Mark scheme Practice paper 2

Mark scheme

Practice paper 2

1 a

V = 10000 × 1.06

¹⁵

= 23965.58 (M1) (A1)

b

23965 58 10000 1

^{0 06}

. = ×

^⎛

+

12^.

⎝⎜

⎞

⎠⎟

n

(M1) (A1) (A1)

= 175.243 = 176 months (A1) [6 marks]

2

ln 2 x − = ⇒ 1 0 2 x − = ⇒ 1 1 2 x = ⇒ = 2 x 1 (A1) y = ln 2 x − ⇒ 1 y ( ) 1 _{= ⇒} 1 m

N

_{= − = −}

¹

1

′

1

(M1) (A1)

N : y  0 = 1 (x  1)  y = x + 1 (M1) (A1) [5 marks]

3

8 5 6 5 64 25 36

5

6 4

6 4 1 04

13

2 2

2

2 2

+ + + +

+ + + +

=

− =

⎧

⎨ ⎪⎪

⎩ ⎪

⎪

⇒ + =

+

a b

a b

a b a b .

. . ==

⎧ ⎨

⎩ 85 (M1) (A1) (A1) (A1)

⇒ = −

+ ( − ) ₌

⎧ ⎨

⎪

⎩⎪ ⇒ = −

− + =

⎧ ⎨

⎩ ⇒ =

=

⎧ ⎨

⎩

b a

a a

b a

a a

b a 13

13 85

13 13 42 0

7 6

2 2 2

(A1) (A1) [6 marks]

4 a

s

₁

= 100 + 5t, s

2

t s s s

2

1 2

1

=

2

⇒ = − (M1) (A1)

s = ⇒ = 0 0 100 + 5 t −

¹

t ⇒ t − 5 t − 100 = 0

2

1 2

2 2

(A1) (AG)

b ¹

2

2

1 2

5 100 0 10 20

t − t − = ⇒ t = − , t = (A1)

^d

d^s

m/s

t

( ) 20 ₌ 15 (A1) [5 marks]

5 a

cos sin arccos sin ln

y x

e x

y x

y x

y

= −

= −

⎧ ⎨

⎪

⎩⎪ ⇒ = ( − )

= ( − )

⎧ ⎨

⎪

⎩⎪

1 1

1

2 1

2

(A1) (A1)

b

(M1) (A1)

x = 1.60, y = 1.57 (A1) (A1) [6 marks]

6 a

Since the probability density function must always be positive and

e

^−x2

> 0 for all the values of x, then k > 0. (R1)

b

2

0

ke

^−x2

d x = 1 (M1) (A1)

k = 1.13 (A1)

c

2

0

kx e

^−x2

d x = 0 556 . (M1) (A1) [6 marks]

7

i i

i i i i i i i i

x y

x y a b c d e f

+ − ( ) ₌ ( + ) _{− =}

⎧ ⎨

⎪

⎩⎪ 1 4 ⇒ = = − = + = − = =

1 2 , 1 , 1 , , 4 , 2 (A1)

x

y

ed bf

x

ad bc

af ce ad bc

=

=

⎧

⎨ ⎪⎪

⎩ ⎪

⎪

⇒

= ( ) ( )

−

( )

−

−

−

− − − ⋅

⋅ − − +

4 1 2

1

i i i

i i

((

i

) ( )

= ( )

( ) ( ) ( )

⎧

⎨

⎪ ⎪

⎩

⎪ ⎪

−

⋅ − + ⋅

⋅ − − + −

1

2 1 4

1 1

i

i i i

i i i i

y

(M2) (A1) (A1)

x x

y

y

= ⇒ = +

= +

− − −

−

= − − −

−

⎧

⎨⎪⎪

⎩⎪

⎪

⎧⎨

⎪

⎩⎪

4 2 2

1 2 2 4 4

1 2

2 6 6 4

i i

i

i

i (A1) (A1) [7 marks]

8 a

The function f is even therefore its antiderivative F is odd,

F (x) = F (x) for all real values. (R1)

a

a

f x ( ) d x _{= ⎡⎣} F a ( ) ₋ F ( ) ₋ a _{⎤⎦ = ×} 2 F a ( ) (M1)

= × ⎡ ( ) ₋ ( )

⎣⎢ ⎤

⎦⎥ = ×

2 0 2

0

F a F N

a

0

f x ( ) d x (A1)

b

The function g is odd therefore its antiderivative G is even,

G(x) = G(x) for all real values. (R1)

a

g x ( ) d x _{= ⎡⎣} G a ( ) ₋ G ( ) ₋ a _{⎤⎦ =} G a ( ) ₋ G a ( ) ₌ 0 (M1) (A1) [6 marks]

9

Given that x

₁

, x

₂

and x

₃

are solutions of the equation

2 x

³

− 3 x

²

+ 4 x − = ⇒ = 5 0 a 2 , b = − 3 , c = 4 , d = − 5 (A1) Using Viete’s formulae

x x x

^b

x x x

a

d a

1

+

2

+

3

= − ,

1 2 3

= − (M1)

x x x

1

x x x x x x x x x x x x

2

2 3 1 2

2

3 1 2 3

2

1 2 3 1 2 3

+ + = ( + + ) ^{(M1) (A1)}

= −

^{− ⎛}

−

⁻

=

⎝⎜

⎞

⎠⎟

5 2

3 2

15

4

(A1) [5 marks]

10 a ^{sinCAB sin}

sin

^sin BC

BCA

AB

CAB

= ⇒ = ⋅ 6

^40°

5

(M1)

∠ = ⋅ = °

( )

^{⎛ °}

⎝⎜

⎞

CAB

1

6

⁴⁰ ⎠⎟

50 5

arcsin

^sin5

. (A1)

∠ = ° − ⋅ = °

( )

^{⎛ °}

⎝⎜

⎞

CAB

2

180 6

⁴⁰ ⎠⎟

129 5

arcsin

^sin5

. (A1)

b ^AC ABC

BC

CAB

AC

( )

( ) ( )

^°

= ⇒ = ⋅

°

=

1 1

1

6

^{89 5}

9 33

40

sin sin

sin .

sin

. (M1) (A1)

AC ABC

BC

CAB

AC

( )

( ) ( )

^°

= ⇒ = ⋅

°

=

2 2

2

6

^{10 5}

1 70

40

sin sin

sin .

sin

. (A1)

c

A =

¹

AB BC ⋅ ⋅ ( ABC )

2

sin

²

(M1)

= ⋅ ⋅ ⋅

¹

( ° ) ₌

2

5 6 sin 10 5 . 2 73 .

(A1) [8 marks]

11

The function

f x x

x

x

( ) sin

cos

,

= −

+

≤ ≤

3 4

3 2

0 2S is given.

a

A vertical asymptote occurs when the denominator equals 0. (R1)

 Ÿ 

³

  Ÿ ‡

3 2cos x 0 cos x

2

1 x (A1)

b

x = ⇒ 0 f ( ) 0 _{= =}

³

0 6

5

. (A1)

c

Ÿ  Ÿ

³ Ÿ ^{0.848, 2.29}

0 3 4 sin 0 sin

4 p q

y x x (M1) (A1) (A1)

d

1 2 3 4 5 6 x

y

(0, 0.6) (0.848, 0)

(1.75, –0.354) (2.29, 0)

(3.61, 3.95)

2 3

–1 –2 4 5 6

1 0

Shape (A1)

Zeroes (A1)

Stationary points (A1)

e

f (x) − g (x) > 0 (M1)

x ∈ [ ^{0 0 221 , .} [ ^∪ ] ^{1 75 6 28 . , .} ] ^{(A1) (A1)}

f

M

₁

(3.41, 3.28), M

₂

(5.16, 2.49) (M1) (A1) Since the function is continuous on the domain y

_max

= 3.28 (A1) [15 marks]

12 a n1

=

n2

=

n1 n2

−

⇒ ≠

⎛

⎝

⎜⎜

⎜

⎞

⎠

⎟⎟

⎟

⎛

⎝

⎜⎜

⎜

⎞

⎠

⎟⎟

⎟ ⋅ ∈

1 1 2

1 2 1

,

k

,k \

(A1) (R1)

b

x y z

x y z

+ + =

− + + =

⎧ ⎨

⎩

2 4

2 2 (M1) (A1)

x = 2 t, y = 2  t, z = t, t  R (A1) (A1) (A1)

c

We take a point P(2, 2, 0) that lies on the line and the direction vector

d =

−

⎛

⎝

⎜ ⎜

⎜

⎞

⎠

⎟ ⎟

⎟ 1 1 1

to fi nd the equation of the plane ∏.

AP

× =

d n

−

⎛

⎝

⎜ ⎜

⎜

⎞

⎠

⎟ ⎟

⎟ ×

−

⎛

⎝

⎜ ⎜

⎜

⎞

⎠

⎟ ⎟

⎟ =

−

−

−

⎛

⎝

⎜ ⎜

⎜

⎞

⎠

⎟ ⎟

⎟ ⇒ = 1

4 2

1 1 1

2 1 3

2 1 3

3

⎛⎛

⎝

⎜ ⎜

⎜

⎞

⎠

⎟ ⎟

⎟ (M1) (A1) (A1)

A  ∏ 2 · 1 + 1 · (2) + 3 · 2 = d  d = 6 (M1) (A1)

∏: 2x + y + 3z = 6 (A1)

d

The direction vector of the line is the normal vector of the plane  and we include the point A.

A   1 · 1 + 1 · (2) + 1 · 2 = d  d = 3 (M1)

: x + y  z = 3 (A1)

2 2 3 7 3

⁷

− + − − = − ⇒ = t t t t ⇒ = t

3

(A1)

⇒ A ′

^⎛_⎝⎜

−

¹

−

^⎞_⎠⎟

3 1 3

7

, ,3

(A1) [17 marks]

13 a

X ~ N (μ,  ) P

P

P P X

X

X X ( < ) ₌

( > ) ₌

⎧ ⎨

⎪

⎩⎪ ⇒ ( < ) ₌

( < ) ₌ 2 0 748

1 7 0 909

2 0 748 1 7 0 011 .

. .

.

. .

⎧⎧ ⎨

⎪

⎩⎪ (M1) (A1) (A1)

2

1 7

1

1

0 748

1

0 011

2 0 748

−

−

= ( )

= ( )

⎧

⎨ ⎪⎪

⎩ ⎪

⎪

⇒ = − ( ) _⋅

=

−

−

−

P V

P V

P V

P Φ

Φ . Φ

.

.

.

1 1 7 . −

¹

( 0 011 . ) _⋅

⎧ ⎨

⎪

⎩⎪ Φ

⁻

V

(A1) (A1)

μ = 1.93, σ = 0.101 (A1) (A1)

b

P(1.75 < X < 2.15) = 0.794 (M1) (A1) (AG)

c

A: “At least one pole satisfi es the standards.”

P(A) = 1  P (A) = 1  (1  0.948)

³

(M1) (A1)

= 0.99986 (A1)

d

B:“All three poles satisfy the standards.”

   

 

P B

P B|A

P A

(M1) (A1)

=

^{0 948}

=

0 99986

3

0 852

.

.

. (A1) [15 marks]

14 a

D ( f ) = {x  : x  0) (A1)

R( f ) = [1, 1] (A1)

b

sin x x x

x x

= ⇒ = ⇒ =

= ⇒ =

⎧ ⎨

⎪

⎩⎪

0

2 4

1 2

2 2

S S

S S (M1) (A1) (A1)

c

A =

4π²

0

sin x d x (M1) (A1) (A1)

= 25.1 (A1)

d

V =

4π²

0

sin x x

( )

²

^d (M1) (A1) (A1)

= 62.0 (A1) [13 marks]