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Orion Tuition Centre Suggested solution for 2016 H2 Maths A Level paper 1 1
Good Luck for paper 2! 1.
4x2 + 4x −14x − 4
− x + 3( )
= 3x2 + 3x − 2x − 4
( )
( )
( )( )( )( )( )
2
2
2
2
4 4 14 34
4 4 14 3 04
3 5 2 04
3 5 2 4 0
3 1 2 4 012 or 43
x x xx
x x xx
x xxx x x
x x x
x x
+ − < +−
+ − − + <−
+ − <−+ − − <
− + − <
∴ < − < <
2. (i) 0
2
0, ln 2 0.693x x
dy dydx dx π= =
= = − = −
(ii) At 0x = , 2y = , 0dydx
=
Equation of tangent is 2y∴ =
At 2
x π= , 1y = , ln 2dydx
= −
( )
1 Equation of tangent is ln 2
21ln 2 ln 2 12
y
x
y x
π
π
−∴ = −−
⇒ = − + +
Equating both equations and solving,
( )ln 2 ln 2 1 22
0.12832
x
xy
π− + + =
⇒ ==
Therefore coordinates of intersection point is (0.128,2)
3.
( ) ( )( )
4
, is a turning point.,
f x k x l m
l ml a m b
= − +
⇒∴ = =
( ) ( )4
4
At 0, , 0c c k l mc bka
= − +−⇒ =
Orion Tuition Centre Suggested solution for 2016 H2 Maths A Level paper 1 2
4. (i)
a +8da + 3d
= r3 ⇒
ad+8
ad+ 3
= r3 ⇒ ad= 3r 2 −8
1− r3 ! 1( )
a +11da +8d
= r7 ⇒
ad+11
ad+8
= r7! 2( )
Sub (1) into (2):
2
3 72
3
3 37
3 3
37
3
10 3
3 8 1113 8 81
3 8 11 113 8 8 8
8 35
5 8 3 0 (shown)
rr r
rrr r rr r
r rrr r
− +− =− +
−− + −⇒ =− + −
− + =−
∴ − + = Solving using GC, 1 (rej) or 0.74r r= = (ii)
( )
( )
1 0.741 0.74 1 0.740.740.26
n
n
n
bbS S
b
∞
−− = −
− −
=
5. (i) ( ) ( )+ × −u v u v
22
2 0 12
2 2 2
a
b
ba ba
= − × + ×= ×
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= × −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎛ ⎞⎜ ⎟= − +⎜ ⎟⎜ ⎟−⎝ ⎠
u v v uv u
(ii) 1
, 2 2 4 2 41
ab a a a
a
−⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= − ∴ × = − = −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠
v u
Unit vector:
2a141
⎛
⎝
⎜⎜⎜
⎞
⎠
⎟⎟⎟= 1⇒ 2 a 18 = 1⇒ a = ± 1
6 3
(iii) ( ) ( ) 2 2. 0 0+ − = ⇒ − =u v u v u v
v2= u
2
= 22 +12 + 22
= 9
v = 3
Orion Tuition Centre Suggested solution for 2016 H2 Maths A Level paper 1 3
6 (ii)
u1 =u0 +2= 4u2 =u1 +10=14u3 =u2 +30= 44
(iii)
ur −ur−1r=1
n
∑=u1 −u0+u2 −u1....+un−1 −un−2+un −un−1=un −u0
ur =ur−1 + r3 + r
ur −ur−1 = r3 + r
From part (i)
ur −ur−1r=1
n
∑ = r3 + rr=1
n
∑
= r(r2 +1)r=1
n
∑ = 14(n)(n+1)(n2 +n+2)
Finally
un −u0 =14(n)(n+1)(n
2 +n+2)
un =14(n)(n+1)(n
2 +n+2)+u0
un =14(n)(n+1)(n
2 +n+2)+2
7. (a) Sub −1+5i into equation given,
(−1+5i)2 +(−1−8i)(−1+5i)+(−17+7i)=1−10i−25+1−5i+8i+40−17+7i=0
Perform Long division with linear factor
w+1−5i we will have,
(w+1−5i)(w−2−3i)=0
Thus the other root is 2+3i (b) Form Quadratic Factor with the conjugate
pairs, we will have,
(z −(1+ai))(z −(1−ai)= z2 −2z +1+a2
The third root must be real, so we are
expecting the equation to be as follows,
(z2 −2z +1+a2)(z +b)=0z3 + z2(−2+b)+ z(1+a2 −2b)+(1+a2)b=0
Comparing Cofficients b=-3, a = 3, k=-30
Orion Tuition Centre Suggested solution for 2016 H2 Maths A Level paper 1 4
8. (i)
y = f (x)= tan(ax +b)f '(x)= asec2(ax +b)= a(1+ tan2(ax +b)= a(1+ y2)= a+ay2
f ''(x)=2ay dy
dx=2ay(a+ay2)
f '''(x)=2a2 dydx
+6a2 y2 dydx
=2a2(a+ay2)+6a2 y2(a+ay2)=2a3 +8a3 y2 +6a3 y4
(ii) ( ) tan4 4
b y f x axπ π⎛ ⎞= ⇒ = = +⎜ ⎟⎝ ⎠
f (0)= tanπ4 =1f '(0)=2af ''(0)= 4a2f '''(0)=16a3
f (x)=1+2ax +2a2x2 + 83a
3x3 + ...
(iii)
b=0,a=2
y = f (x)= tan(2x)f (0)=0f '(0)=2f ''(0)=0f '''(0)=2(23)−16
tan2x =2x + 83 x3
9. (ia)
y = dxdt
⇒ dydt
= d 2xdt2
d 2xdt2 + 2 dx
dt= 10⇒ dy
dt+ 2y = 10
dydt
= 10− 2y
(ib)
110− 2y
dy =∫ dt∫
− 12
ln |10− 2y |= t + c
10− 2y = Ae−2t , A = ±e−2c
y = 5− A2
e−2t
t = 0, dydx
= 0⇒ y = 0
0 = 5− A2⇒ A = 10
y = 5−5e−2t
dydx
= y = 5−5e−2t
x = 5t + 52
e−2t + d
t = 0,x = 0
0 = 52+ d ⇒ d = − 5
2
x = 5t + 52
e−2t − 52
Orion Tuition Centre Suggested solution for 2016 H2 Maths A Level paper 1 5
9. (ii)
d 2xdt2 = 10−5sin
12
t
dxdt
= 10−5sin12
t∫ dt
dxdt
= 10t +10cos12
t +C
t = 0,dxdt
= 0⇒ 0 = 10+C ⇒ C = −10
x = 10t +10cos12
t −10∫x = 5t2 + 20sin
12
t −10t + D
t = 0,x = 0⇒ 0 = D
x = 5t2 + 20sin12
t −10t
(iii) Using GC with equation from (ib) t= 1.47s,
Using GC with equation from (ii) t=1.05s.
10. (ai) y = f (x) = 1+ x x = ( y −1)2
f
−1(x) = (x −1)2,x ≥1 (aii) f (x) = f −1(x)
1+ x = (x −1)2
1+ x = x2 − 2x +1
x = x2 − 2x
x = x4 − 4x3 + 4x2
x3 − 4x2 + 4x −1= 0 x = 2.62, 0 (rej) , 0.382 (rej)
(bi) (b) ( ) ( ) ( )4 6, 7 8, 12 9g g g= = = . No it is
not 1-1 function since g(5)=g(6)=6.
Orion Tuition Centre Suggested solution for 2016 H2 Maths A Level paper 1 6
11. (i)(a) normal vector of p=1 0 22 2 10 1 2
−⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟× =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠ ⎝ ⎠
Since pn is parallel to direction vector of l , l is perpendicular to p
1 1 0 1 23 2 4 0 12 0 2 1 2
1 1 23 2 42 2 1 2
0 2 22 4 30 2 2 1
t
ttt
ttt
λ µ
λλ µµ
λ µλ µ
λ µ
− −⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟− + + = +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
+ = − −⎧⎪⇒ − + + =⎨⎪ − = +⎩
+ + = −⎧⎪⇒ + − =⎨⎪ − − = −⎩
Solving using GC, 8 19 5, ,9 18 9
tλ µ= − = = −
(i)(b) Vector equation of p:
2 1 2
. 1 3 . 1 12 2 2
− −⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟= − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
r
⇒ r.−212
⎛
⎝
⎜⎜⎜
⎞
⎠
⎟⎟⎟
−212
⎛
⎝
⎜⎜⎜
⎞
⎠
⎟⎟⎟= −1
−212
⎛
⎝
⎜⎜⎜
⎞
⎠
⎟⎟⎟
⇒ r.−212
⎛
⎝
⎜⎜⎜
⎞
⎠
⎟⎟⎟
3 = − 13
Let the planes of interest have equation 2
. 1 32
k−⎛ ⎞
⎜ ⎟⇒ =⎜ ⎟⎜ ⎟⎝ ⎠
r
Distance from the planes of interest to p
can be calculated using 13
k ⎛ ⎞− −⎜ ⎟⎝ ⎠
1 123
35 37 or -3 3
k
k
∴ + =
=
Equations are
2 235 37. 1 3 or . 1 33 3
2 2
− −⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= = −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
r r
Which can also be written as 2 2
. 1 35 or . 1 372 2
− −⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= = −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
r r