Paper2Practice [303 marks]

[1 mark]1a.

Consider the expansion of .

Write down the number of terms in this expansion.

Markscheme11 terms A1 N1[1 mark]

(x+ 3)10

[4 marks]1b. Find the term containing .

Markschemeevidence of binomial expansion (M1)

eg , attempt to expand

evidence of choosing correct term (A1)

eg , ,

correct working (A1)

eg , ,

A1 N3[4 marks]

x3

( )nr

an−rbr

term, r = 78th ( )107

(x (3)3 )7

( )107

(x (3)3 )7 ( )103

(x (3)3 )7

262440 (accept 262000 )x3 x3

2a. [2 marks]

The following table shows the average weights ( y kg) for given heights (x cm) in a population of men.

Heights (x cm) 165 170 175 180 185

Weights (y kg) 67.8 70.0 72.7 75.5 77.2

The relationship between the variables is modelled by the regression equation .

Write down the value of and of .

Markscheme (exact) A1 N1

(exact), A1 N1[2 marks]

y = ax+ b

a b

a= 0.486

b = −12.41 −12.4

2b. [2 marks]The relationship between the variables is modelled by the regression equation .

Hence, estimate the weight of a man whose height is 172 cm.

y = ax+ b

Markschemecorrect substitution (A1)eg

(kg) A1 N2[2 marks]

0.486(172) − 12.41

71.182

71.2

[1 mark]2c. Write down the correlation coefficient.

Markscheme

A1 N1[1 mark]

r = 0.997276

r = 0.997

2d. [2 marks]State which two of the following describe the correlation between the variables.strong zero positivenegative no correlation weak

Markschemestrong, positive (must have both correct) A2 N2[2 marks]

[2 marks]3a.

The following diagram shows two perpendicular vectors u and v.

Let . Represent on the diagram above.w= u− v w

MarkschemeMETHOD 1

A1A1 N2

Note: Award A1 for segment connecting endpoints and A1 for direction (must see arrow).

METHOD 2

A1A1 N2

Notes: Award A1 for segment connecting endpoints and A1 for direction (must see arrow).

Additional lines not required.

[2 marks]

3b. [4 marks]Given that and , where , find \(n\).

Markschemeevidence of setting scalar product equal to zero (seen anywhere) R1eg u v correct expression for scalar product (A1)eg

attempt to solve equation (M1)eg

A1 N3[4 marks]

u =⎛⎝⎜

321

⎞⎠⎟ v =

⎛⎝⎜

5n

3

⎞⎠⎟ n ∈ Z

⋅ = 0, 15 + 2n+ 3 = 0

3 × 5 + 2 ×n+ 1 × 3, 2n+ 18 = 0

2n = −18

n = −9

4. [7 marks]Let , where , and . Find the equation of the normal to the graph of at

.

f(x) = g(x)

h(x)g(2) = 18, h(2) = 6, (2) = 5g′ (2) = 2h′ f

x = 2

Markschemerecognizing need to find or (R1)

(seen anywhere) (A1)

correct substitution into the quotient rule (A1)eg

A1

gradient of normal is 6 (A1)attempt to use the point and gradient to find equation of straight line (M1)eg

correct equation in any form A1 N4eg

[7 marks]

f(2) (2)f ′

f(2) = 186

6(5)−18(2)

62

(2) = −f ′ 636

y− f(2) = − (x− 2)1(2)f ′

y− 3 = 6(x− 2), y = 6x− 9

[3 marks]5a.

Let . Part of the graph of is shown in the following diagram.

The graph crosses the -axis at the points and .

Find the -coordinate of and of .

Markschemerecognizing (M1)eg

A1A1 N3[3 marks]

f(x) = 5 −x2 f

x A B

x A B

f(x) = 0

f = 0, = 5x2

x = ±2.23606

x = ± (exact), x = ±2.245√

5b. [3 marks]The region enclosed by the graph of and the -axis is revolved about the -axis.

Find the volume of the solid formed.

Markschemeattempt to substitute either limits or the function into formulainvolving (M1)eg

volume A2 N3[3 marks]

f x 360∘ x

f2

π∫ dx, π ( − 10 + 25), 2π(5 − )x2 2 ∫ 2.24−2.24 x4 x2 ∫ 5√

0 f2

187.328= 187

[2 marks]6a.

The following table shows the amount of fuel ( litres) used by a car to travel certain distances ( km).

Distance (x km) 40 75 120 150 195

Amount of fuel (y litres) 3.6 6.5 9.9 13.1 16.2

This data can be modelled by the regression line with equation .

Write down the value of and of .

Markscheme

A1A1 N2[2 marks]

y x

y = ax+ b

a b

a= 0.0823604, b = 0.306186

a= 0.0824, b = 0.306

[1 mark]6b. Explain what the gradient represents.

Markschemecorrect explanation with reference to number of litres

required for km A1 N1eg represents the (average) amount of fuel (litres) required to drive km, (average) litres per kilometre, (average) rate of change

in fuel used for each km travelled

[1 marks]

a

1

a 1

[2 marks]6c. Use the model to estimate the amount of fuel the car would use if it is driven km.

Markschemevalid approach (M1)eg , sketch

(litres) A1 N2[2 marks]

110

y = 0.0824(110) + 0.306

9.36583

9.37

[2 marks]7a.

Let and be independent events, where and .

Find .

Markschemecorrect substitution (A1)eg

A1 N2[2 marks]

A B P(A) = 0.3 P(B) = 0.6

P(A∩B)

0.3 × 0.6

P(A∩B) = 0.18

[2 marks]7b. Find .P(A∪B)

Markschemecorrect substitution (A1)eg

A1 N2[2 marks]

P(A∪B) = 0.3 + 0.6 − 0.18

P(A∪B) = 0.72

7c. [1 mark]On the following Venn diagram, shade the region that represents .

Markscheme

A1 N1

A∩B′

[2 marks]7d. Find .

Markschemeappropriate approach (M1)eg

(may be seen in Venn diagram) A1 N2[2 marks]

P(A∩ )B′

0.3 − 0.18, P(A) × P( )B′

P(A∩ ) = 0.12B′

[4 marks]8a.

In triangle , and . The area of the triangle is .

Find the two possible values for .

Markschemecorrect substitution into area formula (A1)eg

correct working (A1)eg

;

; A1A1 N3(accept degrees ie ; )

[4 marks]

ABC AB = 6 cm AC = 8 cm 16 cm2

A

(6)(8)sinA = 16, sinA =12

1624

A = arcsin( )23

A = 0.729727656… ,2.41186499… ( , )41.8103149∘ 138.1896851∘

A = 0.730 2.41

41.8∘ 138∘

[3 marks]8b. Given that is obtuse, find .

Markschemeevidence of choosing cosine rule (M1)eg

correct substitution into RHS (angle must be obtuse) (A1)eg ,

A1 N2[3 marks]

A BC

B = A + A − 2(AB)(AC) cosA, + − 2ab cosCC2 B2 C2 a2 b2

B = + − 2(6)(8)cos2.41, + − 2(6)(8)cosC2 62 82 62 82 138∘

BC = 171.55− −−−−√

BC = 13.09786

BC = 13.1 cm

[2 marks]9a.

Let , for . The following diagram shows the graph of .

The graph has a maximum at and a minimum at .

Write down the value of .

Markscheme A2 N2

Note: Award A1 for . [2 marks]

f(x) = pcos(q(x+ r)) + 10 0 ⩽ x ⩽ 20 f

(4,18) (16,2)

r

r = −4

r = 4

[2 marks]9b. Find .

Markschemeevidence of valid approach (M1)eg , distance from

A1 N2[2 marks]

p

maxy value -- y value2

y = 10

p = 8

[2 marks]9c. Find .q

Markschemevalid approach (M1)eg period is , , substitute a point into their

, (do not accept degrees) A1 N2

[2 marks]

24 36024

f(x)

q = ( , exact)2π24

π

120.262

[2 marks]9d. Solve .

Markschemevalid approach (M1)eg line on graph at

(accept ) A1 N2[2 marks] Note: Do not award the final A1 if additional values are given. If an incorrect value of leads to multiple solutions, award the final

A1 only if all solutions within the domain are given.

f(x) = 7

y = 7, 8 cos( (x− 4)) + 10 = 72π24

x = 11.46828

x = 11.5 (11.5,7)

q

10a. [3 marks]

The velocity of a particle in ms is given by , for .

On the grid below, sketch the graph of .

−1 v = − 1esin t 0 ≤ t ≤ 5

v

Markscheme

A1A1A1 N3

Note: Award A1 for approximately correct shape crossing x-axis with .

Only if this A1 is awarded, award the following:

A1 for maximum in circle, A1 for endpoints in circle.

[3 marks]

3 < x < 3.5

[1 mark]10b. Find the total distance travelled by the particle in the first five seconds.

Markscheme (exact), A1 N1

[1 mark]

t = π 3.14

[4 marks]10c. Write down the positive -intercept.

Markschemerecognizing distance is area under velocity curve (M1)

eg , shading on diagram, attempt to integrate

valid approach to find the total area (M1)

eg , , ,

correct working with integration and limits (accept or missing ) (A1)

eg , ,

distance (m) A1 N3

[4 marks]

t

s = ∫ v

area A + area B ∫ vdt− ∫ vdt vdt+ vdt∫ 3.140 ∫ 5

3.14 ∫ |v|

dx dt

vdt+ vdt∫ 3.140 ∫ 3.14

5 3.067… + 0.878… − 1∫ 50 ∣∣esin t ∣∣

= 3.95

11. [7 marks]Consider the expansion of . The constant term is .

Find .

x2(3 + )x2 k

x

816 128

k

Markschemevalid approach (M1)

eg ,

, Pascal’s triangle to line

attempt to find value of r which gives term in (M1)eg exponent in binomial must give

correct working (A1)eg

evidence of correct term (A1)

eg

equating their term and 16128 to solve for M1

eg

A1A1 N2 Note: If no working shown, award N0 for .

Total [7 marks]

( )8r

(3 )x2 8−r( )kx

r

+ ( ) ( ) + ( ) + …(3 )x2 8 81

(3 )x2 7 k

x

82

(3 )x2 6( )kx

29th

x0

, =x−2 x2( )x2 8−r( )kx

r

x0

2(8 − r) − r = −2, 18 − 3r = 0, 2r+ (−8 + r) = −2

( ), ( ) , r = 6, r = 282

86

(3 )x2 2( )kx

2

k

( ) = 16128, =x2 86

(3 )x2 2( )kx

6k6 16128

28(9)

k = ±2

k = 2

[2 marks]12a.

Let .

Find .

Markschemeexpressing as (M1)

A1 N2

[2 marks]

f(x) = −x4−−√3 12

(x)f ′

f x43

(x) = (= )f ′ 43x

13

43x√3

[4 marks]12b. Find .

Markschemeattempt to integrate (M1)

eg

A1A1A1 N4

[4 marks]

∫ f(x)dx

x4−−√3

x+14

3

+143

∫ f(x)dx = − + c37x

73

x

2

[2 marks]13a.

Two events and are such that and .

Given that and are mutually exclusive, find .

Markschemecorrect approach (A1)eg

A1 N2[2 marks]

A B P(A) = 0.2 P(A∪B) = 0.5

A B P(B)

0.5 = 0.2 + P(B), P(A∩B) = 0

P(B) = 0.3

[4 marks]13b. Given that and are independent, find .

MarkschemeCorrect expression for (seen anywhere) A1eg

attempt to substitute into correct formula for (M1)eg

correct working (A1)eg

A1 N3

[4 marks]

A B P(B)

P(A∩B)

P(A∩B) = 0.2P(B), 0.2x

P(A∪B)

P(A∪B) = 0.2 + P(B) − P(A∩B), P(A∪B) = 0.2 +x− 0.2x

0.5 = 0.2 + P(B) − 0.2P(B), 0.8x = 0.3

P(B) = (= 0.375, exact)38

14a. [3 marks]

A particle moves along a straight line such that its velocity, , is given by , for .

On the grid below, sketch the graph of , for .

Markscheme

A1A2 N3

Notes: Award A1 for approximately correct domain .

The shape must be approximately correct, with maximum skewed left. Only if the shape is approximately correct, award A2 for all

the following approximately correct features, in circle of tolerance where drawn (accept seeing correct coordinates for the maximum,

even if point outside circle):

Maximum point, passes through origin, asymptotic to -axis (but must not touch the axis).

If only two of these features are correct, award A1.

[3 marks]

v ms−1 v(t) = 10te−1.7t t ⩾ 0

v 0 ⩽ t ⩽ 4

0 ⩽ t ⩽ 4

t

14b.

Markschemevalid approach (including and ) (M1)eg , area from to (may be shaded in diagram)

A1 N2[2 marks]

0 3

10t dt, f(x)∫ 30 e−1.7t ∫ 3

0 0 3

distance = 3.33 (m)

[3 marks]14c. Find the velocity of the particle when its acceleration is zero.

Markschemerecognizing acceleration is derivative of velocity (R1)eg , attempt to find , reference to maximum on the graph of valid approach to find when (may be seen on graph) (M1)eg

A1 N3 Note: Award R1M1A0 for if velocity is not identified as final answer

[3 marks]

a= dvdt

dvdt

v

v a= 0= 0, 10 − 17t = 0, t = 0.588dv

dte−1.7t e−1.7t

velocity = 2.16 (m )s−1

(0.588,216)

[2 marks]15a.

The time taken for a student to complete a task is normally distributed with a mean of minutes and a standard deviation of

minutes.

A student is selected at random. Find the probability that the student completes the task in less than minutes.

MarkschemeNote: There may be slight differences in answers, depending on whether candidates use tables or GDCs, or their 3 sf answers in

subsequent parts. Do not penalise answers that are consistent with their working and check carefully for FT.

attempt to standardize (M1)eg

A1 N2[2 marks]

20 1.25

21.8

z= , 1.4421.8−201.25

P(T < 21.8) = 0.925

[5 marks]15b. The probability that a student takes between and minutes is . Find the value of .

MarkschemeNote: There may be slight differences in answers, depending on whether candidates use tables or GDCs, or their 3 sf answers insubsequent parts. Do not penalise answers that are consistent with their working and check carefully for FT.

attempt to subtract probabilities (M1)eg

A1EITHERfinding the -value for (A1)eg (from tables), attempt to set up equation using their -value (M1)eg

A1 N3OR

A3 N3[5 marks]

k 21.8 0.3 k

P(T < 21.8) − P(T < k) = 0.3, 0.925 − 0.3P(T < k) = 0.625

z 0.625z= 0.3186 z= 0.3188

z

0.3186 = , − 0.524 × 1.25 = k− 20k−201.25

k = 20.4

k = 20.4

16a.

Consider the graph of the semicircle given by , for . A rectangle is drawn with upper vertices

and on the graph of , and on the -axis, as shown in the following diagram.

Let .(i) Find , giving your answer in terms of .

(ii) Hence, write down an expression for the area of the rectangle, giving your answer in terms of .

Markscheme(i) valid approach (may be seen on diagram) (M1)eg to is

A1 N2(ii) A1 N1[3 marks]

f(x) = 6x−x2− −−−−−√ 0 ⩽ x ⩽ 6 PQRS R

S f PQ x

OP = x

PQ x

x

Q 6 x

PQ = 6 − 2x

A = (6 − 2x) 6x−x2− −−−−−√

[2 marks]16b. Find the rate of change of area when .

Markschemerecognising at needed (must be the derivative of area) (M1)

A1 N2

[2 marks]

x = 2

dAdx

x = 2

= − , − 4.95dAdx

7 2√2

[2 marks]16c. The area is decreasing for . Find the value of and of .

Markscheme A1A1 N2

[4 marks]

a< x < b a b

a= 0.879 b = 3

17a. [3 marks]

The first three terms of an arithmetic sequence are 36, 40, 44,….

(i) Write down the value of d .

(ii) Find .

Markscheme(i) A1 N1

(ii) evidence of valid approach (M1)

e.g. , repeated addition of d from 36

A1 N2

[3 marks]

u8

d = 4

= 36 + 7(4)u8

= 64u8

17b. [3 marks](i) Show that .

(ii) Hence, write down the value of .

Markscheme(i) correct substitution into sum formula A1

e.g. ,

evidence of simplifying

e.g. A1

AG N0

(ii) A1 N1

[3 marks]

= 2 + 34nSn n2

S14

= {2 (36) + (n− 1)(4)}Snn

2{72 + 4n− 4}n

2

{4n+ 68}n

2

= 2 + 34nSn n2

868

18a. [6 marks]

Let and be functions such that .

(a) The graph of is mapped to the graph of under the following transformations:

vertical stretch by a factor of , followed by a translation .

Write down the value of

(i) ;

(ii) ;

(iii) .

(b) Let . The point A( , ) on the graph of is mapped to the point on the graph of . Find .

Markscheme(a) (i) A1 N1

(ii) A1 N1

(iii) A1 N1

[3 marks]

(b) recognizing one transformation (M1)

eg horizontal stretch by , reflection in -axis

is ( , ) A1A1 N3

[3 marks]

Total [6 marks]

f g g(x) = 2f(x+ 1) + 5

f g

k ( )pq

k

p

q

h(x) = −g(3x) 6 5 g A′ h A′

k = 2

p = −1

q = 5

13

x

A′ 2 −5

18b. [3 marks]The graph of is mapped to the graph of under the following transformations:

vertical stretch by a factor of , followed by a translation .

Write down the value of

(i) ;

(ii) ;

(iii) .

f g

k ( )pq

k

p

q

Markscheme(i) A1 N1

(ii) A1 N1

(iii) A1 N1

[3 marks]

k = 2

p = −1

q = 5

[3 marks]18c. Let . The point A( , ) on the graph of is mapped to the point on the graph of . Find .

Markschemerecognizing one transformation (M1)

eg horizontal stretch by , reflection in -axis

is ( , ) A1A1 N3

[3 marks]

Total [6 marks]

h(x) = −g(3x) 6 5 g A′ h A′

13

x

A′ 2 −5

19. [7 marks]A random variable is normally distributed with and .

Find the interquartile range of .

Markschemerecognizing one quartile probability (may be seen in a sketch) (M1)

eg ,

finding standardized value for either quartile (A1)

eg ,

attempt to set up equation (must be with -values) (M1)

eg ,

one correct quartile

eg ,

correct working (A1)

eg other correct quartile,

valid approach for IQR (seen anywhere) (A1)

eg ,

IQR A1 N4

[7 marks]

X μ = 150 σ = 10

X

P(X < ) = 0.75Q3 0.25

z= 0.67448… z= −0.67448…

z

0.67 = −150Q3

10−0.67448 = x−150

10

= 156.74…Q3 = 143.25…Q1

−μ = 6.744…Q3

−Q3 Q1 2( −μ)Q3

= 13.5

[3 marks]20a.

The random variable is normally distributed with mean and standard deviation .

Find .

X 20 5

P(X ≤ 22.9)

Markschemeevidence of appropriate approach (M1)

eg

(A1)

A1 N3

[3 marks]

z= 22.9−205

z= 0.58

P(X ≤ 22.9) = 0.719

[3 marks]20b. Given that , find the value of .

Markscheme-score for is (A1)

valid approach (must be with -values) (M1)

eg using inverse normal,

A1 N3

[3 marks]

P(X < k) = 0.55 k

z 0.55 0.12566…

z

0.1257 = k−205

k = 20.6

[3 marks]21a.

The following diagram shows a triangle ABC.

The area of triangle ABC is cm , AB cm , AC cm and .

Find .

Markschemecorrect substitution into area formula (A1)

eg

setting their area expression equal to (M1)

eg

A1 N2

[3 marks]

80 2 = 18 = x B C =A 50∘

x

(18x)sin 5012

80

9x sin 50 = 80

x = 11.6

[3 marks]21b. Find BC.

Markschemeevidence of choosing cosine rule (M1)

eg

correct substitution into right hand side (may be in terms of ) (A1)

eg

BC A1 N2

[3 marks]

= + + 2ab sinCc2 a2 b2

x

+ − 2(11.6)(18)cos5011.62 182

= 13.8

22. [6 marks]The sum of the first three terms of a geometric sequence is , and the sum of the infinite sequence is . Find thecommon ratio.

Markschemecorrect substitution into sum of a geometric sequence A1

eg ,

correct substitution into sum to infinity A1

eg

attempt to eliminate one variable (M1)

eg substituting

correct equation in one variable (A1)

eg ,

evidence of attempting to solve the equation in a single variable (M1)

eg sketch, setting equation equal to zero,

A1 N4

[6 marks]

62.755 440

62.755 = ( )u11−r3

1−r+ r+ = 62.755u1 u1 u1r

2

= 440u1

1−r

= 440(1 − r)u1

62.755 = 440(1 − r)( )1−r3

1−r440(1 − r)(1 + r+ ) = 62.755r2

62.755 = 440(1 − )r3

r = 0.95 = 1920

[1 mark]23a.

The following diagram shows a circle with centre O and radius cm.

Points A and B are on the circumference of the circle and radians .

The point C is on [OA] such that radians .

Show that .

r

A B = 1.4O

B O =C π

2

OC = rcos1.4

Markschemeuse right triangle trigonometry A1

eg

AG N0

[1 mark]

cos1.4 = OCr

OC = rcos1.4

[7 marks]23b. The area of the shaded region is cm . Find the value of .

Markschemecorrect value for BC

eg , (A1)

area of A1

area of sector A1

attempt to subtract in any order (M1)

eg sector – triangle,

correct equation A1

eg

attempt to solve their equation (M1)

eg sketch, writing as quadratic,

A1 N4

[7 marks]

Note: Exception to FT rule. Award A1FT for a correct FT answer from a quadratic equation involving two trigonometric functions.

25 2 r

BC = rsin 1.4 −r2 (rcos1.4) 2− −−−−−−−−−−−√

ΔOBC = rsin 1.4 × rcos1.412

(= sin 1.4 × cos1.4)12r2

OAB = × 1.412r2

sin 1.4 × cos1.4 − 0.712r2 r2

0.7 − rsin 1.4 × rcos1.4 = 25r2 12

250.616…

r = 6.37

24a. [3 marks]

Consider the points A( , , ) , B( , , ) , and C( , , ) , .

Find

(i) ;

(ii) .

Markscheme(i) appropriate approach (M1)

eg ,

A1 N2

(ii) A1 N1

[3 marks]

5 2 1 6 5 3 7 6 a+ 1 a ∈ R

AB−→−

AC−→−

+AO−→−

OB−→−

B − A

=AB−→− ⎛

⎝⎜132

⎞⎠⎟

=AC−→− ⎛

⎝⎜24a

⎞⎠⎟

[4 marks]24b.

Let be the angle between and .

Find the value of for which .

Markschemevalid reasoning (seen anywhere) R1

eg scalar product is zero,

correct scalar product of their and (may be seen in part (c)) (A1)

eg

correct working for their and (A1)

eg ,

A1 N3

[4 marks]

q AB−→−

AC−→−

a q = π

2

cos =π

2u∙v

|u||v|

AB−→−

AC−→−

1(2) + 3(4) + 2(a)

AB−→−

AC−→−

2a+ 14 2a= −14

a= −7

24c. [8 marks]i. Show that .

ii. Hence, find the value of a for which .

Markschemecorrect magnitudes (may be seen in (b)) (A1)(A1)

,

substitution into formula (M1)

eg ,

simplification leading to required answer A1

eg

AG N0

[4 marks]

correct setup (A1)

eg

valid attempt to solve (M1)

eg sketch, , attempt to square

A2 N3

[4 marks]

cosq = 2a+14

14 +280a2√

q = 1.2

(= )+ +12 32 22− −−−−−−−−−√ 14−−√ (= )+ +22 42 a2− −−−−−−−−−√ 20 + a2

− −−−−−√

cosθ 1×2+3×4+2×a

+ +12 32 22√ + +22 42 a2√

14+2a

14√ 4+16+a2√

cosθ = 14+2a

14√ 20+a2√

cosθ = 2a+14

14 +280a2√

cos1.2 = 2a+14

14 +280a2√

− cos1.2 = 02a+14

14 +280a2√

a= −3.25

[4 marks]24d. Hence, find the value of a for which .q = 1.2

Markschemecorrect setup (A1)

eg

valid attempt to solve (M1)

eg sketch, , attempt to square

A2 N3

[4 marks]

cos1.2 = 2a+14

14 +280a2√

− cos1.2 = 02a+14

14 +280a2√

a= −3.25

25a. [3 marks]The heights of a group of seven-year-old children are normally distributed with mean and standard deviation . Achild is chosen at random from the group.

Find the probability that this child is taller than .

Markschemeevidence of appropriate method (M1)

e.g. , sketch of normal curve showing mean and ,

(A1)

A1 N3

[3 marks]

117 cm 5 cm

122.5 cm

z= 122.5−1175

122.5 1.1

P(Z < 1.1) = 0.8643

0.135666

P(H > 122.5) = 0.136

25b. [3 marks]The heights of a group of seven-year-old children are normally distributed with mean and standard deviation . Achild is chosen at random from the group.

The probability that this child is shorter than is . Find the value of k .

Markscheme (A1)

set up equation (M1)

e.g. , sketch

A1 N3

[3 marks]

117 cm 5 cm

k cm 0.65

z= 0.3853

= 0.3853X−1175

k = 118.926602

k = 199

[3 marks]26a.

A particle moves in a straight line with velocity , for , where v is in centimetres per second and t is inseconds.

Find the acceleration of the particle after 2.7 seconds.

v = 12t− 2 − 1t3 t ≥ 0

Markschemerecognizing that acceleration is the derivative of velocity (seen anywhere) (R1)

e.g.

correctly substituting 2.7 into their expression for a (not into v) (A1)

e.g.

(exact), A1 N3

[3 marks]

a= , ,12 − 6sd2

dt2v′ t2

(2.7)s′′

acceleration = −31.74 −31.7

[3 marks]26b. Find the displacement of the particle after 1.3 seconds.

Markschemerecognizing that displacement is the integral of velocity R1

e.g.

correctly substituting 1.3 (A1)

e.g.

(exact), (cm) A1 N2

[3 marks]

s = ∫ v

vdt∫ 1.30

displacement = 7.41195 7.41

27a. [5 marks]

The following diagram shows the cuboid (rectangular solid) OABCDEFG, where O is the origin, and , ,

.

(i) Find .

(ii) Find .

(iii) Show that .

= 4iOA−→−

= 3jOC−→−

= 2kOD−→−

OB−→−

OF−→

= −4i+ 3j+ 2kAG−→−

Markscheme(i) valid approach (M1)

e.g.

A1 N2

(ii) valid approach (M1)

e.g. ; ;

A1 N2

(iii) correct approach A1

e.g. ; ;

AG N0

[5 marks]

OA + OB

= 4i+ 3jOB−→−

+ +OA−→−

AB−→−

BF−→

+OB−→−

BF−→

+ +OC−→−

CG−→−

GF−→

= 4i+ 3j+ 2kOF−→

+ +AO−→−

OC−→−

CG−→−

+ +AB−→−

BF−→

FG−→

+ +AB−→−

BC−→

CG−→−

= −4i+ 3j+ 2kAG−→−

27b. [4 marks]Write down a vector equation for

(i) the line OF;

(ii) the line AG.

Markscheme(i) any correct equation for (OF) in the form A2 N2

where is 0 or , and is a scalar multiple of

e.g. , ,

(ii) any correct equation for (AG) in the form A2 N2

where is or and is a scalar multiple of

e.g. , ,

[4 marks]

r = a+ tb

a 4i+ 3j+ 2k b 4i+ 3j+ 2k

r = t(4, 3, 2) r =⎛⎝⎜

4t3t2t

⎞⎠⎟ r = 4i+ 3j+ 2k+ t(4i+ 3j+ 2k)

r = a+ sb

a 4i 3j+ 2k b −4i+ 3j+ 2k

r = (4, 0, 0) + s(−4, 3, 2) r =⎛⎝⎜

4 − 4s3s2s

⎞⎠⎟ r = 3j+ 2k+ s(−4i+ 3j+ 2k)

[7 marks]27c. Find the obtuse angle between the lines OF and AG.

Markschemechoosing correct direction vectors, and (A1)(A1)

scalar product (A1)

magnitudes , , (A1)(A1)

substitution into formula M1

e.g.

,

or A1 N4

[7 marks]

OF−→

AG−→−

= −16 + 9 + 4 (= −3)

+ +42 32 22− −−−−−−−−−√ + +(−4)2 32 22− −−−−−−−−−−−−√ ( , )29−−√ 29−−√

cosθ = = (− )−16+9+4

( )×+ +42 32 22√ + +(−4)2 32 22√3

29

95.93777∘ 1.67443 radians

θ = 95.9∘ 1.67

28a. [4 marks]The probability of obtaining “tails” when a biased coin is tossed is . The coin is tossed ten times. Find the probability ofobtaining at least four tails.

Markschemeevidence of recognizing binomial distribution (M1)

e.g. , ,

EITHER

(A1)

evidence of using complement (M1)

e.g. any probability,

A1 N3

OR

summing the probabilities from to (M1)

correct expression or values (A1)

e.g. ,

0.919424

A1 N3

[4 marks]

0.57

X ∼ B(10,0.57) p = 0.57 q = 0.43

P(X ≤ 3) = 2.16 × + 0.00286 + 0.01709 + 0.0604110−4 (= 0.08057)

1− P(X ≥ 4) = 1 − P(X ≤ 3)

0.919423…

P(X ≥ 4) = 0.919

X = 4 X = 10

( )(0.57 (0.43∑r=4

10 10r

)r )10−r 0.14013 + 0.2229 + … + 0.02731 + 0.00362

P(X ≥ 4) = 0.919

28b. [3 marks]The probability of obtaining “tails” when a biased coin is tossed is 0.57. The coin is tossed ten times. Find the probability ofobtaining the fourth tail on the tenth toss.

Markschemeevidence of valid approach (M1)

e.g. three tails in nine tosses,

correct calculation

e.g. , (A1)

A1 N2

[3 marks]

( ) (0.57 (0.4393

)3 )6

( ) (0.57 (0.43 × 0.5793

)3 )6 0.09834 × 0.57

0.05605178…

P(4th tail on 10th toss) = 0.0561

[3 marks]29a.

Consider the expansion of .

Find b.

= 256 + 3072 + … + k + …(2 + )x3 b

x

8x24 x20 x0

Markschemevalid attempt to find term in (M1)

e.g. ,

correct equation A1

e.g.

A1 N2

[3 marks]

x20

( ) ( )(b)81

27 (2 ( ) = 3072x3)7 b

x

( ) ( )(b) = 307281

27

b = 3

[3 marks]29b. Find k.

Markschemeevidence of choosing correct term (M1)

e.g. 7th term,

correct expression A1

e.g.

(accept ) A1 N2

[3 marks]

r = 6

( ) (286

x3)2( )3x

6

k = 81648 81600

[1 mark]30a.

Jose takes medication. After t minutes, the concentration of medication left in his bloodstream is given by ,where A is in milligrams per litre.

Write down .

Markscheme A1 N1

[1 mark]

A(t) = 10(0.5)0.014t

A(0)

A(0) = 10

[2 marks]30b. Find the concentration of medication left in his bloodstream after 50 minutes.

Markschemesubstitution into formula (A1)

e.g. ,

A1 N2

[2 marks]

10(0.5)0.014(50) A(50)

A(50) = 6.16

30c. [5 marks]At 13:00, when there is no medication in Jose’s bloodstream, he takes his first dose of medication. He can take his medicationagain when the concentration of medication reaches 0.395 milligrams per litre. What time will Jose be able to take his medication again?

Markschemeset up equation (M1)

e.g.

attempting to solve (M1)

e.g. graph, use of logs

correct working (A1)

e.g. sketch of intersection,

A1

correct time 18:33 or 18:34 (accept 6:33 or 6:34 but nothing else) A1 N3

[5 marks]

A(t) = 0.395

0.014t log0.5 = log0.0395

t = 333.00025…

[4 marks]31a.

Let , where . The function v is obtained when the graph of f is transformed by

a stretch by a scale factor of parallel to the y-axis,

followed by a translation by the vector .

Find , giving your answer in the form .

Markschemeapplies vertical stretch parallel to the y-axis factor of (M1)

e.g. multiply by , ,

applies horizontal shift 2 units to the right (M1)

e.g. ,

applies a vertical shift 4 units down (M1)

e.g. subtracting 4, ,

A1 N4

[4 marks]

f(t) = 2 + 7t2 t > 0

13

( )2−4

v(t) a(t− b + c)2

13

13

f(t)13

× 213

f(t− 2) t− 2

f(t) − 4 − 473

v(t) = (t− 2 −23

)2 53

31b. [3 marks]A particle moves along a straight line so that its velocity in ms , at time t seconds, is given by v . Find the distance theparticle travels between and .

−1

t = 5.0 t = 6.8

Markschemerecognizing that distance travelled is area under the curve M1

e.g. , sketch

distance = 15.576 (accept 15.6) A2 N2

[3 marks]

∫ v, (t− 2 − t29

)3 53

[3 marks]32a.

A random variable X is distributed normally with a mean of 20 and variance 9.

Find .

Markscheme (A1)

evidence of attempt to find (M1)

e.g. ,

A1 N3

[3 marks]

P(X ≤ 24.5)

σ = 3

P(X ≤ 24.5)

z= 1.5 24.5−203

P(X ≤ 24.5) = 0.933

32b. [5 marks]Let .

(i) Represent this information on the following diagram.

(ii) Find the value of k .

P(X ≤ k) = 0.85

Markscheme

A1A1 N2

Note: Award A1 with shading that clearly extends to right of the mean, A1 for any correct label, either k, area or their value of k.

(ii) (A1)

attempt to set up an equation (M1)

e.g. ,

A1 N3

[5 marks]

z= 1.03(64338)

= 1.0364k−203

= 0.85k−203

k = 23.1

[2 marks]33a.

A box holds 240 eggs. The probability that an egg is brown is 0.05.

Find the expected number of brown eggs in the box.

Markschemecorrect substitution into formula for (A1)

e.g.

A1 N2

[2 marks]

E(X)

0.05 × 240

E(X) = 12

[2 marks]33b. Find the probability that there are 15 brown eggs in the box.

Markschemeevidence of recognizing binomial probability (may be seen in part (a)) (M1)

e.g. ,

A1 N2

[2 marks]

( ) (0.05 (0.9524015

)15 )225 X ∼ B(240,0.05)

P(X = 15) = 0.0733

[3 marks]33c. Find the probability that there are at least 10 brown eggs in the box.

Markscheme (A1)

evidence of valid approach (M1)

e.g. using complement, summing probabilities

A1 N3

[3 marks]

P(X ≤ 9) = 0.236

P(X ≥ 10) = 0.764

[2 marks]34a.

Let , and .

Find .

Markschemeattempt to form composite (M1)

e.g.

A1 N2

[2 marks]

f(x) = 3x g(x) = 2x− 5 h(x) = (f ∘ g)(x)

h(x)

f(2x− 5)

h(x) = 6x− 15

[3 marks]34b. Find .

Markschemeinterchanging x and y (M1)

evidence of correct manipulation (A1)

e.g. ,

A1 N3

[3 marks]

(x)h−1

y+ 15 = 6x = y−x

652

(x) =h−1 x+156

[2 marks]35a.

Let the random variable X be normally distributed with mean 25, as shown in the following diagram.

The shaded region between 25 and 27 represents of the distribution.

Find .

Markschemesymmetry of normal curve (M1)

e.g.

A1 N2

[2 marks]

30%

P(X > 27)

P(X < 25) = 0.5

P(X > 27) = 0.2

[5 marks]35b. Find the standard deviation of X .

MarkschemeMETHOD 1

finding standardized value (A1)

e.g.

evidence of complement (M1)

e.g. , , 0.8

finding z-score (A1)

e.g.

attempt to set up equation involving the standardized value M1

e.g. ,

A1 N3

METHOD 2

set up using normal CDF function and probability (M1)

e.g. ,

correct equation A2

e.g. ,

attempt to solve the equation using GDC (M1)

e.g. solver, graph, trial and error (more than two trials must be shown)

A1 N3

[5 marks]

27−25σ

1 − p P(X < 27)

z= 0.84…

0.84 = 27−25σ

0.84 = X−μσ

σ = 2.38

P(25 <X < 27) = 0.3 P(X < 27) = 0.8

P(25 <X < 27) = 0.3 P(X > 27) = 0.2

σ = 2.38

[2 marks]36a.

A standard die is rolled 36 times. The results are shown in the following table.

Write down the standard deviation.

Markscheme A2 N2

[2 marks]

σ = 1.61

[1 mark]36b. Write down the median score.

Markschememedian A1 N1

[1 mark]

= 4.5

[3 marks]36c. Find the interquartile range.

Markscheme , (may be seen in a box plot) (A1)(A1)

(accept any notation that suggests the interval 3 to 5) A1 N3

[3 marks]

= 3Q1 = 5Q3

IQR = 2

[1 mark]37a.

The n term of an arithmetic sequence is given by .

Write down the common difference.

Markscheme A1 N1

[1 mark]

th = 5 + 2nun

d = 2

37b. [5 marks](i) Given that the n term of this sequence is 115, find the value of n .

(ii) For this value of n , find the sum of the sequence.

Markscheme(i) (A1)

A1 N2

(ii) (may be seen in above) (A1)

correct substitution into formula for sum of arithmetic series (A1)

e.g. , ,

(accept ) A1 N3

[5 marks]

th

5 + 2n = 115

n = 55

= 7u1

= (7 + 115)S55552

= (2(7) + 54(2))S55552

(5 + 2k)∑k=1

55

= 3355S55 3360

[3 marks]38a.

Let .

There are two points of inflexion on the graph of f . Write down the x-coordinates of these points.

Markschemevalid approach R1

e.g. , the max and min of gives the points of inflexion on f

(accept ( ) and ( ) A1A1 N1N1

[3 marks]

(x) = −24 + 9 + 3x+ 1f ′ x3 x2

(x) = 0f ′′ f ′

−0.114, 0.364 −0.114, 0.811 0.364, 2.13)

[2 marks]38b. Let . Explain why the graph of g has no points of inflexion.g(x) = (x)f ′′

MarkschemeMETHOD 1

graph of g is a quadratic function R1 N1

a quadratic function does not have any points of inflexion R1 N1

METHOD 2

graph of g is concave down over entire domain R1 N1

therefore no change in concavity R1 N1

METHOD 3

R1 N1

therefore no points of inflexion as R1 N1

[2 marks]

(x) = −144g′′

(x) ≠ 0g′′

[4 marks]39a.

The following frequency distribution of marks has mean 4.5.

Find the value of x.

Markscheme , (seen anywhere) A1

evidence of substituting into mean (M1)

correct equation A1

e.g. ,

A1 N2

[4 marks]

∑fx = 1(2) + 2(4) + … + 7(4) ∑fx = 146 + 5x

∑ fx

∑ f

= 4.5146+5x34+x

146 + 5x = 4.5(34 +x)

x = 14

[2 marks]39b. Write down the standard deviation.

Markscheme A2 N2

[2 marks]

σ = 1.54

[2 marks]40a.

Let . Part of the graph of f is shown below.

The y-intercept is at (0, 13) .

Show that .

Markschemesubstituting (0, 13) into function M1

e.g.

A1

AG N0

[2 marks]

f(x) = A + 3ekx

A = 10

13 = A + 3e0

13 = A+ 3

A = 10

[3 marks]40b. Given that (correct to 3 significant figures), find the value of k.

Markschemesubstituting into A1

e.g. ,

evidence of solving equation (M1)

e.g. sketch, using

(accept ) A1 N2

[3 marks]

f(15) = 3.49

f(15) = 3.49

3.49 = 10 + 3e15k 0.049 = e15k

ln

k = −0.201 ln 0.04915

40c. [5 marks](i) Using your value of k , find .

(ii) Hence, explain why f is a decreasing function.

(iii) Write down the equation of the horizontal asymptote of the graph f .

(x)f ′

Markscheme(i)

A1A1A1 N3

Note: Award A1 for , A1 for , A1 for the derivative of 3 is zero.

(ii) valid reason with reference to derivative R1 N1

e.g. , derivative always negative

(iii) A1 N1

[5 marks]

f(x) = 10 + 3e−0.201x

f(x) = 10 × −0.201e−0.201x (= −2.01 )e−0.201x

10e−0.201x × − 0.201

(x) < 0f ′

y = 3

40d. [6 marks]Let .

Find the area enclosed by the graphs of f and g .

Markschemefinding limits , (seen anywhere) A1A1

evidence of integrating and subtracting functions (M1)

correct expression A1

e.g. ,

area A2 N4

[6 marks]

g(x) = − + 12x− 24x2

3.8953… 8.6940…

g(x) − f(x)dx∫ 8.693.90 [(− + 12x− 24) − (10 + 3)]dx∫ 8.69

3.90 x2 e−0.201x

= 19.5

[3 marks]41a.

Consider , for and , for . The graph of f is given below.

On the diagram above, sketch the graph of g.

f(x) = 2 −x2 −2 ≤ x ≤ 2 g(x) = sin ex −2 ≤ x ≤ 2

Printed for Atherton High School

© International Baccalaureate Organization 2016 International Baccalaureate® - Baccalauréat International® - Bachillerato Internacional®

Markscheme

A1A1A1 N3

[3 marks]

[2 marks]41b. Solve .

Markscheme , (accept , if working in degrees) A1A1 N2

[2 marks]

f(x) = g(x)

x = −1.32 x = 1.68 x = −1.41 x = 1.39

[2 marks]41c. Write down the set of values of x such that .

Markscheme (accept if working in degrees) A2 N2

[2 marks]

f(x) > g(x)

−1.32 < x < 1.68 −1.41 < x < 1.39