the curve y=9x+ 4x−18

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7 (a) f (−a )=2 a2+5a+17=R

f (−b )=2b2+5 b+17=R

4

Same remainder → 2a2+5a+17=2b2+5b+17

2a2+5 a=2 b2+5b

2(a¿¿2−b2)=5 (b−a)¿

2 (a+b ) ( a−b )=−5(a−b)

a+b=−52

(b) f (2 )=0

6(2)3−7 (2 )2+k (2 )+6=0

k=−13

f ( x )=6 x3−7 x2−13 x+6=(x−2)(6 x2+5 x−3) (the other factor)

( x−2 ) (6 x2+5 x−3 )=0

x=2 , x=−5±√(5)2−4(6)(−3)2(6)

x=2 , x=0.40 ,−1.24

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8 (i) Let C(p , q) length AC = length BC

√( p+3)2+(q+2)2=√( p+1)2+(q+12)2

p2+6 p+9+q2+4 q+4=p2+2 p+1+q2+24q+144

4 p=20 q+132

p=5q+33 ------(1)

(Alternatively, use perpendicular bisector of AB)

Grad of AB = −5

Perpendicular grad AB = 15

Mid point AB = (−2 ,−7)

perpendicular bisector of AB is

y+7=15(x+2)

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5

y=15

x−335

q=15

p−335 -----------(1)

Grad of given line = 15

(Grad of BC )( 15¿=−1

Equation of line BC is q+12=−5( p+1)

−5 p−5=q+12

q=−5 p−17 ----------(2)

Solving simultaneously (1) and (2), p=−2, q=−7Centre (−2 ,−7¿ (shown)

(ii) radius = √(−2+3)2+(−7+2)2=√26

Equation is ¿

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9 (i) s=12(e−2 t−e−t)

when t=0 , s=12 (1−1 )=0 m

The particle is initially at O.

(ii) v=12(−2 e−2 t+e−t)

when t=0 , v=−12m / s

a=12(4 e−2 t−e−t)

when t=0 , a=36m /s2

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10 (a) x2+1=k+x

For real and distinct roots, (−1)2−4 (1)(1−k)>0 B2

6

1−4+4 k>0

k> 34

(b) α +β= p2

α β=−12

αβ+ β

α=−29

2

α2+β2

αβ=−29

2

( p

2)

2

−2(−12

)

−12

=−292

p2

4=−1

2 (−292 )−1

p2=25 p=5∨p=−5(reject ) ( p>0¿

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11 (i) A=x(hx+k )Ax

=hx+k

Plot Ax

against x

x 50 100 150 200 250

Ax

74 110 144 180 214B2

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(ii) from the graph -

y -intercept = k = 40

Gradient = h = 0.70

(iii) A=x2

Ax

=x

Draw the straight line Ax

=x into graph

x=130

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