web view1. 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 x 3 dx=26 . 3 - k...
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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
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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