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GCE A Level H2 Maths

Nov 2011 Paper 1

1) x2

+ x + 1

= (x +1

2)

2

1

4+ 1

= (x +1

2 )2 +3

4

> 0 for all x.

x2 + x 2 < 0(x + 2)(x 1) < 0

+ +

2 1

2 < x < 1

2i) 2.25a 1.5b + c = 4.5

4.41a + 2.1b + c = 3.2

11.56a + 3.4b + c = 4.1

a = 0.215, b = 0.490, c = 3.281

(ii) f(x) = 0.43x 0.49 > 0x > 1.14

3i) dxdt

= 2t

dy

dt=

2

t2

dy

dx=

2/t2

2t=

1

t3

Equation of tangent is

y 2

p=

1

p3 (x p

2)

y =3

p

x

p3

(ii)3

p

x

p3= 0

x = 3p2

Coordinate of Q = (3p2, 0).

Coordinate of R = (0,3

p).

(iii)

xcoord. of midpoint =3

2p2

ycoord. of midpoint =3

2p

x =3

2p2

=3

2(

3

2y)2

Cartesian equation is x =27

8y2

i.e. y2 = 278x

4i) cos6

x

= (1 x2

2+

x4

24...)

6

= 1 6(x2

2

x4

24) + 15(

x2

2

x4

24)

2

+...

= 1 3x2 +x

4

4+

15x4

4+...

= 1 3x2 + 4x4 +...

(ii)(a)

0

a g(x) dx

0a 1 3x2 + 4x4 dx

= [ x x3 +4

5x5 ]

a

0

= a a3

+4

5a

5

When a =

4, 0

a g(x) dx

4 (

4)3 +

4

5(

4)5

= 0.540

(b) From GC,

0/4 g(x) dx = 0.475

Since a =

4is not close to zero,

the graphs of y = cos6 x and y =

1 3x2

+ 4x4

differ significantly

at x =

4. the approximation is

not good.

5i) y = f( | x | )

x

y

-7 -6 -5 - 4 - 3 - 2 - 1 0 1 2 3 4 5 6 7

-3

-2

-1

0

1

2

3

y = | f(x) |

x

y

-7 -6 -5 - 4 - 3 - 2 - 1 0 1 2 3 4 5 6 7

-3

-2

-1

0

1

2

3

(ii) [0, 2]

(iii) 11 f( | x | ) dx

= 2 1 +1

2 2 1

= 3

1

a | f(x) | dx

=1

2 1 1 +

1

2(a 2)(a 2)

=1

2+

1

2(a 2)2 = 3

1

2(a 2)

2=

5

2

(a 2)2 = 5

a 2 = 5 or - 5 (reject since

a > 2)

a = 2 + 5

6i) sin (r +1

2 ) sin (r 1

2 )

= sin r cos2

+ cos r sin2

sin r cos2

+ cos r sin2

= 2 cos r sin2

Proved

(ii)

cos r =sin (r+

1

2) sin (r

1

2)

2 sin2

r=1

ncos r

=1

2 sin2

[ sin32

sin2

+ sin52

sin32

:

(0, 2)

(0, 2)

(2, 0)

(2, 0)

(2, 0)

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2

:

+ sin (n +1

2) sin (n

1

2) ]

=

sin (n +1

2) sin

2

2 sin

2

(iii) Let Pn be the proposition:

r=1

nsin r =

cos2

cos (n +1

2)

2 sin2

When n = 1:

RHS =

cos2

cos32

2 sin2

=

2 sin sin (

2)

2 sin2

= sin = LHSP1 is true.Assume that Pkis true for some

k+ i.e.

r=1

ksin r =

cos2

cos (k +1

2)

2 sin2

Prove that Pk+1 is also true i.e.

r=1

k+1sin r =

cos 2 cos (k +32)

2 sin2

LHS = r=1

k+1sin r

= r=1

ksin r + sin (k + 1)

=

cos2

cos (k+1

2)

2 sin 2

+ sin (k+1)

=

cos2cos(k+

1

2)+2sin

2

sin(k+1)

2 sin2

=

cos2cos(k+

1

2) cos(k+

3

2) +cos(k+

1

2)

2 sin2

=

cos2

cos (k +3

2)

2 sin

2

= RHS

Since P1 is true and Pk is true Pk+1 is true, hence by Math

Induction, Pn is true for all n

+

7i) OM

=OP

+OQ

2

=

1

3a +

3

5b

2

=1

6a +3

10b

Area ofOMP

=1

2area ofOPQ

=1

21

2| OP

| | OQ

| sin OPQ

=1

4| OP

OQ

|

=1

4|1

3a

3

5b |

=1

20| a b |

(ii)(a) | a | = p

2

63

= p 4 + 36 + 9 = 1

p =1

7

(b) | ab | is the length ofprojection ofb on a.

(c) a b

=

2/7

6/73/7

1

12

= 17

12 3(4 3)

2 + 6

=1

7

978

8i) 1

100 v2 dv

=1

20ln

10 + v

10 v+ c

(iia) 1

10 0.1v2dv = dt

1

100 v2 dv =

1

10dt

120 ln 10 + v10 v = 110 t + c

ln

10 + v

10 v= 2t + 20c

10 + v

10 v= e2t+20c

10 + v

10 v= e2t e20c

= Ae2t

where A = e20c

When t = 0, v = 0 A = 110 + v

10 v= e

2t

Let v = 5:15

5 = e2t

2t = ln 3

t =1

2ln 3

(b) When t = 1:10 + v

10 v= e

2

10 + v = 10e2 ve2

v(1 + e2) = 10e

2 10

v = 10e2 1

e2

+ 1

(c)10 + v

10 v= e2t

10 + v = 10e2t ve2t

v(1 + e2t) = 10e

2t 10

v = 10e2t 1

e2t + 1

As t , v 10.

9i) 256 + (n 1)(7) < 107(n 1) >246

n 1 > 351

7

n > 361

7

n = 37.Depth on the 10th day T10

= 256 + 9(7)

= 193Total depth S37

=37

2[ 2(256) + 36(7) ]

= 4810

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(ii) Sn >99

100S

256

1

8

9

n

1 8

9

> 0.99256

1 8

9

1 89 n > 0.99

8

9

n< 0.01

n log8

9< log 0.01

n > 39.099

it takes 40 days.

10i) z2

= 8ei/2

= 8ei(2n

2)

z = 2 2 ei(n 4) , n = 0, 1

= 2 2 ei/4

, 2 2 ei3/4

z1 , z2 = 2(1 i), 2(1 + i)

(ii) w =4 16 4(4 + 2i)

2

=4 8i

2

=4 2(1 i)

2

= 2 (1 i)w1 , w2 = 1 i, 3 + i

(iii)

x

y

-5 -4 - 3 - 2 -1 0 1 2 3 4 5

-2

-1

0

1

2

(iv) The 2 lines are parallel.

11i)

2

5

2

4

1

3=

6

4

5

432

413

=

021

6

4

5

0

2

1=

4 + 10

(6 0)

12 0

=

6

6

12= 6

11

2

4

1

3

1

1

2= 4 1 6 = 3

equation is x + y + 2z = 3.

(ii) l1 : r =

1

23

+

2

41

l2 : r =

2

1

3+

1

5

k

1 + 2 = 2 + 2 = 3 4 2 = 6

2 4 = 1 + 5 4 + 5 = 1 7 = 7 = 1 = 1

3 + = 3 + k 4 = 3 + k k = 7

(iii)

1 + 2

2 43 +

112

= 1 + 2 + 2 4 6 + 2= 3

l1 lies on p. Shown

2 + 1 + 53 7

11

2= 3

2 + + 1 + 5 + 6 14 = 38 = 8 = 1

pt of intersection = (1, 6, 4).(iv)

sin =

1

57

112

1+25+49 1+1+4

=

1 + 5 14

75 6

=8

15 2

= 22.2

GCE A Level H2 Maths

Nov 2011 Paper 2

1i)

x

y

-8 - 7 - 6 -5 - 4 - 3 - 2 - 1 0 1 2 3 4 5 6 7 8 9 10

0

1

2

3

4

5

6

7

8

(ii) 22 + 52 = 29

Max value of | z | = 29 + 3

Min value of | z | = 29 3(iii)

x

y

- 7 - 6 - 5 - 4 - 3 - 2 -1 0 1 2 3 4 5 6 7 8 90

1

2

3

4

5

6

7

Substitute y = x into (x 2)2 +

(y 5)2

= 32:

(x 2)2

+ (x 5)2

= 9

x2

4x + 4 + x2

10x + 25 = 92x

2 14x + 20 = 0

x2

7x + 10 = 0(x 2)(x 5) = 0

x = 2, 5

coord of P = (2, 2) & (5, 5)max value of | z 6 i |

= (5 6)2

+ (5 1)2

= 1 + 16

= 17

2i) Volume V

= (2n 2x)(n 2x)x= (2n2 6nx + 4x2)x

= 2n2x 6nx2 + 4x3

(ii)dV

dx= 2n

2 12nx + 12x

2= 0

6x2

6nx + n2

= 0

x =6n 36n2 24n2

12

=6n 2n 3

12

= (3 + 3 )n

6, (3 3 )

n

6

The answer (3 + 3 )n

6is to be

rejected since to be able to cut

squares of side x from each

51

8

2

(2, 5)

(6, 1)

(2, 5)

P

P

z1

z2

w1

w2 (a)

(b)

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corner, x must be 1

2

ln (2x + 1) = y 3

2x + 1 = ey3

x =ey3 1

2

f1: x ex3 1

2, x

Domain =

Range = (1

2, )

(ii)

x

y

-15 -14 -13 -12 -11 -10 -9 -8 - 7 - 6 - 5 - 4 - 3 - 2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

-7

-6

-5

-4

-3

-2

-1

0

1

2

3

4

5

6

7

(iii) The 2 curves are

symmetrical about the line y = x

they intersect along the line y= x

the xcoord of the points ofintersection satisfies f(x) = x

i.e. ln (2x + 1) + 3 = xi.e. ln (2x + 1) = x 3

x = 0.4847, 5.482

4a)(i) 0n x2 e2x dx

= [x

2

e2x

2 ]

n

0 0n

2x

e2x

2 dx

= n2e2n

2+0

n x e2x

dx

= n2e2n

2+[x

e2x

2]n

00

ne2x

2dx

= n2e2n

2+ n

e2n

2+

1

2[

e2x

2]n

0

= n2e2n

2+ n

e2n

2

1

4[ e2n 1]

= e2n

4(1 + 2n2 + 2n) +

1

4

(ii) 0 x2 e2x dx

=1

4

(b)

x

y

-4 -3 -2 -1 0 1 2 3 4

-2

-1

0

1

2

Volume = 01

16x2

(x2 + 1)2dx

= 0/4

16 tan2

(tan2 + 1)2sec2 d

= 160/4

tan2

sec4

sec2 d

= 160/4 sin

2

cos

2

cos2 d

= 160/4 sin2 d Shown

= 80/4 1 cos 2 d

= 8[ sin 2

2]/4

0

= 8[

4

1

2]

= 22

4

5) P(X < 40) = 0.05

P(Z < 40

) = 0.05

40

= 1.644854

40 = 1.644854P(X < 70) = 0.975

P(Z 5 and nq = 12 > 5, R ~

N(28, 8.4) approximately.

P(R < 25)= P(R < 24.5)

0.114

8i)

(ii) From GC, r = 0.992

Even though | r | is close to 1,

the scatter plot shows that the

relationship between x and y is

quadratic. Therefore the bestmodel between x and y may not

f1

f

y = 1

2

x = 1

2

(3, 0)

(0, 3)

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be y = c + dx.(iii)

The corr coeff calculated using y= a + bx2 is 0.99998, whose

magnitude is even closer to 1 ascompared to r = 0.992.

Therefore y = a + bx2

is a bettermodel.

(iv)

Regression line isy = 22.23049 0.85621x2

Estimate y

= 22.23049 0.85621(3.2)2

= 13.5

9i)

0.05 faulty

0.6 A

0.95 not

faulty

0.07 faulty

0.4 B0.93 not

faulty

(a) P(faulty)= 0.05 0.6 + 0.07 0.4

= 0.03 + 0.028= 0.058

(b) P(A | faulty)

=P(A faulty)

P(faulty)

=0.03

0.058

= 0.517

(ii)(a) 2 0.058 (1 0.058)

= 0.109272= 0.109

(b) P(both by A | exactly 1faulty)

=

P(both by A exactly 1 faulty)P(exactly 1 faulty)

=P(exactly 1 faulty | both by A)P(both by A)

0.109272

=2 0.05 0.95 0.62

0.109272

= 0.313

10i) Let = meaninstallation time of all

components.

Null hypothesis H0: = 38Alternative hypothesis H1: InvNor(0.05)

0.18 n > 1.64485

n < 9.1381n < 83.5

Hence n { 1, 2, 3, ..., 83 }

11i) P(R = 4)

=

18

4

12

6

30

10

=3060 924

30045015

= 0.0941(ii) P(R = r) > P(R = r + 1)

18

r

12

10r

30

10

>

18

r+1

12

9r

30

10

18!

r! (18r)!

12!

(10r)! (2+r)!>

18!

(r+1)! (17r)!

12!

(9r)! (3+r)!

(r+1)!(17r)!(9r)!(3+r)! >r!(18r)!(10r)!(2+r)! Shown

(r + 1)(3 + r) > (18 r)(10 r)

r2

+ 4r + 3 > r2

28r + 18032r > 177

r > 5.53

r = 6

12i) Let X = no. of people

joining the queue in 4 min.X ~ Po(4.8)

P(X 8)= 1 P(X 7)= 0.113

(ii) Let T = no. of peoplejoining the queue in t sec.

T ~ Po(1.2t

60) = Po(0.02t)

P(T 1) = 0.7P(T = 0) + P(T = 1) = 0.7e0.02t + e0.02t 0.02t = 0.7

e0.02t (1 + 0.02t) = 0.7

From GC, t = 54.867

= 55 (to nearest whole number)(iii) Let A, B = no. of people

joining and leaving the queue in

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15 minutes.A ~ Po(1.2 15) = Po(18) ~

N(18, 18) approximately since

18 > 10B ~ Po(1.8 15) = Po(27) ~

N(27, 27) approximately since

27 > 10

A B ~ N(18 27, 18 + 27)= N(9, 45)

P(35 + A B 24)= P(A B 11)= P(A B 11.5) 0.645(iv) The rate of peoplejoining the queue may not be

constant throughout the periodof several hours. There may be

more people queuing up just

before the departure of a flight.