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