paper 2 most likely questions May 2018 [327 marks]

1a. [2 marks]

Let , for .

Find the -intercept of the graph of .

Markschemevalid approach (M1)

eg

0.816496

(exact), 0.816 A1 N2

[2 marks]

f(x) = 6x2−4ex 0 ⩽ x ⩽ 7

x f

f(x) = 0, ± 0.816

x = √ 23

1b. [2 marks]The graph of has a maximum at the point A. Write down the coordinates of A.

Markscheme

A1A1 N2

[2 marks]

f

(2.29099, 2.78124)

A(2.29, 2.78)

1c. [3 marks]On the following grid, sketch the graph of .

Markscheme

A1A1A1 N3

Notes: Award A1 for correct domain and endpoints at and in circles,

A1 for maximum in square,

A1 for approximately correct shape that passes through their -intercept in circle and has changed from concave down to concave upbetween 2.29 and 7.

[3 marks]

f

x = 0 x = 7

x

R

2a. [2 marks]

Let , for . The graph of passes through the point , where .

Find the value of .

Markschemevalid approach (M1)

eg , intersection with

2.32143

(exact), 2.32 A1 N2

[2 marks]

f(x) = 6 − ln(x2 + 2) x ∈ R f (p, 4) p > 0

p

f(p) = 4 y = 4, ± 2.32

p = √e2 − 2

2b. [3 marks]The following diagram shows part of the graph of .

The region enclosed by the graph of , the -axis and the lines and is rotated 360° about the -axis. Find the volume of thesolid formed.

f

f x x = −p x = p x

Markschemeattempt to substitute either their limits or the function into volume formula (must involve , accept reversed limits and absence of and/or , but do not accept any other errors) (M1)

eg

331.989

A2 N3

[3 marks]

f 2 π

dx

∫ 2.32−2.32 f 2, π∫ (6 − ln(x2 + 2))2dx, 105.675

volume = 332

3. [6 marks]In the expansion of , the coefficient of the term in is 11880. Find the value of .

Markschemevalid approach for expansion (must have correct substitution for parameters, but accept an incorrect value for ) (M1)

eg

recognizing need to find term in in binomial expansion (A1)

eg

correct term or coefficient in binomial expansion (may be seen in equation) (A1)

eg

setting up equation in with their coefficient/term (do not accept other powers of ) (M1)

eg

correct equation (A1)

eg

A1 N3

[6 marks]

ax3(2 + ax)11 x5 a

r

( 11r

) (2)11−raxr, ( 113

) (2)8(ax)3, 211 + ( 111

) (2)10(ax)1 + ( 112

) (2)9(ax) + …

x2

r = 2, (ax)2

( 112

) (ax)2(2)9, 55(a2x2)(512), 28160a2

x5 x

ax3( 112

) (ax)2(2)9 = 11880x5

28160a3 = 11880

a = 34

4. [7 marks]The heights of adult males in a country are normally distributed with a mean of 180 cm and a standard deviation of . 17% ofthese men are shorter than 168 cm. 80% of them have heights between and 192 cm.

Find the value of .

σ cm(192 − h) cm

h

Markschemefinding the -value for 0.17 (A1)

eg

setting up equation to find , (M1)

eg

(A1)

EITHER (Properties of the Normal curve)

correct value (seen anywhere) (A1)

eg

correct working (A1)

eg

correct equation in

eg (A1)

35.6536

A1 N3

OR (Trial and error using different values of h)

two correct probabilities whose 2 sf will round up and down, respectively, to 0.8 A2

eg

A2

[7 marks]

z

z = −0.95416

σ

z = , − 0.954 =168−180σ

−12σ

σ = 12.5765

P(X < 192) = 0.83, P(X > 192) = 0.17

P(X < 192 − h) = 0.83 − 0.8, P(X < 192 − h) = 1 − 0.8 − 0.17,

P(X > 192 − h) = 0.8 + 0.17

h

= −1.88079, 192 − h = 156.346(192−h)−180

12.576

h = 35.7

P(192 − 35.6 < X < 192) = 0.799706, P(157 < X < 192) = 0.796284,

P(192 − 36 < X < 192) = 0.801824

h = 35.7

5a. [2 marks]

Consider the graph of , for .

Find the -intercept.

f(x) = + 3ex

5x−10x ≠ 2

y

Markschemevalid approach (M1)

eg ,

-intercept is 2.9 A1 N2

[2 marks]

f(0)

y

5b. [2 marks]Find the equation of the vertical asymptote.

Markschemevalid approach involving equation or inequality (M1)

eg

(must be an equation) A1 N2

[2 marks]

5x − 10 = 0, 2, x ≠ 2

x = 2

5c. [2 marks]Find the minimum value of for .f(x) x > 2

Markscheme7.01710

A2 N2

Note: If candidate gives the minimum point as their final answer, award A1 for .

[2 marks]

min value = 7.02

(3, 7.02)

6a. [2 marks]

In a large university the probability that a student is left handed is 0.08. A sample of 150 students is randomly selected from theuniversity. Let be the expected number of left-handed students in this sample.

Find .

Markschemeevidence of binomial distribution (may be seen in part (b)) (M1)

eg

A1 N2

[2 marks]

k

k

np, 150 × 0.08

k = 12

6b. [2 marks]Hence, find the probability that exactly students are left handed;

Markscheme (A1)

0.119231

probability A1 N2

[2 marks]

k

P (X = 12) = ( 15012

) (0.08)12(0.92)138

= 0.119

6c. [2 marks]Hence, find the probability that fewer than students are left handed.

Markschemerecognition that (M1)

0.456800

A1 N2

[2 marks]

k

X ⩽ 11

P(X < 12) = 0.457

7. [7 marks]The following diagram shows the chord [AB] in a circle of radius 8 cm, where .

Find the area of the shaded segment.

AB = 12 cm

Markschemeattempt to find the central angle or half central angle (M1)

eg , cosine rule, right triangle

correct working (A1)

eg

correct angle (seen anywhere)

eg (A1)

correct sector area

eg (A1)

area of triangle (seen anywhere) (A1)

eg

appropriate approach (seen anywhere) (M1)

eg , their sector-their triangle

22.5269

area of shaded region A1 N4

Note: Award M0A0A0A0A1 then M1A0 (if appropriate) for correct triangle area without any attempt to find an angle in triangle OAB.

[7 marks]

cosθ = , sin−1( ) , 0.722734, 41.4096∘, − sin−1( )82+82−122

2∙8∙868

π

268

AOB

1.69612, 97.1807∘, 2 × sin−1( )68

(8)(8)(1.70), (64π), 54.275912

97.1807360

(8)(8)sin 1.70, (8)(12)sin 0.722, × √64 − 36 × 12, 31.749012

12

12

Atriangle − Asector

= 22.5 (cm2)

8. [7 marks]Let . Find the term in in the expansion of the derivative, .f(x) = (x2 + 3)7 x5 f ′(x)

MarkschemeMETHOD 1

derivative of A2

recognizing need to find term in (seen anywhere) R1

eg

valid approach to find the terms in (M1)

eg , Pascal’s triangle to 6th row

identifying correct term (may be indicated in expansion) (A1)

eg

correct working (may be seen in expansion) (A1)

eg

A1 N3

METHOD 2

recognition of need to find in (seen anywhere) R1

valid approach to find the terms in (M1)

eg , Pascal’s triangle to 7th row

identifying correct term (may be indicated in expansion) (A1)

eg 6th term,

correct working (may be seen in expansion) (A1)

eg

correct term (A1)

differentiating their term in (M1)

eg

A1 N3

[7 marks]

f(x)

7(x2 + 3)6(x2)

x4 (x2 + 3)6

14x (term in x4)

(x2 + 3)6

( 6r

) (x2)6−r(3)r, (x2)6(3)0 + (x2)5(3)1 + …

5th term, r = 2, ( 64

) , (x2)2(3)4

( 64

) (x2)2(3)4, 15 × 34, 14x × 15 × 81(x2)2

17010x5

x6 (x2 + 3)7

(x2 + 3)7

( 7r

) (x2)7−r(3)r, (x2)7(3)0 + (x2)6(3)1 + …

r = 3, ( 73

) , (x2)3(3)4

( 74

) (x2)3(3)4, 35 × 34

2835x6

x6

(2835x6)′, (6)(2835x5)

17010x5

( )

9a. [1 mark]

A particle P moves along a straight line. Its velocity after seconds is given by , for . The following

diagram shows the graph of .

Write down the first value of at which P changes direction.

Markscheme A1 N1

[1 mark]

vP ms−1 t vP = √t sin( t)π

20 ⩽ t ⩽ 8

vP

t

t = 2

9b. [2 marks]Find the total distance travelled by P, for .

Markschemesubstitution of limits or function into formula or correct sum (A1)

eg

9.64782

distance A1 N2

[2 marks]

0 ⩽ t ⩽ 8

∫ 80 |v|dt, ∫ ∣∣vQ∣∣dt, ∫ 2

0 vdt − ∫ 42 vdt + ∫ 6

4 vdt − ∫ 86 vdt

= 9.65 (metres)

⩽ ⩽

9c. [4 marks]A second particle Q also moves along a straight line. Its velocity, after seconds is given by for . After seconds Q has travelled the same total distance as P.

Find .

Markschemecorrect approach (A1)

eg

correct integration (A1)

eg

equating their expression to the distance travelled by their P (M1)

eg

5.93855

5.94 (seconds) A1 N3

[4 marks]

vQ ms−1 t vQ = √t 0 ⩽ t ⩽ 8k

k

s = ∫ √t, ∫ k

0 √tdt, ∫ k

0 ∣∣vQ∣∣dt

∫ √t = t + c, [ x ]k

0, k2

3

32

23

32

23

32

k = 9.65, ∫ k

0 √tdt = 9.6523

32

10. [2 marks]

The following table shows a probability distribution for the random variable , where .

A bag contains white and blue marbles, with at least three of each colour. Three marbles are drawn from the bag, without replacement.The number of blue marbles drawn is given by the random variable .

A game is played in which three marbles are drawn from the bag of ten marbles, without replacement. A player wins a prize if threewhite marbles are drawn.

Jill plays the game nine times. Find the probability that she wins exactly two prizes.

Markschemevalid approach (M1)

eg

0.279081

0.279 A1 N2

[2 marks]

X E(X) = 1.2

X

B(n, p), (n

r)prqn−r, (0.167)2(0.833)7, ( 9

2)

11a. [1 mark]

The following diagram shows a circle, centre O and radius mm. The circle is divided into five equal sectors.

One sector is OAB, and .

Write down the exact value of in radians.

Markscheme A1 N1

[1 mark]

r

AOB = θ

θ

θ = 2π5

11b. [3 marks]

The area of sector AOB is .

Find the value of .

20π mm2

r

Markschemecorrect expression for area (A1)

eg

evidence of equating their expression to (M1)

eg

A1 N2

[3 marks]

A = r2( ) , 12

2π5

πr2

5

20π

r2( ) = 20π, r2 = 100, r = ±1012

2π5

r = 10

11c. [3 marks]Find AB.

MarkschemeMETHOD 1

evidence of choosing cosine rule (M1)

eg

correct substitution of their and into RHS (A1)

eg

11.7557

A1 N2

METHOD 2

evidence of choosing sine rule (M1)

eg

correct substitution of their and (A1)

eg

11.7557

A1 N2

[3 marks]

a2 = b2 + c2 − 2bc cosA

r θ

102 + 102 − 2 × 10 × 10 cos( )2π5

AB = 11.8 (mm)

=sinAa

sinB

b

r θ

=sin 2π

5

AB

sin( (π− ))12

2π

5

10

AB = 11.8 (mm)

12a. [3 marks]

Let and .

The graphs of and intersect at and , where .

Find the value of and of.

Markschemevalid attempt to find the intersection (M1)

eg

, sketch, one correct answer

A1A1 N3

[3 marks]

f(x) = xe−x g(x) = −3f(x) + 1

f g x = p x = q p < q

p

q

f = g

p = 0.357402, q = 2.15329

p = 0.357, q = 2.15

12b. [3 marks]Hence, find the area of the region enclosed by the graphs of and.

Markschemeattempt to set up an integral involving subtraction (in any order) (M1)

eg

0.537667

A2 N3

[3 marks]

f

g

∫ qp [f(x) − g(x)]dx, ∫ q

p f(x)dx−∫ qp g(x)dx

area = 0.538

13a. [3 marks]

The weights, , of newborn babies in Australia are normally distributed with a mean 3.41 kg and standard deviation 0.57 kg. Anewborn baby has a low birth weight if it weighs less than kg.

Given that 5.3% of newborn babies have a low birth weight, find .

Markschemevalid approach (M1)

eg ,

2.48863

A2 N3

[3 marks]

W

w

w

z = −1.61643

w = 2.49 (kg)

13b. [3 marks]A newborn baby has a low birth weight.

Find the probability that the baby weighs at least 2.15 kg.

Markschemecorrect value or expression (seen anywhere)

eg (A1)

evidence of conditional probability (M1)

eg

0.744631

0.745 A1 N2

[3 marks]

0.053 − P(X ⩽ 2.15), 0.039465

, P(2.15⩽X⩽w

P(X⩽w)0.039465

0.053

14a. [3 marks]

A population of rare birds, , can be modelled by the equation , where is the initial population, and is measured in

decades. After one decade, it is estimated that .

(i) Find the value of .

(ii) Interpret the meaning of the value of .

Markscheme(i) valid approach (M1)

eg

A1 N2

(ii) correct interpretation R1 N1

eg population is decreasing, growth rate is negative

[3 marks]

Pt Pt = P0ekt P0 t

= 0.9P1

P0

k

k

0.9 = ek(1)

k = −0.105360

k = ln0.9 (exact), − 0.105

14b. [5 marks]Find the least number of whole years for which .< 0.75Pt

P0

MarkschemeMETHOD 1

valid approach (accept an equality, but do not accept 0.74) (M1)

eg

valid approach to solve their inequality (M1)

eg logs, graph

A1

28 years A2 N2

METHOD 2

valid approach which gives both crossover values accurate to at least 2 sf A2

eg

(A1)

28 years A2 N2

[5 marks]

P < 0.75P0, P0ekt < 0.75P0, 0.75 = et ln0.9

t > 2.73045 (accept t = 2.73045) (2.73982 from − 0.105)

= 0.75241… , = 0.74452…P2.7

P0

P2.8

P0

t = 2.8

15a. [1 mark]

Consider the expansion of .

Write down the number of terms of this expansion.

Markscheme11 terms A1 N1

[1 mark]

(x2 + )102x

15b. [5 marks]Find the coefficient of .x8

Markschemevalid approach (M1)

eg

Pascal’s triangle to row

valid attempt to find value of which gives term in (M1)

eg

identifying required term (may be indicated in expansion) (A1)

eg

correct working (may be seen in expansion) (A1)

eg

3360 A1 N3

[5 marks]

( 10r

) (x2)10−r( )r

, a10b0 + ( 101

)a9b1( 102

)a8b2 + …2x

11th

r x8

(x2)10−r ( ) = x8, x2r( )10−r

= x81xr

2x

r = 6, 5th term, 7th term

( 106

) (x2)6( )4, 210 × 162

x

16. [7 marks]A particle moves in a straight line. Its velocity after seconds is given by

After seconds, the particle is 2 m from its initial position. Find the possible values of .

v ms−1 t

v = 6t − 6, for 0 ⩽ t ⩽ 2.

p p

Markschemecorrect approach (A1)

eg

correct integration (A1)

eg

recognizing that there are two possibilities (M1)

eg 2 correct answers,

two correct equations in A1A1

eg

0.42265, 1.57735

A1A1 N3

[7 marks]

s = ∫ v, ∫ p

0 6t − 6dt

∫ 6t − 6dt = 3t2 − 6t + C, [3t2 − 6t]p

0

s = ±2, c ± 2

p

3p2 − 6p = 2, 3p2 − 6p = −2

p = 0.423 or p = 1.58

17a. [2 marks]

Let and be independent events, with and , where .

Write down an expression for in terms of.

Markscheme (A1)

A1 N2

[2 marks]

C

D P(C) = 2k P(D) = 3k2 0 < k < 0.5

P(C ∩ D)k

P(C ∩ D) = 2k × 3k2

P(C ∩ D) = 6k3

17b. [3 marks]Find .P(C′|D)

MarkschemeMETHOD 1

finding their (seen anywhere) (A1)

eg

correct substitution into conditional probability formula (A1)

eg

A1 N2

METHOD 2

recognizing A1

finding their (only if first line seen) (A1)

eg

A1 N2

[3 marks]

Total [7 marks]

P(C′ ∩ D)

0.4 × 0.27,0.27 − 0.162,0.108

P(C′|D) = , P(C ′∩D)

0.27

(1−2k)(3k2)

3k2

P(C′|D) = 0.4

P(C′|D) = P(C′)

P(C′) = 1 − P(C)

1 − 2k, 1 − 0.6

P(C′|D) = 0.4

18. [7 marks]Let for .

Points and are on the curve of . The tangent to the curve of at is perpendicular to the tangent at . Find the coordinatesof .

f(x) = ln(4x)x 0 < x ≤ 5

P(0.25, 0) Q f f P Q

Q

Markschemerecognizing that the gradient of tangent is the derivative (M1)

eg

finding the gradient of at (A1)

eg

evidence of taking negative reciprocal of their gradient at (M1)

eg

equating derivatives M1

eg

finding the -coordinate of ,

A1 N3

attempt to substitute their into to find the -coordinate of (M1)

eg

A1 N2

[7 marks]

f ′

f P

f ′(0.25) = 16

P

, −−1m

1f′(0.25)

f ′(x) = , f ′ = − , = 16−116

1m

x( )−ln(4x)1x

x2

x Q x = 0.700750

x = 0.701

x f y Q

f(0.7)

y = 1.47083

y = 1.47

19a. [5 marks]

Let and .

Find

(i) ;

(ii) ;

(iii) .

Markscheme(i) correct substitution (A1)

eg

A1 N2

(ii) correct substitution into magnitude formula for or (A1)

eg , correct value for

A1 N2

(iii) A1 N1

[5 marks]

u = 6i + 3j + 6k v = 2i + 2j + k

u ∙ v

|u|

|v|

6 × 2 + 3 × 2 + 6 × 1

u ∙ v = 24

u v

√62 + 32 + 62, √22 + 22 + 12 |v|

|u| = 9

|v| = 3

19b. [2 marks]Find the angle between and .

Markschemecorrect substitution into angle formula (A1)

eg

A1 N2

[2 marks]

Total [7 marks]

u v

, 0.8249×3

0.475882, 27.26604∘

0.476, 27.3∘

20. [8 marks]Let and

. The graphs of and intersect at two distinct points.

Find the possible values of .

f(x) = kx2 + kx

g(x) = x − 0.8 f g

k

Markschemeattempt to set up equation (M1)

eg

rearranging their equation to equal zero M1

eg

evidence of discriminant (if seen explicitly, not just in quadratic formula) (M1)

eg

correct discriminant (A1)

eg

evidence of correct discriminant greater than zero R1

eg , correct answer

both correct values (A1)

eg

correct answer A2 N3

eg

[8 marks]

f = g, kx2 + kx = x − 0.8

kx2 + kx − x + 0.8 = 0, kx2 + x(k − 1) + 0.8 = 0

b2 − 4ac, Δ = (k − 1)2 − 4k × 0.8, D = 0

(k − 1)2 − 4k × 0.8, k2 − 5.2k + 1

k2 − 5.2k + 1 > 0, (k − 1)2 − 4k × 0.8 > 0

0.2, 5

k < 0.2, k ≠ 0, k > 5

[3 marks]21a.

The population of deer in an enclosed game reserve is modelled by the function

, where

is in months, and

corresponds to 1 January 2014.

Find the number of deer in the reserve on 1 May 2014.

Markscheme (A1)

correct substitution into formula (A1)

eg

969 (deer) (must be an integer) A1 N3

[3 marks]

P(t) = 210 sin(0.5t − 2.6) + 990

t

t = 1

t = 5

210 sin(0.5 × 5 − 2.6) + 990, P(5)

969.034982 …

[2 marks]21b. Find the rate of change of the deer population on 1 May 2014.

Markschemeevidence of considering derivative (M1)

eg

(deer per month) A1 N2

[2 marks]

P ′

104.475

104

[1 mark]21c. Interpret the answer to part (i) with reference to the deer population size on 1 May 2014.

Markscheme(the deer population size is) increasing A1 N1

[1 mark]

22. [8 marks]Ramiro and Lautaro are travelling from Buenos Aires to El Moro.

Ramiro travels in a vehicle whose velocity in

is given by

, where

is in seconds.

Lautaro travels in a vehicle whose displacement from Buenos Aires in metres is given by

.

When

, both vehicles are at the same point.

Find Ramiro’s displacement from Buenos Aires when

.

MarkschemeMETHOD 1

(seen anywhere) (A1)

recognizing need to integrate

(M1)

eg

correct expression A1A1

eg

Note: Award A1 for

, and A1 for

.

equate displacements to find C (R1)

eg

A1

attempt to find displacement (M1)

eg

A1 N5

METHOD 2

recognizing need to integrate

(M1)

eg

valid approach involving a definite integral (M1)

eg

ms−1

VR = 40 − t2

t

SL = 2t2 + 60

t = 0

t = 10

SL(0) = 60

VR

SR(t) ∫ VRdt

40t − t3 + C13

40t

− t313

40(0) − (0)3 + C = 60, SL(0) = SR(0)13

C = 60

SR(10), 40(10) − (10)3 + 6013

126.666

126 (exact), 127 (m)23

VR

SR(t) = ∫ VRdt

∫ ba

VRdt

correct expression with limits (A1)

eg

A2

(seen anywhere) (A1)

valid approach to find total displacement (M1)

eg

(exact),

(m) A1 N5

METHOD 3

(seen anywhere) (A1)

recognizing need to integrate

(M1)

eg

correct expression A1A1

eg

Note: Award A1 for

, and A1 for

.

correct expression for Ramiro displacement A1

eg

A1

valid approach to find total displacement (M1)

eg

(exact), 127 (m) A1 N5

[8 marks]

∫ 100 (40 − t2) dt, ∫ 10

0 VRdt, [40t − t3]10

0

13

66.6666

SL(0) = 60

60 + 66.666

126.666

126 23

127

SL(0) = 60

VR

SR(t) = ∫ VRdt

40t − t3 + C13

40t

− t313

SR(10) − SR(0), [40t − t3 + C]10

0

13

66.6666

60 + 66.6666

126 23

23a. [2 marks]

The following table shows the Diploma score and university entrance mark for seven IB Diploma students.

Find the correlation coefficient.

x y

Markschemeevidence of set up (M1)

eg correct value for (or for or , seen in (b))

A1 N2

[2 marks]

r a r

0.996010

r = 0.996 [0.996, 0.997]

23b. [2 marks]The relationship can be modelled by the regression line with equation .

Write down the value of and of .

Markscheme

A1A1 N2

[2 marks]

y = ax + b

a b

a = 3.15037, b = −15.4393

a = 3.15 [3.15, 3.16], b = −15.4 [−15.5, − 15.4]

23c. [2 marks]Rita scored a total of in her IB Diploma.

Use your regression line to estimate Rita’s university entrance mark.

Markschemesubstituting into their equation (M1)

eg

A1 N2

[2 marks]

Total [6 marks]

26

26

y = 3.15(26) − 15.4

66.4704

66.5 [66.4, 66.5]

24. [2 marks]

A particle starts from point and moves along a straight line. Its velocity, , after seconds is given by , for . The particle is at rest when .

The following diagram shows the graph of .

Find the distance travelled by the particle for .

Markschemecorrect substitution of function and/or limits into formula (A1)

(accept absence of d , but do not accept any errors)

eg

distance is A1 N2

[2 marks]

A v ms−1 t v(t) = e cos t − 112

0 ≤ t ≤ 4 t = π

2

v

0 ≤ t ≤ π

2

t

∫0 v, ∫ ∣∣e

cos t − 1∣∣dt, ∫ (e cos t − 1)π

212

12

0.613747

0.614 [0.613, 0.614] (m)

[3 marks]25a.

The following diagram shows triangle ABC.

Find AC.

Markschemeevidence of choosing cosine rule (M1)

eg

correct substitution into the right-hand side (A1)

eg

A1 N2

[3 marks]

AC2 = AB2 + BC2 − 2(AB)(BC) cos(ABC)

62 + 102 − 2(6)(10) cos 100∘

AC = 12.5234

AC = 12.5 (cm)

25b. [3 marks]Find.

Markschemeevidence of choosing a valid approach (M1)

eg sine rule, cosine rule

correct substitution (A1)

eg

A1 N2

[3 marks]

BCA

= , cos(BCA) =sin(BCA)

6sin 100∘

12.5(AC)2+102−62

2(AC)(10)

BCA = 28.1525

BCA = 28.2∘

26a. [4 marks]

The weights of fish in a lake are normally distributed with a mean of g and standard deviation . It is known that of the fishhave weights between g and g.

(i) Write down the probability that a fish weighs more than g.

(ii) Find the probability that a fish weighs less than g.

760 σ 78.87%705 815

760

815

MarkschemeNote: There may be slight differences in answers, depending on which values candidates carry through in subsequent parts. Acceptanswers that are consistent with their working.

(i) A1 N1

(ii) evidence of valid approach (M1)

recognising symmetry,

correct working (A1)

eg

A1 N2

[4 marks]

P(X > 760) = 0.5 (exact), [0.499, 0.500]

, 1 − P(W < 815), + 78.87%0.78872

21.132

0.5 + 0.39435, 1 − 0.10565,

0.89435 (exact), 0.894 [0.894, 0.895]

26b. [4 marks](i) Write down the standardized value for g.

(ii) Hence or otherwise, find.

Markscheme(i) A1 N1

(ii) evidence of appropriate approach (M1)

eg

correct substitution (A1)

eg

A1 N2

[4 marks]

815

σ

1.24999

z = 1.25 [1.24, 1.25]

σ = , x−μ

1.25815−760

σ

1.25 = , 815−760σ

815−7601.24999

44.0003

σ = 44.0 [44.0, 44.1] (g)

26c. [2 marks]A fishing contest takes place in the lake. Small fish, called tiddlers, are thrown back into the lake. The maximum weight of a tiddleris standard deviations below the mean.

Find the maximum weight of a tiddler.

Markschemecorrect working (A1)

eg

A1 N2

[2 marks]

1.5

760 − 1.5 × 44

693.999

694 [693, 694] (g)

26d. [2 marks]A fish is caught at random. Find the probability that it is a tiddler.

Markscheme

A2 N2

[2 marks]

0.0668056

P(X < 694) = 0.0668 [0.0668, 0.0669]

26e. [2 marks] of the fish in the lake are salmon. of the salmon are tiddlers. Given that a fish caught at random is a tiddler, find theprobability that it is a salmon.

Markschemerecognizing conditional probability (seen anywhere) (M1)

eg

appropriate approach involving conditional probability (M1)

eg ,

correct working

eg P (salmon and tiddler) (A1)

A1 N2

[4 marks]

Total [16 marks]

25% 10%

P(A|B), 0.0250.0668

P(S|T) = P(S and T)

P(T)

= 0.25 × 0.1, 0.25×0.10.0668

0.374220

0.374 [0.374, 0.375]

27a. [5 marks]

The first two terms of a geometric sequence are and .

(i) Find the common ratio.

(ii) Hence or otherwise, find .

un u1 = 4 u2 = 4.2

u5

Markscheme(i) valid approach (M1)

eg

A1 N2

(ii) attempt to substitute into formula, with their (M1)

eg

correct substitution (A1)

eg

A1 N2

[5 marks]

r = , u2

u1

44.2

r = 1.05 (exact)

r

4 × 1.05n, 4 × 1.05 × 1.05…

4 × 1.054, 4 × 1.05 × 1.05 × 1.05 × 1.05

u5 = 4.862025 (exact), 4.86 [4.86, 4.87]

27b. [5 marks]Another sequence is defined by , where , and , such that and .

(i) Find the value of .

(ii) Find the value of .

Markscheme(i) attempt to substitute (M1)

eg

A1 N2

(ii) correct substitution of into A1

eg

correct work (A1)

eg finding intersection point,

A1 N2

[5 marks]

vn vn = ank a, k ∈ R n ∈ Z+ v1 = 0.05 v2 = 0.25

a

k

n = 1

0.05 = a × 1k

a = 0.05

n = 2 v2

0.25 = a × 2k

k = log2( ) , 0.250.05

log5log2

2.32192

k = log25 (exact), 2.32 [2.32, 2.33]

27c. [5 marks]Find the smallest value of for which .

Markschemecorrect expression for (A1)

eg

EITHER

correct substitution into inequality (accept equation) (A1)

eg

valid approach to solve inequality (accept equation) (M1)

eg finding point of intersection,

(must be an integer) A1 N2

OR

table of values

when A1

when A1

(must be an integer) A1 N2

[4 marks]

Total [14 marks]

n vn > un

un

4 × 1.05n−1

0.05 × nk > 4 × 1.05n−1

n = 7.57994 (7.59508 from 2.32)

n = 8

n = 7, u7 = 5.3604, v7 = 4.5836

n = 8, u8 = 5.6284, v8 = 6.2496

n = 8

28a. [3 marks]

Adam is a beekeeper who collected data about monthly honey production in his bee hives. The data for six of his hives is shown in thefollowing table.

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

Write down the value of and of .

P = aN + b

a b

Markschemeevidence of setup (M1)

eg correct value for or

A1A1 N3

[3 marks]

a b

a = 6.96103, b = −454.805

a = 6.96, b = −455 (accept 6.96x − 455)

28b. [2 marks]Use this regression line to estimate the monthly honey production from a hive that has 270 bees.

Markschemesubstituting into their equation (M1)

eg

1424.67

A1 N2

[2 marks]

N = 270

6.96(270) − 455

P = 1420 (g)

28c. [1 mark]

Adam has 200 hives in total. He collects data on the monthly honey production of all the hives. This data is shown in the followingcumulative frequency graph.

Adam’s hives are labelled as low, regular or high production, as defined in the following table.

Write down the number of low production hives.

Markscheme40 (hives) A1 N1

[1 mark]

28d. [3 marks]

Adam knows that 128 of his hives have a regular production.

Find the value of ;

Markschemevalid approach (M1)

eg

168 hives have a production less than (A1)

A1 N3

[3 marks]

k

128 + 40

k

k = 1640

28e. [2 marks]Find the number of hives that have a high production.

Markschemevalid approach (M1)

eg

32 (hives) A1 N2

[2 marks]

200 − 168

28f. [3 marks]Adam decides to increase the number of bees in each low production hive. Research suggests that there is a probability of 0.75that a low production hive becomes a regular production hive. Calculate the probability that 30 low production hives become regularproduction hives.

Markschemerecognize binomial distribution (seen anywhere) (M1)

eg

correct values (A1)

eg (check FT) and and

0.144364

0.144 A1 N2

[3 marks]

X ∼ B(n, p), (n

r)pr(1 − p)n−r

n = 40 p = 0.75 r = 30, ( 4030

) 0.7530(1 − 0.75)10

29a. [2 marks]

Note: In this question, distance is in metres and time is in seconds.

A particle P moves in a straight line for five seconds. Its acceleration at time is given by , for .

Write down the values of when .

Markscheme A1A1 N2

[2 marks]

t a = 3t2 − 14t + 8 0 ⩽ t ⩽ 5

t a = 0

t = (exact), 0.667, t = 423

29b. [2 marks]Hence or otherwise, find all possible values of for which the velocity of P is decreasing.

Markschemerecognizing that is decreasing when is negative (M1)

eg , sketch of

correct interval A1 N2

eg

[2 marks]

t

v a

a < 0, 3t2 − 14t + 8 ⩽ 0 a

< t < 423

29c. [6 marks]

When , the velocity of P is .

Find an expression for the velocity of P at time .

Markschemevalid approach (do not accept a definite integral) (M1)

eg

correct integration (accept missing ) (A1)(A1)(A1)

substituting , (must have ) (M1)

eg

A1 N6

[6 marks]

t = 0 3 ms−1

t

v ∫ a

c

t3 − 7t2 + 8t + c

t = 0, v = 3 c

3 = 03 − 7(02) + 8(0) + c, c = 3

v = t3 − 7t2 + 8t + 3

29d. [4 marks]Find the total distance travelled by P when its velocity is increasing.

Markschemerecognizing that increases outside the interval found in part (b) (M1)

eg , diagram

one correct substitution into distance formula (A1)

eg

one correct pair (A1)

eg 3.13580 and 11.0833, 20.9906 and 35.2097

14.2191 A1 N2

[4 marks]

v

0 < t < , 4 < t < 523

∫0 |v| , ∫ 54 |v|, ∫ 4 |v|, ∫ 5

0 |v|23

23

d = 14.2 (m)

30a. [1 mark]

Let and , for .

The graph of can be obtained from the graph of by two transformations:

Write down the value of ;

f(x) = lnx g(x) = 3 + ln( )x

2x > 0

g f

a horizontal stretch of scale factor q followed by

a translation of (h

k) .

q

Markscheme A1 N1

Note: Accept , , and , 2.31 as candidate may have rewritten as equal to .

[1 mark]

q = 2

q = 1 h = 0 k = 3 − ln(2) g(x) 3 + ln(x) − ln(2)

30b. [1 mark]Write down the value of ;

Markscheme A1 N1

Note: Accept , , and , 2.31 as candidate may have rewritten as equal to .

[1 mark]

h

h = 0

q = 1 h = 0 k = 3 − ln(2) g(x) 3 + ln(x) − ln(2)

30c. [1 mark]Write down the value of .

Markscheme A1 N1

Note: Accept , , and , 2.31 as candidate may have rewritten as equal to .

[1 mark]

k

k = 3

q = 1 h = 0 k = 3 − ln(2) g(x) 3 + ln(x) − ln(2)

30d. [2 marks]

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

The graph of intersects the graph of at two points. These points have coordinates 0.111 and 3.31 correct to three significantfigures.

Find .

Markscheme2.72409

2.72 A2 N2

[2 marks]

h(x) = g(x) × cos(0.1x) 0 < x < 4 h y = x

h h−1 x

∫ 3.310.111 (h(x) − x)dx

30e. [3 marks]Hence, find the area of the region enclosed by the graphs of and .

Markschemerecognizing area between and equals 2.72 (M1)

eg

recognizing graphs of and are reflections of each other in (M1)

eg area between and equals between and

5.44819

5.45 A1 N3

[??? marks]

h h−1

y = x h

h h−1 y = x

y = x h y = x h−1

2 × 2.72∫ 3.310.111 (x − h−1(x))dx = 2.72

30f. [7 marks]Let be the vertical distance from a point on the graph of to the line . There is a point on the graph of where isa maximum.

Find the coordinates of P, where .

Markschemevalid attempt to find (M1)

eg difference in -coordinates,

correct expression for (A1)

eg

valid approach to find when is a maximum (M1)

eg max on sketch of , attempt to solve

0.973679

A2 N4

substituting their value into (M1)

2.26938

A1 N2

[7 marks]

d h y = x P(a, b) h d

0.111 < a < 3.31

d

y d = h(x) − x

d

(ln x + 3) (cos0.1x) − x12

d

d d′ = 0

x = 0.974

x h(x)

y = 2.27

31a. [2 marks]

A random variable is normally distributed with mean, . In the following diagram, the shaded region between 9 and represents 30%of the distribution.

Find .

Markschemevalid approach (M1)

eg

(exact) A1 N2

[2 marks]

X μ μ

P(X < 9)

P(X < μ) = 0.5, 0.5 − 0.3

P(X < 9) = 0.2

31b. [3 marks]

The standard deviation of is 2.1.

Find the value of .

X

μ

Markscheme (may be seen in equation) (A1)

valid attempt to set up an equation with their (M1)

eg

10.7674

A1 N3

[3 marks]

z = −0.841621

z

−0.842 = , − 0.842 = , z =μ−X

σ

X−μ

σ

9−μ

2.1

μ = 10.8

31c. [5 marks]

The random variable is normally distributed with mean and standard deviation 3.5. The events and are independent,and .

Find .

Markscheme (seen anywhere) (A1)

valid approach (M1)

eg

correct equation (A1)

eg

A1

A1 N3

[5 marks]

Y λ X > 9 Y > 9P ((X > 9) ∩ (Y > 9)) = 0.4

λ

P(X > 9) = 0.8

P(A) × P(B)

0.8 × P(Y > 9) = 0.4

P(Y > 9) = 0.5

λ = 9

31d. [5 marks]Given that , find .

Markschemefinding (seen anywhere) (A2)

recognizing conditional probability (M1)

eg

correct working (A1)

eg

0.746901

0.747 A1 N3

[5 marks]

Y > 9 P(Y < 13)

P(9 < Y < 13) = 0.373450

P(A|B), P(Y < 13|Y > 9)

0.3730.5

32a. [6 marks]

The following diagram shows the graph of , for .

The graph of has a minimum point at and a maximum point at .

(i) Find the value of .

(ii) Show that .

(iii) Find the value of .

f(x) = asin bx + c 0 ⩽ x ⩽ 12

f (3, 5) (9, 17)

c

b = π

6

a

Markscheme(i) valid approach (M1)

eg

A1 N2

(ii) valid approach (M1)

eg period is 12, per

A1

AG N0

(iii) METHOD 1

valid approach (M1)

eg

, substitution of points

A1 N2

METHOD 2

valid approach (M1)

eg

, amplitude is 6

A1 N2

[6 marks]

5+172

c = 11

= , 9 − 32πb

b = 2π12

b = π

6

5 = asin( × 3) + 11π

6

a = −6

17−52

a = −6

32b. [3 marks]

The graph of is obtained from the graph of by a translation of . The maximum point on the graph of has coordinates

.

(i) Write down the value of .

(ii) Find .

Markscheme(i)

A1 N1

(ii)

A2 N2

[3 marks]

g f (k

0) g

(11.5, 17)

k

g(x)

k = 2.5

g(x) = −6 sin( (x − 2.5)) + 11π

6

32c. [6 marks]

The graph of changes from concave-up to concave-down when .

(i) Find .

(ii) Hence or otherwise, find the maximum positive rate of change of .

g x = w

w

g

Markscheme(i) METHOD 1 Using

recognizing that a point of inflexion is required M1

egsketch, recognizing change in concavity

evidence of valid approach (M1)

eg

, sketch, coordinates of max/min on

(exact) A1 N2

METHOD 2 Using

recognizing that a point of inflexion is required M1

eg sketch, recognizing change in concavity

evidence of valid approach involving translation (M1)

eg

, sketch,

(exact) A1 N2

(ii) valid approach involving the derivative of or (seen anywhere) (M1)

eg

, max on derivative, sketch of derivative

attempt to find max value on derivative M1

eg

, dot on max of sketch

3.14159

max rate of change (exact), 3.14 A1 N2

[6 marks]

g

g′′(x) = 0 g′

w = 8.5

f

x = w − k 6 + 2.5

w = 8.5

g f

g′(w), − πcos( x)π

6

−πcos( (8.5 − 2.5)), f ′(6)π

6

= π

33a. [2 marks]

A particle P starts from a point A and moves along a horizontal straight line. Its velocity after seconds is given by

The following diagram shows the graph of .

Find the initial velocity of .

Markschemevalid attempt to substitute into the correct function (M1)

eg

2 A1 N2

[2 marks]

v cms−1 t

v(t) = { −2t + 2, for 0 ⩽ t ⩽ 1

3√t + − 7, for 1 ⩽ t ⩽ 124t2

v

P

t = 0

−2(0) + 2

33b. [2 marks]

P is at rest when and .

Find the value of .

t = 1 t = p

p

Markschemerecognizing when P is at rest (M1)

5.21834

A1 N2

[2 marks]

v = 0

p = 5.22 (seconds)

33c. [4 marks]

When , the acceleration of P is zero.

(i) Find the value of .

(ii) Hence, find the speed of P when .

Markscheme(i) recognizing that (M1)

eg

, minimum on graph

1.95343

A1 N2

(ii) valid approach to find their minimum (M1)

eg

, reference to min on graph

1.75879

speed A1 N2

[4 marks]

t = q

q

t = q

a = v′

v′ = 0

q = 1.95

v(q), − 1.75879

= 1.76 (cms−1)

33d. [6 marks](i) Find the total distance travelled by P between and .

(ii) Hence or otherwise, find the displacement of P from A when .

Markscheme(i) substitution of correct into distance formula, (A1)

eg

4.45368

distance A1 N2

(ii) displacement from to (seen anywhere) (A1)

eg

displacement from to (A1)

eg

valid approach to find displacement for M1

eg

displacement A1 N2

[6 marks]

t = 1 t = p

t = p

v(t)

∫ p

1∣∣3√t + − 7∣∣dt, ∣∣∫ 3√t + − 7dt∣∣

4t2

4t2

= 4.45 (cm)

t = 1 t = p

−4.45368, ∫ p

1 (3√t + − 7) dt4t2

t = 0 t = 1

∫ 10 (−2t + 2)dt, 0.5 × 1 × 2, 1

0 ⩽ t ⩽ p

∫ 10 (−2t + 2)dt + ∫ p

1 (3√t + − 7) dt, ∫ 10 (−2t + 2)dt − 4.454

t2

−3.45368

= −3.45 (cm)

34a. [2 marks]

The points A and B lie on a line

, and have position vectors and respectively. Let O be the origin. This is shown on the following diagram.

Find .

Markschemevalid approach (addition or subtraction) (M1)

eg

A1 N2

[2 marks]

L⎛⎜⎝

−3−22

⎞⎟⎠⎛⎜⎝

64

−1

⎞⎟⎠

−−→AB

AO + OB, B − A

−−→AB =

⎛⎜⎝96

−3

⎞⎟⎠

34b.

The point C also lies on

, such that .

Show that .

L−−→AC = 2

−−→CB

−−→OC =

⎛⎜⎝320

⎞⎟⎠

MarkschemeMETHOD 1

valid approach using (M1)

eg

correct working A1

eg

all three equations A1

eg ,

AG N0

METHOD 2

valid approach (M1)

eg

correct working A1

eg

correct substitution of and

A1

eg

AG N0

METHOD 3

valid approach (M1)

eg , diagram,

correct working A1

eg

correct working involving A1

eg

AG N0

[3 marks]

−−→OC =

⎛⎜⎝x

y

z

⎞⎟⎠−−→AC =

⎛⎜⎝x + 3y + 2z − 2

⎞⎟⎠ , −−→CB =

⎛⎜⎝6 − x

4 − y

−1 − z

⎞⎟⎠

⎛⎜⎝x + 3y + 2z − 2

⎞⎟⎠ =⎛⎜⎝

12 − 2x

8 − 2y

−2 − 2z

⎞⎟⎠

x + 3 = 12 − 2x, y + 2 = 8 − 2y, z − 2 = −2 − 2z

−−→OC =

⎛⎜⎝320

⎞⎟⎠

−−→OC −

−−→OA = 2 (−−→

OB −−−→OC)

3−−→OC = 2

−−→OB +

−−→OA

−−→OB

−−→OA

3−−→OC = 2

⎛⎜⎝64

−1

⎞⎟⎠ +⎛⎜⎝

−3−22

⎞⎟⎠ , 3−−→OC =

⎛⎜⎝960

⎞⎟⎠−−→OC =

⎛⎜⎝320

⎞⎟⎠

−−→AC =

−−→AB2

3

−−→CB =

−−→AB1

3

−−→AC =

⎛⎜⎝64

−2

⎞⎟⎠ , −−→CB =

⎛⎜⎝32

−1

⎞⎟⎠−−→OC

−−→OC =

⎛⎜⎝−3−22

⎞⎟⎠ +⎛⎜⎝

64

−2

⎞⎟⎠ , ⎛⎜⎝

64

−1

⎞⎟⎠ −⎛⎜⎝

32

−1

⎞⎟⎠−−→OC =

⎛⎜⎝320

⎞⎟⎠

34c. [5 marks]

Let

be the angle between and .

Find .

Markschemefinding scalar product and magnitudes (A1)(A1)(A1)

scalar product

magnitudes

substitution into formula M1

eg

A1 N4

[5 marks]

θ−−→AB

−−→OC

θ

= (9 × 3) + (6 × 2) + (−3 × 0) (= 39)

√81 + 36 + 9 (= 11.22), √9 + 4 (= 3.605)

cosθ =(9×3)+12

√126×√13

θ = 0.270549 (accept 15.50135∘)

θ = 0.271 (accept 15.5∘)

34d. [6 marks]

Let D be a point such that

, where . Let E be a point on such that is a right angle. This is shown on the following diagram.

(i) Show that .

(ii) The distance from D to line is less than 3 units. Find the possible values of .

Markscheme(i) attempt to use a trig ratio M1

eg

attempt to express in terms of M1

eg

correct working A1

eg

AG N0

(ii) valid approach involving the segment DE (M1)

eg

recognizing

correct working (accept equation) (A1)

eg

A1 N2

[6 marks]

−−→OD = k

−−→OC k > 1 L

CED

∣∣∣−−→DE∣∣∣ = (k − 1)∣

∣∣−−→OC∣

∣∣sin θ

L k

sin θ = , ∣∣∣−−→CE∣

∣∣ = ∣∣∣−−→CD∣

∣∣cosθDECD

−−→CD

−−→OC

−−→OC +

−−→CD =

−−→OD, OC + CD = OD

∣∣∣k

−−→OC −

−−→OC∣

∣∣sin θ

∣∣∣−−→DE∣∣∣ = (k − 1)∣

∣∣−−→OC∣

∣∣sin θ

∣∣∣−−→DE∣∣∣ < 3, DE = 3

(k − 1)(√13)sin 0.271 < 3, k − 1 = 3.11324

1 < k < 4.11 (accept k < 4.11 but not k = 4.11)

35a. [2 marks]

A factory has two machines, A and B. The number of breakdowns of each machine is independent from day to day.

Let be the number of breakdowns of Machine A on any given day. The probability distribution for can be modelled by the followingtable.

Find .

Markschemeevidence of summing to 1 (M1)

eg

A1 N2

[2 marks]

A A

k

0.55 + 0.3 + 0.1 + k = 1

k = 0.05 (exact)

35b. [3 marks](i) A day is chosen at random. Write down the probability that Machine A has no breakdowns.

(ii) Five days are chosen at random. Find the probability that Machine A has no breakdowns on exactly four of these days.

Markscheme(i) 0.55 A1 N1

(ii) recognizing binomial probability (M1)

eg

A1 N2

[3 marks]

X : B(n, p), ( 54

) , (0.55)4(1 − 0.55), (n

r)prqn−r

P(X = 4) = 0.205889

P(X = 4) = 0.206

35c. [2 marks]

Let be the number of breakdowns of Machine B on any given day. The probability distribution for can be modelled by the followingtable.

Find .

Markschemecorrect substitution into formula for (A1)

eg

A1 N2

[2 marks]

B B

E(B)

E(X)

0.2 + (2 × 0.08) + (3 × 0.02)

E(B) = 0.42 (exact)

35d. [8 marks]

On Tuesday, the factory uses both Machine A and Machine B. The variables and are independent.

(i) Find the probability that there are exactly two breakdowns on Tuesday.

(ii) Given that there are exactly two breakdowns on Tuesday, find the probability that both breakdowns are of Machine A.

Markscheme(i) valid attempt to find one possible way of having 2 breakdowns (M1)

eg and, tree diagram

one correct calculation for 1 way (seen anywhere) (A1)

eg

recognizing there are 3 ways of having 2 breakdowns (M1)

eg A twice or B twice or one breakdown each

correct working (A1)

eg

A1 N3

(ii) recognizing conditional probability (M1)

eg

correct working (A1)

eg

A1 N2

[8 marks]

A B

2A, 2B, 1A

1B

0.1 × 0.7, 0.55 × 0.08, 0.3 × 0.2

(0.1 × 0.7) + (0.55 × 0.08) + (0.3 × 0.2)

P(2 breakdowns) = 0.174 (exact)

P(A|B), P(2A|2breakdowns)

0.1×0.70.174

P(A = 2|two breakdowns) = 0.402298

P(A = 2|two breakdowns) = 0.402

36a. [5 marks]

A particle P moves along a straight line so that its velocity, , after seconds, is given by , for. The initial displacement of P from a fixed point O is 4 metres.

Find the displacement of P from O after 5 seconds.

vms−1 t v = cos3t − 2 sin t − 0.50 ⩽ t ⩽ 5

MarkschemeMETHOD 1

recognizing (M1)

recognizing displacement of P in first 5 seconds (seen anywhere) A1

(accept missing )

eg

valid approach to find total displacement (M1)

eg

0.284086

0.284 (m) A2 N3

METHOD 2

recognizing (M1)

correct integration A1

eg (do not penalize missing “ ”)

attempt to find (M1)

eg

attempt to substitute into their expression with (M1)

eg

0.284086

0.284 (m) A1 N3

[5 marks]

s = ∫ v

dt

∫ 50 vdt, − 3.71591

4 + (−3.7159), s = 4 + ∫ 50 v

s = ∫ v

sin 3t + 2 cost − + c13

t

2c

c

4 = sin(0) + 2 cos(0) − − + c, 4 = sin 3t + 2 cost − + c, 2 + c = 413

02

13

t

2

t = 5 c

s(5), sin(15) + 2 cos(5)5 − − + 213

52

36b. [2 marks]

The following sketch shows the graph of .

Find when P is first at rest.

Markschemerecognizing that at rest, (M1)

A1 N2

[2 marks]

v

v = 0

t = 0.179900

t = 0.180 (secs)

36c. [2 marks]Write down the number of times P changes direction.

Markschemerecognizing when change of direction occurs (M1)

eg crosses axis

2 (times) A1 N2

[2 marks]

v t

36d. [2 marks]Find the acceleration of P after 3 seconds.

Markschemeacceleration is (seen anywhere) (M1)

eg

0.743631

A1 N2

[2 marks]

v′

v′(3)

0.744 (ms−2)

36e. [3 marks]Find the maximum speed of P.

Markschemevalid approach involving max or min of (M1)

eg

, graph

one correct co-ordinate for min (A1)

eg

A1 N2

[3 marks]

v

v′ = 0, a = 0

1.14102, − 3.27876

3.28 (ms−1)

37a. [2 marks]

Let , for .

Write down the equation of the horizontal asymptote of the graph of .

Markscheme (correct equation only) A2 N2

[2 marks]

f(x) = + 21x−1

x > 1

f

y = 2

37b. [2 marks]Find .

Markschemevalid approach (M1)

eg

A1 N2

[2 marks]

f ′(x)

(x − 1)−1 + 2, f ′(x) =0(x−1)−1

(x−1)2

−(x − 1)−2, f ′(x) = −1

(x−1)2

37c. [2 marks]

Let , for . The graphs of and have the same horizontal asymptote.

Write down the value of .

g(x) = ae−x + b x ⩾ 1 f g

b

Markschemecorrect equation for the asymptote of

eg (A1)

A1 N2

[2 marks]

g

y = b

b = 2

37d. [4 marks]Given that , find the value of .

Markschemecorrect derivative of g (seen anywhere) (A2)

eg

correct equation (A1)

eg

7.38905

A1 N2

[4 marks]

g′(1) = −e a

g′(x) = −ae−x

−e = −ae−1

a = e2 (exact), 7.39

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© International Baccalaureate Organization 2018

International Baccalaureate® - Baccalauréat International® - Bachillerato Internacional®

37e. [4 marks]There is a value of , for, for which the graphs of and have the same gradient. Find this gradient.

Markschemeattempt to equate their derivatives (M1)

eg

valid attempt to solve their equation (M1)

eg correct value outside the domain of such as 0.522 or 4.51,

correct solution (may be seen in sketch) (A1)

eg

gradient is A1 N3

[4 marks]

x

1 < x < 4 f g

f ′(x) = g′(x), = −ae−x−1

(x−1)2

f

x = 2, (2, − 1)

−1