IB Questionbank

Updates to Questionbank

User interface language: English | Español

SL Paper 2

Let f(x) = ln x − 5x , for x > 0 .

Solve f '(x) = f "(x).

Markscheme

METHOD 1 (using GDC)

valid approach (M1)

0.558257

x = 0.558 A1 N2

Note: Do not award A1 if additional answers given.

METHOD 2 (analytical)

attempt to solve their equation f '(x) = f "(x) (do not accept $\frac{1}{x} - 5 = - \frac{1}{{{x^2}}}$ ) (M1)

eg $5{x^2} - x - 1 = 0,\,\,\frac{{1 \pm \sqrt {21} }}{{10}},\,\,\frac{1}{x} = \frac{{ - 1 \pm \sqrt {21} }}{2},\,\, - 0.358$

0.558257

x = 0.558 A1 N2

Note: Do not award A1 if additional answers given.

[2 marks]

Examiners report

[N/A]

Let $f(x) = - 0.5{x^4} + 3{x^2} + 2x$ . The following diagram shows part of the graph of $f$ .

M17/5/MATME/SP2/ENG/TZ2/08

There are $x$ -intercepts at $x = 0$ and at $x = p$ . There is a maximum at A where $x = a$ , and a point of inflexion at B where $x = b$ .

Find the value of $p$ .

[2]

Write down the coordinates of A.

[2]

b.i.

Write down the rate of change of $f$ at A.

[1]

b.ii.

Find the coordinates of B.

[4]

c.i.

Find the the rate of change of $f$ at B.

[3]

c.ii.

Let $R$ be the region enclosed by the graph of $f$ , the $x$ -axis, the line $x = b$ and the line $x = a$ . The region $R$ is rotated 360° about the $x$ -axis. Find the volume of the solid formed.

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

evidence of valid approach (M1)

eg $\,\,\,\,\,$ $f(x) = 0,{\text{ }}y = 0$

2.73205

$p = 2.73$ A1 N2

[2 marks]

1.87938, 8.11721

$(1.88,{\text{ }}8.12)$ A2 N2

[2 marks]

b.i.

rate of change is 0 (do not accept decimals) A1 N1

[1 marks]

b.ii.

METHOD 1 (using GDC)

valid approach M1

eg $\,\,\,\,\,$ $f’’ = 0$ , max/min on $f’,{\text{ }}x = - 1$

sketch of either $f^{'}$ or $f^{″}$ , with max/min or root (respectively) (A1)

$x = 1$ A1 N1

Substituting their $x$ value into $f$ (M1)

eg $\,\,\,\,\,$ $f(1)$

$y = 4.5$ A1 N1

METHOD 2 (analytical)

$f’’ = - 6{x^2} + 6$ A1

setting $f’’ = 0$ (M1)

$x = 1$ A1 N1

substituting their $x$ value into $f$ (M1)

eg $\,\,\,\,\,$ $f(1)$

$y = 4.5$ A1 N1

[4 marks]

c.i.

recognizing rate of change is $f^{'}$ (M1)

eg $\,\,\,\,\,$ $y’,{\text{ }}f’(1)$

rate of change is 6 A1 N2

[3 marks]

c.ii.

attempt to substitute either limits or the function into formula (M1)

involving ${f^2}$ (accept absence of $\pi$ and/or ${\text{d}}x$ )

eg $\,\,\,\,\,$ $\pi \int {{{( - 0.5{x^4} + 3{x^2} + 2x)}^2}{\text{d}}x,{\text{ }}\int_1^{1.88} {{f^2}} }$

128.890

${\text{volume}} = 129$ A2 N3

[3 marks]

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

Consider the function $f(x) = - {x^4} + a{x^2} + 5$ , where $a$ is a constant. Part of the graph of $y = f(x)$ is shown below.

M17/5/MATSD/SP2/ENG/TZ2/06

It is known that at the point where $x = 2$ the tangent to the graph of $y = f(x)$ is horizontal.

There are two other points on the graph of $y = f(x)$ at which the tangent is horizontal.

Write down the $y$ -intercept of the graph.

[1]

Find $f'(x)$ .

[2]

Show that $a = 8$ .

[2]

c.i.

Find $f(2)$ .

[2]

c.ii.

Write down the $x$ -coordinates of these two points;

[2]

d.i.

Write down the intervals where the gradient of the graph of $y = f(x)$ is positive.

[2]

d.ii.

Write down the range of $f(x)$ .

[2]

Write down the number of possible solutions to the equation $f(x) = 5$ .

[1]

The equation $f(x) = m$ , where $m \in \mathbb{R}$ , has four solutions. Find the possible values of $m$ .

[2]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

5 (A1)

Note: Accept an answer of $(0,{\text{ }}5)$ .

[1 mark]

$\left( {f'(x) = } \right) - 4{x^3} + 2ax$ (A1)(A1)

Note: Award (A1) for $- 4{x^3}$ and (A1) for $+ 2ax$ . Award at most (A1)(A0) if extra terms are seen.

[2 marks]

$- 4 \times {2^3} + 2a \times 2 = 0$ (M1)(M1)

Note: Award (M1) for substitution of $x = 2$ into their derivative, (M1) for equating their derivative, written in terms of $a$ , to 0 leading to a correct answer (note, the 8 does not need to be seen).

$a = 8$ (AG)

[2 marks]

c.i.

$\left( {f(2) = } \right) - {2^4} + 8 \times {2^2} + 5$ (M1)

Note: Award (M1) for correct substitution of $x = 2$ and $a = 8$ into the formula of the function.

21 (A1)(G2)

[2 marks]

c.ii.

$(x = ){\text{ }} - 2,{\text{ }}(x = ){\text{ 0}}$ (A1)(A1)

Note: Award (A1) for each correct solution. Award at most (A0)(A1)(ft) if answers are given as $( - 2{\text{ }},21)$ and $(0,{\text{ }}5)$ or $( - 2,{\text{ }}0)$ and $(0,{\text{ }}0)$ .

[2 marks]

d.i.

$x < - 2,{\text{ }}0 < x < 2$ (A1)(ft)(A1)(ft)

Note: Award (A1)(ft) for $x < - 2$ , follow through from part (d)(i) provided their value is negative.

Award (A1)(ft) for $0 < x < 2$ , follow through only from their 0 from part (d)(i); 2 must be the upper limit.

Accept interval notation.

[2 marks]

d.ii.

$y \leqslant 21$ (A1)(ft)(A1)

Notes: Award (A1)(ft) for 21 seen in an interval or an inequality, (A1) for “ $y \leqslant$ ”.

Accept interval notation.

Accept $- \infty < y \leqslant 21$ or $f(x) \leqslant 21$ .

Follow through from their answer to part (c)(ii). Award at most (A1)(ft)(A0) if $x$ is seen instead of $y$ . Do not award the second (A1) if a (finite) lower limit is seen.

[2 marks]

3 (solutions) (A1)

[1 mark]

$5 < m < 21$ or equivalent (A1)(ft)(A1)

Note: Award (A1)(ft) for 5 and 21 seen in an interval or an inequality, (A1) for correct strict inequalities. Follow through from their answers to parts (a) and (c)(ii).

Accept interval notation.

[2 marks]

Examiners report

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

d.i.

[N/A]

d.ii.

[N/A]

Contestants in a TV gameshow try to get through three walls by passing through doors without falling into a trap. Contestants choose doors at random.
If they avoid a trap they progress to the next wall.
If a contestant falls into a trap they exit the game before the next contestant plays.
Contestants are not allowed to watch each other attempt the game.

The first wall has four doors with a trap behind one door.

Ayako is a contestant.

Natsuko is the second contestant.

The second wall has five doors with a trap behind two of the doors.

The third wall has six doors with a trap behind three of the doors.

The following diagram shows the branches of a probability tree diagram for a contestant in the game.

Write down the probability that Ayako avoids the trap in this wall.

[1]

Find the probability that only one of Ayako and Natsuko falls into a trap while attempting to pass through a door in the first wall.

[3]

Copy the probability tree diagram and write down the relevant probabilities along the branches.

[3]

A contestant is chosen at random. Find the probability that this contestant fell into a trap while attempting to pass through a door in the second wall.

[2]

d.i.

A contestant is chosen at random. Find the probability that this contestant fell into a trap.

[3]

d.ii.

120 contestants attempted this game.

Find the expected number of contestants who fell into a trap while attempting to pass through a door in the third wall.

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\frac{3}{4}$ (0.75, 75%) (A1)

[1 mark]

$\frac{3}{4} \times \frac{1}{4} + \frac{1}{4} \times \frac{3}{4}$ OR $2 \times \frac{3}{4} \times \frac{1}{4}$ (M1)(M1)

Note: Award (M1) for their product $\frac{1}{4} \times \frac{3}{4}$ seen, and (M1) for adding their two products or multiplying their product by 2.

$= \frac{3}{8}\,\,\,\,\left( {\frac{6}{{16}},\,\,0.375,\,\,37.5{\text{% }}} \right)$ (A1)(ft) (G3)

Note: Follow through from part (a), but only if the sum of their two fractions is 1.

[3 marks]

(A1)(ft)(A1)(A1)

Note: Award (A1) for each correct pair of branches. Follow through from part (a).

[3 marks]

$\frac{3}{4} \times \frac{2}{5}$ (M1)

Note: Award (M1) for correct probabilities multiplied together.

$= \frac{3}{{10}}\,\,\,\left( {0.3,\,\,30{\text{% }}} \right)$ (A1)(ft) (G2)

Note: Follow through from their tree diagram or part (a).

[2 marks]

d.i.

$1 - \frac{3}{4} \times \frac{2}{5} \times \frac{3}{6}$ OR $\frac{1}{4} + \frac{3}{4} \times \frac{2}{5} + \frac{3}{4} \times \frac{3}{5} \times \frac{3}{6}$ (M1)(M1)

Note: Award (M1) for $\frac{3}{4} \times \frac{3}{5} \times \frac{3}{6}$ and (M1) for subtracting their correct probability from 1, or adding to their $\frac{1}{4} + \frac{3}{4} \times \frac{2}{5}$ .

$= \frac{{93}}{{120}}\,\,\,\,\left( {\frac{{31}}{{40}},\,\,0.775,\,\,77.5{\text{% }}} \right)$ (A1)(ft) (G2)

Note: Follow through from their tree diagram.

[3 marks]

d.ii.

$\frac{3}{4} \times \frac{3}{5} \times \frac{3}{6} \times 120$ (M1)(M1)

Note: Award (M1) for $\frac{3}{4} \times \frac{3}{5} \times \frac{3}{6}\,\,\,\,\left( {\frac{3}{4} \times \frac{3}{5} \times \frac{3}{6}\,\,{\text{OR}}\,\,\frac{{27}}{{120}}\,\,{\text{OR}}\,\,\frac{9}{{40}}} \right)$ and (M1) for multiplying by 120.

= 27 (A1)(ft) (G3)

Note: Follow through from their tree diagram or their $\frac{3}{4} \times \frac{3}{5} \times \frac{3}{6}$ from their calculation in part (d)(ii).

[3 marks]

Examiners report

[N/A]

d.i.

[N/A]

d.ii.

[N/A]

All lengths in this question are in centimetres.

A solid metal ornament is in the shape of a right pyramid, with vertex $V$ and square base $ABCD$ . The centre of the base is $X$ . Point $V$ has coordinates $(1, 5, 0)$ and point $A$ has coordinates $(- 1, 1, 6)$ .

The volume of the pyramid is $57.2 {cm}^{3}$ , correct to three significant figures.

Find $AV$ .

[2]

Given that $A \hat{V} B = 40 °$ , find $AB$ .

[3]

Find the height of the pyramid, $VX$ .

[3]

A second ornament is in the shape of a cuboid with a rectangular base of length $2 x cm$ , width $x cm$ and height $y cm$ . The cuboid has the same volume as the pyramid.

The cuboid has a minimum surface area of $S {cm}^{2}$ . Find the value of $S$ .

[5]

Markscheme

attempt to use the distance formula to find $AV$ (M1)

$\sqrt{{(1 - (- 1))}^{2} + {(5 - 1)}^{2} + {(0 - 6)}^{2}}$

$= 7.48331 \dots$

$= 7.48 (cm) (= \sqrt{56} or 2 \sqrt{14})$ A1

[2 marks]

METHOD 1

attempt to apply cosine rule OR sine rule to find $AB$ (M1)

$(AB =) \sqrt{7.48 \dots^{2} + 7.48 \dots^{2} - 2 \times 7.48 \dots \times 7.48 \dots \cos (40 °)}$ OR $\frac{AB}{\sin 40 °} = \frac{\sqrt{56}}{\sin 70 °}$ (A1)

$= 5.11888 \dots$

$= 5.12 (cm)$ A1

METHOD 2

Let $M$ be the midpoint of $[AB]$

attempt to apply right-angled trigonometry on triangle $AVM$ (M1)

$= 2 \times 7.48 \dots \times \sin (20 °)$ (A1)

$= 5.11888 \dots$

$= 5.12 (cm)$ A1

[3 marks]

METHOD 1

equating volume of pyramid formula to $57.2$ (M1)

$\frac{1}{3} \times 5.11 \dots^{2} \times h = 57.2$ (A1)

$h = 6.54886 \dots$

$h = 6.55 (cm)$ A1

METHOD 2

Let $M$ be the midpoint of $[AB]$

${AV}^{2} = {AM}^{2} + {MX}^{2} + {XV}^{2}$ (M1)

$\Rightarrow XV = \sqrt{7.48 \dots^{2} - {(\frac{5.11 \dots}{2})}^{2} - {(\frac{5.11 \dots}{2})}^{2}}$ (A1)

$h = 6.54886 \dots$

$h = 6.55 (cm)$ A1

[3 marks]

$V = x \times 2 x \times y = 57.2$ (A1)

$S = 2 (2 x^{2} + x y + 2 x y)$ A1

Note: Condone use of $A$ .

attempt to substitute $y = \frac{57.2}{2 x^{2}}$ into their expression for surface area (M1)

$(S (x) =) 4 x^{2} + 6 x (\frac{57.2}{2 x^{2}})$

EITHER

attempt to find minimum turning point on graph of area function (M1)

$\frac{d S}{d x} = 8 x - 171.6 x^{- 2} = 0$ OR $x = 2.77849 \dots$ (M1)

THEN

$92.6401 \dots$

minimum surface area $= 92.6 ({cm}^{2})$ A1

[5 marks]

Examiners report

Parts (a), (b) and (c) were completed well by many candidates, but few were able to make any significant progress in part (d).

In part (a), many candidates were able to apply the distance formula and successfully find AV. However, a common error was to work in two-dimensions and to apply Pythagoras' Theorem once, neglecting completely the z-coordinates. Many recognised the need to use either the sine or cosine rule in part (b) to find the length AB. Common errors in this part included: the GDC being set incorrectly in radians; applying right-angled trigonometry on a 40°, 40°, 70° triangle; or using $\frac{1}{2} AB$ as the length of AX in triangle AVX.

Despite the formula for the volume of a pyramid being in the formula booklet, a common error in part (c) was to omit the factor of $\frac{1}{3}$ from the volume formula, or not to recognise that the area of the base of the pyramid was ${AB}^{2}$ .

The most challenging part of this question proved to be the optimization of the surface area of the cuboid in part (d). Although some candidates were able to form an equation involving the volume of the cuboid, an expression for the surface area eluded most. A common error was to gain a surface area which involved eight sides rather than six. It was surprising that few who were able to find both the equation and an expression were able to progress any further. Of those that did, few used their GDC to find the minimum surface area directly, with most preferring the more time consuming analytical approach.

[N/A]

160 students attend a dual language school in which the students are taught only in Spanish or taught only in English.

A survey was conducted in order to analyse the number of students studying Biology or Mathematics. The results are shown in the Venn diagram.

Set S represents those students who are taught in Spanish.

Set B represents those students who study Biology.

Set M represents those students who study Mathematics.

A student from the school is chosen at random.

Find the number of students in the school that are taught in Spanish.

[2]

a.i.

Find the number of students in the school that study Mathematics in English.

[2]

a.ii.

Find the number of students in the school that study both Biology and Mathematics.

[2]

a.iii.

Write down $n\left( {S \cap \left( {M \cup B} \right)} \right)$ .

[1]

b.i.

Write down $n\left( {B \cap M \cap S'} \right)$ .

[1]

b.ii.

Find the probability that this student studies Mathematics.

[2]

c.i.

Find the probability that this student studies neither Biology nor Mathematics.

[2]

c.ii.

Find the probability that this student is taught in Spanish, given that the student studies Biology.

[2]

c.iii.

Markscheme

10 + 40 + 28 + 17 (M1)

= 95 (A1)(G2)

Note: Award (M1) for each correct sum (for example: 10 + 40 + 28 + 17) seen.

[2 marks]

a.i.

20 + 12 (M1)

= 32 (A1)(G2)

Note: Award (M1) for each correct sum (for example: 10 + 40 + 28 + 17) seen.

[2 marks]

a.ii.

12 + 40 (M1)

= 52 (A1)(G2)

Note: Award (M1) for each correct sum (for example: 10 + 40 + 28 + 17) seen.

[2 marks]

a.iii.

78 (A1)

[1 mark]

b.i.

12 (A1)

[1 mark]

b.ii.

$\frac{{100}}{{160}}$ $\left( {\frac{5}{8}{\text{,}}\,\,0.625{\text{,}}\,\,62.5\,{\text{% }}} \right)$ (A1)(A1) (G2)

Note: Throughout part (c), award (A1) for correct numerator, (A1) for correct denominator. All answers must be probabilities to award (A1).

[2 marks]

c.i.

$\frac{{42}}{{160}}$ $\left( {\frac{{21}}{{80}}{\text{,}}\,\,0.263\,\,\left( {0.2625} \right){\text{,}}\,\,26.3\,{\text{% }}\,\,\left( {26.25\,{\text{% }}} \right)} \right)$ (A1)(A1) (G2)

Note: Throughout part (c), award (A1) for correct numerator, (A1) for correct denominator. All answers must be probabilities to award (A1).

[2 marks]

c.ii.

$\frac{{50}}{{70}}$ $\left( {\frac{5}{7}{\text{,}}\,\,0.714\,\,\left( {0.714285 \ldots } \right){\text{,}}\,\,71.4\,{\text{% }}\,\,\left( {71.4285 \ldots \,{\text{% }}} \right)} \right)$ (A1)(A1) (G2)

Note: Throughout part (c), award (A1) for correct numerator, (A1) for correct denominator. All answers must be probabilities to award (A1).

[2 marks]

c.iii.

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

a.iii.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

c.iii.

Let $f(x) = 6 - \ln ({x^2} + 2)$ , for $x \in \mathbb{R}$ . The graph of $f$ passes through the point $(p,{\text{ }}4)$ , where $p > 0$ .

Find the value of $p$ .

[2]

The following diagram shows part of the graph of $f$ .

N17/5/MATME/SP2/ENG/TZ0/05.b

The region enclosed by the graph of $f$ , the $x$ -axis and the lines $x = - p$ and $x = p$ is rotated 360° about the $x$ -axis. Find the volume of the solid formed.

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

valid approach (M1)

eg $\,\,\,\,\,$ $f(p) = 4$ , intersection with $y = 4,{\text{ }} \pm 2.32$

2.32143

$p = \sqrt {{{\text{e}}^2} - 2}$ (exact), 2.32 A1 N2

[2 marks]

attempt to substitute either their limits or the function into volume formula (must involve ${f^2}$ , accept reversed limits and absence of $\pi$ and/or ${\text{d}}x$ , but do not accept any other errors) (M1)

eg $\,\,\,\,\,$ $\int_{ - 2.32}^{2.32} {{f^2},{\text{ }}\pi \int {{{\left( {6 - \ln ({x^2} + 2)} \right)}^2}{\text{d}}x,{\text{ 105.675}}} }$

331.989

${\text{volume}} = 332$ A2 N3

[3 marks]

Examiners report

[N/A]

On one day 180 flights arrived at a particular airport. The distance travelled and the arrival status for each incoming flight was recorded. The flight was then classified as on time, slightly delayed, or heavily delayed.

The results are shown in the following table.

A χ² test is carried out at the 10 % significance level to determine whether the arrival status of incoming flights is independent of the distance travelled.

The critical value for this test is 7.779.

A flight is chosen at random from the 180 recorded flights.

State the alternative hypothesis.

[1]

Calculate the expected frequency of flights travelling at most 500 km and arriving slightly delayed.

[2]

Write down the number of degrees of freedom.

[1]

Write down the χ² statistic.

[2]

d.i.

Write down the associated p-value.

[1]

d.ii.

State, with a reason, whether you would reject the null hypothesis.

[2]

Write down the probability that this flight arrived on time.

[2]

Given that this flight was not heavily delayed, find the probability that it travelled between 500 km and 5000 km.

[2]

Two flights are chosen at random from those which were slightly delayed.

Find the probability that each of these flights travelled at least 5000 km.

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

The arrival status is dependent on the distance travelled by the incoming flight (A1)

Note: Accept “associated” or “not independent”.

[1 mark]

$\frac{{60 \times 45}}{{180}}$ OR $\frac{{60}}{{180}} \times \frac{{45}}{{180}} \times 180$ (M1)

Note: Award (M1) for correct substitution into expected value formula.

= 15 (A1) (G2)

[2 marks]

4 (A1)

Note: Award (A0) if “2 + 2 = 4” is seen.

[1 mark]

9.55 (9.54671…) (G2)

Note: Award (G1) for an answer of 9.54.

[2 marks]

d.i.

0.0488 (0.0487961…) (G1)

[1 mark]

d.ii.

Reject the Null Hypothesis (A1)(ft)

Note: Follow through from their hypothesis in part (a).

9.55 (9.54671…) > 7.779 (R1)(ft)

0.0488 (0.0487961…) < 0.1 (R1)(ft)

Note: Do not award (A1)(ft)(R0)(ft). Follow through from part (d). Award (R1)(ft) for a correct comparison, (A1)(ft) for a consistent conclusion with the answers to parts (a) and (d). Award (R1)(ft) for χ²_calc > χ²_crit, provided the calculated value is explicitly seen in part (d)(i).

[2 marks]

$\frac{{52}}{{180}}\,\,\left( {0.289,\,\,\frac{{13}}{{45}},\,\,28.9\,{\text{% }}} \right)$ (A1)(A1) (G2)

Note: Award (A1) for correct numerator, (A1) for correct denominator.

[2 marks]

$\frac{{35}}{{97}}\,\,\left( {0.361,\,\,36.1\,{\text{% }}} \right)$ (A1)(A1) (G2)

Note: Award (A1) for correct numerator, (A1) for correct denominator.

[2 marks]

$\frac{{14}}{{45}} \times \frac{{13}}{{44}}$ (A1)(M1)

Note: Award (A1) for two correct fractions and (M1) for multiplying their two fractions.

$= \frac{{182}}{{1980}}\,\,\left( {0.0919,\,\,\frac{{91}}{{990}},\,0.091919 \ldots ,\,9.19\,{\text{% }}} \right)$ (A1) (G2)

[3 marks]

Examiners report

[N/A]

d.i.

[N/A]

d.ii.

[N/A]

Consider the function $f\left( x \right) = {x^2}{{\text{e}}^{3x}}$ , $x \in \mathbb{R}$ .

The graph of $f$ has a horizontal tangent line at $x = 0$ and at $x = a$ . Find $a$ .

Markscheme

valid method (M1)

eg $f'\left( x \right) = 0$ , ,

$a = - 0.667\left( { = - \frac{2}{3}} \right)$ (accept $x = - 0.667$ ) A1 N2

[2 marks]

Examiners report

[N/A]

Sila High School has 110 students. They each take exactly one language class from a choice of English, Spanish or Chinese. The following table shows the number of female and male students in the three different language classes.

A ${\chi ^2}$ test was carried out at the 5 % significance level to analyse the relationship between gender and student choice of language class.

Use your graphic display calculator to write down

The critical value at the 5 % significance level for this test is 5.99.

One student is chosen at random from this school.

Another student is chosen at random from this school.

Write down the null hypothesis, H₀, for this test.

[1]

State the number of degrees of freedom.

[1]

the expected frequency of female students who chose to take the Chinese class.

[1]

c.i.

the ${\chi ^2}$ statistic.

[2]

c.ii.

State whether or not H₀ should be rejected. Justify your statement.

[2]

Find the probability that the student does not take the Spanish class.

[2]

e.i.

Find the probability that neither of the two students take the Spanish class.

[3]

e.ii.

Find the probability that at least one of the two students is female.

[3]

e.iii.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

(H₀:) (choice of) language is independent of gender (A1)

Note: Accept “there is no association between language (choice) and gender”. Accept “language (choice) is not dependent on gender”. Do not accept “not related” or “not correlated” or “not influenced”.

[1 mark]

2 (AG)

[1 mark]

16.4 (16.4181…) (G1)

[1 mark]

c.i.

$\chi _{{\text{calc}}}^2 = 8.69$ (8.68507…) (G2)

[2 marks]

c.ii.

(we) reject the null hypothesis (A1)(ft)

8.68507… > 5.99 (R1)(ft)

Note: Follow through from part (c)(ii). Accept “do not accept” in place of “reject.” Do not award (A1)(ft)(R0).

(we) reject the null hypothesis (A1)

0.0130034 < 0.05 (R1)

Note: Accept “do not accept” in place of “reject.” Do not award (A1)(ft)(R0).

[2 marks]

$\frac{{88}}{{110}}\,\,\,\left( {\frac{4}{5}{\text{,}}\,\,0.8{\text{,}}\,\,80{\text{% }}} \right)$ (A1)(A1)(G2)

Note: Award (A1) for correct numerator, (A1) for correct denominator.

[2 marks]

e.i.

$\frac{{88}}{{110}} \times \frac{{87}}{{109}}$ (M1)(M1)

Note: Award (M1) for multiplying two fractions. Award (M1) for multiplying their correct fractions.

$\left( {\frac{{46}}{{110}}} \right)\left( {\frac{{45}}{{109}}} \right) + 2\left( {\frac{{46}}{{110}}} \right)\left( {\frac{{42}}{{109}}} \right) + \left( {\frac{{42}}{{110}}} \right)\left( {\frac{{41}}{{109}}} \right)$ (M1)(M1)

Note: Award (M1) for correct products; (M1) for adding 4 products.

$0.639\,\,\,\left( {0.638532 \ldots {\text{,}}\,\,\frac{{348}}{{545}}{\text{,}}\,\,63.9{\text{% }}} \right)$ (A1)(ft)(G2)

Note: Follow through from their answer to part (e)(i).

[3 marks]

e.ii.

$1 - \frac{{67}}{{110}} \times \frac{{66}}{{109}}$ (M1)(M1)

Note: Award (M1) for multiplying two correct fractions. Award (M1) for subtracting their product of two fractions from 1.

$\frac{{43}}{{110}} \times \frac{{42}}{{109}} + \frac{{43}}{{110}} \times \frac{{67}}{{109}} + \frac{{67}}{{110}} \times \frac{{43}}{{109}}$ (M1)(M1)

Note: Award (M1) for correct products; (M1) for adding three products.

$0.631\,\,\,\left( {0.631192 \ldots {\text{,}}\,\,63.1{\text{% ,}}\,\,\frac{{344}}{{545}}} \right)$ (A1)(G2)

[3 marks]

e.iii.

Examiners report

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

e.i.

[N/A]

e.ii.

[N/A]

e.iii.

In a company it is found that 25 % of the employees encountered traffic on their way to work. From those who encountered traffic the probability of being late for work is 80 %.

From those who did not encounter traffic, the probability of being late for work is 15 %.

The tree diagram illustrates the information.

The company investigates the different means of transport used by their employees in the past year to travel to work. It was found that the three most common means of transport used to travel to work were public transportation (P ), car (C ) and bicycle (B ).

The company finds that 20 employees travelled by car, 28 travelled by bicycle and 19 travelled by public transportation in the last year.

Some of the information is shown in the Venn diagram.

There are 54 employees in the company.

Write down the value of a.

[1]

a.i.

Write down the value of b.

[1]

a.ii.

Use the tree diagram to find the probability that an employee encountered traffic and was late for work.

[2]

b.i.

Use the tree diagram to find the probability that an employee was late for work.

[3]

b.ii.

Use the tree diagram to find the probability that an employee encountered traffic given that they were late for work.

[3]

b.iii.

Find the value of x.

[1]

c.i.

Find the value of y.

[1]

c.ii.

Find the number of employees who, in the last year, did not travel to work by car, bicycle or public transportation.

[2]

Find $n\left( {\left( {C \cup B} \right) \cap P'} \right)$ .

[2]

Markscheme

a = 0.2 (A1)

[1 mark]

a.i.

b = 0.85 (A1)

[1 mark]

a.ii.

0.25 × 0.8 (M1)

Note: Award (M1) for a correct product.

$= 0.2\,\,\,\left( {\,\frac{1}{5},\,\,\,20\% } \right)$ (A1)(G2)

[2 marks]

b.i.

0.25 × 0.8 + 0.75 × 0.15 (A1)(ft)(M1)

Note: Award (A1)(ft) for their (0.25 × 0.8) and (0.75 × 0.15), (M1) for adding two products.

$= 0.313\,\,\,\left( {0.3125,\,\,\,\frac{5}{{16}},\,\,\,31.3\% } \right)$ (A1)(ft)(G3)

Note: Award the final (A1)(ft) only if answer does not exceed 1. Follow through from part (b)(i).

[3 marks]

b.ii.

$\frac{{0.25 \times 0.8}}{{0.25 \times 0.8 + 0.75 \times 0.15}}$ (A1)(ft)(A1)(ft)

Note: Award (A1)(ft) for a correct numerator (their part (b)(i)), (A1)(ft) for a correct denominator (their part (b)(ii)). Follow through from parts (b)(i) and (b)(ii).

$= 0.64\,\,\,\left( {\frac{{16}}{{25}},\,\,64{\text{% }}} \right)$ (A1)(ft)(G3)

Note: Award final (A1)(ft) only if answer does not exceed 1.

[3 marks]

b.iii.

(x =) 3 (A1)

[1 Mark]

c.i.

(y =) 10 (A1)(ft)

Note: Following through from part (c)(i) but only if their x is less than or equal to 13.

[1 Mark]

c.ii.

54 − (10 + 3 + 4 + 2 + 6 + 8 + 13) (M1)

Note: Award (M1) for subtracting their correct sum from 54. Follow through from their part (c).

= 8 (A1)(ft)(G2)

Note: Award (A1)(ft) only if their sum does not exceed 54. Follow through from their part (c).

[2 marks]

6 + 8 + 13 (M1)

Note: Award (M1) for summing 6, 8 and 13.

27 (A1)(G2)

[2 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

b.iii.

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

Consider the function $f\left( x \right) = \frac{{27}}{{{x^2}}} - 16x,\,\,\,x \ne 0$ .

Sketch the graph of y = f (x), for −4 ≤ x ≤ 3 and −50 ≤ y ≤ 100.

[4]

Use your graphic display calculator to find the equation of the tangent to the graph of y = f (x) at the point (–2, 38.75).

Give your answer in the form y = mx + c.

[2]

b.iii.

Sketch the graph of the function g (x) = 10x + 40 on the same axes.

[2]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

(A1)(A1)(A1)(A1)

Note: Award (A1) for axis labels and some indication of scale; accept y or f(x).

Use of graph paper is not required. If no scale is given, assume the given window for zero and minimum point.

Award (A1) for smooth curve with correct general shape.

Award (A1) for x-intercept closer to y-axis than to end of sketch.

Award (A1) for correct local minimum with x-coordinate closer to y-axis than end of sketch and y-coordinate less than half way to top of sketch.

Award at most (A1)(A0)(A1)(A1) if the sketch intersects the y-axis or if the sketch curves away from the y-axis as x approaches zero.

[4 marks]

y = −9.25x + 20.3 (y = −9.25x + 20.25) (A1)(A1)

Note: Award (A1) for −9.25x, award (A1) for +20.25, award a maximum of (A0)(A1) if answer is not an equation.

[2 marks]

b.iii.

correct line, y = 10x + 40, seen on sketch (A1)(A1)

Note: Award (A1) for straight line with positive gradient, award (A1) for x-intercept and y-intercept in approximately the correct positions. Award at most (A0)(A1) if ruler not used. If the straight line is drawn on different axes to part (a), award at most (A0)(A1).

[2 marks]

Examiners report

[N/A]

b.iii.

[N/A]

Consider the curve y = 2x³ − 9x² + 12x + 2, for −1 < x < 3

Sketch the curve for −1 < x < 3 and −2 < y < 12.

[4]

A teacher asks her students to make some observations about the curve.

Three students responded.
Nadia said “The x-intercept of the curve is between −1 and zero”.
Rick said “The curve is decreasing when x < 1 ”.
Paula said “The gradient of the curve is less than zero between x = 1 and x = 2 ”.

State the name of the student who made an incorrect observation.

[1]

Find $\frac{{{\text{dy}}}}{{{\text{dx}}}}$ .

[3]

Show that the stationary points of the curve are at x = 1 and x = 2.

[2]

Given that y = 2x³ − 9x² + 12x + 2 = k has three solutions, find the possible values of k.

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

(A1)(A1)(A1)(A1)

Note: Award (A1) for correct window (condone a window which is slightly off) and axes labels. An indication of window is necessary. −1 to 3 on the x-axis and −2 to 12 on the y-axis and a graph in that window.
(A1) for correct shape (curve having cubic shape and must be smooth).
(A1) for both stationary points in the 1^st quadrant with approximate correct position,
(A1) for intercepts (negative x-intercept and positive y intercept) with approximate correct position.

[4 marks]

Rick (A1)

Note: Award (A0) if extra names stated.

[1 mark]

6x² − 18x + 12 (A1)(A1)(A1)

Note: Award (A1) for each correct term. Award at most (A1)(A1)(A0) if extra terms seen.

[3 marks]

6x² − 18x + 12 = 0 (M1)

Note: Award (M1) for equating their derivative to 0. If the derivative is not explicitly equated to 0, but a subsequent solving of their correct equation is seen, award (M1).

6(x − 1)(x − 2) = 0 (or equivalent) (M1)

Note: Award (M1) for correct factorization. The final (M1) is awarded only if answers are clearly stated.

Award (M0)(M0) for substitution of 1 and of 2 in their derivative.

x = 1, x = 2 (AG)

[2 marks]

6 < k < 7 (A1)(A1)(ft)(A1)

Note: Award (A1) for an inequality with 6, award (A1)(ft) for an inequality with 7 from their part (c) provided it is greater than 6, (A1) for their correct strict inequalities. Accept ]6, 7[ or (6, 7).

[3 marks]

Examiners report

[N/A]

Let $f\left( x \right) = 12\,\,{\text{cos}}\,x - 5\,\,{\text{sin}}\,x,\,\, - \pi \leqslant x \leqslant 2\pi$ , be a periodic function with $f\left( x \right) = f\left( {x + 2\pi } \right)$

The following diagram shows the graph of $f$ .

There is a maximum point at A. The minimum value of $f$ is −13 .

A ball on a spring is attached to a fixed point O. The ball is then pulled down and released, so that it moves back and forth vertically.

The distance, d centimetres, of the centre of the ball from O at time t seconds, is given by

$d\left( t \right) = f\left( t \right) + 17,\,\,0 \leqslant t \leqslant 5.$

Find the coordinates of A.

[2]

For the graph of $f$ , write down the amplitude.

[1]

b.i.

For the graph of $f$ , write down the period.

[1]

b.ii.

Hence, write $f\left( x \right)$ in the form $p\,\,{\text{cos}}\,\left( {x + r} \right)$ .

[3]

Find the maximum speed of the ball.

[3]

Find the first time when the ball’s speed is changing at a rate of 2 cm s⁻².

[5]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

−0.394791,13

A(−0.395, 13) A1A1 N2

[2 marks]

13 A1 N1

[1 mark]

b.i.

${2\pi }$ , 6.28 A1 N1

[1 mark]

b.ii.

valid approach (M1)

eg recognizing that amplitude is p or shift is r

$f\left( x \right) = 13\,\,{\text{cos}}\,\left( {x + 0.395} \right)$ (accept p = 13, r = 0.395) A1A1 N3

Note: Accept any value of r of the form $0.395 + 2\pi k,\,\,k \in \mathbb{Z}$

[3 marks]

recognizing need for d ′(t) (M1)

eg −12 sin(t) − 5 cos(t)

correct approach (accept any variable for t) (A1)

eg −13 sin(t + 0.395), sketch of d′, (1.18, −13), t = 4.32

maximum speed = 13 (cms⁻¹) A1 N2

[3 marks]

recognizing that acceleration is needed (M1)

eg a(t), d "(t)

correct equation (accept any variable for t) (A1)

eg $a\left( t \right) = - 2,\,\,\left| {\frac{{\text{d}}}{{{\text{d}}t}}\left( {d'\left( t \right)} \right)} \right| = 2,\,\, - 12\,\,{\text{cos}}\,\left( t \right) + 5\,\,{\text{sin}}\,\left( t \right) = - 2$

valid attempt to solve their equation (M1)

eg sketch, 1.33

1.02154

1.02 A2 N3

[5 marks]

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

Consider the function $f\left( x \right) = \frac{1}{3}{x^3} + \frac{3}{4}{x^2} - x - 1$ .

The function has one local maximum at $x = p$ and one local minimum at $x = q$ .

Write down the $y$ -intercept of the graph of $y = f\left( x \right)$ .

[1]

Sketch the graph of $y = f\left( x \right)$ for −3 ≤ $x$ ≤ 3 and −4 ≤ $y$ ≤ 12.

[4]

Determine the range of $f\left( x \right)$ for $p$ ≤ $x$ ≤ $q$ .

[3]

Markscheme

−1 (A1)

Note: Accept (0, −1).

[1 mark]

(A1)(A1)(A1)(A1)

Note: Award (A1) for correct window and axes labels, −3 to 3 should be indicated on the $x$ -axis and −4 to 12 on the $y$ -axis.
(A1)) for smooth curve with correct cubic shape;
(A1) for $x$ -intercepts: one close to −3, the second between −1 and 0, and third between 1 and 2; and $y$ -intercept at approximately −1;
(A1) for local minimum in the 4th quadrant and maximum in the 2nd quadrant, in approximately correct positions.
Graph paper does not need to be used. If window not given award at most (A0)(A1)(A0)(A1).

[4 marks]

$- 1.27 \leqslant {\text{ }}f\left( x \right) \leqslant 1.33\,\,\,\left( { - 1.27083 \ldots \leqslant {\text{ }}f\left( x \right) \leqslant 1.33333 \ldots {\text{,}}\,\, - \frac{{61}}{{48}} \leqslant {\text{ }}f\left( x \right) \leqslant \frac{4}{3}} \right)$ (A1)(ft)(A1)(ft)(A1)

Note: Award (A1) for −1.27 seen, (A1) for 1.33 seen, and (A1) for correct weak inequalities with their endpoints in the correct order. For example, award (A0)(A0)(A0) for answers like $5 \leqslant {\text{ }}f\left( x \right) \leqslant 2$ . Accept $y$ in place of $f\left( x \right)$ . Accept alternative correct notation such as [−1.27, 1.33].

Follow through from their $p$ and $q$ values from part (g) only if their $f\left( p \right)$ and $f\left( q \right)$ values are between −4 and 12. Award (A0)(A0)(A0) if their values from (g) are given as the endpoints.

[3 marks]

Examiners report

[N/A]

A rocket is travelling in a straight line, with an initial velocity of $140$ m s⁻¹. It accelerates to a new velocity of $500$ m s⁻¹ in two stages.

During the first stage its acceleration, $a$ m s⁻², after $t$ seconds is given by $a\left( t \right) = 240\,{\text{sin}}\left( {2t} \right)$ , where $0 \leqslant t \leqslant k$ .

The first stage continues for $k$ seconds until the velocity of the rocket reaches $375$ m s⁻¹.

Find an expression for the velocity, $v$ m s⁻¹, of the rocket during the first stage.

[4]

Find the distance that the rocket travels during the first stage.

[4]

During the second stage, the rocket accelerates at a constant rate. The distance which the rocket travels during the second stage is the same as the distance it travels during the first stage.

Find the total time taken for the two stages.

[6]

Markscheme

recognizing that $v = \int a$ (M1)

correct integration A1

eg $- 120\,{\text{cos}}\left( {2t} \right) + c$

attempt to find $c$ using their $v\left( t \right)$ (M1)

eg $- 120\,{\text{cos}}\left( 0 \right) + c = 140$

$v\left( t \right) = - 120\,{\text{cos}}\left( {2t} \right) + 260$ A1 N3

[4 marks]

evidence of valid approach to find time taken in first stage (M1)

eg graph, $- 120\,{\text{cos}}\left( {2t} \right) + 260 = 375$

$k = 1.42595$ A1

attempt to substitute their $v$ and/or their limits into distance formula (M1)

eg $\int_0^{1.42595} {\left| v \right|}$ , $\int {260 - 120} \,{\text{cos}}\left( {2t} \right)$ , $\int_0^k {\left( {260 - 120\,{\text{cos}}\left( {2t} \right)} \right)} \,{\text{d}}t$

$353.608$

distance is $354$ (m) A1 N3

[4 marks]

recognizing velocity of second stage is linear (seen anywhere) R1

eg graph, $s = \frac{1}{2}h\left( {a + b} \right)$ , $v = mt + c$

valid approach (M1)

eg $\int {v = 353.608}$

correct equation (A1)

eg $\frac{1}{2}h\left( {375 + 500} \right) = 353.608$

time for stage two $=0.808248$ ( $0.809142$ from 3 sf) A2

$2.23420$ ( $2.23914$ from 3 sf)

$2.23$ seconds ( $2.24$ from 3 sf) A1 N3

[6 marks]

Examiners report

[N/A]

A company performs an experiment on the efficiency of a liquid that is used to detect a nut allergy.

A group of 60 people took part in the experiment. In this group 26 are allergic to nuts. One person from the group is chosen at random.

A second person is chosen from the group.

When the liquid is added to a person’s blood sample, it is expected to turn blue if the person is allergic to nuts and to turn red if the person is not allergic to nuts.

The company claims that the probability that the test result is correct is 98% for people who are allergic to nuts and 95% for people who are not allergic to nuts.

It is known that 6 in every 1000 adults are allergic to nuts.

This information can be represented in a tree diagram.

N17/5/MATSD/SP2/ENG/TZ0/04.c.d.e.f.g

An adult, who was not part of the original group of 60, is chosen at random and tested using this liquid.

The liquid is used in an office to identify employees who might be allergic to nuts. The liquid turned blue for 38 employees.

Find the probability that both people chosen are not allergic to nuts.

[2]

Copy and complete the tree diagram.

[3]

Find the probability that this adult is allergic to nuts and the liquid turns blue.

[2]

Find the probability that the liquid turns blue.

[3]

Find the probability that the tested adult is allergic to nuts given that the liquid turned blue.

[3]

Estimate the number of employees, from this 38, who are allergic to nuts.

[2]

Markscheme

$\frac{{34}}{{60}} \times \frac{{33}}{{59}}$ (M1)

Note: Award (M1) for their correct product.

$= 0.317{\text{ }}\left( {\frac{{187}}{{590}},{\text{ }}0.316949 \ldots ,{\text{ }}31.7\% } \right)$ (A1)(ft)(G2)

Note: Follow through from part (a).

[2 marks]

N17/5/MATSD/SP2/ENG/TZ0/04.c/M (A1)(A1)(A1)

Note: Award (A1) for each correct pair of branches.

[3 marks]

$0.006 \times 0.98$ (M1)

Note: Award (M1) for multiplying 0.006 by 0.98.

$= 0.00588{\text{ }}\left( {\frac{{147}}{{25000}},{\text{ }}0.588\% } \right)$ (A1)(G2)

[2 marks]

$0.006 \times 0.98 + 0.994 \times 0.05{\text{ }}(0.00588 + 0.994 \times 0.05)$ (A1)(ft)(M1)

Note: Award (A1)(ft) for their two correct products, (M1) for adding two products.

$= 0.0556{\text{ }}\left( {0.05558,{\text{ }}5.56\% ,{\text{ }}\frac{{2779}}{{50000}}} \right)$ (A1)(ft)(G3)

Note: Follow through from parts (c) and (d).

[3 marks]

$\frac{{0.006 \times 0.98}}{{0.05558}}$ (M1)(M1)

Note: Award (M1) for their correct numerator, (M1) for their correct denominator.

$= 0.106{\text{ }}\left( {0.105793 \ldots ,{\text{ }}10.6\% ,{\text{ }}\frac{{42}}{{397}}} \right)$ (A1)(ft)(G3)

Note: Follow through from parts (d) and (e).

[3 marks]

$0.105793 \ldots \times 38$ (M1)

Note: Award (M1) for multiplying 38 by their answer to part (f).

$= 4.02{\text{ }}(4.02015 \ldots )$ (A1)(ft)(G2)

Notes: Follow through from part (f). Use of 3 sf result from part (f) results in an answer of 4.03 (4.028).

[2 marks]

Examiners report

[N/A]

Let $f\left( x \right) = x - 8$ , $g\left( x \right) = {x^4} - 3$ and $h\left( x \right) = f\left( {g\left( x \right)} \right)$ .

Find $h\left( x \right)$ .

[2]

Let ${\text{C}}$ be a point on the graph of $h$ . The tangent to the graph of $h$ at ${\text{C}}$ is parallel to the graph of $f$ .

Find the $x$ -coordinate of ${\text{C}}$ .

[5]

Markscheme

attempt to form composite (in any order) (M1)

eg $f\left( {{x^4} - 3} \right)$ , ${\left( {x - 8} \right)^4} - 3$

$h\left( x \right) = {x^4} - 11$ A1 N2

[2 marks]

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

eg ${h'}$

correct derivative (seen anywhere) (A1)

$h'\left( x \right) = 4{x^3}$

correct value for gradient of $f$ (seen anywhere) (A1)

$f'\left( x \right) = 1$ , $m = 1$

setting their derivative equal to $1$ (M1)

$4{x^3} = 1$

$0.629960$

$x = \sqrt[3]{{\frac{1}{4}}}$ (exact), $0.630$ A1 N3

[5 marks]

Examiners report

[N/A]

Let $f\left( x \right) = {{\text{e}}^{2\,{\text{sin}}\left( {\frac{{\pi x}}{2}} \right)}}$ , for x > 0.

The k th maximum point on the graph of f has x-coordinate x_k where $k \in {\mathbb{Z}^ + }$ .

Given that x_k_{+ 1} = x_k + a, find a.

[4]

Hence find the value of n such that $\sum\limits_{k = 1}^n {{x_k} = 861}$ .

[4]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

valid approach to find maxima (M1)

eg one correct value of x_k, sketch of f

any two correct consecutive values of x_k (A1)(A1)

eg x₁ = 1, x₂ = 5

a = 4 A1 N3

[4 marks]

recognizing the sequence x_1, x₂_, x_{3, …,} x_n is arithmetic (M1)

eg d = 4

correct expression for sum (A1)

eg $\frac{n}{2}\left( {2\left( 1 \right) + 4\left( {n - 1} \right)} \right)$

valid attempt to solve for n (M1)

eg graph, 2n² − n − 861 = 0

n = 21 A1 N2

[4 marks]

Examiners report

[N/A]

The population of fish in a lake is modelled by the function

$f\left( t \right) = \frac{{1000}}{{1 + 24{{\text{e}}^{ - 0.2t}}}}$ , 0 ≤ $t$ ≤ 30 , where $t$ is measured in months.

Find the population of fish at $t$ = 10.

[2]

Find the rate at which the population of fish is increasing at $t$ = 10.

[2]

Find the value of $t$ for which the population of fish is increasing most rapidly.

[2]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

valid approach (M1)

eg $f$ (10)

235.402

235 (fish) (must be an integer) A1 N2

[2 marks]

recognizing rate of change is derivative (M1)

eg rate = ${f'}$ , ${f'}$ (10) , sketch of ${f'}$ , 35 (fish per month)

35.9976

36.0 (fish per month) A1 N2

[2 marks]

valid approach (M1)

eg maximum of ${f'}$ , ${f''}$ = 0

15.890

15.9 (months) A1 N2

[2 marks]

Examiners report

[N/A]

Let $f\left( x \right) = {x^4} - 54{x^2} + 60x$ , for $- 1 \leqslant x \leqslant 6$ . The following diagram shows the graph of $f$ .

There are $x$ -intercepts at $x = 0$ and at $x = p$ . There is a maximum at point ${\text{A}}$ where $x = a$ , and a point of inflexion at point ${\text{B}}$ where $x = b$ .

Find the value of $p$ .

[2]

Write down the coordinates of ${\text{A}}$ .

[2]

b.i.

Find the equation of the tangent to the graph of $f$ at ${\text{A}}$ .

[2]

b.ii.

Find the coordinates of ${\text{B}}$ .

[5]

c.i.

Find the rate of change of $f$ at ${\text{B}}$ .

[2]

c.ii.

Let $R$ be the region enclosed by the graph of $f$ , the $x$ -axis and the lines $x = p$ and $x = b$ . The region $R$ is rotated 360º about the $x$ -axis. Find the volume of the solid formed.

[3]

Markscheme

evidence of valid approach (M1)

eg $f\left( x \right) = 0$ , $y = 0$

$1.13843$

$p = 1.14$ A1 N2

[2 marks]

$0.562134$ , $16.7641$

$\left( {0.562{\text{, }}16.8} \right)$ A2 N2

[2 marks]

b.i.

valid approach (M1)

eg tangent at maximum point is horizontal, $f' = 0$

$y = 16.8$ (must be an equation) A1 N2

[2 marks]

b.ii.

METHOD 1 (using GDC)

valid approach M1

eg $f'' = 0$ , max/min on ${f'}$ , $x = - 3$

sketch of either ${f'}$ or ${f''}$ , with max/min or root (respectively) (A1)

$x = 3$ A1 N1

substituting their $x$ value into $f$ (M1)

eg $f\left( 3 \right)$

$y = - 225$ (exact) (accept $\left( {3{\text{, }} - 225} \right)$ ) A1 N1

METHOD 2 (analytical)

$f'' = 12{x^2} - 108$ A1

valid approach (M1)

eg $f'' = 0$ , $x = \pm 3$

$x = 3$ A1 N1

substituting their $x$ value into $f$ (M1)

eg $f\left( 3 \right)$

$y = - 225$ (exact) (accept $\left( {3{\text{, }} - 225} \right)$ ) A1 N1

[5 marks]

c.i.

recognizing rate of change is ${f'}$ (M1)

eg ${y'}$ , $f'\left( 3 \right)$

rate of change is $-156$ (exact) A1 N2

[2 marks]

c.ii.

attempt to substitute either their limits or the function into volume formula (M1)

eg $\int_{1.14}^3 {{f^2}}$ , $\pi \int {{{\left( {{x^4} - 54{x^2} + 60x} \right)}^2}} {\text{d}}x$ , $25\,752.0$

$80\,902.3$

volume $= 80\,900$ A2 N3

[3 marks]

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

Let $f(x) = \ln x$ and $g(x) = 3 + \ln \left( {\frac{x}{2}} \right)$ , for $x > 0$ .

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

$\begin{array}{*{20}{l}} {{\text{a horizontal stretch of scale factor }}q{\text{ followed by}}} \\ {{\text{a translation of }}\left( {\begin{array}{*{20}{c}} h \\ k \end{array}} \right).} \end{array}$

Let $h(x) = g(x) \times \cos (0.1x)$ , for $0 < x < 4$ . The following diagram shows the graph of $h$ and the line $y = x$ .

M17/5/MATME/SP2/ENG/TZ1/10.b.c

The graph of $h$ intersects the graph of ${h^{ - 1}}$ at two points. These points have $x$ coordinates 0.111 and 3.31 correct to three significant figures.

Write down the value of $q$ ;

[1]

a.i.

Write down the value of $h$ ;

[1]

a.ii.

Write down the value of $k$ .

[1]

a.iii.

Find $\int_{0.111}^{3.31} {\left( {h(x) - x} \right){\text{d}}x}$ .

[2]

b.i.

Hence, find the area of the region enclosed by the graphs of $h$ and ${h^{ - 1}}$ .

[3]

b.ii.

Let $d$ be the vertical distance from a point on the graph of $h$ to the line $y = x$ . There is a point ${\text{P}}(a,{\text{ }}b)$ on the graph of $h$ where $d$ is a maximum.

Find the coordinates of P, where $0.111 < a < 3.31$ .

[7]

Markscheme

$q = 2$ A1 N1

Note: Accept $q = 1$ , $h = 0$ , and $k = 3 - \ln (2)$ , 2.31 as candidate may have rewritten $g(x)$ as equal to $3 + \ln (x) - \ln (2)$ .

[1 mark]

a.i.

$h = 0$ A1 N1

Note: Accept $q = 1$ , $h = 0$ , and $k = 3 - \ln (2)$ , 2.31 as candidate may have rewritten $g(x)$ as equal to $3 + \ln (x) - \ln (2)$ .

[1 mark]

a.ii.

$k = 3$ A1 N1

Note: Accept $q = 1$ , $h = 0$ , and $k = 3 - \ln (2)$ , 2.31 as candidate may have rewritten $g(x)$ as equal to $3 + \ln (x) - \ln (2)$ .

[1 mark]

a.iii.

2.72409

2.72 A2 N2

[2 marks]

b.i.

recognizing area between $y = x$ and $h$ equals 2.72 (M1)

eg $\,\,\,\,\,$ M17/5/MATME/SP2/ENG/TZ1/10.b.ii/M

recognizing graphs of $h$ and ${h^{ - 1}}$ are reflections of each other in $y = x$ (M1)

eg $\,\,\,\,\,$ area between $y = x$ and $h$ equals between $y = x$ and ${h^{ - 1}}$

$2 \times 2.72\int_{0.111}^{3.31} {\left( {x - {h^{ - 1}}(x)} \right){\text{d}}x = 2.72}$

5.44819

5.45 A1 N3

[??? marks]

b.ii.

valid attempt to find $d$ (M1)

eg $\,\,\,\,\,$ difference in $y$ -coordinates, $d = h(x) - x$

correct expression for $d$ (A1)

eg $\,\,\,\,\,$ $\left( {\ln \frac{1}{2}x + 3} \right)(\cos 0.1x) - x$

valid approach to find when $d$ is a maximum (M1)

eg $\,\,\,\,\,$ max on sketch of $d$ , attempt to solve $d’ = 0$

0.973679

$x = 0.974$ A2 N4

substituting their $x$ value into $h(x)$ (M1)

2.26938

$y = 2.27$ A1 N2

[7 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

a.iii.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

A particle moves in a straight line such that its velocity, $v m s^{- 1}$ , at time $t$ seconds is given by $v = \frac{(t^{2} + 1) \cos t}{4}, 0 \leq t \leq 3$ .

Determine when the particle changes its direction of motion.

[2]

Find the times when the particle’s acceleration is $- 1.9 {m s}^{- 2}$ .

[3]

Find the particle’s acceleration when its speed is at its greatest.

[2]

Markscheme

recognises the need to find the value of $t$ when $v = 0$ (M1)

$t = 1.57079 \dots (= \frac{π}{2})$

$t = 1.57 (= \frac{π}{2})$ (s) A1

[2 marks]

recognises that $a (t) = v' (t)$ (M1)

$t_{1} = 2.26277 \dots, t_{2} = 2.95736 \dots$

$t_{1} = 2.26, t_{2} = 2.96$ (s) A1A1

Note: Award M1A1A0 if the two correct answers are given with additional values outside $0 \leq t \leq 3$ .

[3 marks]

speed is greatest at $t = 3$ (A1)

$a = - 1.83778 \dots$

$a = - 1.84 ({m s}^{- 2})$ A1

[2 marks]

Examiners report

In part (a) many did not realize the change of motion occurred when $v = 0$ . A common error was finding $v (0)$ or thinking that it was at the maximum of $v$ .

In part (b), most candidates knew to differentiate but some tried to substitute in $- 1.9$ for $t$ , while others struggled to differentiate the function by hand rather than using the GDC. Many candidates tried to solve the equation analytically and did not use their technology. Of those who did, many had their calculators in degree mode.

Almost all candidates who attempted part (c) thought the greatest speed was the same as the maximum of $v$ .

[N/A]

A scientist conducted a nine-week experiment on two plants, $A$ and $B$ , of the same species. He wanted to determine the effect of using a new plant fertilizer. Plant $A$ was given fertilizer regularly, while Plant $B$ was not.

The scientist found that the height of Plant $A, h_{A} cm$ , at time $t$ weeks can be modelled by the function $h_{A} (t) = \sin (2 t + 6) + 9 t + 27$ , where $0 \leq t \leq 9$ .

The scientist found that the height of Plant $B, h_{B} cm$ , at time $t$ weeks can be modelled by the function $h_{B} (t) = 8 t + 32$ , where $0 \leq t \leq 9$ .

Use the scientist’s models to find the initial height of

Plant $B$ .

[1]

a.i.

Plant $A$ correct to three significant figures.

[2]

a.ii.

Find the values of $t$ when $h_{A} (t) = h_{B} (t)$ .

[3]

For $0 \leq t \leq 9$ , find the total amount of time when the rate of growth of Plant $B$ was greater than the rate of growth of Plant $A$ .

[6]

Markscheme

$32$ (cm) A1

[1 mark]

a.i.

$h_{A} (0) = \sin (6) + 27$ (M1)

$= 26.7205 \dots$

$= 26.7$ (cm) A1

[2 marks]

a.ii.

attempts to solve $h_{A} (t) = h_{B} (t)$ for $t$ (M1)

$t = 4.00746 \dots, 4.70343 \dots, 5.88332 \dots$

$t = 4.01, 4.70, 5.88$ (weeks) A2

[3 marks]

recognises that $h_{A}' (t)$ and $h_{B}' (t)$ are required (M1)

attempts to solve $h_{A}' (t) = h_{B}' (t)$ for $t$ (M1)

$t = 1.18879 \dots$ and $2.23598 \dots$ OR $4.33038 \dots$ and $5.37758 \dots$ OR $7.47197 \dots$ and $8.51917 \dots$ (A1)

Note: Award full marks for $t = \frac{4 π}{3} - 3, \frac{5 π}{3} - 3, (\frac{7 π}{3} - 3, \frac{8 π}{3} - 3 \frac{10 π}{3} - 3, \frac{11 π}{3} - 3)$ .

Award subsequent marks for correct use of these exact values.

$1.18879 \dots < t < 2.23598 \dots$ OR $4.33038 \dots < t < 5.37758 \dots$ OR

$7.47197 \dots < t < 8.51917 \dots$ (A1)

attempts to calculate the total amount of time (M1)

$3 (2.2359 \dots - 1.1887 \dots) (= 3 ((\frac{5 π}{3} - 3) - (\frac{4 π}{3} - 3)))$

$= 3.14 (= π)$ (weeks) A1

[6 marks]

Examiners report

Many students did not change their calculators back to radian mode. This meant they had no chance of correctly answering parts (c) and (d), since even if follow through was given, there were not enough intersections on the graphs.

Most managed part (a) and some attempted to equate the functions in part b) but few recognised that 'rate of growth' was the derivatives of the given functions, and of those who did, most were unable to find them.

Almost all the candidates who did solve part (c) gave the answer $3 \times 1.05 = 3.15$ , when working with more significant figures would have given them 3.14. They lost the last mark.

a.i.

[N/A]

a.ii.

[N/A]

A manufacturer makes trash cans in the form of a cylinder with a hemispherical top. The trash can has a height of 70 cm. The base radius of both the cylinder and the hemispherical top is 20 cm.

A designer is asked to produce a new trash can.

The new trash can will also be in the form of a cylinder with a hemispherical top.

This trash can will have a height of H cm and a base radius of r cm.

There is a design constraint such that H + 2r = 110 cm.

The designer has to maximize the volume of the trash can.

Write down the height of the cylinder.

[1]

Find the total volume of the trash can.

[4]

Find the height of the cylinder, h , of the new trash can, in terms of r.

[2]

Show that the volume, V cm³ , of the new trash can is given by

$V = 110\pi {r^3}$ .

[3]

Using your graphic display calculator, find the value of r which maximizes the value of V.

[2]

The designer claims that the new trash can has a capacity that is at least 40% greater than the capacity of the original trash can.

State whether the designer’s claim is correct. Justify your answer.

[4]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

50 (cm) (A1)

[1 mark]

$\pi \times 50 \times {20^2} + \frac{1}{2} \times \frac{4}{3} \times \pi \times {20^3}$ (M1)(M1)(M1)

Note: Award (M1) for their correctly substituted volume of cylinder, (M1) for correctly substituted volume of sphere formula, (M1) for halving the substituted volume of sphere formula. Award at most (M1)(M1)(M0) if there is no addition of the volumes.

$= 79600\,\,\left( {{\text{c}}{{\text{m}}^3}} \right)\,\,\left( {79587.0 \ldots \left( {{\text{c}}{{\text{m}}^3}} \right)\,,\,\,\frac{{76000}}{3}\pi } \right)$ (A1)(ft) (G3)

Note: Follow through from part (a).

[4 marks]

h = H − r (or equivalent) OR H = 110 − 2r (M1)

Note: Award (M1) for writing h in terms of H and r or for writing H in terms of r.

(h =) 110 − 3r (A1) (G2)

[2 marks]

$\left( {V = } \right)\,\,\,\,\frac{2}{3}\pi {r^3} + \pi {r^2} \times \left( {110 - 3r} \right)$ (M1)(M1)(M1)

Note: Award (M1) for volume of hemisphere, (M1) for correct substitution of their h into the volume of a cylinder, (M1) for addition of two correctly substituted volumes leading to the given answer. Award at most (M1)(M1)(M0) for subsequent working that does not lead to the given answer. Award at most (M1)(M1)(M0) for substituting H = 110 − 2r as their h.

$V = 110\pi {r^2} - \frac{7}{3}\pi {r^3}$ (AG)

[3 marks]

(r =) 31.4 (cm) (31.4285… (cm)) (G2)

$\left( \pi \right)\left( {220r - 7{r^2}} \right) = 0$ (M1)

Note: Award (M1) for setting the correct derivative equal to zero.

(r =) 31.4 (cm) (31.4285… (cm)) (A1)

[2 marks]

$\left( {V = } \right)\,\,\,\,110\pi {\left( {31.4285 \ldots } \right)^3} - \frac{7}{3}\pi {\left( {31.4285 \ldots } \right)^3}$ (M1)

Note: Award (M1) for correct substitution of their 31.4285… into the given equation.

= 114000 (113781…) (A1)(ft)

Note: Follow through from part (e).

(increase in capacity =) $\frac{{113.781 \ldots - 79587.0 \ldots }}{{79587.0 \ldots }} \times 100 = 43.0\,\,\left( {\text{% }} \right)$ (R1)(ft)

Note: Award (R1)(ft) for finding the correct percentage increase from their two volumes.

1.4 × 79587.0… = 111421.81… (R1)(ft)

Note: Award (R1)(ft) for finding the capacity of a trash can 40% larger than the original.

Claim is correct (A1)(ft)

Note: Follow through from parts (b), (e) and within part (f). The final (R1)(A1)(ft) can be awarded for their correct reason and conclusion. Do not award (R0)(A1)(ft).

[4 marks]

Examiners report

[N/A]

A particle P moves along a straight line. The velocity v m s⁻¹ of P after t seconds is given by v (t) = 7 cos t − 5t ^{cos t}, for 0 ≤ t ≤ 7.

The following diagram shows the graph of v.

Find the initial velocity of P.

[2]

Find the maximum speed of P.

[3]

Write down the number of times that the acceleration of P is 0 m s⁻² .

[3]

Find the acceleration of P when it changes direction.

[4]

Find the total distance travelled by P.

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

initial velocity when t = 0 (M1)

eg v(0)

v = 7 (m s⁻¹) A1 N2

[2 marks]

recognizing maximum speed when $\left| v \right|$ is greatest (M1)

eg minimum, maximum, v' = 0

one correct coordinate for minimum (A1)

eg 6.37896, −24.6571

24.7 (ms⁻¹) A1 N2

[3 marks]

recognizing a = v ′ (M1)

eg $a = \frac{{{\text{d}}v}}{{{\text{d}}t}}$ , correct derivative of first term

identifying when a = 0 (M1)

eg turning points of v, t-intercepts of v ′

3 A1 N3

[3 marks]

recognizing P changes direction when v = 0 (M1)

t = 0.863851 (A1)

−9.24689

a = −9.25 (ms⁻²) A2 N3

[4 marks]

correct substitution of limits or function into formula (A1)
eg $\int_0^7 {\left| {\,v\,} \right|,\,\int_0^{0.8638} {v{\text{d}}t - \int_{0.8638}^7 {v{\text{d}}t} } ,\,\,\int {\left| {\,7\,{\text{cos}}\,x - 5{x^{{\text{cos}}\,x}}\,} \right|} \,dx,\,\,3.32 = 60.6}$

63.8874

63.9 (metres) A2 N3

[3 marks]

Examiners report

[N/A]

A particle P moves along a straight line. Its velocity ${v_{\text{P}}}{\text{ m}}\,{{\text{s}}^{ - 1}}$ after $t$ seconds is given by ${v_{\text{P}}} = \sqrt t \sin \left( {\frac{\pi }{2}t} \right)$ , for $0 \leqslant t \leqslant 8$ . The following diagram shows the graph of ${v_{\text{P}}}$ .

M17/5/MATME/SP2/ENG/TZ1/07

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

[1]

a.i.

Find the total distance travelled by P, for $0 \leqslant t \leqslant 8$ .

[2]

a.ii.

A second particle Q also moves along a straight line. Its velocity, ${v_{\text{Q}}}{\text{ m}}\,{{\text{s}}^{ - 1}}$ after $t$ seconds is given by ${v_{\text{Q}}} = \sqrt t$ for $0 \leqslant t \leqslant 8$ . After $k$ seconds Q has travelled the same total distance as P.

Find $k$ .

[4]

Markscheme

$t = 2$ A1 N1

[1 mark]

a.i.

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

eg $\,\,\,\,\,$ $\int_0^8 {\left| v \right|{\text{d}}t,{\text{ }}\int {\left| {{v_Q}} \right|{\text{d}}t,{\text{ }}\int_0^2 {v{\text{d}}t - \int_2^4 {v{\text{d}}t + \int_4^6 {v{\text{d}}t - \int_6^8 {v{\text{d}}t} } } } } }$

9.64782

distance $= 9.65{\text{ (metres)}}$ A1 N2

[2 marks]

a.ii.

correct approach (A1)

eg $\,\,\,\,\,$ $s = \int {\sqrt t ,{\text{ }}\int_0^k {\sqrt t } } {\text{d}}t,{\text{ }}\int_0^k {\left| {{v_{\text{Q}}}} \right|{\text{d}}t}$

correct integration (A1)

eg $\,\,\,\,\,$ $\int {\sqrt t = \frac{2}{3}{t^{\frac{3}{2}}} + c,{\text{ }}\left[ {\frac{2}{3}{x^{\frac{3}{2}}}} \right]_0^k,{\text{ }}\frac{2}{3}{k^{\frac{3}{2}}}}$

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

eg $\,\,\,\,\,$ $\frac{2}{3}{k^{\frac{3}{2}}} = 9.65,{\text{ }}\int_0^k {\sqrt t {\text{d}}t = 9.65}$

5.93855

5.94 (seconds) A1 N3

[4 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

Let $f\left( x \right) = \frac{{16}}{x}$ . The line $L$ is tangent to the graph of $f$ at $x = 8$ .

$L$ can be expressed in the form r $= \left( {\begin{array}{*{20}{c}} 8 \\ 2 \end{array}} \right) + t$ u.

The direction vector of $y = x$ is $\left( {\begin{array}{*{20}{c}} 1 \\ 1 \end{array}} \right)$ .

Find the gradient of $L$ .

[2]

Find u.

[2]

Find the acute angle between $y = x$ and $L$ .

[5]

Find $\left( {f \circ f} \right)\left( x \right)$ .

[3]

d.i.

Hence, write down ${f^{ - 1}}\left( x \right)$ .

[1]

d.ii.

Hence or otherwise, find the obtuse angle formed by the tangent line to $f$ at $x = 8$ and the tangent line to $f$ at $x = 2$ .

[3]

d.iii.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

attempt to find $f'\left( 8 \right)$ (M1)

eg $f'\left( x \right)$ , ${y'}$ , $- 16{x^{ - 2}}$

−0.25 (exact) A1 N2

[2 marks]

u null or any scalar multiple A2 N2

[2 marks]

correct scalar product and magnitudes (A1)(A1)(A1)

scalar product $= 1 \times 4 + 1 \times - 1\,\,\,\left( { = 3} \right)$

magnitudes $= \sqrt {{1^2} + {1^2}}$ , $\sqrt {{4^2} + {{\left( { - 1} \right)}^2}}$ $\left( { = \sqrt 2 {\text{,}}\,\,\sqrt {17} } \right)$

substitution of their values into correct formula (M1)

eg $\frac{{4 - 1}}{{\sqrt {{1^2} + {1^2}} \sqrt {{4^2} + {{\left( { - 1} \right)}^2}} }}$ , $\frac{{ - 3}}{{\sqrt 2 \sqrt {17} }}$ , 2.1112, 120.96°

1.03037 , 59.0362°

angle = 1.03 , 59.0° A1 N4

[5 marks]

attempt to form composite $\left( {f \circ f} \right)\left( x \right)$ (M1)

eg $f\left( {f\left( x \right)} \right)$ , $f\left( {\frac{{16}}{x}} \right)$ , $\frac{{16}}{{f\left( x \right)}}$

correct working (A1)

eg $\frac{{16}}{{\frac{{16}}{x}}}$ , $16 \times \frac{x}{{16}}$

$\left( {f \circ f} \right)\left( x \right) = x$ A1 N2

[3 marks]

d.i.

${f^{ - 1}}\left( x \right) = \frac{{16}}{x}$ (accept $y = \frac{{16}}{x}$ , $\frac{{16}}{x}$ ) A1 N1

Note: Award A0 in part (ii) if part (i) is incorrect.
Award A0 in part (ii) if the candidate has found ${f^{ - 1}}\left( x \right) = \frac{{16}}{x}$ by interchanging $x$ and $y$ .

[1 mark]

d.ii.

METHOD 1

recognition of symmetry about $y = x$ (M1)

eg (2, 8) ⇔ (8, 2)

evidence of doubling their angle (M1)

eg $2 \times 1.03$ , $2 \times 59.0$

2.06075, 118.072°

2.06 (radians) (118 degrees) A1 N2

METHOD 2

finding direction vector for tangent line at $x = 2$ (A1)

eg $\left( {\begin{array}{*{20}{c}} { - 1} \\ 4 \end{array}} \right)$ , $\left( {\begin{array}{*{20}{c}} 1 \\ { - 4} \end{array}} \right)$

substitution of their values into correct formula (must be from vectors) (M1)

eg $\frac{{ - 4 - 4}}{{\sqrt {{1^2} + {4^2}} \sqrt {{4^2} + {{\left( { - 1} \right)}^2}} }}$ , $\frac{8}{{\sqrt {17} \sqrt {17} }}$

2.06075, 118.072°

2.06 (radians) (118 degrees) A1 N2

METHOD 3

using trigonometry to find an angle with the horizontal (M1)

eg ${\text{tan}}\,\theta = - \frac{1}{4}$ , ${\text{tan}}\,\theta = - 4$

finding both angles of rotation (A1)

eg ${\theta _1} = 0.244978{\text{, 14}}{\text{.0362}}^\circ$ , ${\theta _1} = 1.81577{\text{, 104}}{\text{.036}}^\circ$

2.06075, 118.072°

2.06 (radians) (118 degrees) A1 N2

[3 marks]

d.iii.

Examiners report

[N/A]

d.i.

[N/A]

d.ii.

[N/A]

d.iii.

Let $f''\left( x \right) = \left( {{\text{cos}}\,2x} \right)\left( {{\text{sin}}\,6x} \right)$ , for 0 ≤ $x$ ≤ 1.

Sketch the graph of ${f''}$ on the grid below:

[3]

Find the $x$ -coordinates of the points of inflexion of the graph of $f$ .

[3]

Hence find the values of $x$ for which the graph of $f$ is concave-down.

[2]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

A1A1A1 N3

Note: Only if the shape is approximately correct with exactly 2 maximums and 1 minimum on the interval 0 ≤ $x$ ≤ 0, award the following:
A1 for correct domain with both endpoints within circle and oval.
A1 for passing through the other $x$ -intercepts within the circles.
A1 for passing through the three turning points within circles (ignore $x$ -intercepts and extrema outside of the domain).

[3 marks]

evidence of reasoning (may be seen on graph) (M1)

eg $f'' = 0$ , (0.524, 0), (0.785, 0)

0.523598, 0.785398

$x = 0.524\,\,\left( { = \frac{\pi }{6}} \right)$ , $x = 0.785\,\,\left( { = \frac{\pi }{4}} \right)$ A1A1 N3

Note: Award M1A1A0 if any solution outside domain (eg $x = 0$ ) is also included.

[3 marks]

$0.524 < x < 0.785\,\,\,\left( {\frac{\pi }{6} < x < \frac{\pi }{4}} \right)$ A2 N2

Note: Award A1 if any correct interval outside domain also included, unless additional solutions already penalized in (b).
Award A0 if any incorrect intervals are also included.

[2 marks]

Examiners report

[N/A]

All lengths in this question are in metres.

Consider the function $f\left( x \right) = \sqrt {\frac{{4 - {x^2}}}{8}}$ , for −2 ≤ $x$ ≤ 2. In the following diagram, the shaded region is enclosed by the graph of $f$ and the $x$ -axis.

A container can be modelled by rotating this region by 360˚ about the $x$ -axis.

Water can flow in and out of the container.

The volume of water in the container is given by the function $g\left( t \right)$ , for 0 ≤ $t$ ≤ 4 , where $t$ is measured in hours and $g\left( t \right)$ is measured in m³. The rate of change of the volume of water in the container is given by $g'\left( t \right) = 0.9 - 2.5\,{\text{cos}}\left( {0.4{t^2}} \right)$ .

The volume of water in the container is increasing only when $p$ < $t$ < $q$ .

Find the volume of the container.

[3]

Find the value of $p$ and of $q$ .

[3]

b.i.

During the interval $p$ < $t$ < $q$ , he volume of water in the container increases by $k$ m³. Find the value of $k$ .

[3]

b.ii.

When $t$ = 0, the volume of water in the container is 2.3 m³. It is known that the container is never completely full of water during the 4 hour period.

Find the minimum volume of empty space in the container during the 4 hour period.

[5]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

attempt to substitute correct limits or the function into formula involving ${f^2}$ (M1)

eg $\pi \int_{ - 2}^2 {{y^2}\,{\text{d}}y}$ , $\pi {\int {\left( {\sqrt {\frac{{4 - {x^2}}}{8}} } \right)} ^2}{\text{d}}x$

4.18879

volume = 4.19, $\frac{4}{3}\pi$ (exact) (m³) A2 N3

Note: If candidates have their GDC incorrectly set in degrees, award M marks where appropriate, but no A marks may be awarded. Answers from degrees are p = 13.1243 and q = 26.9768 in (b)(i) and 12.3130 or 28.3505 in (b)(ii).

[3 marks]

recognizing the volume increases when ${g'}$ is positive (M1)

eg $g'\left( t \right)$ > 0, sketch of graph of ${g'}$ indicating correct interval

1.73387, 3.56393

$p$ = 1.73, $p$ = 3.56 A1A1 N3

[3 marks]

b.i.

valid approach to find change in volume (M1)

eg $g\left( q \right) - g\left( p \right)$ , $\int_p^q {g'\left( t \right){\text{d}}t}$

3.74541

total amount = 3.75 (m³) A2 N3

[3 marks]

b.ii.

Note: There may be slight differences in the final answer, depending on which values candidates carry through from previous parts. Accept answers that are consistent with correct working.

recognizing when the volume of water is a maximum (M1)

eg maximum when $t = q$ , $\int_0^q {g'\left( t \right){\text{d}}t}$

valid approach to find maximum volume of water (M1)

eg $2.3 + \int_0^q {g'\left( t \right){\text{d}}t}$ , $2.3 + \int_0^p {g'\left( t \right){\text{d}}t} + 3.74541$ , 3.85745

correct expression for the difference between volume of container and maximum value (A1)

eg $4.18879 - \left( {2.3 + \int_0^q {g'\left( t \right){\text{d}}t} } \right)$ , 4.19 − 3.85745

0.331334

0.331 (m³) A2 N3

[5 marks]

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

A particle moves along a straight line so that its velocity, $v$ m s⁻¹, after $t$ seconds is given by $v\left( t \right) = {1.4^t} - 2.7$ , for 0 ≤ $t$ ≤ 5.

Find when the particle is at rest.

[2]

Find the acceleration of the particle when $t = 2$ .

[2]

Find the total distance travelled by the particle.

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

valid approach (M1)

eg $v\left( t \right) = 0$ , sketch of graph

2.95195

$t = {\text{lo}}{{\text{g}}_{1.4}}2.7$ (exact), $t = 2.95$ (s) A1 N2

[2 marks]

valid approach (M1)

eg $a\left( t \right) = v'\left( t \right)$ , $v'\left( 2 \right)$

0.659485

$a\left( 2 \right)$ = 1.96 ln 1.4 (exact), $a\left( 2 \right)$ = 0.659 (m s⁻²) A1 N2

[2 marks]

correct approach (A1)

eg $\int_0^5 {\left| {v\left( t \right)} \right|} \,{\text{d}}t$ , $\int_0^{2.95} {\left( { - v\left( t \right)} \right)} \,{\text{d}}t + \int_{295}^5 {v\left( t \right)} \,{\text{d}}t$

5.3479

distance = 5.35 (m) A2 N3

[3 marks]

Examiners report

[N/A]

Let $f(x) = x{{\text{e}}^{ - x}}$ and $g(x) = - 3f(x) + 1$ .

The graphs of $f$ and $g$ intersect at $x = p$ and $x = q$ , where $p < q$ .

Find the value of $p$ and of $q$ .

[3]

Hence, find the area of the region enclosed by the graphs of $f$ and $g$ .

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

valid attempt to find the intersection (M1)

eg $\,\,\,\,\,$ $f = g$ , sketch, one correct answer

$p = 0.357402,{\text{ }}q = 2.15329$

$p = 0.357,{\text{ }}q = 2.15$ A1A1 N3

[3 marks]

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

eg $\,\,\,\,\,$ $\int_p^q {\left[ {f(x) - g(x)} \right]{\text{d}}x,{\text{ }}} \int_p^q {f(x){\text{d}}x - } \int_p^q {g(x){\text{d}}x}$

0.537667

${\text{area}} = 0.538$ A2 N3

[3 marks]

Examiners report

[N/A]

The displacement, in centimetres, of a particle from an origin, O, at time t seconds, is given by s(t) = t ² cos t + 2t sin t, 0 ≤ t ≤ 5.

Find the maximum distance of the particle from O.

[3]

Find the acceleration of the particle at the instant it first changes direction.

[4]

Markscheme

use of a graph to find the coordinates of the local minimum (M1)

s = −16.513... (A1)

maximum distance is 16.5 cm (to the left of O) A1

[3 marks]

attempt to find time when particle changes direction eg considering the first maximum on the graph of s or the first t – intercept on the graph of s'. (M1)

t = 1.51986... (A1)

attempt to find the gradient of s' for their value of t, s" (1.51986...) (M1)

=–8.92 (cm/s²) A1

[4 marks]

Examiners report

[N/A]

A function $f$ is given by $f(x) = (2x + 2)(5 - {x^2})$ .

The graph of the function $g(x) = {5^x} + 6x - 6$ intersects the graph of $f$ .

Expand the expression for $f(x)$ .

[1]

b.i.

Find $f’(x)$ .

[3]

b.ii.

Draw the graph of $f$ for $- 3 \leqslant x \leqslant 3$ and $- 40 \leqslant y \leqslant 20$ . Use a scale of 2 cm to represent 1 unit on the $x$ -axis and 1 cm to represent 5 units on the $y$ -axis.

[4]

Write down the coordinates of the point of intersection.

[2]

Markscheme

$10x - 2{x^3} + 10 - 2{x^2}$ (A1)

Notes: The expansion may be seen in part (b)(ii).

[1 mark]

b.i.

$10 - 6{x^2} - 4x$ (A1)(ft)(A1)(ft)(A1)(ft)

Notes: Follow through from part (b)(i). Award (A1)(ft) for each correct term. Award at most (A1)(ft)(A1)(ft)(A0) if extra terms are seen.

[3 marks]

b.ii.

N17/5/MATSD/SP2/ENG/TZ0/05.d/M (A1)(A1)(ft)(A1)(ft)(A1)

Notes: Award (A1) for correct scale; axes labelled and drawn with a ruler.

Award (A1)(ft) for their correct $x$ -intercepts in approximately correct location.

Award (A1) for correct minimum and maximum points in approximately correct location.

Award (A1) for a smooth continuous curve with approximate correct shape. The curve should be in the given domain.

Follow through from part (a) for the $x$ -intercepts.

[4 marks]

$(1.49,{\text{ }}13.9){\text{ }}\left( {(1.48702 \ldots ,{\text{ }}13.8714 \ldots )} \right)$ (G1)(ft)(G1)(ft)

Notes: Award (G1) for 1.49 and (G1) for 13.9 written as a coordinate pair. Award at most (G0)(G1) if parentheses are missing. Accept $x = 1.49$ and $y = 13.9$ . Follow through from part (b)(i).

[2 marks]

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

In this question distance is in centimetres and time is in seconds.

Particle A is moving along a straight line such that its displacement from a point P, after $t$ seconds, is given by ${s_{\text{A}}} = 15 - t - 6{t^3}{{\text{e}}^{ - 0.8t}}$ , 0 ≤ $t$ ≤ 25. This is shown in the following diagram.

Another particle, B, moves along the same line, starting at the same time as particle A. The velocity of particle B is given by ${v_{\text{B}}} = 8 - 2t$ , 0 ≤ $t$ ≤ 25.

Find the initial displacement of particle A from point P.

[2]

Find the value of $t$ when particle A first reaches point P.

[2]

Find the value of $t$ when particle A first changes direction.

[2]

Find the total distance travelled by particle A in the first 3 seconds.

[3]

Given that particles A and B start at the same point, find the displacement function ${s_{\text{B}}}$ for particle B.

[5]

e.i.

Find the other value of $t$ when particles A and B meet.

[2]

e.ii.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

valid approach (M1)

eg ${s_{\text{A}}}\left( 0 \right){\text{,}}\,\,s\left( 0 \right){\text{,}}\,\,t = 0$

15 (cm) A1 N2

[2 marks]

valid approach (M1)

eg ${s_{\text{A}}} = 0{\text{,}}\,\,s = 0{\text{,}}\,\,6.79321{\text{,}}\,\,14.8651$

2.46941

$t$ = 2.47 (seconds) A1 N2

[2 marks]

recognizing when change in direction occurs (M1)

eg slope of $s$ changes sign, $s' = 0$ , minimum point, 10.0144, (4.08, −4.66)

4.07702

$t$ = 4.08 (seconds) A1 N2

[2 marks]

METHOD 1 (using displacement)

correct displacement or distance from P at $t = 3$ (seen anywhere) (A1)

eg −2.69630, 2.69630

valid approach (M1)

eg 15 + 2.69630, $s\left( 3 \right) - s\left( 0 \right)$ , −17.6963

17.6963

17.7 (cm) A1 N2

METHOD 2 (using velocity)

attempt to substitute either limits or the velocity function into distance formula involving $\left| v \right|$ (M1)

eg $\int_0^3 {\left| v \right|{\text{d}}t}$ , $\int {\left| { - 1 - 18{t^2}{{\text{e}}^{ - 0.8t}} + 4.8{t^3}{{\text{e}}^{ - 0.8t}}} \right|}$

17.6963

17.7 (cm) A1 N2

[3 marks]

recognize the need to integrate velocity (M1)

eg $\int {v\left( t \right)}$

$8t - \frac{{2{t^2}}}{2} + c$ (accept $x$ instead of $t$ and missing $c$ ) (A2)

substituting initial condition into their integrated expression (must have $c$ ) (M1)

eg $15 = 8\left( 0 \right) - \frac{{2{{\left( 0 \right)}^2}}}{2} + c$ , $c = 15$

${s_{\text{B}}}\left( t \right) = 8t - {t^2} + 15$ A1 N3

[5 marks]

e.i.

valid approach (M1)

eg ${s_{\text{A}}} = {s_{\text{B}}}$ , sketch, (9.30404, 2.86710)

9.30404

$t = 9.30$ (seconds) A1 N2

Note: If candidates obtain ${s_{\text{B}}}\left( t \right) = 8t - {t^2}$ in part (e)(i), there are 2 solutions for part (e)(ii), 1.32463 and 7.79009. Award the last A1 in part (e)(ii) only if both solutions are given.

[2 marks]

e.ii.

Examiners report

[N/A]

e.i.

[N/A]

e.ii.

A particle moves along a straight line so that its velocity, $v {m s}^{- 1}$ , after $t$ seconds is given by $v (t) = e^{\sin t} + 4 \sin t$ for $0 \leq t \leq 6$ .

Find the value of $t$ when the particle is at rest.

[2]

Find the acceleration of the particle when it changes direction.

[3]

Find the total distance travelled by the particle.

[2]

Markscheme

recognizing at rest $v = 0$ (M1)

$t = 3.34692 \dots$

$t = 3.35$ (seconds) A1

Note: Award (M1)A0 for additional solutions to $v = 0$ eg $t = - 0.205$ or $t = 6.08$ .

[2 marks]

recognizing particle changes direction when $v = 0$ OR when $t = 3.34692 \dots$ (M1)

$a = - 4.71439 \dots$

$a = - 4.71 ({ms}^{- 2})$ A2

[3 marks]

distance travelled $= \int_{0}^{6} |v| d t$ OR

$\int_{0}^{3.34 \dots} (e^{\sin (t)} + 4 \sin (t)) d t - \int_{3.34 \dots}^{6} (e^{\sin (t)} + 4 \sin (t)) d t (= 14.3104 \dots + 6.44300 \dots)$ (A1)

$= 20.7534 \dots$

$= 20.8$ (metres) A1

[2 marks]

Examiners report

The majority of candidates found this question challenging but were often able to gain some of the marks in each part. However, it was not uncommon to see candidates manage either all of this question, or none of it, which unfortunately suggested that not all candidates had covered this content.

In part (a), while many recognized $v = 0$ when the particle is at rest, a common error was to assume that $t = 0$ . It was pleasing to see the majority of those who had the correct equation, manage to progress to the correct value of t, having recognised the domain and the angle measure. Inevitably, a few candidates ignored the domain and obtained $t = - 0.205$ , or found $t = 0$ .

Part (b) was not well done. The most successful approach was to use the GDC to find the gradient of the curve at the value of t obtained in part (a). This was well communicated, concise and generally accurate, although some either rounded incorrectly, or obtained $a = 4.71$ rather than $a = - 4.71$ . Of those that did not use the GDC, many were aware that $a (t) = v' (t)$ and made an attempt to find an expression for $v' (t)$ . However, the majority appeared not to recognise when the particle would change direction and went on to substitute an incorrect value of $t$ , not realising that they had obtained the required value earlier.

Those that had been successful in parts (a) and (b), particularly if they had been using their GDC, were generally able to complete part (c). However, it was disappointing that many candidates who understood that an integral was required, did not refer to the formula booklet and omitted the absolute value from the integrand.

[N/A]

A particle moves in a straight line such that its velocity, $v {ms}^{- 1}$ , at time $t$ seconds is given by

$v = 4 t^{2} - 6 t + 9 - 2 \sin (4 t), 0 \leq t \leq 1$ .

The particle’s acceleration is zero at $t = T$ .

Find the value of $T$ .

[2]

Let $s_{1}$ be the distance travelled by the particle from $t = 0$ to $t = T$ and let $s_{2}$ be the distance travelled by the particle from $t = T$ to $t = 1$ .

Show that $s_{2} > s_{1}$ .

[3]

Markscheme

* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.

attempts either graphical or symbolic means to find the value of $t$ when $\frac{d v}{d t} = 0$ (M1)

$T = 0.465 (s)$ A1

[2 marks]

attempts to find the value of either $s_{1} = \int_{0}^{0.46494 \dots} v d t$ or $s_{2} = \int_{0.46494 \dots}^{1} v d t$ (M1)

$s_{1} = 3.02758 \dots$ and $s_{2} = 3.47892 \dots$ A1A1

Note: Award as above for obtaining, for example, $s_{2} - s_{1} = 0.45133 \dots$ or $\frac{s_{2}}{s_{1}} = 1.14907 \dots$

Note: Award a maximum of M1A1A0FT for use of an incorrect value of $T$ from part (a).

so $s_{2} > s_{1}$ AG

[3 marks]

Examiners report

[N/A]

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 $t$ is given by $a = 3{t^2} - 14t + 8$ , for $0 \leqslant t \leqslant 5$ .

When $t = 0$ , the velocity of P is $3{\text{ m}}\,{{\text{s}}^{ - 1}}$ .

Write down the values of $t$ when $a = 0$ .

[2]

Hence or otherwise, find all possible values of $t$ for which the velocity of P is decreasing.

[2]

Find an expression for the velocity of P at time $t$ .

[6]

Find the total distance travelled by P when its velocity is increasing.

[4]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$t = \frac{2}{3}{\text{ (exact), }}0.667,{\text{ }}t = 4$ A1A1 N2

[2 marks]

recognizing that $v$ is decreasing when $a$ is negative (M1)

eg $\,\,\,\,\,$ $a < 0,{\text{ }}3{t^2} - 14t + 8 \leqslant 0$ , sketch of $a$

correct interval A1 N2

eg $\,\,\,\,\,$ $\frac{2}{3} < t < 4$

[2 marks]

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

eg $\,\,\,\,\,$ $v\int a$

correct integration (accept missing $c$ ) (A1)(A1)(A1)

${t^3} - 7{t^2} + 8t + c$

substituting $t = 0,{\text{ }}v = 3$ , (must have $c$ ) (M1)

eg $\,\,\,\,\,$ $3 = {0^3} - 7({0^2}) + 8(0) + c,{\text{ }}c = 3$

$v = {t^3} - 7{t^2} + 8t + 3$ A1 N6

[6 marks]

recognizing that $v$ increases outside the interval found in part (b) (M1)

eg $\,\,\,\,\,$ $0 < t < \frac{2}{3},{\text{ }}4 < t < 5$ , diagram

one correct substitution into distance formula (A1)

eg $\,\,\,\,\,$ $\int_0^{\frac{2}{3}} {\left| v \right|,{\text{ }}\int_4^5 {\left| v \right|} ,{\text{ }}\int_{\frac{2}{3}}^4 {\left| v \right|} ,{\text{ }}\int_0^5 {\left| v \right|} }$

one correct pair (A1)

eg $\,\,\,\,\,$ 3.13580 and 11.0833, 20.9906 and 35.2097

14.2191 A1 N2

$d = 14.2{\text{ (m)}}$

[4 marks]

Examiners report

[N/A]

Consider a function $f (x)$ , for $x \geq 0$ . The derivative of $f$ is given by $f' (x) = \frac{6 x}{x^{2} + 4}$ .

The graph of $f$ is concave-down when $x > n$ .

Show that $f'' (x) = \frac{24 - 6 x^{2}}{{(x^{2} + 4)}^{2}}$ .

[4]

Find the least value of $n$ .

[2]

Find $\int \frac{6 x}{x^{2} + 4} d x$ .

[3]

Let $R$ be the region enclosed by the graph of $f$ , the $x$ -axis and the lines $x = 1$ and $x = 3$ . The area of $R$ is $19.6$ , correct to three significant figures.

Find $f (x)$ .

[7]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

evidence of choosing the quotient rule (M1)

eg $\frac{v u' - u v'}{v^{2}}$

derivative of $6 x$ is $6$ (must be seen in rule) (A1)

derivative of $x^{2} + 4$ is $2 x$ (must be seen in rule) (A1)

correct substitution into the quotient rule A1

eg $\frac{6 (x^{2} + 4) - (6 x) (2 x)}{{(x^{2} + 4)}^{2}}$

$f'' (x) = \frac{24 - 6 x^{2}}{{(x^{2} + 4)}^{2}}$ AG N0

METHOD 2

evidence of choosing the product rule (M1)

eg $v u' + u v'$

derivative of $6 x$ is $6$ (must be seen in rule) (A1)

derivative of ${(x^{2} + 4)}^{- 1}$ is $- 2 x {(x^{2} + 4)}^{- 2}$ (must be seen in rule) (A1)

correct substitution into the product rule A1

eg $6 {(x^{2} + 4)}^{- 1} + (- 1) (6 x) (2 x) {(x^{2} + 4)}^{- 2}$

$f'' (x) = \frac{24 - 6 x^{2}}{{(x^{2} + 4)}^{2}}$ AG N0

[4 marks]

METHOD 1 (2^nd derivative) (M1)

valid approach

eg $f'' < 0, 24 - 6 x^{2} < 0, n = \pm 2, x = 2$

$n = 2$ (exact) A1 N2

METHOD 2 (1^st derivative)

valid attempt to find local maximum on $f'$ (M1)

eg sketch with max indicated, $(2, 1.5), x = 2$

$n = 2$ (exact) A1 N2

[2 marks]

evidence of valid approach using substitution or inspection (M1)

eg $\int 3 (2 x) \frac{1}{u} d x, u = x^{2} + 4, d u = 2 x d x, \int 3 \times (\frac{1}{u}) d u$

$\int \frac{6 x}{(x^{2} + 4)} d x = 3 \ln (x^{2} + 4) + c$ A2 N3

[3 marks]

recognizing that area $= \int_{1}^{3} f (x) d x$ (seen anywhere) (M1)

recognizing that their answer to (c) is their $f (x)$ (accept absence of $c$ ) (M1)

eg $f (x) = 3 \ln (x^{2} + 4) + c, f (x) = 3 \ln (x^{2} + 4)$

correct value for $\int_{1}^{3} 3 \ln (x^{2} + 4) d x$ (seen anywhere) (A1)

eg $12.4859$

correct integration for $\int_{1}^{3} c d x$ (seen anywhere) (A1)

${[c x]}_{1}^{3}, 2 c$

adding their integrated expressions and equating to $19.6$ (do not accept an expression which involves an integral) (M1)

eg $12.4859 + 2 c = 19.6, 2 c = 7.114$

$c = 3.55700$ (A1)

$f (x) = 3 \ln (x^{2} + 4) + 3.56$ A1 N4

[7 marks]

Examiners report

[N/A]

A particle P starts from a point A and moves along a horizontal straight line. Its velocity $v{\text{ cm}}\,{{\text{s}}^{ - 1}}$ after $t$ seconds is given by

$v(t) = \left\{ {\begin{array}{*{20}{l}} { - 2t + 2,}&{{\text{for }}0 \leqslant t \leqslant 1} \\ {3\sqrt t + \frac{4}{{{t^2}}} - 7,}&{{\text{for }}1 \leqslant t \leqslant 12} \end{array}} \right.$

The following diagram shows the graph of $v$ .

N16/5/MATME/SP2/ENG/TZ0/09

P is at rest when $t = 1$ and $t = p$ .

When $t = q$ , the acceleration of P is zero.

Find the initial velocity of $P$ .

[2]

Find the value of $p$ .

[2]

(i) Find the value of $q$ .

(ii) Hence, find the speed of P when $t = q$ .

[4]

(i) Find the total distance travelled by P between $t = 1$ and $t = p$ .

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

[6]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

valid attempt to substitute $t = 0$ into the correct function (M1)

eg $\,\,\,\,\,$ $- 2(0) + 2$

2 A1 N2

[2 marks]

recognizing $v = 0$ when P is at rest (M1)

5.21834

$p = 5.22{\text{ }}({\text{seconds}})$ A1 N2

[2 marks]

(i) recognizing that $a = v'$ (M1)

eg $\,\,\,\,\,$ $v' = 0$ , minimum on graph

1.95343

$q = 1.95$ A1 N2

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

eg $\,\,\,\,\,$ $v(q),{\text{ }} - 1.75879$ , reference to min on graph

1.75879

speed $= 1.76{\text{ }}(c\,{\text{m}}\,{{\text{s}}^{ - 1}})$ A1 N2

[4 marks]

(i) substitution of correct $v(t)$ into distance formula, (A1)

eg $\,\,\,\,\,$ $\int_1^p {\left| {3\sqrt t + \frac{4}{{{t^2}}} - 7} \right|{\text{d}}t,{\text{ }}\left| {\int {3\sqrt t + \frac{4}{{{t^2}}} - 7{\text{d}}t} } \right|}$

4.45368

distance $= 4.45{\text{ }}({\text{cm}})$ A1 N2

(ii) displacement from $t = 1$ to $t = p$ (seen anywhere) (A1)

eg $\,\,\,\,\,$ $- 4.45368,{\text{ }}\int_1^p {\left( {3\sqrt t + \frac{4}{{{t^2}}} - 7} \right){\text{d}}t}$

displacement from $t = 0$ to $t = 1$ (A1)

eg $\,\,\,\,\,$ $\int_0^1 {( - 2t + 2){\text{d}}t,{\text{ }}0.5 \times 1 \times 2,{\text{ 1}}}$

valid approach to find displacement for $0 \leqslant t \leqslant p$ M1

eg $\,\,\,\,\,$ $\int_0^1 {( - 2t + 2){\text{d}}t + \int_1^p {\left( {3\sqrt t + \frac{4}{{{t^2}}} - 7} \right){\text{d}}t,{\text{ }}\int_0^1 {( - 2t + 2){\text{d}}t - 4.45} } }$

$- 3.45368$

displacement $= - 3.45{\text{ }}({\text{cm}})$ A1 N2

[6 marks]

Examiners report

[N/A]

Let $f(x) = {({x^2} + 3)^7}$ . Find the term in ${x^5}$ in the expansion of the derivative, $f’(x)$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

derivative of $f(x)$ A2

$7{({x^2} + 3)^6}(x2)$

recognizing need to find ${x^4}$ term in ${({x^2} + 3)^6}$ (seen anywhere) R1

eg $\,\,\,\,\,$ $14x{\text{ (term in }}{x^4})$

valid approach to find the terms in ${({x^2} + 3)^6}$ (M1)

eg $\,\,\,\,\,$ $\left( {\begin{array}{*{20}{c}} 6 \\ r \end{array}} \right){({x^2})^{6 - r}}{(3)^r},{\text{ }}{({x^2})^6}{(3)^0} + {({x^2})^5}{(3)^1} + \ldots$ , Pascal’s triangle to 6th row

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

eg $\,\,\,\,\,$ ${\text{5th term, }}r = 2,{\text{ }}\left( {\begin{array}{*{20}{c}} 6 \\ 4 \end{array}} \right),{\text{ }}{({x^2})^2}{(3)^4}$

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

eg $\,\,\,\,\,$ $\left( {\begin{array}{*{20}{c}} 6 \\ 4 \end{array}} \right){({x^2})^2}{(3)^4},{\text{ }}15 \times {3^4},{\text{ }}14x \times 15 \times 81{({x^2})^2}$

$17010{x^5}$ A1 N3

METHOD 2

recognition of need to find ${x^6}$ in ${({x^2} + 3)^7}$ (seen anywhere) R1

valid approach to find the terms in ${({x^2} + 3)^7}$ (M1)

eg $\,\,\,\,\,$ $\left( {\begin{array}{*{20}{c}} 7 \\ r \end{array}} \right){({x^2})^{7 - r}}{(3)^r},{\text{ }}{({x^2})^7}{(3)^0} + {({x^2})^6}{(3)^1} + \ldots$ , Pascal’s triangle to 7th row

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

eg $\,\,\,\,\,$ 6th term, $r = 3,{\text{ }}\left( {\begin{array}{*{20}{c}} 7 \\ 3 \end{array}} \right),{\text{ (}}{{\text{x}}^2}{)^3}{(3)^4}$

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

eg $\,\,\,\,\,$ $\left( {\begin{array}{*{20}{c}} 7 \\ 4 \end{array}} \right){{\text{(}}{{\text{x}}^2})^3}{(3)^4},{\text{ }}35 \times {3^4}$

correct term (A1)

$2835{x^6}$

differentiating their term in ${x^6}$ (M1)

eg $\,\,\,\,\,$ $(2835{x^6})',{\text{ (6)(2835}}{{\text{x}}^5})$

$17010{x^5}$ A1 N3

[7 marks]

Examiners report

[N/A]

Consider the function $f$ defined by $f (x) = 90 e^{- 0.5 x}$ for $x \in ℝ^{+}$ .

The graph of $f$ and the line $y = x$ intersect at point $P$ .

The line $L$ has a gradient of $- 1$ and is a tangent to the graph of $f$ at the point $Q$ .

The shaded region $A$ is enclosed by the graph of $f$ and the lines $y = x$ and $L$ .

Find the $x$ -coordinate of $P$ .

[2]

Find the exact coordinates of $Q$ .

[4]

Show that the equation of $L$ is $y = - x + 2 \ln 45 + 2$ .

[2]

Find the $x$ -coordinate of the point where $L$ intersects the line $y = x$ .

[1]

d.i.

Hence, find the area of $A$ .

[4]

d.ii.

The line $L$ is tangent to the graphs of both $f$ and the inverse function $f^{- 1}$ .

Find the shaded area enclosed by the graphs of $f$ and $f^{- 1}$ and the line $L$ .

[2]

Markscheme

Attempt to find the point of intersection of the graph of $f$ and the line $y = x$ (M1)

$x = 5.56619 \dots$

$= 5.57$ A1

[2 marks]

$f' (x) = - 45 e^{- 0.5 x}$ A1

attempt to set the gradient of $f$ equal to $- 1$ (M1)

$- 45 e^{- 0.5 x} = -1$

$Q$ has coordinates $(2 \ln 45, 2)$ (accept ( $- 2 \ln \frac{1}{45}, 2$ ) A1A1

Note: Award A1 for each value, even if the answer is not given as a coordinate pair.

Do not accept $\frac{\ln \frac{1}{45}}{- 0.5}$ or $\frac{\ln 45}{0.5}$ as a final value for $x$ . Do not accept $2.0$ or $2.00$ as a final value for $y$ .

[4 marks]

attempt to substitute coordinates of $Q$ (in any order) into an appropriate equation (M1)

$y - 2 = - (x - 2 \ln 45)$ OR $2 = - 2 \ln 45 + c$ A1

equation of $L$ is $y = - x + 2 \ln 45 + 2$ AG

[2 marks]

$x = \ln 45 + 1 (= 4.81)$ A1

[1 mark]

d.i.

appropriate method to find the sum of two areas using integrals of the difference of two functions (M1)

Note: Allow absence of incorrect limits.

$\int_{4.806 \dots}^{5.566 \dots} (x - (- x + 2 \ln 45 + 2)) d x + \int_{5.566 \dots}^{7.613 \dots} (90 e^{- 0.5 x} - (- x + 2 \ln 45 + 2)) d x$ (A1)(A1)

Note: Award A1 for one correct integral expression including correct limits and integrand.
Award A1 for a second correct integral expression including correct limits and integrand.

$= 1.52196 \dots$

$= 1.52$ A1

[4 marks]

d.ii.

by symmetry $2 \times 1.52 \dots$ (M1)

$= 3.04$ A1

Note: Accept any answer that rounds to $3.0$ (but do not accept $3$ ).

[2 marks]

Examiners report

[N/A]

d.i.

[N/A]

d.ii.

[N/A]

A particle moves in a straight line. The velocity, $v {ms}^{- 1}$ , of the particle at time $t$ seconds is given by $v (t) = t \sin t - 3$ , for $0 \leq t \leq 10$ .

The following diagram shows the graph of $v$ .

Find the smallest value of $t$ for which the particle is at rest.

[2]

Find the total distance travelled by the particle.

[2]

Find the acceleration of the particle when $t = 7$ .

[2]

Markscheme

recognising $v = 0$ (M1)

$t = 6.74416 \dots$

$= 6.74$ (sec) A1

Note: Do not award A1 if additional values are given.

[2 marks]

$\int_{0}^{10} |v (t)| d t$ OR $- \int_{0}^{6.74416 \dots} v (t) d t + \int_{6.74416 \dots}^{9.08837 \dots} v (t) d t - \int_{9.08837 \dots}^{10} v (t) d t$ (A1)

$= 37.0968 \dots$

$= 37.1 (m)$ A1

[2 marks]

recognizing acceleration at $t = 7$ is given by $v' (7)$ (M1)

acceleration $= 5.93430 \dots$

$= 5.93 ({ms}^{- 2})$ A1

[2 marks]

Examiners report

[N/A]

Hyungmin designs a concrete bird bath. The bird bath is supported by a pedestal. This is shown in the diagram.

The interior of the bird bath is in the shape of a cone with radius $r$ , height $h$ and a constant slant height of $50 cm$ .

Let $V$ be the volume of the bird bath.

Hyungmin wants the bird bath to have maximum volume.

Write down an equation in $r$ and $h$ that shows this information.

[1]

Show that $V = \frac{2500 π h}{3} - \frac{π h^{3}}{3}$ .

[1]

Find $\frac{d V}{d h}$ .

[2]

Using your answer to part (c), find the value of $h$ for which $V$ is a maximum.

[2]

Find the maximum volume of the bird bath.

[2]

To prevent leaks, a sealant is applied to the interior surface of the bird bath.

Find the surface area to be covered by the sealant, given that the bird bath has maximum volume.

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$h^{2} + r^{2} = 50^{2}$ (or equivalent) (A1)

Note: Accept equivalent expressions such as $r = \sqrt{2500 - h^{2}}$ or $h = \sqrt{2500 - r^{2}}$ . Award (A0) for a final answer of $\pm \sqrt{2500 - h^{2}}$ or $\pm \sqrt{2500 - r^{2}}$ , or any further incorrect working.

[1 mark]

$\frac{1}{3} \times π \times (2500 - h^{2}) \times h$ OR $\frac{1}{3} \times π \times {(\sqrt{2500 - h^{2}})}^{2} \times h$ (M1)

Note: Award (M1) for correct substitution in the volume of cone formula.

$V = \frac{2500 π h}{3} - \frac{π h^{3}}{3}$ (AG)

Note: The final line must be seen, with no incorrect working, for the (M1) to be awarded.

[1 mark]

$(\frac{d V}{d h} =) \frac{2500 π}{3} - π h^{2}$ (A1)(A1)

Note: Award (A1) for $\frac{2500 π}{3}$ , (A1) for $- π h^{2}$ . Award at most (A1)(A0) if extra terms are seen. Award (A0) for the term $- \frac{3 π h^{2}}{3}$ .

[2 marks]

$0 = \frac{2500 π}{3} - π h^{2}$ (M1)

Note: Award (M1) for equating their derivative to zero. Follow through from part (c).

sketch of $\frac{d V}{d h}$ (M1)

Note: Award (M1) for a labelled sketch of $\frac{d V}{d h}$ with the curve/axes correctly labelled or the $x$ -intercept explicitly indicated.

$(h =) 28.9 (cm) (\sqrt{\frac{2500}{3}}, \frac{50}{\sqrt{3}}, \frac{50 \sqrt{3}}{3}, 28.8675 \dots)$ (A1)(ft)

Note: An unsupported $28.9 cm$ is awarded no marks. Graphing the function $V (h)$ is not an acceptable method and (M0)(A0) should be awarded. Follow through from part (c). Given the restraints of the question, $h \geq 50$ is not possible.

[2 marks]

$(V =) \frac{2500 \times π \times 28.8675 \dots}{3} - \frac{π {(28.8675 \dots)}^{3}}{3}$ (M1)

$\frac{1}{3} π {(40.828 \dots)}^{2} \times 28.8675 \dots$ (M1)

Note: Award (M1) for substituting their $28.8675 \dots$ in the volume formula.

$(V =) 50400 ({cm}^{3}) (50383.3 \dots)$ (A1)(ft)(G2)

Note: Follow through from part (d).

[2 marks]

$(S =) π \times \sqrt{2500 - {(28.8675 \dots)}^{2}} \times 50$ (A1)(ft)(M1)

Note: Award (A1) for their correct radius seen $(40.8248 \dots, \sqrt{2500 - {(28.8675 \dots)}^{2}})$ .
Award (M1) for correctly substituted curved surface area formula for a cone.

$(S =) 6410 ({cm}^{2}) (6412.74 \dots)$ (A1)(ft)(G2)

Note: Follow through from parts (a) and (d).

[3 marks]

Examiners report

[N/A]

All lengths in this question are in metres.

Let $f(x) = - 0.8{x^2} + 0.5$ , for $- 0.5 \leqslant x \leqslant 0.5$ . Mark uses $f(x)$ as a model to create a barrel. The region enclosed by the graph of $f$ , the $x$ -axis, the line $x = - 0.5$ and the line $x = 0.5$ is rotated 360° about the $x$ -axis. This is shown in the following diagram.

N16/5/MATME/SP2/ENG/TZ0/06

Use the model to find the volume of the barrel.

[3]

The empty barrel is being filled with water. The volume $V{\text{ }}{{\text{m}}^3}$ of water in the barrel after $t$ minutes is given by $V = 0.8(1 - {{\text{e}}^{ - 0.1t}})$ . How long will it take for the barrel to be half-full?

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

attempt to substitute correct limits or the function into the formula involving

${y^2}$

eg $\,\,\,\,\,$ $\pi \int_{ - 0.5}^{0.5} {{y^2}{\text{d}}x,{\text{ }}\pi \int {{{( - 0.8{x^2} + 0.5)}^2}{\text{d}}x} }$

0.601091

volume $= 0.601{\text{ }}({{\text{m}}^3})$ A2 N3

[3 marks]

attempt to equate half their volume to $V$ (M1)

eg $\,\,\,\,\,$ $0.30055 = 0.8(1 - {{\text{e}}^{ - 0.1t}})$ , graph

4.71104

4.71 (minutes) A2 N3

[3 marks]

Examiners report

[N/A]

Let g(x) = −(x − 1)² + 5.

Let f(x) = x². The following diagram shows part of the graph of f.

The graph of g intersects the graph of f at x = −1 and x = 2.

Write down the coordinates of the vertex of the graph of g.

[1]

On the grid above, sketch the graph of g for −2 ≤ x ≤ 4.

[3]

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

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

(1,5) (exact) A1 N1

[1 mark]

A1A1A1 N3

Note: The shape must be a concave-down parabola.
Only if the shape is correct, award the following for points in circles:
A1 for vertex,
A1 for correct intersection points,
A1 for correct endpoints.

[3 marks]

integrating and subtracting functions (in any order) (M1)
eg $\int {f - g}$

correct substitution of limits or functions (accept missing dx, but do not accept any errors, including extra bits) (A1)
eg $\int_{ - 1}^2 {g - f,\,\,\int { - {{\left( {x - 1} \right)}^2}} } + 5 - {x^2}$

area = 9 (exact) A1 N2

[3 marks]

Examiners report

[N/A]

Let $f\left( x \right) = \,\,{\text{sin}}\,\left( {{e^x}} \right)$ for 0 ≤ $x$ ≤ 1.5. The following diagram shows the graph of $f$ .

Find the x-intercept of the graph of $f$ .

[2]

The region enclosed by the graph of $f$ , the y-axis and the x-axis is rotated 360° about the x-axis.

Find the volume of the solid formed.

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

valid approach (M1)
eg $f\left( x \right) = 0,\,\,\,\,{e^x} = 180$ or 0…

1.14472

$x = {\text{ln}}\,\pi$ (exact), 1.14 A1 N2

[2 marks]

attempt to substitute either their limits or the function into formula involving ${f^2}$ . (M1)

eg ${\int_0^{1.14} {{f^2},\,\,\pi \int {\left( {{\text{sin}}\,\left( {{e^x}} \right)} \right)} } ^2}dx,\,\,0.795135$

2.49799

volume = 2.50 A2 N3

[3 marks]

Examiners report

[N/A]

A water container is made in the shape of a cylinder with internal height $h$ cm and internal base radius $r$ cm.

N16/5/MATSD/SP2/ENG/TZ0/06

The water container has no top. The inner surfaces of the container are to be coated with a water-resistant material.

The volume of the water container is $0.5{\text{ }}{{\text{m}}^3}$ .

The water container is designed so that the area to be coated is minimized.

One can of water-resistant material coats a surface area of $2000{\text{ c}}{{\text{m}}^2}$ .

Write down a formula for $A$ , the surface area to be coated.

[2]

Express this volume in ${\text{c}}{{\text{m}}^3}$ .

[1]

Write down, in terms of $r$ and $h$ , an equation for the volume of this water container.

[1]

Show that $A = \pi {r^2} + \frac{{1\,000\,000}}{r}$ .

[2]

Find $\frac{{{\text{d}}A}}{{{\text{d}}r}}$ .

[3]

Using your answer to part (e), find the value of $r$ which minimizes $A$ .

[3]

Find the value of this minimum area.

[2]

Find the least number of cans of water-resistant material that will coat the area in part (g).

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$(A = ){\text{ }}\pi {r^2} + 2\pi rh$ (A1)(A1)

Note: Award (A1) for either $\pi {r^2}$ OR $2\pi rh$ seen. Award (A1) for two correct terms added together.

[2 marks]

$500\,000$ (A1)

Notes: Units not required.

[1 mark]

$500\,000 = \pi {r^2}h$ (A1)(ft)

Notes: Award (A1)(ft) for $\pi {r^2}h$ equating to their part (b).

Do not accept unless $V = \pi {r^2}h$ is explicitly defined as their part (b).

[1 mark]

$A = \pi {r^2} + 2\pi r\left( {\frac{{500\,000}}{{\pi {r^2}}}} \right)$ (A1)(ft)(M1)

Note: Award (A1)(ft) for their ${\frac{{500\,000}}{{\pi {r^2}}}}$ seen.

Award (M1) for correctly substituting only ${\frac{{500\,000}}{{\pi {r^2}}}}$ into a correct part (a).

Award (A1)(ft)(M1) for rearranging part (c) to $\pi rh = \frac{{500\,000}}{r}$ and substituting for $\pi rh$ in expression for $A$ .

$A = \pi {r^2} + \frac{{1\,000\,000}}{r}$ (AG)

Notes: The conclusion, $A = \pi {r^2} + \frac{{1\,000\,000}}{r}$ , must be consistent with their working seen for the (A1) to be awarded.

Accept ${10^6}$ as equivalent to ${1\,000\,000}$ .

[2 marks]

$2\pi r - \frac{{{\text{1}}\,{\text{000}}\,{\text{000}}}}{{{r^2}}}$ (A1)(A1)(A1)

Note: Award (A1) for $2\pi r$ , (A1) for $\frac{1}{{{r^2}}}$ or ${r^{ - 2}}$ , (A1) for $- {\text{1}}\,{\text{000}}\,{\text{000}}$ .

[3 marks]

$2\pi r - \frac{{1\,000\,000}}{{{r^2}}} = 0$ (M1)

Note: Award (M1) for equating their part (e) to zero.

${r^3} = \frac{{1\,000\,000}}{{2\pi }}$ OR $r = \sqrt[3]{{\frac{{1\,000\,000}}{{2\pi }}}}$ (M1)

Note: Award (M1) for isolating $r$ .

sketch of derivative function (M1)

with its zero indicated (M1)

$(r = ){\text{ }}54.2{\text{ }}({\text{cm}}){\text{ }}(54.1926 \ldots )$ (A1)(ft)(G2)

[3 marks]

$\pi {(54.1926 \ldots )^2} + \frac{{1\,000\,000}}{{(54.1926 \ldots )}}$ (M1)

Note: Award (M1) for correct substitution of their part (f) into the given equation.

$= 27\,700{\text{ }}({\text{c}}{{\text{m}}^2}){\text{ }}(27\,679.0 \ldots )$ (A1)(ft)(G2)

[2 marks]

$\frac{{27\,679.0 \ldots }}{{2000}}$ (M1)

Note: Award (M1) for dividing their part (g) by 2000.

$= 13.8395 \ldots$ (A1)(ft)

Notes: Follow through from part (g).

14 (cans) (A1)(ft)(G3)

Notes: Final (A1) awarded for rounding up their $13.8395 \ldots$ to the next integer.

[3 marks]

Examiners report

[N/A]

A group of 66 people went on holiday to Hawaii. During their stay, three trips were arranged: a boat trip ( $B$ ), a coach trip ( $C$ ) and a helicopter trip ( $H$ ).

From this group of people:

3	went on all three trips;
16	went on the coach trip only;
13	went on the boat trip only;
5	went on the helicopter trip only;
x	went on the coach trip and the helicopter trip but not the boat trip;
2x	went on the boat trip and the helicopter trip but not the coach trip;
4x	went on the boat trip and the coach trip but not the helicopter trip;
8	did not go on any of the trips.

One person in the group is selected at random.

Draw a Venn diagram to represent the given information, using sets labelled $B$ , $C$ and $H$ .

[5]

Show that $x = 3$ .

[2]

Write down the value of $n(B \cap C)$ .

[1]

Find the probability that this person

(i) went on at most one trip;

(ii) went on the coach trip, given that this person also went on both the helicopter trip and the boat trip.

[4]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

N16/5/MATSD/SP2/ENG/TZ0/02.a/M (A5)

Notes: Award (A1) for rectangle and three labelled intersecting circles (U need not be seen),

(A1) for 3 in the correct region,

(A1) for 8 in the correct region,

(A1) for 5, 13 and 16 in the correct regions,

(A1) for $x$ , $2x$ and $4x$ in the correct regions.

[5 marks]

$8 + 13 + 16 + 3 + 5 + x + 2x + 4x = 66$ (M1)

Note: Award (M1) for either a completely correct equation or adding all the terms from their diagram in part (a) and equating to 66.

Award (M0)(A0) if their equation has no $x$ .

$7x = 66 - 45$ OR $7x + 45 = 66$ (A1)

Note: Award (A1) for adding their like terms correctly, but only when the solution to their equation is equal to 3 and is consistent with their original equation.

$x = 3$ (AG)

Note: The conclusion $x = 3$ must be seen for the (A1) to be awarded.

[2 marks]

15 (A1)(ft)

Note: Follow through from part (a). The answer must be an integer.

[1 mark]

(i) $\frac{{42}}{{66}}{\text{ }}\left( {\frac{7}{{11}},{\text{ }}0.636,{\text{ }}63.6\% } \right)$ (A1)(ft)(A1)(G2)

Note: Award (A1)(ft) for numerator, (A1) for denominator. Follow through from their Venn diagram.

(ii) $\frac{3}{9}{\text{ }}\left( {\frac{1}{3},{\text{ }}0.333,{\text{ }}33.3\% } \right)$ (A1)(A1)(ft)(G2)

Note: Award (A1) for numerator, (A1)(ft) for denominator. Follow through from their Venn diagram.

[4 marks]

Examiners report

[N/A]

Consider the curves $y = x^{2} \sin x$ and $y = - 1 - \sqrt{1 + 4 {(x + 2)}^{2}}$ for $- π \leq x \leq 0$ .

Find the $x$ -coordinates of the points of intersection of the two curves.

[3]

Find the area, $A$ , of the region enclosed by the two curves.

[4]

Markscheme

attempts to solve $x^{2} \sin x = - 1 - \sqrt{1 + 4 {(x + 2)}^{2}}$ (M1)

$x = - 2.76, - 1.54$ A1A1

Note: Award A1A0 if additional solutions outside the domain are given.

[3 marks]

$A = \int_{- 2.762 \dots}^{- 1.537 \dots} (- 1 - \sqrt{1 + 4 {(x + 2)}^{2}} - x^{2} \sin x) d x$ (or equivalent) (M1)(A1)

Note: Award M1 for attempting to form an integrand involving “top curve” − “bottom curve”.

so $A = 1.47$ A2

[4 marks]

Examiners report

[N/A]

Let $f\left( x \right) = 4 - 2{{\text{e}}^x}$ . The following diagram shows part of the graph of $f$ .

Find the $x$ -intercept of the graph of $f$ .

[2]

The region enclosed by the graph of $f$ , the $x$ -axis and the $y$ -axis is rotated 360º about the $x$ -axis. Find the volume of the solid formed.

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

valid approach (M1)

eg $f\left( x \right) = 0$ , $4 - 2{{\text{e}}^x} = 0$

0.693147

$x$ = ln 2 (exact), 0.693 A1 N2

[2 marks]

attempt to substitute either their correct limits or the function into formula (M1)

involving ${f^2}$

eg $\int_0^{0.693} {{f^2}}$ , $\pi {\int {\left( {4 - 2{{\text{e}}^x}} \right)} ^2}{\text{d}}x$ , $\int_0^{{\text{ln}}\,2} {{{\left( {4 - 2{{\text{e}}^x}} \right)}^2}}$

3.42545

volume = 3.43 A2 N3

[3 marks]

Examiners report

[N/A]

Find $\int (6 x + 7) d x$ .

[3]

Given $f' (x) = 6 x + 7$ and $f (1.2) = 7.32$ , find $f (x)$ .

[3]

Markscheme

correct integration $3 x^{2} + 7 x + c$ A1A1A1

Note: Award A1 for $3 x^{2}$ , A1 for $7 x$ and A1 for $+ c$

[3 marks]

recognition that $f (x) = \int f' (x) d x$ (M1)

$3 {(1.2)}^{2} + 7 (1.2) + c = 7.32$ (A1)

$c = - 5.4$

$f (x) = 3 x^{2} + 7 x - 5.4$ A1

[3 marks]

Examiners report

[N/A]

Violeta plans to grow flowers in a rectangular plot. She places a fence to mark out the perimeter of the plot and uses 200 metres of fence. The length of the plot is $x$ metres.

M17/5/MATSD/SP2/ENG/TZ2/05

Violeta places the fence so that the area of the plot is maximized.

By selling her flowers, Violeta earns 2 Bulgarian Levs (BGN) per square metre of the plot.

Show that the width of the plot, in metres, is given by $100 - x$ .

[1]

Write down the area of the plot in terms of $x$ .

[1]

Find the value of $x$ that maximizes the area of the plot.

[2]

Show that Violeta earns 5000 BGN from selling the flowers grown on the plot.

[2]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\frac{{200 - 2x}}{2}$ (or equivalent) (M1)

$2x + 2y = 200$ (or equivalent) (M1)

Note: Award (M1) for a correct expression leading to $100 - x$ (the $100 - x$ does not need to be seen). The 200 must be seen for the (M1) to be awarded. Do not accept $100 - x$ substituted in the perimeter of the rectangle formula.

$100 - x$ (AG)

[1 mark]

$({\text{area}} = ){\text{ }}x(100 - x)$ $\,\,\,$ OR $\,\,\,$ $- {x^2} + 100x$ (or equivalent) (A1)

[1 mark]

$x = \frac{{ - 100}}{{ - 2}}$ $\,\,\,$ OR $\,\,\,$ $- 2x + 100 = 0$ $\,\,\,$ OR $\,\,\,$ graphical method (M1)

Note: Award (M1) for use of axis of symmetry formula or first derivative equated to zero or a sketch graph.

$x = 50$ (A1)(ft)(G2)

Note: Follow through from part (b), provided x is positive and less than 100.

[2 marks]

$50(100 - 50) \times 2$ (M1)(M1)

Note: Award (M1) for substituting their $x$ into their formula for area (accept “ $50 \times 50$ ” for the substituted formula), and (M1) for multiplying by 2. Award at most (M0)(M1) if their calculation does not lead to 5000 (BGN), although the 5000 (BGN) does not need to be seen explicitly.

Substitution of 50 into area formula may be seen in part (c).

5000 (BGN) (AG)

[2 marks]

Examiners report

[N/A]

Consider the function $f\left( x \right) = \frac{{48}}{x} + k{x^2} - 58$ , where x > 0 and k is a constant.

The graph of the function passes through the point with coordinates (4 , 2).

P is the minimum point of the graph of f (x).

Sketch the graph of y = f (x) for 0 < x ≤ 6 and −30 ≤ y ≤ 60.
Clearly indicate the minimum point P and the x-intercepts on your graph.

Markscheme

(A1)(A1)(ft)(A1)(ft)(A1)(ft)

Note: Award (A1) for correct window. Axes must be labelled.
(A1)(ft) for a smooth curve with correct shape and zeros in approximately correct positions relative to each other.
(A1)(ft) for point P indicated in approximately the correct position. Follow through from their x-coordinate in part (c). (A1)(ft) for two x-intercepts identified on the graph and curve reflecting asymptotic properties.

[4 marks]

Examiners report

[N/A]

The function $f$ is defined by $f (x) = \frac{4 x + 1}{x + 4}$ , where $x \in ℝ, x \neq - 4$ .

For the graph of $f$

The graphs of $f$ and $f^{- 1}$ intersect at $x = p$ and $x = q$ , where $p < q$ .

write down the equation of the vertical asymptote.

[1]

a.i.

find the equation of the horizontal asymptote.

[2]

a.ii.

Find $f^{- 1} (x)$ .

[4]

b.i.

Using an algebraic approach, show that the graph of $f^{- 1}$ is obtained by a reflection of the graph of $f$ in the $y$ -axis followed by a reflection in the $x$ -axis.

[4]

b.ii.

Find the value of $p$ and the value of $q$ .

[2]

c.i.

Hence, find the area enclosed by the graph of $f$ and the graph of $f^{- 1}$ .

[3]

c.ii.

Markscheme

$x = - 4$ A1

[1 mark]

a.i.

attempt to substitute into $y = \frac{a}{c}$ OR table with large values of $x$ OR sketch of $f$ showing asymptotic behaviour (M1)

$y = 4$ A1

[2 marks]

a.ii.

$y = \frac{4 x + 1}{x + 4}$

attempt to interchange $x$ and $y$ (seen anywhere) M1

$x y + 4 y = 4 x + 1$ OR $x y + 4 x = 4 y + 1$ (A1)

$x y - 4 x = 1 - 4 y$ OR $x y - 4 y = 1 - 4 x$ (A1)

$f^{- 1} (x) = \frac{1 - 4 x}{x - 4}$ (accept $y = \frac{1 - 4 x}{x - 4}$ ) A1

[4 marks]

b.i.

reflection in $y$ -axis given by $f (- x)$ (M1)

$f (- x) = \frac{- 4 x + 1}{- x + 4}$ (A1)

reflection of their $f (- x)$ in $x$ -axis given by $- f (- x)$ accept "now $- f (x)$ " M1

$(- f (- x) =) - \frac{- 4 x + 1}{- x + 4}$

$= \frac{- 4 x + 1}{x - 4}$ OR $\frac{4 x - 1}{- x + 4}$ A1

$= \frac{1 - 4 x}{x - 4} (= f^{- 1} (x))$ AG

Note: If the candidate attempts to show the result using a particular coordinate on the graph of $f$ rather than a general coordinate on the graph of $f$ , where appropriate, award marks as follows:
M0A0 for eg $(2, 3) \to (- 2, 3)$
M0A0 for $(- 2, 3) \to (- 2, - 3)$

[4 marks]

b.ii.

attempt to solve $f (x) = f^{- 1} (x)$ using graph or algebraically (M1)

$p = - 1$ AND $q = 1$ A1

Note: Award (M1)A0 if only one correct value seen.

[2 marks]

c.i.

attempt to set up an integral to find area between $f$ and $f^{- 1}$ (M1)

$\int_{- 1}^{1} (\frac{4 x + 1}{x + 4} - \frac{1 - 4 x}{x - 4}) d x$ (A1)

$= 0.675231 \dots$

$= 0.675$ A1

[3 marks]

c.ii.

Examiners report

Candidates mostly found the first part of this question accessible, with many knowing how to find the equation of both asymptotes. Common errors included transposing the asymptotes, or finding where an asymptote occurred but not giving it as an equation.

Candidates knew how to start part (b)(i), with most attempting to find the inverse function by firstly interchanging $x$ and $y$ . However, many struggled with the algebra required to change the subject, and were not awarded all the marks. A common error in this part was for candidates to attempt to find an expression for $f' (x)$ , rather than one for $f^{- 1} (x)$ . Few candidates were able to answer part (b)(ii). Many appeared not to know that a reflection in the $y$ -axis is given by $f (- x)$ , or that a reflection in the $x$ -axis is given by $- f (x)$ . Many of those that did, multiplied both the numerator and denominator by $- 1$ when taking the negative of their $f (- x)$ , i.e. $- (\frac{- 4 x + 1}{- x + 4})$ was often simplified as $\frac{4 x - 1}{x - 4}$ . However, the majority of candidates either did not attempt this question part or attempted to describe a graphical approach often involving a specific point, rather than an algebraic approach.

Those that attempted part (c), and had the correct expression for $f^{- 1} (x)$ , were usually able to gain all the marks. However, those that had an incorrect expression, or had found $f' (x)$ , often proceeded to find an area, even when there was not an area enclosed by their two curves.

a.i.

[N/A]

a.ii.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

c.i.

[N/A]

c.ii.

Haruka has an eco-friendly bag in the shape of a cuboid with width 12 cm, length 36 cm and height of 9 cm. The bag is made from five rectangular pieces of cloth and is open at the top.

Nanako decides to make her own eco-friendly bag in the shape of a cuboid such that the surface area is minimized.

The width of Nanako’s bag is x cm, its length is three times its width and its height is y cm.

The volume of Nanako’s bag is 3888 cm³.

Calculate the area of cloth, in cm², needed to make Haruka’s bag.

[2]

Calculate the volume, in cm³, of the bag.

[2]

Use this value to write down, and simplify, the equation in x and y for the volume of Nanako’s bag.

[2]

Write down and simplify an expression in x and y for the area of cloth, A, used to make Nanako’s bag.

[2]

Use your answers to parts (c) and (d) to show that

$A = 3{x^2} + \frac{{10368}}{x}$ .

[2]

Find $\frac{{{\text{d}}A}}{{{\text{d}}x}}$ .

[3]

Use your answer to part (f) to show that the width of Nanako’s bag is 12 cm.

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

36 × 12 + 2(9 ×12) + 2(9 × 36) (M1)

Note: Award (M1) for correct substitution into surface area of cuboid formula.

= 1300 (cm²) (1296 (cm²)) (A1)(G2)

[2 marks]

36 × 9 ×12 (M1)

Note: Award (M1) for correct substitution into volume of cuboid formula.

= 3890 (cm³) (3888 (cm³)) (A1)(G2)

[2 marks]

3 $x$ × $x$ × $y$ = 3888 (M1)

Note: Award (M1) for correct substitution into volume of cuboid formula and equated to 3888.

$x$ ² $y$ = 1296 (A1)(G2)

Note: Award (A1) for correct fully simplified volume of cuboid.

Accept $y = \frac{{1296}}{{{x^2}}}$ .

[2 marks]

(A =) 3x² + 2(xy) + 2(3xy) (M1)

Note: Award (M1) for correct substitution into surface area of cuboid formula.

(A =) 3x² + 8xy (A1)(G2)

Note: Award (A1) for correct simplified surface area of cuboid formula.

[2 marks]

$A = 3{x^2} + 8x\left( {\frac{{1296}}{{{x^2}}}} \right)$ (A1)(ft)(M1)

Note: Award (A1)(ft) for correct rearrangement of their part (c) seen (rearrangement may be seen in part(c)), award (M1) for substitution of their part (c) into their part (d) but only if this leads to the given answer, which must be shown.

$A = 3{x^2} + \frac{{10368}}{x}$ (AG)

[2 marks]

$\left( {\frac{{{\text{d}}A}}{{{\text{d}}x}}} \right) = 6x - \frac{{10368}}{{{x^2}}}$ (A1)(A1)(A1)

Note: Award (A1) for $6x$ , (A1) for −10368, (A1) for ${x^{ - 2}}$ . Award a maximum of (A1)(A1)(A0) if any extra terms seen.

[3 marks]

$6x - \frac{{10368}}{{{x^2}}} = 0$ (M1)

Note: Award (M1) for equating their ${\frac{{{\text{d}}A}}{{{\text{d}}x}}}$ to zero.

$6{x^3} = 10368$ OR $6{x^3} - 10368 = 0$ OR ${x^3} - 1728 = 0$ (M1)

Note: Award (M1) for correctly rearranging their equation so that fractions are removed.

$x = \sqrt[3]{{1728}}$ (A1)

$x = 12$ (cm) (AG)

Note: The (AG) line must be seen for the final (A1) to be awarded. Substituting $x = 12$ invalidates the method, award a maximum of (M1)(M0)(A0).

[3 marks]

Examiners report

[N/A]

The derivative of a function $g$ is given by $g' (x) = 3 x^{2} + 5 e^{x}$ , where $x \in ℝ$ . The graph of $g$ passes through the point $(0, 4)$ . Find $g (x)$ .

Markscheme

METHOD 1

recognises that $g (x) = \int (3 x^{2} + 5 e^{x}) d x$ (M1)

$g (x) = x^{3} + 5 e^{x} (+ C)$ (A1)(A1)

Note: Award A1 for each integrated term.

substitutes $x = 0$ and $y = 4$ into their integrated function (must involve $+ C$ ) (M1)

$4 = 0 + 5 + C \Rightarrow C = - 1$

$g (x) = x^{3} + 5 e^{x} - 1$ A1

METHOD 2

attempts to write both sides in the form of a definite integral (M1)

$\int_{0}^{x} g' (t) d t = \int_{0}^{x} (3 t^{2} + 5 e^{t}) d t$ (A1)

$g (x) - 4 = x^{3} + 5 e^{x} - 5 e^{0}$ (A1)(A1)

Note: Award A1 for $g (x) - 4$ and A1 for $x^{3} + 5 e^{x} - 5 e^{0}$ .

$g (x) = x^{3} + 5 e^{x} - 1$ A1

[5 marks]

Examiners report

While many students were successful in solving this question, some did not consider the constant of integration or struggled to integrate the exponential term. A few students lost the final mark for stopping at $C = - 1$ and not giving the formula for $g (x)$ .

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

A particle moves along a horizontal line starting at a fixed point A. The velocity $v$ of the particle, at time $t$ , is given by $v(t) = \frac{{2{t^2} - 4t}}{{{t^2} - 2t + 2}}$ , for $0 \leqslant t \leqslant 5$ . The following diagram shows the graph of $v$

M17/5/MATME/SP2/ENG/TZ2/07

There are $t$ -intercepts at $(0,{\text{ }}0)$ and $(2,{\text{ }}0)$ .

Find the maximum distance of the particle from A during the time $0 \leqslant t \leqslant 5$ and justify your answer.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1 (displacement)

recognizing $s = \int {v{\text{d}}t}$ (M1)

consideration of displacement at $t = 2$ and $t = 5$ (seen anywhere) M1

eg $\,\,\,\,\,$ $\int_0^2 v$ and $\int_0^5 v$

Note: Must have both for any further marks.

correct displacement at $t = 2$ and $t = 5$ (seen anywhere) A1A1

$- 2.28318$ (accept 2.28318), 1.55513

valid reasoning comparing correct displacements R1

eg $\,\,\,\,\,$ $\left| { - 2.28} \right| > \left| {1.56} \right|$ , more left than right

2.28 (m) A1 N1

Note: Do not award the final A1 without the R1.

METHOD 2 (distance travelled)

recognizing distance $= \int {\left| v \right|{\text{d}}t}$ (M1)

consideration of distance travelled from $t = 0$ to 2 and $t = 2$ to 5 (seen anywhere) M1

eg $\,\,\,\,\,$ $\int_0^2 v$ and $\int_2^5 v$

Note: Must have both for any further marks

correct distances travelled (seen anywhere) A1A1

2.28318, (accept $- 2.28318$ ), 3.83832

valid reasoning comparing correct distance values R1

eg $\,\,\,\,\,$ $3.84 - 2.28 < 2.28,{\text{ }}3.84 < 2 \times 2.28$

2.28 (m) A1 N1

Note: Do not award the final A1 without the R1.

[6 marks]

Examiners report

[N/A]