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SL Paper 2

The rate of change of the height $(h)$ of a ball above horizontal ground, measured in metres, $t$ seconds after it has been thrown and until it hits the ground, can be modelled by the equation

$\frac{d h}{d t} = 11.4 - 9.8 t$

The height of the ball when $t = 0$ is $1.2 m$ .

Find an expression for the height $h$ of the ball at time $t$ .

[6]

Find the value of $t$ at which the ball hits the ground.

[2]

b.i.

Hence write down the domain of $h$ .

[1]

b.ii.

Find the range of $h$ .

[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.

$h = \int (11.4 - 9.8 t) d t$ M1

$h = 11.4 t - 4.9 t^{2} (+ c)$ A1A1

When $t = 0, h = 1.2$ (M1)

$c = 1.2$ (A1)

$(h =) 1.2 + 11.4 t - 4.9 t^{2}$ A1

[6 marks]

$2.43 (2.42741 \dots)$ seconds (M1)A1

[2 marks]

b.i.

$0 \leq t \leq 2.43$ A1

Note: Accept $0 \leq t < 2.43$ .

[1 mark]

b.ii.

Maximum value is $7.83061 \dots$ (M1)

Range is $0 \leq h \leq 7.83$ A1A1

Note: Accept $0 \leq h < 7.83$ .

[3 marks]

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[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]

The cross-sectional view of a tunnel is shown on the axes below. The line $[AB]$ represents a vertical wall located at the left side of the tunnel. The height, in metres, of the tunnel above the horizontal ground is modelled by $y = - 0.1 x^{3} + 0.8 x^{2}, 2 \leq x \leq 8$ , relative to an origin $O$ .

Point $A$ has coordinates $(2, 0)$ , point $B$ has coordinates $(2, 2.4)$ , and point $C$ has coordinates $(8, 0)$ .

When $x = 4$ the height of the tunnel is $6.4 m$ and when $x = 6$ the height of the tunnel is $7.2 m$ . These points are shown as $D$ and $E$ on the diagram, respectively.

Find $\frac{d y}{d x}$ .

[2]

a.i.

Hence find the maximum height of the tunnel.

[4]

a.ii.

Use the trapezoidal rule, with three intervals, to estimate the cross-sectional area of the tunnel.

[3]

Write down the integral which can be used to find the cross-sectional area of the tunnel.

[2]

c.i.

Hence find the cross-sectional area of the tunnel.

[2]

c.ii.

Markscheme

evidence of power rule (at least one correct term seen) (M1)

$\frac{d y}{d x} = - 0.3 x^{2} + 1.6 x$ A1

[2 marks]

a.i.

$- 0.3 x^{2} + 1.6 x = 0$ M1

$x = 5.33 (5.33333 \dots, \frac{16}{3})$ A1

$y = - 0.1 \times 5.33333 \dots^{3} + 0.8 \times 5.33333 \dots^{2}$ (M1)

Note: Award M1 for substituting their zero for $\frac{d y}{d x} (5.333 \dots)$ into $y$ .

$7.59 m (7.58519 \dots)$ A1

Note: Award M0A0M0A0 for an unsupported $7.59$ .
Award at most M0A0M1A0 if only the last two lines in the solution are seen.
Award at most M1A0M1A1 if their $x = 5.33$ is not seen.

[6 marks]

a.ii.

$A = \frac{1}{2} \times 2 ((2.4 + 0) + 2 (6.4 + 7.2))$ (A1)(M1)

Note: Award A1 for $h = 2$ seen. Award M1 for correct substitution into the trapezoidal rule (the zero can be omitted in working).

$= 29.6 m^{2}$ A1

[3 marks]

$A = \int_{2}^{8} - 0.1 x^{3} + 0.8 x^{2} d x$ OR $A = \int_{2}^{8} y d x$ A1A1

Note: Award A1 for a correct integral, A1 for correct limits in the correct location. Award at most A0A1 if $d x$ is omitted.

[2 marks]

c.i.

$A = 32.4 m^{2}$ A2

Note: As per the marking instructions, FT from their integral in part (c)(i). Award at most A1FTA0 if their area is $> 48$ , this is outside the constraints of the question (a $6 \times 8$ rectangle).

[2 marks]

c.ii.

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

c.i.

[N/A]

c.ii.

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]

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 $f$ .

Find the exact value of each of the zeros of $f$ .

[3]

Expand the expression for $f(x)$ .

[1]

b.i.

Find $f’(x)$ .

[3]

b.ii.

Use your answer to part (b)(ii) to find the values of $x$ for which $f$ is increasing.

[3]

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

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

$- 1,{\text{ }}\sqrt 5 ,{\text{ }} - \sqrt 5$ (A1)(A1)(A1)

Note: Award (A1) for –1 and each exact value seen. Award at most (A1)(A0)(A1) for use of 2.23606… instead of $\sqrt 5$ .

[3 marks]

$10x - 2{x^3} + 10 - 2{x^2}$ (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.

$10 - 6{x^2} - 4x > 0$ (M1)

Notes: Award (M1) for their $f’(x) > 0$ . Accept equality or weak inequality.

$- 1.67 < x < 1{\text{ }}\left( { - \frac{5}{3} < x < 1,{\text{ }} - 1.66666 \ldots < x < 1} \right)$ (A1)(ft)(A1)(ft)(G2)

Notes: Award (A1)(ft) for correct endpoints, (A1)(ft) for correct weak or strict inequalities. Follow through from part (b)(ii). Do not award any marks if there is no answer in part (b)(ii).

[3 marks]

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 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]

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.

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.

(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]

e.i.

[N/A]

e.ii.

[N/A]

e.iii.

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

Find $f'\left( x \right)$ .

[3]

Find the gradient of the graph of $y = f\left( x \right)$ at $x = 2$ .

[2]

Find the equation of the tangent line to the graph of $y = f\left( x \right)$ at $x = 2$ . Give the equation in the form $ax + by + d = 0$ where, $a$ , $b$ , and $d \in \mathbb{Z}$ .

[2]

Markscheme

${x^2} + \frac{3}{2}x - 1$ (A1)(A1)(A1)

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

[3 marks]

${2^2} + \frac{3}{2} \times 2 - 1$ (M1)

Note: Award (M1) for correct substitution of 2 in their derivative of the function.

6 (A1)(ft)(G2)

Note: Follow through from part (d).

[2 marks]

$\frac{8}{3} = 6\left( 2 \right) + c$ (M1)

Note: Award (M1) for 2, their part (a) and their part (e) substituted into equation of a straight line.

$c = - \frac{{28}}{3}$

$\left( {y - \frac{8}{3}} \right) = 6\left( {x - 2} \right)$ (M1)

Note: Award (M1) for 2, their part (a) and their part (e) substituted into equation of a straight line.

$y = 6x - \frac{{28}}{3}\,\,\left( {y = 6x - 9.33333 \ldots } \right)$ (M1)

Note: Award (M1) for their answer to (e) and intercept $- \frac{{28}}{3}$ substituted in the gradient-intercept line equation.

$- 18x + 3y + 28 = 0$ (accept integer multiples) (A1)(ft)(G2)

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

[2 marks]

Examiners report

[N/A]

The following diagram shows the graph of $f(x) = a\sin bx + c$ , for $0 \leqslant x \leqslant 12$ .

N16/5/MATME/SP2/ENG/TZ0/10

The graph of $f$ has a minimum point at $(3,{\text{ }}5)$ and a maximum point at $(9,{\text{ }}17)$ .

The graph of $g$ is obtained from the graph of $f$ by a translation of $\left( {\begin{array}{*{20}{c}} k \\ 0 \end{array}} \right)$ . The maximum point on the graph of $g$ has coordinates $(11.5,{\text{ }}17)$ .

The graph of $g$ changes from concave-up to concave-down when $x = w$ .

(i) Find the value of $c$ .

(ii) Show that $b = \frac{\pi }{6}$ .

(iii) Find the value of $a$ .

[6]

(i) Write down the value of $k$ .

(ii) Find $g(x)$ .

[3]

(i) Find $w$ .

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

[6]

Markscheme

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

(i) valid approach (M1)

eg $\,\,\,\,\,$ $\frac{{5 + 17}}{2}$

$c = 11$ A1 N2

(ii) valid approach (M1)

eg $\,\,\,\,\,$ period is 12, per $= \frac{{2\pi }}{b},{\text{ }}9 - 3$

$b = \frac{{2\pi }}{{12}}$ A1

$b = \frac{\pi }{6}$ AG N0

(iii) METHOD 1

valid approach (M1)

eg $\,\,\,\,\,$ $5 = a\sin \left( {\frac{\pi }{6} \times 3} \right) + 11$ , substitution of points

$a = - 6$ A1 N2

METHOD 2

valid approach (M1)

eg $\,\,\,\,\,$ $\frac{{17 - 5}}{2}$ , amplitude is 6

$a = - 6$ A1 N2

[6 marks]

(i) $k = 2.5$ A1 N1

(ii) $g(x) = - 6\sin \left( {\frac{\pi }{6}(x - 2.5)} \right) + 11$ A2 N2

[3 marks]

(i) METHOD 1 Using $g$

recognizing that a point of inflexion is required M1

eg $\,\,\,\,\,$ sketch, recognizing change in concavity

evidence of valid approach (M1)

eg $\,\,\,\,\,$ $g''(x) = 0$ , sketch, coordinates of max/min on ${g'}$

$w = 8.5$ (exact) A1 N2

METHOD 2 Using $f$

recognizing that a point of inflexion is required M1

eg $\,\,\,\,\,$ sketch, recognizing change in concavity

evidence of valid approach involving translation (M1)

eg $\,\,\,\,\,$ $x = w - k$ , sketch, $6 + 2.5$

$w = 8.5$ (exact) A1 N2

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

eg $\,\,\,\,\,$ $g'(w),{\text{ }} - \pi \cos \left( {\frac{\pi }{6}x} \right)$ , max on derivative, sketch of derivative

attempt to find max value on derivative M1

eg $\,\,\,\,\,$ $- \pi \cos \left( {\frac{\pi }{6}(8.5 - 2.5)} \right),{\text{ }}f'(6)$ , dot on max of sketch

3.14159

max rate of change $= \pi$ (exact), 3.14 A1 N2

[6 marks]

Examiners report

[N/A]

A hollow chocolate box is manufactured in the form of a right prism with a regular hexagonal base. The height of the prism is $h cm$ , and the top and base of the prism have sides of length $x cm$ .

Given that $\sin 60 ° = \frac{\sqrt{3}}{2}$ , show that the area of the base of the box is equal to $\frac{3 \sqrt{3} x^{2}}{2}$ .

[2]

Given that the total external surface area of the box is $1200 {cm}^{2}$ , show that the volume of the box may be expressed as $V = 300 \sqrt{3} x - \frac{9}{4} x^{3}$ .

[5]

Sketch the graph of $V = 300 \sqrt{3} x - \frac{9}{4} x^{3}$ , for $0 \leq x \leq 16$ .

[2]

Find an expression for $\frac{d V}{d x}$ .

[2]

Find the value of $x$ which maximizes the volume of the box.

[2]

Hence, or otherwise, find the maximum possible volume of the box.

[2]

The box will contain spherical chocolates. The production manager assumes that they can calculate the exact number of chocolates in each box by dividing the volume of the box by the volume of a single chocolate and then rounding down to the nearest integer.

Explain why the production manager is incorrect.

[1]

Markscheme

evidence of splitting diagram into equilateral triangles M1

area $= 6 (\frac{1}{2} x^{2} \sin 60 °)$ A1

$= \frac{3 \sqrt{3} x^{2}}{2}$ AG

Note: The AG line must be seen for the final A1 to be awarded.

[2 marks]

total surface area of prism $1200 = 2 (3 x^{2} \frac{\sqrt{3}}{2}) + 6 x h$ M1A1

Note: Award M1 for expressing total surface areas as a sum of areas of rectangles and hexagons, and A1 for a correctly substituted formula, equated to $1200$ .

$h = \frac{400 - \sqrt{3} x^{2}}{2 x}$ A1

volume of prism $= \frac{3 \sqrt{3}}{2} x^{2} \times h$ (M1)

$= \frac{3 \sqrt{3}}{2} x^{2} (\frac{400 - \sqrt{3} x^{2}}{2 x})$ A1

$= 300 \sqrt{3} x - \frac{9}{4} x^{3}$ AG

Note: The AG line must be seen for the final A1 to be awarded.

[5 marks]

A1A1

Note: Award A1 for correct shape, A1 for roots in correct place with some indication of scale (indicated by a labelled point).

[2 marks]

$\frac{d V}{d x} = 300 \sqrt{3} - \frac{27}{4} x^{2}$ A1A1

Note: Award A1 for a correct term.

[2 marks]

from the graph of $V$ or $\frac{d V}{d x}$ OR solving $\frac{d V}{d x} = 0$ (M1)

$x = 8.88 (8.877382 \dots)$ A1

[2 marks]

from the graph of $V$ OR substituting their value for $x$ into $V$ (M1)

$V_{max} = 3040 {cm}^{3} (3039.34 \dots)$ A1

[2 marks]

EITHER
wasted space / spheres do not pack densely (tesselate) A1

OR
the model uses exterior values / assumes infinite thinness of materials and hence the modelled volume is not the true volume A1

[1 mark]

Examiners report

[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 zero of f (x).

[1]

b.i.

Use your graphic display calculator to find the coordinates of the local minimum point.

[2]

b.ii.

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.

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]

1.19 (1.19055…) (A1)

Note: Accept an answer of (1.19, 0).

Do not follow through from an incorrect sketch.

[1 mark]

b.i.

(−1.5, 36) (A1)(A1)

Note: Award (A0)(A1) if parentheses are omitted.

Accept x = −1.5, y = 36.

[2 marks]

b.ii.

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.

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

b.iii.

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]

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).

Find the value of k.

[2]

Using your value of k , find f ′(x).

[3]

Use your answer to part (b) to show that the minimum value of f(x) is −22 .

[3]

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.

[4]

Markscheme

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

$\frac{{48}}{4} + k \times {4^2} - 58 = 2$ (M1)
Note: Award (M1) for correct substitution of x = 4 and y = 2 into the function.

k = 3 (A1) (G2)

[2 marks]

$\frac{{ - 48}}{{{x^2}}} + 6x$ (A1)(A1)(A1)(ft) (G3)

Note: Award (A1) for −48 , (A1) for x⁻², (A1)(ft) for their 6x. Follow through from part (a). Award at most (A1)(A1)(A0) if additional terms are seen.

[3 marks]

$\frac{{ - 48}}{{{x^2}}} + 6x = 0$ (M1)

Note: Award (M1) for equating their part (b) to zero.

x = 2 (A1)(ft)

Note: Follow through from part (b). Award (M1)(A1) for $\frac{{ - 48}}{{{{\left( 2 \right)}^2}}} + 6\left( 2 \right) = 0$ seen.

Award (M0)(A0) for x = 2 seen either from a graphical method or without working.

$\frac{{48}}{2} + 3 \times {2^2} - 58\,\,\,\left( { = - 22} \right)$ (M1)

Note: Award (M1) for substituting their 2 into their function, but only if the final answer is −22. Substitution of the known result invalidates the process; award (M0)(A0)(M0).

−22 (AG)

[3 marks]

(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]

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 this person is not allergic to nuts.

[2]

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

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

$\frac{{34}}{{60}}{\text{ }}\left( {\frac{{17}}{{30}},{\text{ }}0.567,{\text{ }}0.566666 \ldots ,{\text{ }}56.7\% } \right)$ (A1)(A1)

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

[2 marks]

$\frac{{34}}{{60}} \times \frac{{33}}{{59}}$ (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]

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]

A box of chocolates is to have a ribbon tied around it as shown in the diagram below.

The box is in the shape of a cuboid with a height of $3$ cm. The length and width of the box are $x$ and $y$ cm.

After going around the box an extra $10$ cm of ribbon is needed to form the bow.

The volume of the box is $450 {cm}^{3}$ .

Find an expression for the total length of the ribbon $L$ in terms of $x$ and $y$ .

[2]

Show that $L = 2 x + \frac{300}{x} + 22$

[3]

Find $\frac{d L}{d x}$

[3]

Solve $\frac{d L}{d x} = 0$

[2]

Hence or otherwise find the minimum length of ribbon required.

[2]

Markscheme

$L = 2 x + 2 y + 12 + 10 = 2 x + 2 y + 22$ A1A1

Note: A1 for $2 x + 2 y$ and A1 for $12 + 10$ or $22$ .

[2 marks]

$V = 3 x y = 450$ A1

$y = \frac{150}{x}$ A1

$L = 2 x + 2 (\frac{150}{x}) + 22$ M1

$L = 2 x + \frac{300}{x} + 22$ AG

[3 marks]

$L = 2 x + 300 x^{- 1} + 22$ (M1)

$\frac{d L}{d x} = 2 - \frac{300}{x^{2}}$ A1A1

Note: A1 for $2$ (and $0$ ), A1 for $\frac{300}{x^{2}}$ .

[3 marks]

$\frac{300}{x^{2}} = 2$ (M1)

$x = \sqrt{150} = 12.2 (12.2474 \dots)$ A1

[2 marks]

$L = 2 \sqrt{150} + \frac{300}{\sqrt{150}} + 22 = 71.0 (70.9897 \dots) cm$ (M1)A1

[2 marks]

Examiners report

[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]

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]

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]

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]

A theatre set designer is designing a piece of flat scenery in the shape of a hill. The scenery is formed by a curve between two vertical edges of unequal height. One edge is $2$ metres high and the other is $1$ metre high. The width of the scenery is $6$ metres.

A coordinate system is formed with the origin at the foot of the $2$ metres high edge. In this coordinate system the highest point of the cross‐section is at $(2, 3.5)$ .

A set designer wishes to work out an approximate value for the area of the scenery $(A m^{2})$ .

In order to obtain a more accurate measure for the area the designer decides to model the curved edge with the polynomial $h (x) = a x^{3} + b x^{2} + c x + d a, b, c, d \in ℝ$ where $h$ metres is the height of the curved edge a horizontal distance $x m$ from the origin.

Explain why $A < 21$ .

[1]

By dividing the area between the curve and the $x$ ‐axis into two trapezoids of unequal width show that $A > 14.5$ , justifying the direction of the inequality.

[4]

Write down the value of $d$ .

[1]

Use differentiation to show that $12 a + 4 b + c = 0$ .

[2]

Determine two other linear equations in $a$ , $b$ and $c$ .

[3]

Hence find an expression for $h (x)$ .

[3]

Use the expression found in (f) to calculate a value for $A$ .

[2]

Markscheme

The area $A$ is less than the rectangle containing the cross-section which is equal to $6 \times 3.5 = 21$ R1

Note: $6 \times 3.5 = 21$ is not sufficient for R1.

[1 mark]

$\frac{1}{2} \times 2 \times (2 + 3.5) + \frac{1}{2} \times 4 \times (3.5 + 1)$ (M1)(A1)

$= 14.5$ A1

This is an underestimate as the trapezoids are enclosed by (are under) the curve. R1

Note: This can be shown in a diagram.

[4 marks]

$h (0) = 2 \Rightarrow d = 2$ A1

[1 mark]

$h' (x) = 3 a x^{2} + 2 b x + c$ A1

$h' (2) = 0$ M1

hence $12 a + 4 b + c = 0$ AG

[2 marks]

Substitute the points $(2, 3.5)$ and $(6, 1)$ (M1)

$8 a + 4 b + 2 c + 2 = 3.5 (8 a + 4 b + 2 c = 1.5)$

and

$216 a + 36 b + 6 c + 2 = 1 (216 a + 36 b + 6 c = - 1)$ A1A1

[3 marks]

Solve on a GDC (M1)

$h (x) = 0.0365 x^{3} - 0.521 x^{2} + 1.65 x + 2$ A2

$(h (x) = 0.0364583 \dots x^{3} - 0.520833 \dots x^{2} + 1.64583 \dots x + 2)$

[3 marks]

$\int_{0}^{6} 0.0364583 \dots x^{3} - 0.520833 \dots x^{2} + 1.64583 \dots x + 2 d x$

$= 15.9 (15.9374 \dots) (m^{2})$ (M1)A1

Note: Accept $16.0 (16.014)$ from the three significant figure answer to part (g).

[2 marks]

Examiners report

[N/A]

The Happy Straw Company manufactures drinking straws.

The straws are packaged in small closed rectangular boxes, each with length 8 cm, width 4 cm and height 3 cm. The information is shown in the diagram.

Each week, the Happy Straw Company sells $x$ boxes of straws. It is known that $\frac{{{\text{d}}P}}{{{\text{d}}x}} = - 2x + 220$ , $x$ ≥ 0, where $P$ is the weekly profit, in dollars, from the sale of $x$ thousand boxes.

Calculate the surface area of the box in cm².

[2]

Calculate the length AG.

[2]

Find the number of boxes that should be sold each week to maximize the profit.

[3]

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

[5]

Find the least number of boxes which must be sold each week in order to make a profit.

[3]

Markscheme

2(8 × 4 + 3 × 4 + 3 × 8) M1

= 136 (cm²) A1

[2 marks]

$\sqrt {{8^2} + {4^2} + {3^2}}$ M1

(AG =) 9.43 (cm) (9.4339…, $\sqrt {89}$ ) A1

[2 marks]

$- 2x + 220 = 0$ M1

$x = 110$ A1

110 000 (boxes) A1

[3 marks]

$P\left( x \right) = \int { - 2x + 220} \,{\text{d}}x$ M1

Note: Award M1 for evidence of integration.

$P\left( x \right) = - {x^2} + 220x + c$ A1A1

Note: Award A1 for either $- {x^2}$ or $220x$ award A1 for both correct terms and constant of integration.

$1700 = - {\left( {20} \right)^2} + 220\left( {20} \right) + c$ M1

$c = - 2300$

$P\left( x \right) = - {x^2} + 220x - 2300$ A1

[5 marks]

$- {x^2} + 220x - 2300 = 0$ M1

$x = 11.005$ A1

11 006 (boxes) A1

Note: Award M1 for their $P\left( x \right) = 0$ , award A1 for their correct solution to $x$ .
Award the final A1 for expressing their solution to the minimum number of boxes. Do not accept 11 005, the nearest integer, nor 11 000, the answer expressed to 3 significant figures, as these will not satisfy the demand of the question.

[3 marks]

Examiners report

[N/A]

A cafe makes $x$ litres of coffee each morning. The cafe’s profit each morning, $C$ , measured in dollars, is modelled by the following equation

$C = \frac{x}{10} (k^{2} - \frac{3}{100} x^{2})$

where $k$ is a positive constant.

The cafe’s manager knows that the cafe makes a profit of $$ 426$ when $20$ litres of coffee are made in a morning.

The manager of the cafe wishes to serve as many customers as possible.

Find an expression for $\frac{d C}{d x}$ in terms of $k$ and $x$ .

[3]

Hence find the maximum value of $C$ in terms of $k$ . Give your answer in the form $p k^{3}$ , where $p$ is a constant.

[4]

Find the value of $k$ .

[2]

c.i.

Use the model to find how much coffee the cafe should make each morning to maximize its profit.

[1]

c.ii.

Sketch the graph of $C$ against $x$ , labelling the maximum point and the $x$ -intercepts with their coordinates.

[3]

Determine the maximum amount of coffee the cafe can make that will not result in a loss of money for the morning.

[2]

Markscheme

attempt to expand given expression (M1)

$C = \frac{x k^{2}}{10} - \frac{3 x^{3}}{1000}$

$\frac{d C}{d x} = \frac{k^{2}}{10} - \frac{9 x^{2}}{1000}$ M1A1

Note: Award M1 for power rule correctly applied to at least one term and A1 for correct answer.

[3 marks]

equating their $\frac{d C}{d x}$ to zero (M1)

$\frac{k^{2}}{10} - \frac{9 x^{2}}{1000} = 0$

$x^{2} = \frac{100 k^{2}}{9}$

$x = \frac{10 k}{3}$ (A1)

substituting their $x$ back into given expression (M1)

$C_{max} = \frac{10 k}{30} (k^{2} - \frac{300 k^{2}}{900})$

$C_{max} = \frac{2 k^{3}}{9} (0.222 \dots k^{3})$ A1

[4 marks]

substituting $20$ into given expression and equating to $426$ M1

$426 = \frac{20}{10} (k^{2} - \frac{3}{100} {(20)}^{2})$

$k = 15$ A1

[2 marks]

c.i.

$50$ A1

[1 mark]

c.ii.

A1A1A1

Note: Award A1 for graph drawn for positive $x$ indicating an increasing and then decreasing function, A1 for maximum labelled and A1 for graph passing through the origin and $86.6$ , marked on the $x$ -axis or whose coordinates are given.

[3 marks]

setting their expression for $C$ to zero OR choosing correct $x$ -intercept on their graph of $C$ (M1)

$x_{max} = 86.6 (86.6025 \dots)$ litres A1

[2 marks]

Examiners report

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

A sector of a circle, centre $O$ and radius $4.5 m$ , is shown in the following diagram.

A square field with side $8 m$ has a goat tied to a post in the centre by a rope such that the goat can reach all parts of the field up to $4.5 m$ from the post.

^{[Source: mynamepong, n.d. Goat [image online] Available at: https://thenounproject.com/term/goat/1761571/}
^{This file is licensed under the Creative Commons Attribution-ShareAlike 3.0 Unported (CC BY-SA 3.0)}
^{https://creativecommons.org/licenses/by-sa/3.0/deed.en [Accessed 22 April 2010] Source adapted.]}

Let $V$ be the volume of grass eaten by the goat, in cubic metres, and $t$ be the length of time, in hours, that the goat has been in the field.

The goat eats grass at the rate of $\frac{d V}{d t} = 0.3 t e^{- t}$ .

Find the angle $AÔB$ .

[3]

a.i.

Find the area of the shaded segment.

[5]

a.ii.

Find the area of a circle with radius $4.5 m$ .

[2]

b.i.

Find the area of the field that can be reached by the goat.

[3]

b.ii.

Find the value of $t$ at which the goat is eating grass at the greatest rate.

[2]

Markscheme

$(\frac{1}{2} AÔB =) arccos (\frac{4}{4.5}) = 27.266 \dots$ (M1)(A1)

$AÔB = 54.532 \dots \approx 54.5 °$ ( $0.951764 \dots \approx 0.952$ radians) A1

Note: Other methods may be seen; award (M1)(A1) for use of a correct trigonometric method to find an appropriate angle and then A1 for the correct answer.

[3 marks]

a.i.

finding area of triangle

EITHER

area of triangle $= \frac{1}{2} \times 4 . 5^{2} \times \sin (54.532 \dots)$ (M1)

Note: Award M1 for correct substitution into formula.

$= 8.24621 \dots \approx 8.25 m^{2}$ (A1)

$AB = 2 \times \sqrt{4 . 5^{2} - 4^{2}} = 4.1231 \dots$

area triangle $= \frac{4.1231 \dots \times 4}{2}$ (M1)

$= 8.24621 \dots \approx 8.25 m^{2}$ (A1)

finding area of sector

EITHER

area of sector $= \frac{54.532 \dots}{360} \times π \times 4 . 5^{2}$ (M1)

$= 9.63661 \dots \approx 9.64 m^{2}$ (A1)

area of sector $= \frac{1}{2} \times 0.9517641 \dots \times 4 . 5^{2}$ (M1)

$= 9.63661 \dots \approx 9.64 m^{2}$ (A1)

THEN

area of segment $= 9.63661 \dots - 8.24621 \dots$

$= 1.39 m^{2} (1.39040 \dots)$ A1

[5 marks]

a.ii.

$π \times 4 . 5^{2}$ (M1)

$63.6 m^{2} (63.6172 \dots m^{2})$ A1

[2 marks]

b.i.

METHOD 1

$4 \times 1.39040 . . . (5.56160)$ (A1)

subtraction of four segments from area of circle (M1)

$= 58.1 m^{2} (58.055 \dots)$ A1

METHOD 2

$4 (0.5 \times 4 . 5^{2} \times \sin 54.532 \dots) + 4 (\frac{35.4679}{360} \times π \times 4 . 5^{2})$ (M1)

$= 32.9845 \dots + 25.0707$ (A1)

$= 58.1 m^{2} (58.055 \dots)$ A1

[3 marks]

b.ii.

sketch of $\frac{d V}{d t}$ OR $\frac{d V}{d t} = 0.110363 \dots$ OR attempt to find where $\frac{d^{2} V}{d t^{2}} = 0$ (M1)

$t = 1$ hour A1

[2 marks]

Examiners report

Part (a)(i) proved to be difficult for many candidates. About half of the candidates managed to correctly find the angle $A \hat{O} B$ . A variety of methods were used: cosine to find half of $A \hat{O} B$ then double it; sine to find angle $A \hat{O} B$ , then find half of A $A \hat{O} B$ and double it; Pythagoras to find half of AB and then sine rule to find half of angle $A \hat{O} B$ then double it; Pythagoras to find half of AB, then double it and use cosine rule to find angle $A \hat{O} B$ . Many candidates lost a mark here due to premature rounding of an intermediate value and hence the final answer was not correct (to three significant figures).

In part (a)(ii) very few candidates managed to find the correct area of the shaded segment and include the correct units. Some only found the area of the triangle or the area of the sector and then stopped.

In part (b)(i), nearly all candidates managed to find the area of a circle.

In part (b)(ii), finding the area of the field reached by the goat proved troublesome for most of the candidates. It appeared as if the candidates did not fully understand the problem. Very few candidates realized the connection to part (a)(ii).

Part (c) was accessed by only a handful of candidates. The candidates could simply have graphed the function on their GDC to find the greatest value, but most did not realize this.

a.i.

[N/A]

a.ii.

[N/A]

b.i.

[N/A]

b.ii.

[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]

Consider the function $g(x) = {x^3} + k{x^2} - 15x + 5$ .

The tangent to the graph of $y = g(x)$ at $x = 2$ is parallel to the line $y = 21x + 7$ .

Find $g'(x)$ .

[3]

Show that $k = 6$ .

[2]

b.i.

Find the equation of the tangent to the graph of $y = g(x)$ at $x = 2$ . Give your answer in the form $y = mx + c$ .

[3]

b.ii.

Use your answer to part (a) and the value of $k$ , to find the $x$ -coordinates of the stationary points of the graph of $y = g(x)$ .

[3]

Find $g’( - 1)$ .

[2]

d.i.

Hence justify that $g$ is decreasing at $x = - 1$ .

[1]

d.ii.

Find the $y$ -coordinate of the local minimum.

[2]

Markscheme

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

$3{x^2} + 2kx - 15$ (A1)(A1)(A1)

Note: Award (A1) for $3{x^2}$ , (A1) for $2kx$ and (A1) for $- 15$ . Award at most (A1)(A1)(A0) if additional terms are seen.

[3 marks]

$21 = 3{(2)^2} + 2k(2) - 15$ (M1)(M1)

Note: Award (M1) for equating their derivative to 21. Award (M1) for substituting 2 into their derivative. The second (M1) should only be awarded if correct working leads to the final answer of $k = 6$ .

Substituting in the known value, $k = 6$ , invalidates the process; award (M0)(M0).

$k = 6$ (AG)

[2 marks]

b.i.

$g(2) = {(2)^3} + (6){(2)^2} - 15(2) + 5{\text{ }}( = 7)$ (M1)

Note: Award (M1) for substituting 2 into $g$ .

$7 = 21(2) + c$ (M1)

Note: Award (M1) for correct substitution of 21, 2 and their 7 into gradient intercept form.

$y - 7 = 21(x - 2)$ (M1)

Note: Award (M1) for correct substitution of 21, 2 and their 7 into gradient point form.

$y = 21x - 35$ (A1) (G2)

[3 marks]

b.ii.

$3{x^2} + 12x - 15 = 0$ (or equivalent) (M1)

Note: Award (M1) for equating their part (a) (with $k = 6$ substituted) to zero.

$x = - 5,{\text{ }}x = 1$ (A1)(ft)(A1)(ft)

Note: Follow through from part (a).

[3 marks]

$3{( - 1)^2} + 12( - 1) - 15$ (M1)

Note: Award (M1) for substituting $- 1$ into their derivative, with $k = 6$ substituted. Follow through from part (a).

$= - 24$ (A1)(ft) (G2)

[2 marks]

d.i.

$g’( - 1) < 0$ (therefore $g$ is decreasing when $x = - 1$ ) (R1)

[1 marks]

d.ii.

$g(1) = {(1)^3} + (6){(1)^2} - 15(1) + 5$ (M1)

Note: Award (M1) for correctly substituting 6 and their 1 into $g$ .

$= - 3$ (A1)(ft) (G2)

Note: Award, at most, (M1)(A0) or (G1) if answer is given as a coordinate pair. Follow through from part (c).

[2 marks]

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

d.i.

[N/A]

d.ii.

[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]

The Maxwell Ohm Company is designing a portable Bluetooth speaker. The speaker is in the shape of a cylinder with a hemisphere at each end of the cylinder.

The dimensions of the speaker, in centimetres, are illustrated in the following diagram where $r$ is the radius of the hemisphere, and $l$ is the length of the cylinder, with $r > 0$ and $l \geq 0$ .

The Maxwell Ohm Company has decided that the speaker will have a surface area of $300 {cm}^{2}$ .

The quality of sound from the speaker will improve as $V$ increases.

Write down an expression for $V$ , the volume (cm³) of the speaker, in terms of $r$ , $l$ and $π$ .

[2]

Write down an equation for the surface area of the speaker in terms of $r$ , $l$ and $π$ .

[3]

Given the design constraint that $l = \frac{150 - 2 π r^{2}}{π r}$ , show that $V = 150 r - \frac{2 π r^{3}}{3}$ .

[2]

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

[2]

Using your answer to part (d), show that $V$ is a maximum when $r$ is equal to $\sqrt{\frac{75}{π}} cm$ .

[2]

Find the length of the cylinder for which $V$ is a maximum.

[2]

Calculate the maximum value of $V$ .

[2]

Use your answer to part (f) to identify the shape of the speaker with the best quality of sound.

[1]

Markscheme

$(V =) \frac{4 π r^{3}}{3} + π r^{2} l$ (or equivalent) (A1)(A1)

Note: Award (A1) for either the volume of a hemisphere formula multiplied by $2$ or the volume of a cylinder formula, and (A1) for completely correct expression. Accept equivalent expressions. Award at most (A1)(A0) if $h$ is used instead of $l$ .

[2 marks]

$300 = 4 π r^{2} + 2 π r l$ (A1)(A1)(A1)

Note: Award (A1) for the surface area of a hemisphere multiplied by $2$ . Award (A1) for the surface area of a cylinder. Award (A1) for the addition of their formulas equated to $300$ . Award at most (A1)(A1)(A0) if $h$ is used instead of $l$ , unless already penalized in part (a).

[3 marks]

$V = \frac{4 π r^{3}}{3} + π r^{2} (\frac{150 - 2 π r^{2}}{π r})$ (M1)

Note: Award (M1) for their correctly substituted formula for $V$ .

$V = \frac{4 π r^{3}}{3} + 150 r - 2 π r^{3}$ (M1)

Note: Award (M1) for correct expansion of brackets and simplification of the cylinder expression in $V$ leading to the final answer.

$V = 150 r - \frac{2 π r^{3}}{3}$ (AG)

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

[2 marks]

$(\frac{d V}{d r} =) 150 - 2 π r^{2}$ (A1)(A1)

Note: Award (A1) for $150$ . Award (A1) for $- 2 π r^{2}$ . Award maximum (A1)(A0) if extra terms seen.

[2 marks]

$150 - 2 π r^{2} = 0$ OR $\frac{d V}{d r} = 0$ OR sketch of $\frac{d V}{d r}$ with $x$ -intercept indicated (M1)

Note: Award (M1) for equating their derivative to zero or a sketch of their derivative with $x$ -intercept indicated.

$r = \sqrt{\frac{150}{2 π}}$ OR $r^{2} = \frac{150}{2 π}$ (A1)

$r = \sqrt{\frac{75}{π}}$ (AG)

Note: The (AG) line must be seen for the preceding (A1) to be awarded.

[2 marks]

$(l =) \frac{150 - 2 π {(\sqrt{\frac{75}{π}})}^{2}}{π (\sqrt{\frac{75}{π}})}$ (M1)

Note: Award (M1) for correct substitution in the given formula for the length of the cylinder.

$(l =) 0 (cm)$ (A1)(G2)

Note: Award (M1)(A1) for correct substitution of the $3$ sf approximation $4.89$ leading to a correct answer of zero.

[2 marks]

$V = 150 (\sqrt{\frac{75}{π}}) - \frac{2 π {(\sqrt{\frac{75}{π}})}^{3}}{3}$ OR $V = \frac{4 π {(\sqrt{\frac{75}{π}})}^{3}}{3}$ (M1)

Note: Award (M1) for correct substitution in the formula for the volume of the speaker or the volume of a sphere.

$489 (488.602 \dots, 100 \sqrt{\frac{75}{π}}) ({cm}^{3})$ (A1)(G2)

Note: Accept $489.795 \dots$ from use of $3$ sf value of $\sqrt{\frac{75}{π}}$ . Award (M1)(A1)(ft) for correct substitution in their volume of speaker. Follow through from parts (a) and (f).

[2 marks]

sphere (spherical) (A1)(ft)

Note: Question requires the use of part (f) so if there is no answer to part (f), part (h) is awarded (A0). Follow through from their $l > 0$ .

[1 mark]

Examiners report

[N/A]

The graph of the quadratic function $f (x) = \frac{1}{2} (x - 2) (x + 8)$ intersects the $y$ -axis at $(0, c)$ .

The vertex of the function is $(- 3, - 12.5)$ .

The equation $f (x) = 12$ has two solutions. The first solution is $x = - 10$ .

Let $T$ be the tangent at $x = - 3$ .

Find the value of $c$ .

[2]

Write down the equation for the axis of symmetry of the graph.

[2]

Use the symmetry of the graph to show that the second solution is $x = 4$ .

[1]

Write down the $x$ -intercepts of the graph.

[2]

On graph paper, draw the graph of $y = f (x)$ for $- 10 \leq x \leq 4$ and $- 14 \leq y \leq 14$ . Use a scale of $1 cm$ to represent $1$ unit on the $x$ -axis and $1 cm$ to represent $2$ units on the $y$ -axis.

[4]

Write down the equation of $T$ .

[2]

f.i.

Draw the tangent $T$ on your graph.

[1]

f.ii.

Given $f (a) = 5.5$ and $f' (a) = - 6$ , state whether the function, $f$ , is increasing or decreasing at $x = a$ . Give a reason for your answer.

[2]

Markscheme

$\frac{1}{2} (0 - 2) (0 + 8)$ OR $\frac{1}{2} (0^{2} + 6 (0) - 16)$ (or equivalent) (M1)

Note: Award (M1) for evaluating $f (0)$ .

$(c =) - 8$ (A1)(G2)

Note: Award (G2) if $- 8$ or $(0, - 8)$ seen.

[2 marks]

$x = - 3$ (A1)(A1)

Note: Award (A1) for “ $x =$ constant”, (A1) for the constant being $- 3$ . The answer must be an equation.

[2 marks]

$(- 3 - - 10) + - 3$ (M1)

$(- 8 - - 10) + 2$ (M1)

$\frac{- 10 + x}{2} = - 3$ (M1)

diagram showing axis of symmetry and given points ( $x$ -values labels, $- 10$ , $- 3$ and $4$ , are sufficient) and an indication that the horizontal distances between the axis of symmetry and the given points are $7$ . (M1)

Note: Award (M1) for correct working using the symmetry between $x = - 10$ and $x = - 3$ . Award (M0) if candidate has used $x = - 10$ and $x = 4$ to show the axis of symmetry is $x = - 3$ . Award (M0) if candidate solved $f (x) = 12$ or evaluated $f (- 10)$ and $f (4)$ .

$(x =) 4$ (AG)

[1 mark]

$- 8$ and $2$ (A1)(A1)

Note: Accept $x = - 8, y = 0$ and $x = 2, y = 0$ or $(- 8, 0)$ and $(2, 0)$ , award at most (A0)(A1) if parentheses are omitted.

[2 marks]

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

Note: Award (A1) for labelled axes with correct scale, correct window. Award (A1) for the vertex, $(- 3, - 12.5)$ , in correct location.
Award (A1) for a smooth continuous curve symmetric about their vertex. Award (A1)(ft) for the curve passing through their $x$ and $y$ intercepts in correct location. Follow through from their parts (a) and (d).

If graph paper is not used:
Award at most (A0)(A0)(A1)(A1)(ft). Their graph should go through their $- 8$ and $2$ for the last (A1)(ft) to be awarded.

[4 marks]

$y = - 12.5$ OR $y = 0 x - 12.5$ (A1)(A1)

Note: Award (A1) for " $y =$ constant", (A1) for the constant being $- 12.5$ . The answer must be an equation.

[2 marks]

f.i.

tangent to the graph drawn at $x = - 3$ (A1)(ft)

Note: Award (A1) for a horizontal straight-line tangent to curve at approximately $x = - 3$ . Award (A0) if a ruler is not used. Follow through from their part (e).

[1 mark]

f.ii.

decreasing (A1)

gradient (of tangent line) is negative (at $x = a$ ) OR $f' (a) < 0$ (R1)

Note: Do not accept "gradient (of tangent line) is $- 6$ ". Do not award (A1)(R0).

[2 marks]

Examiners report

[N/A]

f.i.

[N/A]

f.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 $= \left( {\begin{array}{*{20}{c}} 4 \\ { - 1} \end{array}} \right)$ 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 $f\left( x \right) = 4 - 2{{\text{e}}^x}$ . 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]

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]

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]

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]

The cloth used to make Nanako’s bag costs 4 Japanese Yen (JPY) per cm².

Find the cost of the cloth used to make Nanako’s bag.

[2]

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]

$\left( {3{{\left( {12} \right)}^2} + \frac{{10368}}{{12}}} \right) \times 4$ (M1)

Note: Award (M1) for substituting 12 into the area formula and for multiplying the area formula by 4.

= 5180 (JPY) (5184 (JPY)) (A1)(G2)

[2 marks]

Examiners report

[N/A]

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

Find $f'\left( x \right)$ .

[4]

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

[2]

Markscheme

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

choosing product rule (M1)

eg $uv' + vu'$ , ${\left( {{x^2}} \right)^\prime }\left( {{{\text{e}}^{3x}}} \right) + {\left( {{{\text{e}}^{3x}}} \right)^\prime }{x^2}$

correct derivatives (must be seen in the rule) A1A1

eg $2x$ , ${\text{3}}{{\text{e}}^{3x}}$

$f'\left( x \right) = 2x{{\text{e}}^{3x}} + 3{x^2}{{\text{e}}^{3x}}$ A1 N4

[4 marks]

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]

Let f(x) = ln x − 5x , for x > 0 .

Find f '(x).

[2]

Find f "(x).

[1]

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

[2]

Markscheme

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

$f'\left( x \right) = \frac{1}{x} - 5$ A1A1 N2

[2 marks]

f "(x) = −x⁻² A1 N1

[1 mark]

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]

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.

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]

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]

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]

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]

The function $f\left( x \right) = \frac{1}{3}{x^3} + \frac{1}{2}{x^2} + kx + 5$ has a local maximum and a local minimum. The local maximum is at $x = - 3$ .

Show that $k = - 6$ .

[5]

Find the coordinates of the local minimum.

[2]

Write down the interval where the gradient of the graph of $f\left( x \right)$ is negative.

[2]

Determine the equation of the normal at $x = - 2$ in the form $y = mx + c$ .

[5]

Markscheme

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

${x^2} + x + k$ (A1)(A1)(A1)

Note: Award (A1) for each correct term. Award at most (A1)(A1)(A0) if additional terms are seen or for an answer ${x^2} + x - 6$ . If their derivative is seen in parts (b), (c) or (d) and not in part (a), award at most (A1)(A1)(A0).

${\left( { - 3} \right)^2} + \left( { - 3} \right) + k = 0$ (M1)(M1)

Note: Award (M1) for substituting in $x = - 3$ into their derivative and (M1) for setting it equal to zero. Substituting $k = - 6$ invalidates the process, award at most (A1)(A1)(A1)(M0)(M0).

$\left( {k = } \right) - 6$ (AG)

Note: For the final (M1) to be awarded, no incorrect working must be seen, and must lead to the conclusion $k = - 6$ . The final (AG) must be seen.

[5 marks]

(2, −2.33) OR $\left( {2{\text{,}}\,\, - \frac{7}{3}} \right)$ (A1)(A1)

Note: Award (A1) for each correct coordinate. Award (A0)(A1) if parentheses are missing. Accept $x = 2$ , $y = - 2.33$ . Award (M1)(A0) for their derivative, a quadratic expression with –6 substituted for $k$ , equated to zero but leading to an incorrect answer.

[2 marks]

$- 3 < x < 2$ (A1)(ft)(A1)

Note: Award (A1) for $x > - 3$ , (A1)(ft) for $x < 2$ . Follow through for their "2" in part (b). It is possible to award (A0)(A1). For $- 3 < y < 2$ award (A1)(A0). Accept equivalent notation such as (−3, 2). Award (A0)(A1)(ft) for $- 3 \leqslant x \leqslant 2$ .

[2 marks]

−4 (A1)(ft)

Note: Award (A1)(ft) for the gradient of the tangent seen. If an incorrect derivative was used in part (a), then working for their $f'\left( { - 2} \right)$ must be seen. Follow through from their derivative in part (a).

gradient of normal is $\frac{1}{4}$ (A1)(ft)

Note: Award (A1)(ft) for the negative reciprocal of their gradient of tangent. Follow through within this part. Award (G2) for an unsupported gradient of the normal.

$\frac{{49}}{3}$ $\left( {f\left( { - 2} \right) = \frac{1}{3}{{\left( { - 2} \right)}^3} + \frac{1}{2}{{\left( { - 2} \right)}^2} - 6\left( { - 2} \right) + 5 = \frac{{49}}{3}} \right)$ (A1)

Note: Award (A1) for $\frac{{49}}{3}$ (16.3333…) seen.

$\frac{{49}}{3} = \frac{1}{4}\left( { - 2} \right) + c$ OR $y - \frac{{49}}{3} = \frac{1}{4}\left( {x - - 2} \right)$ (M1)

Note: Award (M1) for substituting their normal gradient into equation of line formula.

$y = \frac{1}{4}x + \frac{{101}}{6}$ OR $y = 0.25x + 16.8333 \ldots$ (A1)(ft)(G4)

Note: Award (G4) for the correct equation of line in correct form without any prior working. The final (A1)(ft) is contingent on $y = \frac{{49}}{3}$ and $x = - 2$ .

[5 marks]

Examiners report

[N/A]

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]

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]

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 = 17 (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]

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]

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]

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]