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SL Paper 1

The function $f$ is defined for all $x \in ℝ$ . The line with equation $y = 6 x - 1$ is the tangent to the graph of $f$ at $x = 4$ .

The function $g$ is defined for all $x \in ℝ$ where $g (x) = x^{2} - 3 x$ and $h (x) = f (g (x))$ .

Write down the value of $f' (4)$ .

[1]

Find $f (4)$ .

[1]

Find $h (4)$ .

[2]

Hence find the equation of the tangent to the graph of $h$ at $x = 4$ .

[3]

Markscheme

$f' (4) = 6$ A1

[1 mark]

$f (4) = 6 \times 4 - 1 = 23$ A1

[1 mark]

$h (4) = f (g (4))$ (M1)

$h (4) = f (4^{2} - 3 \times 4) = f (4)$

$h (4) = 23$ A1

[2 marks]

attempt to use chain rule to find $h'$ (M1)

$f' (g (x)) \times g' (x)$ OR $(x^{2} - 3 x)' \times f' (x^{2} - 3 x)$

$h' (4) = (2 \times 4 - 3) f' (4^{2} - 3 \times 4)$ A1

$= 30$

$y - 23 = 30 (x - 4)$ OR $y = 30 x - 97$ A1

[3 marks]

Examiners report

[N/A]

Let $g\left( x \right) = {p^x} + q$ , for $x{\text{, }}p{\text{, }}q \in \mathbb{R}{\text{, }}p > 1$ . The point ${\text{A}}\left( {0{\text{, }}a} \right)$ lies on the graph of $g$ .

Let $f\left( x \right) = {g^{ - 1}}\left( x \right)$ . The point ${\text{B}}$ lies on the graph of $f$ and is the reflection of point ${\text{A}}$ in the line $y = x$ .

The line ${L_1}$ is tangent to the graph of $f$ at ${\text{B}}$ .

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

[2]

Given that $f'\left( a \right) = \frac{1}{{{\text{ln}}\,p}}$ , find the equation of ${L_1}$ in terms of $x$ , $p$ and $q$ .

[5]

The line ${L_2}$ is tangent to the graph of $g$ at ${\text{A}}$ and has equation $y = \left( {{\text{ln}}\,p} \right)x + q + 1$ .

The line ${L_2}$ passes through the point $\left( { - 2{\text{, }} - 2} \right)$ .

The gradient of the normal to $g$ at ${\text{A}}$ is $\frac{1}{{{\text{ln}}\left( {\frac{1}{3}} \right)}}$ .

Find the equation of ${L_1}$ in terms of $x$ .

[7]

Markscheme

${\text{B}}\left( {a{\text{, }}0} \right)$ (accept ${\text{B}}\left( {q + 1{\text{, }}0} \right)$ ) A2 N2

[2 marks]

Note: There are many approaches to this part, and the steps may be done in any order. Please check working and award marks in line with the markscheme, noting that candidates may work with the equation of the line before finding $a$ .

FINDING $a$

valid attempt to find an expression for $a$ in terms of $q$ (M1)

$g\left( 0 \right) = a{\text{, }}\,{p^0} + q = a$

$a = q + 1$ (A1)

FINDING THE EQUATION OF ${L_1}$

EITHER

attempt to substitute tangent gradient and coordinates into equation of straight line (M1)

eg $y - 0 = f'\left( a \right)\left( {x - a} \right){\text{, }}\,y = f'\left( a \right)\left( {x - \left( {q + 1} \right)} \right)$

correct equation in terms of $a$ and $p$ (A1)

eg $y - 0 = \frac{1}{{{\text{ln}}\left( p \right)}}\left( {x - a} \right)$

attempt to substitute tangent gradient and coordinates to find $b$

eg $0 = \frac{1}{{{\text{ln}}\left( p \right)}}\left( a \right) + b$

$b = \frac{{ - a}}{{{\text{ln}}\left( p \right)}}$ (A1)

THEN (must be in terms of both $p$ and $q$ )

$y = \frac{1}{{{\text{ln}}\,p}}\left( {x - q - 1} \right){\text{, }}\,y = \frac{1}{{{\text{ln}}\,p}}x - \frac{{q + 1}}{{{\text{ln}}\,p}}$ A1 N3

Note: Award A0 for final answers in the form ${L_1} = \frac{1}{{{\text{ln}}\,p}}\left( {x - q - 1} \right)$

[5 marks]

Note: There are many approaches to this part, and the steps may be done in any order. Please check working and award marks in line with the markscheme, noting that candidates may find $q$ in terms of $p$ before finding a value for $p$ .

FINDING $p$

valid approach to find the gradient of the tangent (M1)

eg ${m_1}{m_2} = - 1{\text{, }}\,\, - \frac{1}{{\frac{1}{{{\text{ln}}\left( {\frac{1}{3}} \right)}}}}{\text{, }}\,\, - {\text{ln}}\left( {\frac{1}{3}} \right){\text{, }}\,\, - \frac{1}{{{\text{ln}}\,p}} = \frac{1}{{{\text{ln}}\left( {\frac{1}{3}} \right)}}$

correct application of log rule (seen anywhere) (A1)

eg ${\text{ln}}{\left( {\frac{1}{3}} \right)^{ - 1}}{\text{, }}\,\, - \left( {{\text{ln}}\left( 1 \right) - {\text{ln}}\left( 3 \right)} \right)$

correct equation (seen anywhere) A1

eg ${\text{ln}}\,p = {\text{ln}}\,3{\text{, }}\,\,p = 3$

FINDING $q$

correct substitution of $\left( { - 2{\text{, }} - 2} \right)$ into ${L_2}$ equation (A1)

eg $- 2 = \left( {{\text{ln}}\,p} \right)\left( { - 2} \right) + q + 1$

$q = 2\,{\text{ln}}\,p - 3{\text{, }}\,\,q = 2\,{\text{ln}}\,3 - 3$ (seen anywhere) A1

FINDING ${L_1}$

correct substitution of their $p$ and $q$ into their ${L_1}$ (A1)

eg $y = \frac{1}{{{\text{ln}}\,3}}\left( {x - \left( {2\,{\text{ln}}\,3 - 3} \right) - 1} \right)$

$y = \frac{1}{{{\text{ln}}\,3}}\left( {x - 2\,{\text{ln}}\,3 + 2} \right){\text{, }}\,\,y = \frac{1}{{{\text{ln}}\,3}}x - \frac{{2\,{\text{ln}}\,3 - 2}}{{{\text{ln}}\,3}}$ A1 N2

Note: Award A0 for final answers in the form ${L_1} = \frac{1}{{{\text{ln}}\,3}}\left( {x - 2\,{\text{ln}}\,3 + 2} \right)$ .

[7 marks]

Examiners report

[N/A]

Consider the functions $f (x) = \frac{1}{x - 4} + 1$ , for $x \neq 4$ , and $g (x) = x - 3$ for $x \in ℝ$ .

The following diagram shows the graphs of $f$ and $g$ .

The graphs of $f$ and $g$ intersect at points $A$ and $B$ . The coordinates of $A$ are $(3, 0)$ .

In the following diagram, the shaded region is enclosed by the graph of $f$ , the graph of $g$ , the $x$ -axis, and the line $x = k$ , where $k \in ℤ$ .

The area of the shaded region can be written as $\ln (p) + 8$ , where $p \in ℤ$ .

Find the coordinates of $B$ .

[5]

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

[10]

Markscheme

$\frac{1}{x - 4} + 1 = x - 3$ (M1)

$x^{2} - 8 x + 15 = 0$ OR ${(x - 4)}^{2} = 1$ (A1)

valid attempt to solve their quadratic (M1)

$(x - 3) (x - 5) = 0$ OR $x = \frac{8 \pm \sqrt{8^{2} - 4 (1) (15)}}{2 (1)}$ OR $(x - 4) = \pm 1$

$x = 5 (x = 3, x = 5)$ (may be seen in answer) A1

$B (5, 2)$ (accept $x = 5, y = 2$ ) A1

[5 marks]

recognizing two correct regions from $x = 3$ to $x = 5$ and from $x = 5$ to $x = k$ (R1)

triangle $+ \int_{5}^{k} f (x) d x$ OR $\int_{3}^{5} g (x) d x + \int_{5}^{k} f (x) d x$ OR $\int_{3}^{5} (x - 3) d x + \int_{5}^{k} (\frac{1}{x - 4} + 1) d x$

area of triangle is $2$ OR $\frac{2 \cdot 2}{2}$ OR $(\frac{5^{2}}{2} - 3 (5)) - (\frac{3^{2}}{2} - 3 (3))$ (A1)

correct integration (A1)(A1)

$\int (\frac{1}{x - 4} + 1) d x = \ln (x - 4) + x (+ C)$

Note: Award A1 for $\ln (x - 4)$ and A1 for $x$ .
Note: The first three A marks may be awarded independently of the R mark.

substitution of their limits (for $x$ ) into their integrated function (in terms of $x$ ) (M1)

$\ln (k - 4) + k - (\ln 1 + 5)$

${[\ln (x - 4) + x]}_{5}^{k} = \ln (k - 4) + k - 5$ A1

adding their two areas (in terms of $k$ ) and equating to $\ln p + 8$ (M1)

$2 + \ln (k - 4) + k - 5 = \ln p + 8$

equating their non-log terms to $8$ (equation must be in terms of $k$ ) (M1)

$k - 3 = 8$

$k = 11$ A1

$11 - 4 = p$

$p = 7$ A1

[10 marks]

Examiners report

Nearly all candidates knew to set up an equation with $f (x) = g (x)$ in order to find the intersection of the two graphs, and most were able to solve the resulting quadratic equation. Candidates were not as successful in part (b), however. While some candidates recognized that there were two regions to be added together, very few were able to determine the correct boundaries of these regions, with many candidates integrating one or both functions from $x = 3$ to $x = k$ . While a good number of candidates were able to correctly integrate the function(s), without the correct bounds the values of $k$ and $p$ were unattainable.

[N/A]

The following table shows the probability distribution of a discrete random variable $A$ , in terms of an angle $\theta$ .

M17/5/MATME/SP1/ENG/TZ1/10

Show that $\cos \theta = \frac{3}{4}$ .

[6]

Given that $\tan \theta > 0$ , find $\tan \theta$ .

[3]

Let $y = \frac{1}{{\cos x}}$ , for $0 < x < \frac{\pi }{2}$ . The graph of $y$ between $x = \theta$ and $x = \frac{\pi }{4}$ is rotated 360° about the $x$ -axis. Find the volume of the solid formed.

[6]

Markscheme

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

evidence of summing to 1 (M1)

eg $\,\,\,\,\,$ $\sum {p = 1}$

correct equation A1

eg $\,\,\,\,\,$ $\cos \theta + 2\cos 2\theta = 1$

correct equation in $\cos \theta$ A1

eg $\,\,\,\,\,$ $\cos \theta + 2(2{\cos ^2}\theta - 1) = 1,{\text{ }}4{\cos ^2}\theta + \cos \theta - 3 = 0$

evidence of valid approach to solve quadratic (M1)

eg $\,\,\,\,\,$ factorizing equation set equal to $0,{\text{ }}\frac{{ - 1 \pm \sqrt {1 - 4 \times 4 \times ( - 3)} }}{8}$

correct working, clearly leading to required answer A1

eg $\,\,\,\,\,$ $(4\cos \theta - 3)(\cos \theta + 1),{\text{ }}\frac{{ - 1 \pm 7}}{8}$

correct reason for rejecting $\cos \theta \ne - 1$ R1

eg $\,\,\,\,\,$ $\cos \theta$ is a probability (value must lie between 0 and 1), $\cos \theta > 0$

Note: Award R0 for $\cos \theta \ne - 1$ without a reason.

$\cos \theta = \frac{3}{4}$ AG N0

valid approach (M1)

eg $\,\,\,\,\,$ sketch of right triangle with sides 3 and 4, ${\sin ^2}x + {\cos ^2}x = 1$

correct working

(A1)

eg $\,\,\,\,\,$ missing side $= \sqrt 7 ,{\text{ }}\frac{{\frac{{\sqrt 7 }}{4}}}{{\frac{3}{4}}}$

$\tan \theta = \frac{{\sqrt 7 }}{3}$ A1 N2

[3 marks]

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

eg $\,\,\,\,\,$ $\pi \int_\theta ^{\frac{\pi }{4}} {{f^2},{\text{ }}\int {{{\left( {\frac{1}{{\cos x}}} \right)}^2}} }$

correct substitution of both limits and function (A1)

eg $\,\,\,\,\,$ $\pi \int_\theta ^{\frac{\pi }{4}} {{{\left( {\frac{1}{{\cos x}}} \right)}^2}{\text{d}}x}$

correct integration (A1)

eg $\,\,\,\,\,$ $\tan x$

substituting their limits into their integrated function and subtracting (M1)

eg $\,\,\,\,\,$ $\tan \frac{\pi }{4} - \tan \theta$

Note: Award M0 if they substitute into original or differentiated function.

$\tan \frac{\pi }{4} = 1$ (A1)

eg $\,\,\,\,\,$ $1 - \tan \theta$

$V = \pi - \frac{{\pi \sqrt 7 }}{3}$ A1 N3

[6 marks]

Examiners report

[N/A]

Let $\theta$ be an obtuse angle such that ${\text{sin}}\,\theta = \frac{3}{5}$ .

Let $f\left( x \right) = {{\text{e}}^x}\,{\text{sin}}\,x - \frac{{3x}}{4}$ .

Find the value of ${\text{tan}}\,\theta$ .

[4]

Line $L$ passes through the origin and has a gradient of ${\text{tan}}\,\theta$ . Find the equation of $L$ .

[2]

The following diagram shows the graph of $f$ for 0 ≤ $x$ ≤ 3. Line $M$ is a tangent to the graph of $f$ at point P.

Given that $M$ is parallel to $L$ , find the $x$ -coordinate of P.

[4]

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 sketch of triangle with sides 3 and 5, ${\text{co}}{{\text{s}}^2}\,\theta = 1 - {\text{si}}{{\text{n}}^2}\,\theta$

correct working (A1)

eg missing side is 4 (may be seen in sketch), ${\text{cos}}\,\theta = \frac{4}{5}$ , ${\text{cos}}\,\theta = - \frac{4}{5}$

${\text{tan}}\,\theta = - \frac{3}{4}$ A2 N4

[4 marks]

correct substitution of either gradient or origin into equation of line (A1)

(do not accept $y = mx + b$ )

eg $y = x\,{\text{tan}}\,\theta$ , $y - 0 = m\left( {x - 0} \right)$ , $y = mx$

$y = - \frac{3}{4}x$ A2 N4

Note: Award A1A0 for $L = - \frac{3}{4}x$ .

[2 marks]

valid approach to equate their gradients (M1)

eg $f' = {\text{tan}}\,\theta$ , $f' = - \frac{3}{4}$ , ${{\text{e}}^x}\,{\text{cos}}\,x + {{\text{e}}^x}\,{\text{sin}}\,x - \frac{3}{4} = - \frac{3}{4}$ , ${{\text{e}}^x}\,\left( {{\text{cos}}\,x + {\text{sin}}\,x} \right) - \frac{3}{4} = - \frac{3}{4}$

correct equation without ${{\text{e}}^x}$ (A1)

eg ${\text{sin}}\,x = - {\text{cos}}\,x$ , ${\text{cos}}\,x + {\text{sin}}\,x = 0$ , $\frac{{ - {\text{sin}}\,x}}{{{\text{cos}}\,x}} = 1$

correct working (A1)

eg ${\text{tan}}\,\theta = - 1$ , $x = 135^\circ$

$x = \frac{{3\pi }}{4}$ (do not accept $135^\circ$ ) A1 N1

Note: Do not award the final A1 if additional answers are given.

[4 marks]

Examiners report

[N/A]

A school café sells three flavours of smoothies: mango ( $M$ ), kiwi fruit ( $K$ ) and banana ( $B$ ).
85 students were surveyed about which of these three flavours they like.

35 students liked mango, 37 liked banana, and 26 liked kiwi fruit
2 liked all three flavours
20 liked both mango and banana
14 liked mango and kiwi fruit
3 liked banana and kiwi fruit

Using the given information, complete the following Venn diagram.

[2]

Find the number of surveyed students who did not like any of the three flavours.

[2]

A student is chosen at random from the surveyed students.

Find the probability that this student likes kiwi fruit smoothies given that they like mango smoothies.

[2]

Markscheme

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

(A1)(A1) (C2)

Note: Award (A1) for 18, 12 and 1 in correct place on Venn diagram, (A1) for 3, 16 and 11 in correct place on Venn diagram.

[2 marks]

85 − (3 + 16 + 11 + 18 + 12 + 1 + 2) (M1)

Note: Award (M1) for subtracting the sum of their values from 85.

22 (A1)(ft) (C2)

Note: Follow through from their Venn diagram in part (a).
If any numbers that are being subtracted are negative award (M1)(A0).

[2 marks]

$\frac{{14}}{{35}}\,\,\,\left( {\frac{2}{5}{\text{,}}\,\,0.4{\text{,}}\,\,40{\text{% }}} \right)$ (A1)(ft)(A1)(ft) (C2)

Note: Award (A1) for correct numerator; (A1) for correct denominator. Follow through from their Venn diagram.

[2 marks]

Examiners report

[N/A]

The diagram shows a circular horizontal board divided into six equal sectors. The sectors are labelled white (W), yellow (Y) and blue (B).

A pointer is pinned to the centre of the board. The pointer is to be spun and when it stops the colour of the sector on which the pointer stops is recorded. The pointer is equally likely to stop on any of the six sectors.

Eva will spin the pointer twice. The following tree diagram shows all the possible outcomes.

Find the probability that both spins are yellow.

[2]

Find the probability that at least one of the spins is yellow.

[3]

Write down the probability that the second spin is yellow, given that the first spin is blue.

[1]

Markscheme

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

$\frac{1}{3} \times \frac{1}{3}$ OR ${\left( {\frac{1}{3}} \right)^2}$ (M1)

Note: Award (M1) for multiplying correct probabilities.

$\frac{1}{9}$ (0.111, 0.111111…, 11.1%) (A1) (C2)

[2 marks]

$\left( {\frac{1}{2} \times \frac{1}{3}} \right) + \left( {\frac{1}{6} \times \frac{1}{3}} \right) + \frac{1}{3}$ (M1)(M1)

Note: Award (M1) for $\left( {\frac{1}{2} \times \frac{1}{3}} \right)$ and $\left( {\frac{1}{6} \times \frac{1}{3}} \right)$ or equivalent, and (M1) for $\frac{1}{3}$ and adding only the three correct probabilities.

$1 - {\left( {\frac{2}{3}} \right)^2}$ (M1)(M1)

Note: Award (M1) for ${\frac{2}{3}}$ seen and (M1) for subtracting ${\left( {\frac{2}{3}} \right)^2}$ from 1. This may be shown in a tree diagram with “yellow” and “not yellow” branches.

$\frac{5}{9}$ (0.556, 0.555555…, 55.6%) (A1)(ft) (C3)

Note: Follow through marks may be awarded if their answer to part (a) is used in a correct calculation.

[3 marks]

$\frac{1}{3}$ (0.333, 0.333333…, 33.3%) (A1) (C1)

[1 mark]

Examiners report

[N/A]

A particle $P$ moves along the $x$ -axis. The velocity of $P$ is $v m s^{- 1}$ at time $t$ seconds, where $v (t) = 4 + 4 t - 3 t^{2}$ for $0 \leq t \leq 3$ . When $t = 0, P$ is at the origin $O$ .

Find the value of $t$ when $P$ reaches its maximum velocity.

[2]

a.i.

Show that the distance of $P$ from $O$ at this time is $\frac{88}{27}$ metres.

[5]

a.ii.

Sketch a graph of $v$ against $t$ , clearly showing any points of intersection with the axes.

[4]

Find the total distance travelled by $P$ .

[5]

Markscheme

valid approach to find turning point ( $v' = 0, - \frac{b}{2 a}$ , average of roots) (M1)

$4 - 6 t = 0$ OR $- \frac{4}{2 (- 3)}$ OR $\frac{- \frac{2}{3} + 2}{2}$

$t = \frac{2}{3}$ (s) A1

[2 marks]

a.i.

attempt to integrate $v$ (M1)

$\int v d t = \int (4 + 4 t - 3 t^{2}) d t = 4 t + 2 t^{2} - t^{3} (+ c)$ A1A1

Note: Award A1 for $4 t + 2 t^{2}$ , A1 for $- t^{3}$ .

attempt to substitute their $t$ into their solution for the integral (M1)

distance $= 4 (\frac{2}{3}) + 2 {(\frac{2}{3})}^{2} - {(\frac{2}{3})}^{3}$

$= \frac{8}{3} + \frac{8}{9} - \frac{8}{27}$ (or equivalent) A1

$= \frac{88}{27}$ (m) AG

[5 marks]

a.ii.

valid approach to solve $4 + 4 t - 3 t^{2} = 0$ (may be seen in part (a)) (M1)

$(2 - t) (2 + 3 t)$ OR $\frac{- 4 \pm \sqrt{16 + 48}}{- 6}$

correct $x$ - intercept on the graph at $t = 2$ A1

Note: The following two A marks may only be awarded if the shape is a concave down parabola. These two marks are independent of each other and the (M1).

correct domain from $0$ to $3$ starting at $(0, 4)$ A1

Note: The $3$ must be clearly indicated.

vertex in approximately correct place for $t = \frac{2}{3}$ and $v > 4$ A1

[4 marks]

recognising to integrate between $0$ and $2$ , or $2$ and $3$ OR $\int_{0}^{3} |4 + 4 t - 3 t^{2}| d t$ (M1)

$\int_{0}^{2} (4 + 4 t - 3 t^{2}) d t$

$= 8$ A1

$\int_{2}^{3} (4 + 4 t - 3 t^{2}) d t$

$= - 5$ A1

valid approach to sum the two areas (seen anywhere) (M1)

$\int_{0}^{2} v d t - \int_{2}^{3} v d t$ OR $\int_{0}^{2} v d t + |\int_{2}^{3} v d t|$

total distance travelled $= 13$ (m) A1

[5 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

Rosewood College has 120 students. The students can join the sports club ( $S$ ) and the music club ( $M$ ).

For a student chosen at random from these 120, the probability that they joined both clubs is $\frac{1}{4}$ and the probability that they joined the music club is $\frac{1}{3}$ .

There are 20 students that did not join either club.

Complete the Venn diagram for these students.

N17/5/MATSD/SP1/ENG/TZ0/07.a

[2]

One of the students who joined the sports club is chosen at random. Find the probability that this student joined both clubs.

[2]

Determine whether the events $S$ and $M$ are independent.

[2]

Markscheme

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

N17/5/MATSD/SP1/ENG/TZ0/07.a/M (A1)(A1) (C2)

Note: Award (A1) for 30 in correct area, (A1) for 60 and 10 in the correct areas.

[2 marks]

$\frac{{30}}{{90}}{\text{ }}\left( {\frac{1}{3},{\text{ }}0.333333 \ldots ,{\text{ }}33.3333 \ldots \% } \right)$ (A1)(ft)(A1)(ft) (C2)

Note: Award (A1)(ft) for correct numerator of 30, (A1)(ft) for correct denominator of 90. Follow through from their Venn diagram.

[2 marks]

${\text{P}}(S) \times {\text{P}}(M) = \frac{3}{4} \times \frac{1}{3} = \frac{1}{4}$ (R1)

Note: Award (R1) for multiplying their by $\frac{1}{3}$ .

therefore the events are independent $\left( {{\text{as P}}(S \cap M) = \frac{1}{4}} \right)$ (A1)(ft) (C2)

Note: Award (R1)(A1)(ft) for an answer which is consistent with their Venn diagram.

Do not award (R0)(A1)(ft).

Do not award final (A1) if ${\text{P}}(S) \times {\text{P}}(M)$ is not calculated. Follow through from part (a).

[2 marks]

Examiners report

[N/A]

Consider $f (x) = 4 \cos x (1 - 3 \cos 2 x + 3 \cos^{2} 2 x - \cos^{3} 2 x)$ .

Expand and simplify ${(1 - a)}^{3}$ in ascending powers of $a$ .

[2]

a.i.

By using a suitable substitution for $a$ , show that $1 - 3 \cos 2 x + 3 \cos^{2} 2 x - \cos^{3} 2 x = 8 \sin^{6} x$ .

[4]

a.ii.

Show that $\int_{0}^{m} f (x) d x = \frac{32}{7} \sin^{7} m$ , where $m$ is a positive real constant.

[4]

b.i.

It is given that $\int_{m}^{\frac{π}{2}} f (x) d x = \frac{127}{28}$ , where $0 \leq m \leq \frac{π}{2}$ . Find the value of $m$ .

[5]

b.ii.

Markscheme

EITHER

attempt to use binomial expansion (M1)

$1 + {}_{3}C_{1} \times 1 \times (- a) + {}_{3}C_{2} \times 1 \times {(- a)}^{2} + 1 \times {(- a)}^{3}$

$(1 - a) (1 - a) (1 - a)$

$= (1 - a) (1 - 2 a + a^{2})$ (M1)

THEN

$= 1 - 3 a + 3 a^{2} - a^{3}$ A1

[2 marks]

a.i.

$a = \cos 2 x$ (A1)

So, $1 - 3 \cos 2 x + 3 \cos^{2} 2 x - \cos^{3} 2 x =$

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

attempt to substitute any double angle rule for $\cos 2 x$ into ${(1 - \cos 2 x)}^{3}$ (M1)

$= {(2 \sin^{2} x)}^{3}$ A1

$= 8 \sin^{6} x$ AG

Note: Allow working RHS to LHS.

[4 marks]

a.ii.

recognizing to integrate $\int (4 \cos x \times 8 \sin^{6} x) d x$ (M1)

EITHER

applies integration by inspection (M1)

$32 \int (\cos x \times {(\sin x)}^{6}) d x$

$= \frac{32}{7} \sin^{7} x (+ c)$ A1

${[\frac{32}{7} \sin^{7} x]}_{0}^{m} (= \frac{32}{7} \sin^{7} m - \frac{32}{7} \sin^{7} 0)$ A1

$u = \sin x \Rightarrow \frac{d u}{d x} = \cos x$ (M1)

$\int 32 \cos x (\sin^{6} x) d x = \int 32 u^{6} d u$

$= \frac{32}{7} u^{7} (+ c)$ A1

${[\frac{32}{7} \sin^{7} x]}_{0}^{m}$ OR ${[\frac{32}{7} u^{7}]}_{0}^{\sin m} (= \frac{32}{7} \sin^{7} m - \frac{32}{7} \sin^{7} 0)$ A1

THEN

$= \frac{32}{7} \sin^{7} m$ AG

[4 marks]

b.i.

EITHER

$\int_{m}^{\frac{π}{2}} f (x) d x (= {[\frac{32}{7} \sin^{7} x]}_{m}^{\frac{π}{2}}) = \frac{32}{7} \sin^{7} \frac{π}{2} - \frac{32}{7} \sin^{7} m$ M1

$\frac{32}{7} \sin^{7} \frac{π}{2} - \frac{32}{7} \sin^{7} m = \frac{127}{28}$ OR $\frac{32}{7} (1 - \sin^{7} m) = \frac{127}{28}$ (M1)

$\int_{0}^{\frac{π}{2}} f (x) d x = \int_{0}^{m} f (x) d x + \int_{m}^{\frac{π}{2}} f (x) d x$ M1

$\frac{32}{7} = \frac{32}{7} \sin^{7} m + \frac{127}{28}$ (M1)

THEN

$\sin^{7} m = \frac{1}{128} (= \frac{1}{2^{7}})$ (A1)

$\sin m = \frac{1}{2}$ (A1)

$m = \frac{π}{6}$ A1

[5 marks]

b.ii.

Examiners report

Many candidates successfully expanded the binomial, with the most common error being to omit the negative sign with a. The connection between (a)(i) and (ii) was often noted but not fully utilised with candidates embarking on unnecessary complex algebraic expansions of expressions involving double angle rules. Candidates often struggled to apply inspection or substitution when integrating. As a 'show that' question, b(i) provided a useful result to be utilised in (ii). So even without successfully completing (i) candidates could apply it in part (ii). Not many managed to do so.

a.i.

[N/A]

a.ii.

[N/A]

b.i.

[N/A]

b.ii.

In an international competition, participants can answer questions in only one of the three following languages: Portuguese, Mandarin or Hindi. 80 participants took part in the competition. The number of participants answering in Portuguese, Mandarin or Hindi is shown in the table.

A boy is chosen at random.

State the number of boys who answered questions in Portuguese.

[1]

Find the probability that the boy answered questions in Hindi.

[2]

Two girls are selected at random.

Calculate the probability that one girl answered questions in Mandarin and the other answered questions in Hindi.

[3]

Markscheme

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

20 (A1) (C1)

[1 mark]

$\frac{5}{{43}}\,\,\,\left( {0.11627 \ldots ,\,\,11.6279 \ldots {\text{% }}} \right)$ (A1)(A1) (C2)

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

[2 marks]

$\frac{7}{{37}} \times \frac{{12}}{{36}} + \frac{{12}}{{37}} \times \frac{7}{{36}}$ (A1)(M1)

Note: Award (A1) for first or second correct product seen, (M1) for adding their two products or for multiplying their product by two.

$= \frac{{14}}{{111}}\,\,\left( {\,0.12612 \ldots ,\,\,12.6126\,{\text{% }}} \right)$ (A1) (C3)

[3 marks]

Examiners report

[N/A]

A small cuboid box has a rectangular base of length $3x$ cm and width $x$ cm, where $x > 0$ . The height is $y$ cm, where $y > 0$ .

The sum of the length, width and height is $12$ cm.

The volume of the box is $V$ cm³.

Write down an expression for $y$ in terms of $x$ .

[1]

Find an expression for $V$ in terms of $x$ .

[2]

Find $\frac{{{\text{d}}V}}{{{\text{d}}x}}$ .

[2]

Find the value of $x$ for which $V$ is a maximum.

[4]

d.i.

Justify your answer.

[3]

d.ii.

Find the maximum volume.

[2]

Markscheme

$y = 12 - 4x$ A1 N1

[1 mark]

correct substitution into volume formula (A1)

eg $3x \times x \times y{\text{, }}x \times 3x \times \left( {12 - x - 3x} \right){\text{, }}\left( {12 - 4x} \right)\left( x \right)\left( {3x} \right)$

$V = 3{x^2}\left( {12 - 4x} \right)\,\,\left( { = 36{x^2} - 12{x^3}} \right)$ A1 N2

Note: Award A0 for unfinished answers such as $3{x^2}\left( {12 - x - 3x} \right)$ .

[2 marks]

$\frac{{{\text{d}}V}}{{{\text{d}}x}} = 72x - 36{x^2}$ A1A1 N2

Note: Award A1 for $72x$ and A1 for $- 36{x^2}$ .

[2 marks]

valid approach to find maximum (M1)

eg $V' = 0{\text{, }}72x - 36{x^2} = 0$

correct working (A1)

eg $x\left( {72 - 36x} \right){\text{, }}\frac{{ - 72 \pm \sqrt {{{72}^2} - 4 \cdot \left( { - 36} \right) \cdot 0} }}{{2\left( { - 36} \right)}}{\text{, }}36x = 72{\text{, }}36x\left( {2 - x} \right) = 0$

$x = 2$ A2 N2

Note: Award A1 for $x = 2$ and $x = 0$ .

[4 marks]

d.i.

valid approach to explain that $V$ is maximum when $x = 2$ (M1)

eg attempt to find ${V''}$ , sign chart (must be labelled ${V'}$ )

correct value/s A1

eg $V''\left( 2 \right) = 72 - 72 \times 2$ , $V'\left( a \right)$ where $a < 2$ and $V'\left( b \right)$ where $b > 2$

correct reasoning R1

eg $V''\left( 2 \right) < 0$ , ${V'}$ is positive for $x < 2$ and negative for $x > 2$

Note: Do not award R1 unless A1 has been awarded.

$V$ is maximum when $x = 2$ AG N0

[3 marks]

d.ii.

correct substitution into their expression for volume A1

eg $3 \times {2^2}\left( {12 - 4 \times 2} \right)$ , $36\left( {{2^2}} \right) - 12\left( {{2^3}} \right)$

$V = 48$ (cm³) A1 N1

[2 marks]

Examiners report

[N/A]

d.i.

[N/A]

d.ii.

[N/A]

Let $f\left( x \right) = \frac{1}{3}{x^3} + {x^2} - 15x + 17$ .

The graph of $f$ has horizontal tangents at the points where $x$ = $a$ and $x$ = $b$ , $a$ < $b$ .

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

[2]

Find the value of $a$ and the value of $b$ .

[3]

Sketch the graph of $y = f'\left( x \right)$ .

[1]

c.i.

Hence explain why the graph of $f$ has a local maximum point at $x = a$ .

[1]

c.ii.

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

[3]

d.i.

Hence, use your answer to part (d)(i) to show that the graph of $f$ has a local minimum point at $x = b$ .

[1]

d.ii.

The normal to the graph of $f$ at $x = a$ and the tangent to the graph of $f$ at $x = b$ intersect at the point ( $p$ , $q$ ) .

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

[5]

Markscheme

$f'\left( x \right) = {x^2} + 2x - 15$ (M1)A1

[2 marks]

correct reasoning that $f'\left( x \right) = 0$ (seen anywhere) (M1)

${x^2} + 2x - 15 = 0$

valid approach to solve quadratic M1

$\left( {x - 3} \right)\left( {x + 5} \right)$ , quadratic formula

correct values for $x$

3, −5

correct values for $a$ and $b$

$a$ = −5 and $b$ = 3 A1

[3 marks]

[1 mark]

c.i.

first derivative changes from positive to negative at $x = a$ A1

so local maximum at $x = a$ AG

[1 mark]

c.ii.

$f''\left( x \right) = 2x + 2$ A1

substituting their $b$ into their second derivative (M1)

$f''\left( 3 \right) = 2 \times 3 + 2$

$f''\left( b \right) = 8$ (A1)

[3 marks]

d.i.

$f''\left( b \right)$ is positive so graph is concave up R1

so local minimum at $x = b$ AG

[1 mark]

d.ii.

normal to $f$ at $x = a$ is $x$ = −5 (seen anywhere) (A1)

attempt to find $y$ -coordinate at their value of $b$ (M1)

$f\left( 3 \right) =$ −10 (A1)

tangent at $x = b$ has equation $y$ = −10 (seen anywhere) A1

intersection at (−5, −10)

$p$ = −5 and $q$ = −10 A1

[5 marks]

Examiners report

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

d.i.

[N/A]

d.ii.

[N/A]

The following diagram shows part of the graph of a quadratic function $f$ .

The graph of $f$ has its vertex at $(3, 4)$ , and it passes through point $Q$ as shown.

The function can be written in the form $f (x) = a {(x - h)}^{2} + k$ .

The line $L$ is tangent to the graph of $f$ at $Q$ .

Now consider another function $y = g (x)$ . The derivative of $g$ is given by $g' (x) = f (x) - d$ , where $d \in ℝ$ .

Write down the equation of the axis of symmetry.

[1]

Write down the values of $h$ and $k$ .

[2]

b.i.

Point $Q$ has coordinates $(5, 12)$ . Find the value of $a$ .

[2]

b.ii.

Find the equation of $L$ .

[4]

Find the values of $d$ for which $g$ is an increasing function.

[3]

Find the values of $x$ for which the graph of $g$ is concave-up.

[3]

Markscheme

$x = 3$ A1

Note: Must be an equation in the form “ $x =$ ”. Do not accept $3$ or $\frac{- b}{2 a} = 3$ .

[1 mark]

$h = 3, k = 4$ (accept $a {(x - 3)}^{2} + 4$ ) A1A1

[2 marks]

b.i.

attempt to substitute coordinates of $Q$ (M1)

$12 = a {(5 - 3)}^{2} + 4, 4 a + 4 = 12$

$a = 2$ A1

[2 marks]

b.ii.

recognize need to find derivative of $f$ (M1)

$f' (x) = 4 (x - 3)$ or $f' (x) = 4 x - 12$ A1

$f' (5) = 8$ (may be seen as gradient in their equation) (A1)

$y - 12 = 8 (x - 5)$ or $y = 8 x - 28$ A1

Note: Award A0 for $L = 8 x - 28$ .

[4 marks]

METHOD 1

Recognizing that for $g$ to be increasing, $f (x) - d > 0$ , or $g' > 0$ (M1)

The vertex must be above the $x$ -axis, $4 - d > 0, d - 4 < 0$ (R1)

$d < 4$ A1

METHOD 2

attempting to find discriminant of $g'$ (M1)

${(- 12)}^{2} - 4 (2) (22 - d)$

recognizing discriminant must be negative (R1)

$- 32 + 8 d < 0$ OR $Δ < 0$

$d < 4$ A1

[3 marks]

recognizing that for $g$ to be concave up, $g'' > 0$ (M1)

$g'' > 0$ when $f' > 0, 4 x - 12 > 0, x - 3 > 0$ (R1)

$x > 3$ A1

[3 marks]

Examiners report

In parts (a) and (b) of this question, a majority of candidates recognized the connection between the coordinates of the vertex and the axis of symmetry and the values of $h$ and $k$ , and most candidates were able to successfully substitute the coordinates of point Q to find the value of $a$ . In part (c), the candidates who recognized the need to use the derivative to find the gradient of the tangent were generally successful in finding the equation of the line, although many did not give their equation in the proper form in terms of $x$ and $y$ , and instead wrote $L = 8 x - 28$ , thus losing the final mark. Parts (d) and (e) were much more challenging for candidates. Although a good number of candidates recognized that $g' (x) > 0$ in part (d), and $g'' (x) > 0$ in part (e), very few were able to proceed beyond this point and find the correct inequalities for their final answers.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

A function, $f$ , has its derivative given by $f' (x) = 3 x^{2} - 12 x + p$ , where $p \in ℝ$ . The following diagram shows part of the graph of $f'$ .

The graph of $f'$ has an axis of symmetry $x = q$ .

The vertex of the graph of $f'$ lies on the $x$ -axis.

The graph of $f$ has a point of inflexion at $x = a$ .

Find the value of $q$ .

[2]

Write down the value of the discriminant of $f'$ .

[1]

b.i.

Hence or otherwise, find the value of $p$ .

[3]

b.ii.

Find the value of the gradient of the graph of $f'$ at $x = 0$ .

[3]

Sketch the graph of $f ″$ , the second derivative of $f$ . Indicate clearly the $x$ -intercept and the $y$ -intercept.

[2]

Write down the value of $a$ .

[1]

e.i.

Find the values of $x$ for which the graph of $f$ is concave-down. Justify your answer.

[2]

e.ii.

Markscheme

EITHER

attempt to use $x = - \frac{b}{2 a}$ (M1)

$q = - \frac{- 12}{2 \times 3}$

attempt to complete the square (M1)

$3 {(x - 2)}^{2} - 12 + p$

attempt to differentiate and equate to $0$ (M1)

$f'' (x) = 6 x - 12 = 0$

THEN

$q = 2$ A1

[2 marks]

discriminant $= 0$ A1

[1 mark]

b.i.

EITHER

attempt to substitute into $b^{2} - 4 a c$ (M1)

${(- 12)}^{2} - 4 \times 3 \times p = 0$ A1

$f' (2) = 0$ (M1)

$- 12 + p = 0$ A1

THEN

$p = 12$ A1

[3 marks]

b.ii.

$f'' (x) = 6 x - 12$ A1

attempt to find $f'' (0)$ (M1)

$= 6 \times 0 - 12$

gradient $= - 12$ A1

[3 marks]

A1A1

Note: Award A1 for line with positive gradient, A1 for correct intercepts.

[2 marks]

$a = 2$ A1

[1 mark]

e.i.

$x < 2$ A1

$f'' (x) < 0$ (for $x < 2$ ) OR the $f''$ is below the $x$ -axis (for $x < 2$ )

OR $f''$ (sign diagram must be labelled $f''$ ) R1

[2 marks]

e.ii.

Examiners report

Candidates did score well on this question. As always, candidates are encouraged to read the questions carefully for key words such as 'value' as opposed to 'expression'. So, if asked for the value of the discriminant, their answer should be a number and not an expression found from $b^{2} - 4 a c$ . As such the value of the discriminant in (b)(i) was often seen in (b)(ii). Please ask students to use a straight edge when sketching a straight line! Overall, the reasoning mark for determining where the graph of f is concave-down, was an improvement on previous years. Sign diagrams were typically well labelled, and the description contained clarity regarding which function was being referred to.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

e.i.

[N/A]

e.ii.

Consider $f(x) = \log k(6x - 3{x^2})$ , for $0 < x < 2$ , where $k > 0$ .

The equation $f(x) = 2$ has exactly one solution. Find the value of $k$ .

Markscheme

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

METHOD 1 – using discriminant

correct equation without logs (A1)

eg $\,\,\,\,\,$ $6x - 3{x^2} = {k^2}$

valid approach (M1)

eg $\,\,\,\,\,$ $- 3{x^2} + 6x - {k^2} = 0,{\text{ }}3{x^2} - 6x + {k^2} = 0$

recognizing discriminant must be zero (seen anywhere) M1

eg $\,\,\,\,\,$ $\Delta = 0$

correct discriminant (A1)

eg $\,\,\,\,\,$ ${6^2} - 4( - 3)( - {k^2}),{\text{ }}36 - 12{k^2} = 0$

correct working (A1)

eg $\,\,\,\,\,$ $12{k^2} = 36,{\text{ }}{k^2} = 3$

$k = \sqrt 3$ A2 N2

METHOD 2 – completing the square

correct equation without logs (A1)

eg $\,\,\,\,\,$ $6x - 3{x^2} = {k^2}$

valid approach to complete the square (M1)

eg $\,\,\,\,\,$ $3({x^2} - 2x + 1) = - {k^2} + 3,{\text{ }}{x^2} - 2x + 1 - 1 + \frac{{{k^2}}}{3} = 0$

correct working (A1)

eg $\,\,\,\,\,$ $3{(x - 1)^2} = - {k^2} + 3,{\text{ }}{(x - 1)^2} - 1 + \frac{{{k^2}}}{3} = 0$

recognizing conditions for one solution M1

eg $\,\,\,\,\,$ ${(x - 1)^2} = 0,{\text{ }} - 1 + \frac{{{k^2}}}{3} = 0$

correct working (A1)

eg $\,\,\,\,\,$ $\frac{{{k^2}}}{3} = 1,{\text{ }}{k^2} = 3$

$k = \sqrt 3$ A2 N2

[7 marks]

Examiners report

[N/A]

A factory produces shirts. The cost, C, in Fijian dollars (FJD), of producing x shirts can be modelled by

C(x) = (x − 75)² + 100.

The cost of production should not exceed 500 FJD. To do this the factory needs to produce at least 55 shirts and at most s shirts.

Find the cost of producing 70 shirts.

[2]

Find the value of s.

[2]

Find the number of shirts produced when the cost of production is lowest.

[2]

Markscheme

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

(70 − 75)² + 100 (M1)

Note: Award (M1) for substituting in x = 70.

125 (A1) (C2)

[2 marks]

(s − 75)² + 100 = 500 (M1)

Note: Award (M1) for equating C(x) to 500. Accept an inequality instead of =.

(M1)

Note: Award (M1) for sketching correct graph(s).

(s =) 95 (A1) (C2)

[2 marks]

(M1)

Note: Award (M1) for an attempt at finding the minimum point using graph.

$\frac{{95 + 55}}{2}$ (M1)

Note: Award (M1) for attempting to find the mid-point between their part (b) and 55.

(C'(x) =) 2x − 150 = 0 (M1)

Note: Award (M1) for an attempt at differentiation that is correctly equated to zero.

75 (A1) (C2)

[2 marks]

Examiners report

[N/A]

The equation of a curve is $y = \frac{1}{2}{x^4} - \frac{3}{2}{x^2} + 7$ .

The gradient of the tangent to the curve at a point P is $- 10$ .

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

[2]

Find the coordinates of P.

[4]

Markscheme

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

$2{x^3} - 3x$ (A1)(A1) (C2)

Note: Award (A1) for $2{x^3}$ , award (A1) for $- 3x$ .

Award at most (A1)(A0) if there are any extra terms.

[2 marks]

$2{x^3} - 3x = - 10$ (M1)

Note: Award (M1) for equating their answer to part (a) to $- 10$ .

$x = - 2$ (A1)(ft)

Note: Follow through from part (a). Award (M0)(A0) for $- 2$ seen without working.

$y = \frac{1}{2}{( - 2)^4} - \frac{3}{2}{( - 2)^2} + 7$ (M1)

Note: Award (M1) substituting their $- 2$ into the original function.

$y = 9$ (A1)(ft) (C4)

Note: Accept $( - 2,{\text{ }}9)$ .

[4 marks]

Examiners report

[N/A]

Consider a function $f$ with domain $a < x < b$ . The following diagram shows the graph of $f'$ , the derivative of $f$ .

The graph of $f'$ , the derivative of $f$ , has $x$ -intercepts at $x = p, x = 0$ and $x = t$ . There are local maximum points at $x = q$ and $x = t$ and a local minimum point at $x = r$ .

Find all the values of $x$ where the graph of $f$ is increasing. Justify your answer.

[2]

Find the value of $x$ where the graph of $f$ has a local maximum.

[1]

Find the value of $x$ where the graph of $f$ has a local minimum. Justify your answer.

[2]

c.i.

Find the values of $x$ where the graph of $f$ has points of inflexion. Justify your answer.

[3]

c.ii.

The total area of the region enclosed by the graph of $f'$ , the derivative of $f$ , and the $x$ -axis is $20$ .

Given that $f (p) + f (t) = 4$ , find the value of $f (0)$ .

[6]

Markscheme

Special note: In this question if candidates use the word 'gradient' in their reasoning. e.g. gradient is positive, it must be clear whether this is the gradient of $f$ or the gradient of $f'$ to earn the R mark.

$f$ increases when $p < x < 0$ A1

$f$ increases when $f' (x) > 0$ OR $f'$ is above the $x$ -axis R1

Note: Do not award A0R1.

[2 marks]

$x = 0$ A1

[1 mark]

$f$ is minimum when $x = p$ A1

because $f' (p) = 0, f' (x) < 0$ when $x < p$ and $f' (x) > 0$ when $x > p$

(may be seen in a sign diagram clearly labelled as $f'$ )

OR because $f'$ changes from negative to positive at $x = p$

OR $f' (p) = 0$ and slope of $f'$ is positive at $x = p$ R1

Note: Do not award A0 R1

[2 marks]

c.i.

$f$ has points of inflexion when $x = q, x = r$ and $x = t$ A2

$f'$ has turning points at $x = q, x = r$ and $x = t$

$f'' (q) = 0, f'' (r) = 0$ and $f'' (t) = 0$ and $f'$ changes from increasing to decreasing or vice versa at each of these $x$ -values (may be seen in a sign diagram clearly labelled as $f''$ and $f'$ ) R1

Note: Award A0 if any incorrect answers are given. Do not award A0R1

[3 marks]

c.ii.

recognizing area from $p$ to $t$ (seen anywhere) M1

$\int_{p}^{t} |f' (x)| d x$

recognizing to negate integral for area below $x$ -axis (M1)

$\int_{p}^{0} f' (x) d x - \int_{0}^{t} f' (x) d x$ OR $\int_{p}^{0} f' (x) d x + \int_{t}^{0} f' (x) d x$

$\int_{m}^{n} f' (x) d x = f (n) - f (m)$ (for any integral) (M1)

$f (0) - f (p) - [f (t) - f (0)]$ OR $f (0) - f (p) + f (0) - f (t)$ (A1)

$2 f (0) - [f (t) + f (p)] = 20, 2 f (0) - 4 = 20$ (A1)

$f (0) = 12$ A1

[6 marks]

Examiners report

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

Let $f(x) = {x^2} - x$ , for $x \in \mathbb{R}$ . The following diagram shows part of the graph of $f$ .

N17/5/MATME/SP1/ENG/TZ0/08

The graph of $f$ crosses the $x$ -axis at the origin and at the point ${\text{P}}(1,{\text{ }}0)$ .

The line $L$ intersects the graph of $f$ at another point Q, as shown in the following diagram.

N17/5/MATME/SP1/ENG/TZ0/08.c.d

Find the area of the region enclosed by the graph of $f$ and the line $L$ .

Markscheme

valid approach (M1)

eg $\,\,\,\,\,$ $\int {L - f,{\text{ }}\int_{ - 1}^1 {(1 - {x^2}){\text{d}}x} }$ , splitting area into triangles and integrals

correct integration (A1)(A1)

eg $\,\,\,\,\,$ $\left[ {x - \frac{{{x^3}}}{3}} \right]_{ - 1}^1,{\text{ }} - \frac{{{x^3}}}{3} - \frac{{{x^2}}}{2} + \frac{{{x^2}}}{2} + x$

substituting their limits into their integrated function and subtracting (in any order) (M1)

eg $\,\,\,\,\,$ $1 - \frac{1}{3} - \left( { - 1 - \frac{{ - 1}}{3}} \right)$

Note: Award M0 for substituting into original or differentiated function.

area $= \frac{4}{3}$ A2 N3

[6 marks]

Examiners report

[N/A]

Let $f\left( x \right) = \frac{1}{{\sqrt {2x - 1} }}$ , for $x > \frac{1}{2}$ .

Find $\int {{{\left( {f\left( x \right)} \right)}^2}{\text{d}}x}$ .

[3]

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

The shaded region R is enclosed by the graph of f, the x-axis, and the lines x = 1 and x = 9 . Find the volume of the solid formed when R is revolved 360° about the x-axis.

[4]

Markscheme

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

correct working (A1)

eg $\int {\frac{1}{{2x - 1}}{\text{d}}x,\,\,\int {{{\left( {2x - 1} \right)}^{ - 1}},\,\,\frac{1}{{2x - 1}},\,\,\int {{{\left( {\frac{1}{{\sqrt u }}} \right)}^2}\frac{{{\text{d}}u}}{2}} } }$

${\int {\left( {f\left( x \right)} \right)} ^2}{\text{d}}x = \frac{1}{2}{\text{ln}}\left( {2x - 1} \right) + c$ A2 N3

Note: Award A1 for $\frac{1}{2}{\text{ln}}\left( {2x - 1} \right)$ .

[3 marks]

attempt to substitute either limits or the function into formula involving f ² (accept absence of $\pi$ / dx) (M1)

eg $\int_1^9 {{y^2}{\text{d}}x,\,\,} \pi {\int {\left( {\frac{1}{{\sqrt {2x - 1} }}} \right)} ^2}{\text{d}}x,\,\,\left[ {\frac{1}{2}{\text{ln}}\left( {2x - 1} \right)} \right]_1^9$

substituting limits into their integral and subtracting (in any order) (M1)

eg $\frac{\pi }{2}\left( {{\text{ln}}\left( {17} \right) - {\text{ln}}\left( 1 \right)} \right),\,\,\pi \left( {0 - \frac{1}{2}{\text{ln}}\left( {2 \times 9 - 1} \right)} \right)$

correct working involving calculating a log value or using log law (A1)

eg ${\text{ln}}\left( 1 \right) = 0,\,\,{\text{ln}}\left( {\frac{{17}}{1}} \right)$

$\frac{\pi }{2}{\text{ln}}17\,\,\,\,\left( {{\text{accept }}\pi {\text{ln}}\sqrt {17} } \right)$ A1 N3

Note: Full FT may be awarded as normal, from their incorrect answer in part (a), however, do not award the final two A marks unless they involve logarithms.

[4 marks]

Examiners report

[N/A]

Let $y = {\left( {{x^3} + x} \right)^{\frac{3}{2}}}$ .

Consider the functions $f\left( x \right) = \sqrt {{x^3} + x}$ and $g\left( x \right) = 6 - 3{x^2}\sqrt {{x^3} + x}$ , for $x$ ≥ 0.

The graphs of $f$ and $g$ are shown in the following diagram.

The shaded region $R$ is enclosed by the graphs of $f$ , $g$ , the $y$ -axis and $x = 1$ .

Hence find $\int {\left( {3{x^2} + 1} \right)\sqrt {{x^3} + x} } \,{\text{d}}x$ .

[3]

Write down an expression for the area of $R$ .

[2]

Hence find the exact area of $R$ .

[6]

Markscheme

integrating by inspection from (a) or by substitution (M1)

eg $\frac{2}{3}\int {\frac{3}{2}} \left( {3{x^2} + 1} \right)\sqrt {{x^3} + x} \,{\text{d}}x$ , $u = {x^3} + x$ , $\frac{{{\text{d}}u}}{{{\text{d}}x}} = 3{x^2} + 1$ , $\int {{u^{\frac{1}{2}}}}$ , $\frac{{{u^{\frac{3}{2}}}}}{{1.5}}$

correct integrated expression in terms of $x$ A2 N3

eg $\frac{2}{3}{\left( {{x^3} + x} \right)^{\frac{3}{2}}} + C$ , $\frac{{{{\left( {{x^3} + x} \right)}^{1.5}}}}{{1.5}} + C$

[3 marks]

integrating and subtracting functions (in any order) (M1)

eg $\int {g - f}$ , $\int {f - \int g }$

correct integral (including limits, accept absence of ${{\text{d}}x}$ ) A1 N2

eg $\int_0^1 {\left( {g - f} \right)} \,{\text{d}}x$ , $\int_0^1 {6 - 3{x^2}\sqrt {{x^3} + x} - \sqrt {{x^3} + x} } \,{\text{d}}x$ , $\int_0^1 {g\left( x \right) - } \int_0^1 {f\left( x \right)}$

[2 marks]

recognizing $\sqrt {{x^3} + x}$ is a common factor (seen anywhere, may be seen in part (c)) (M1)

eg $\left( { - 3{x^2} - 1} \right)\sqrt {{x^3} + x}$ , $\int {6 - \left( {3{x^2} + 1} \right)} \sqrt {{x^3} + x}$ , $\left( {3{x^2} - 1} \right)\sqrt {{x^3} + x}$

correct integration (A1)(A1)

eg $6x - \frac{2}{3}{\left( {{x^3} + x} \right)^{\frac{3}{2}}}$

Note: Award A1 for $6x$ and award A1 for $- \frac{2}{3}{\left( {{x^3} + x} \right)^{\frac{3}{2}}}$ .

substituting limits into their integrated function and subtracting (in any order) (M1)

eg $6 - \frac{2}{3}{\left( {{1^3} + 1} \right)^{\frac{3}{2}}}$ , $0 - \left[ {6 - \frac{2}{3}{{\left( {{1^3} + 1} \right)}^{\frac{3}{2}}}} \right]$

correct working (A1)

eg $6 - \frac{2}{3} \times 2\sqrt 2$ , $6 - \frac{2}{3} \times \sqrt 4 \times \sqrt 2$

area of $R = 6 - \frac{{4\sqrt 2 }}{3}\,\,\,\left( { = 6 - \frac{2}{3}\sqrt 8 {\text{,}}\,\,\,6 - \frac{2}{3} \times {2^{\frac{3}{2}}}{\text{,}}\,\,\,\frac{{18 - 4\sqrt 2 }}{3}} \right)$ A1 N3

[6 marks]

Examiners report

[N/A]

Let $f\left( x \right) = {x^3} - 2{x^2} + ax + 6$ . Part of the graph of $f$ is shown in the following diagram.

The graph of $f$ crosses the $y$ -axis at the point P. The line L is tangent to the graph of $f$ at P.

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

[2]

b.i.

Hence, find the equation of L in terms of $a$ .

[4]

b.ii.

The graph of $f$ has a local minimum at the point Q. The line L passes through Q.

Find the value of $a$ .

[8]

Markscheme

$f' = 3{x^2} - 4x + a$ A2 N2

[2 marks]

b.i.

valid approach (M1)

eg $f'\left( 0 \right)$

correct working (A1)

eg $3{\left( 0 \right)^2} - 4\left( 0 \right) + a$ , slope = $a$ , $f'\left( 0 \right) = a$

attempt to substitute gradient and coordinates into linear equation (M1)

eg $y - 6 = a\left( {x - 0} \right)$ , $y - 0 = a\left( {x - 6} \right)$ , $6 = a\left( 0 \right) + c$ , L $= ax + 6$

correct equation A1 N3

eg $y = ax + 6$ , $y - 6 = ax$ , $y - 6 = a\left( {x - 0} \right)$

[4 marks]

b.ii.

valid approach to find intersection (M1)

eg $f\left( x \right) = L$

correct equation (A1)

eg ${x^3} - 2{x^2} + ax + 6 = ax + 6$

correct working (A1)

eg ${x^3} - 2{x^2} = 0$ , ${x^2}(x - 2) = 0$

$x = 2$ at Q (A1)

valid approach to find minimum (M1)

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

correct equation (A1)

eg $3{x^2} - 4x + a = 0$

substitution of their value of $x$ at Q into their $f'\left( x \right) = 0$ equation (M1)

eg $3{\left( 2 \right)^2} - 4\left( 2 \right) + a = 0$ , $12 - 8 + a = 0$

$a$ = −4 A1 N0

[8 marks]

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

A closed cylindrical can with radius r centimetres and height h centimetres has a volume of 20 $\pi$ cm³.

The material for the base and top of the can costs 10 cents per cm² and the material for the curved side costs 8 cents per cm². The total cost of the material, in cents, is C.

Express h in terms of r.

[2]

Show that $C = 20\pi {r^2} + \frac{{320\pi }}{r}$ .

[4]

Given that there is a minimum value for C, find this minimum value in terms of $\pi$ .

[9]

Markscheme

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

correct equation for volume (A1)
eg $\pi {r^2}h = 20\pi$

$h = \frac{{20}}{{{r^2}}}$ A1 N2

[2 marks]

attempt to find formula for cost of parts (M1)
eg 10 × two circles, 8 × curved side

correct expression for cost of two circles in terms of r (seen anywhere) A1
eg $2\pi {r^2} \times 10$

correct expression for cost of curved side (seen anywhere) (A1)
eg $2\pi r \times h \times 8$

correct expression for cost of curved side in terms of r A1
eg $8 \times 2\pi r \times \frac{{20}}{{{r^2}}},\,\,\frac{{320\pi }}{{{r^2}}}$

$C = 20\pi {r^2} + \frac{{320\pi }}{r}$ AG N0

[4 marks]

recognize $C' = 0$ at minimum (R1)
eg $C' = 0,\,\,\frac{{{\text{d}}C}}{{{\text{d}}r}} = 0$

correct differentiation (may be seen in equation)

$C' = 40\pi r - \frac{{320\pi }}{{{r^2}}}$ A1A1

correct equation A1
eg $40\pi r - \frac{{320\pi }}{{{r^2}}} = 0,\,\,40\pi r\frac{{320\pi }}{{{r^2}}}$

correct working (A1)
eg $40{r^3} = 320,\,\,{r^3} = 8$

r = 2 (m) A1

attempt to substitute their value of r into C
eg $20\pi \times 4 + 320 \times \frac{\pi }{2}$ (M1)

correct working
eg $80\pi + 160\pi$ (A1)

$240\pi$ (cents) A1 N3

Note: Do not accept 753.6, 753.98 or 754, even if 240 $\pi$ is seen.

[9 marks]

Examiners report

[N/A]

Let $f(x) = \cos x$ .

Let $g(x) = {x^k}$ , where $k \in {\mathbb{Z}^ + }$ .

Let $k = 21$ and $h(x) = \left( {{f^{(19)}}(x) \times {g^{(19)}}(x)} \right)$ .

(i) Find the first four derivatives of $f(x)$ .

(ii) Find ${f^{(19)}}(x)$ .

[4]

(i) Find the first three derivatives of $g(x)$ .

(ii) Given that ${g^{(19)}}(x) = \frac{{k!}}{{(k - p)!}}({x^{k - 19}})$ , find $p$ .

[5]

(i) Find $h'(x)$ .

(ii) Hence, show that $h'(\pi ) = \frac{{ - 21!}}{2}{\pi ^2}$ .

[7]

Markscheme

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

(i) $f'(x) = - \sin x,{\text{ }}f''(x) = - \cos x,{\text{ }}{f^{(3)}}(x) = \sin x,{\text{ }}{f^{(4)}}(x) = \cos x$ A2 N2

(ii) valid approach (M1)

eg $\,\,\,\,\,$ recognizing that 19 is one less than a multiple of 4, ${f^{(19)}}(x) = {f^{(3)}}(x)$

${f^{(19)}}(x) = \sin x$ A1 N2

[4 marks]

(i) $g'(x) = k{x^{k - 1}}$

$g''(x) = k(k - 1){x^{k - 2}},{\text{ }}{g^{(3)}}(x) = k(k - 1)(k - 2){x^{k - 3}}$ A1A1 N2

(ii) METHOD 1

correct working that leads to the correct answer, involving the correct expression for the 19th derivative A2

eg $\,\,\,\,\,$ $k(k - 1)(k - 2) \ldots (k - 18) \times \frac{{(k - 19)!}}{{(k - 19)!}},{{\text{ }}_k}{P_{19}}$

$p = 19$ (accept $\frac{{k!}}{{(k - 19)!}}{x^{k - 19}}$ ) A1 N1

METHOD 2

correct working involving recognizing patterns in coefficients of first three derivatives (may be seen in part (b)(i)) leading to a general rule for 19th coefficient A2

eg $\,\,\,\,\,$ $g'' = 2!\left( {\begin{array}{*{20}{c}} k \\ 2 \end{array}} \right),{\text{ }}k(k - 1)(k - 2) = \frac{{k!}}{{(k - 3)!}},{\text{ }}{g^{(3)}}(x){ = _k}{P_3}({x^{k - 3}})$

${g^{(19)}}(x) = 19!\left( {\begin{array}{*{20}{c}} k \\ {19} \end{array}} \right),{\text{ }}19! \times \frac{{k!}}{{(k - 19)! \times 19!}},{{\text{ }}_k}{P_{19}}$

$p = 19$ (accept $\frac{{k!}}{{(k - 19)!}}{x^{k - 19}}$ ) A1 N1

[5 marks]

(i) valid approach using product rule (M1)

eg $\,\,\,\,\,$ $uv' + vu',{\text{ }}{f^{(19)}}{g^{(20)}} + {f^{(20)}}{g^{(19)}}$

correct 20th derivatives (must be seen in product rule) (A1)(A1)

eg $\,\,\,\,\,$ ${g^{(20)}}(x) = \frac{{21!}}{{(21 - 20)!}}x,{\text{ }}{f^{(20)}}(x) = \cos x$

$h'(x) = \sin x(21!x) + \cos x\left( {\frac{{21!}}{2}{x^2}} \right){\text{ }}\left( {{\text{accept }}\sin x\left( {\frac{{21!}}{{1!}}x} \right) + \cos x\left( {\frac{{21!}}{{2!}}{x^2}} \right)} \right)$ A1 N3

(ii) substituting $x = \pi$ (seen anywhere) (A1)

eg $\,\,\,\,\,$ ${f^{(19)}}(\pi ){g^{(20)}}(\pi ) + {f^{(20)}}(\pi ){g^{(19)}}(\pi ),{\text{ }}\sin \pi \frac{{21!}}{{1!}}\pi + \cos \pi \frac{{21!}}{{2!}}{\pi ^2}$

evidence of one correct value for $\sin \pi$ or $\cos \pi$ (seen anywhere) (A1)

eg $\,\,\,\,\,$ $\sin \pi = 0,{\text{ }}\cos \pi = - 1$

evidence of correct values substituted into $h'(\pi )$ A1

eg $\,\,\,\,\,$ $21!(\pi )\left( {0 - \frac{\pi }{{2!}}} \right),{\text{ }}21!(\pi )\left( { - \frac{\pi }{2}} \right),{\text{ }}0 + ( - 1)\frac{{21!}}{2}{\pi ^2}$

Note: If candidates write only the first line followed by the answer, award A1A0A0.

$\frac{{ - 21!}}{2}{\pi ^2}$ AG N0

[7 marks]

Examiners report

[N/A]

The expression $\frac{3 \sqrt{x} - 5}{\sqrt{x}}$ can be written as $3 - 5 x^{p}$ . Write down the value of $p$ .

[1]

Hence, find the value of $\int_{1}^{9} (\frac{3 \sqrt{x} - 5}{\sqrt{x}}) d x$ .

[4]

Markscheme

$\frac{3 \sqrt{x} - 5}{\sqrt{x}} = 3 - 5 x^{- \frac{1}{2}}$ A1

$p = - \frac{1}{2}$

[1 mark]

$\int \frac{3 \sqrt{x} - 5}{\sqrt{x}} d x = 3 x - 10 x^{\frac{1}{2}} (+ c)$ A1A1

substituting limits into their integrated function and subtracting (M1)

$3 (9) - 10 {(9)}^{\frac{1}{2}} - (3 (1) - 10 {(1)}^{\frac{1}{2}})$ OR $27 - 10 \times 3 - (3 - 10)$

$= 4$ A1

[4 marks]

Examiners report

Many candidates could give the value of p correctly. However, many did struggle with the integration, including substituting limits into the integrand, without integrating at all. An incorrect value of p often resulted in arithmetic of greater complexity.

[N/A]

In this question, all lengths are in metres and time is in seconds.

Consider two particles, $P_{1}$ and $P_{2}$ , which start to move at the same time.

Particle $P_{1}$ moves in a straight line such that its displacement from a fixed-point is given by $s (t) = 10 - \frac{7}{4} t^{2}$ , for $t \geq 0$ .

Find an expression for the velocity of $P_{1}$ at time $t$ .

[2]

Particle $P_{2}$ also moves in a straight line. The position of $P_{2}$ is given by $r = (\begin{matrix} - 1 \\ 6 \end{matrix}) + t (\begin{matrix} 4 \\ - 3 \end{matrix})$ .

The speed of $P_{1}$ is greater than the speed of $P_{2}$ when $t > q$ .

Find the value of $q$ .

[5]

Markscheme

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

recognizing velocity is derivative of displacement (M1)

eg $v = \frac{d s}{d t}, \frac{d}{d t} (10 - \frac{7}{4} t^{2})$

velocity $= - \frac{14}{4} t (= - \frac{7}{2} t)$ A1 N2

[2 marks]

valid approach to find speed of $P_{2}$ (M1)

eg $|(\begin{matrix} 4 \\ - 3 \end{matrix})|, \sqrt{4^{2} + {(- 3)}^{2}}$ , velocity $= \sqrt{4^{2} + {(- 3)}^{2}}$

correct speed (A1)

eg $5 {m s}^{- 1}$

recognizing relationship between speed and velocity (may be seen in inequality/equation) R1

eg $|- \frac{7}{2} t|$ , speed = | velocity | , graph of $P_{1}$ speed , $P_{1}$ speed $= \frac{7}{2} t, P_{2}$ velocity $= - 5$

correct inequality or equation that compares speed or velocity (accept any variable for $q$ ) A1

eg $|- \frac{7}{2} t| > 5, - \frac{7}{2} q < - 5, \frac{7}{2} q > 5, \frac{7}{2} q = 5$

$q = \frac{10}{7}$ (seconds) (accept $t > \frac{10}{7}$ , do not accept $t = \frac{10}{7}$ ) A1 N2

Note: Do not award the last two A1 marks without the R1.

[5 marks]

Examiners report

[N/A]

The following diagram shows part of the graph of $f (x) = \frac{k}{x}$ , for $x > 0, k > 0$ .

Let $P (p, \frac{k}{p})$ be any point on the graph of $f$ . Line $L_{1}$ is the tangent to the graph of $f$ at $P$ .

Line $L_{1}$ intersects the $x$ -axis at point $A (2 p, 0)$ and the $y$ -axis at point B.

Find $f' (p)$ in terms of $k$ and $p$ .

[2]

a.i.

Show that the equation of $L_{1}$ is $k x + p^{2} y - 2 p k = 0$ .

[2]

a.ii.

Find the area of triangle $AOB$ in terms of $k$ .

[5]

The graph of $f$ is translated by $(\begin{matrix} 4 \\ 3 \end{matrix})$ to give the graph of $g$ .
In the following diagram:

point $Q$ lies on the graph of $g$
points $C$ , $D$ and $E$ lie on the vertical asymptote of $g$
points $D$ and $F$ lie on the horizontal asymptote of $g$
point $G$ lies on the $x$ -axis such that $FG$ is parallel to $DC$ .

Line $L_{2}$ is the tangent to the graph of $g$ at $Q$ , and passes through $E$ and $F$ .

Given that triangle $EDF$ and rectangle $CDFG$ have equal areas, find the gradient of $L_{2}$ in terms of $p$ .

[6]

Markscheme

$f' (x) = - k x^{- 2}$ (A1)

$f' (p) = - k p^{- 2} (= - \frac{k}{p^{2}})$ A1 N2

[2 marks]

a.i.

attempt to use point and gradient to find equation of $L_{1}$ M1

eg $y - \frac{k}{p} = - k p^{- 2} (x - p), \frac{k}{p} = - \frac{k}{p^{2}} (p) + b$

correct working leading to answer A1

eg $p^{2} y - k p = - k x + k p, y - \frac{k}{p} = - \frac{k}{p^{2}} x + \frac{k}{p}, y = - \frac{k}{p^{2}} x + \frac{2 k}{p}$

$k x + p^{2} y - 2 p k = 0$ AG N0

[2 marks]

a.ii.

METHOD 1 – area of a triangle

recognizing $x = 0$ at $B$ (M1)

correct working to find $y$ -coordinate of null (A1)

eg $p^{2} y - 2 p k = 0$

$y$ -coordinate of null at $y = \frac{2 k}{p}$ (may be seen in area formula) A1

correct substitution to find area of triangle (A1)

eg $\frac{1}{2} (2 p) (\frac{2 k}{p}), p \times (\frac{2 k}{p})$

area of triangle $AOB = 2 k$ A1 N3

METHOD 2 – integration

recognizing to integrate $L_{1}$ between $0$ and $2 p$ (M1)

eg $\int_{0}^{2 p} L_{1} d x, \int_{0}^{2 p} - \frac{k}{p^{2}} x + \frac{2 k}{p}$

correct integration of both terms A1

eg $- \frac{k x^{2}}{2 p^{2}} + \frac{2 k x}{p}, - \frac{k}{2 p^{2}} x^{2} + \frac{2 k}{p} x + c, {[- \frac{k}{2 p^{2}} x^{2} + \frac{2 k}{p} x]}_{0}^{2 p}$

substituting limits into their integrated function and subtracting (in either order) (M1)

eg $- \frac{k {(2 p)}^{2}}{2 p^{2}} + \frac{2 k (2 p)}{p} - (0), - \frac{4 k p^{2}}{2 p^{2}} + \frac{4 k p}{p}$

correct working (A1)

eg $- 2 k + 4 k$

area of triangle $AOB = 2 k$ A1 N3

[5 marks]

Note: In this question, the second M mark may be awarded independently of the other marks, so it is possible to award (M0)(A0)M1(A0)(A0)A0.

recognizing use of transformation (M1)

eg area of triangle $AOB$ = area of triangle $DEF$ , $g (x) = \frac{k}{x - 4} + 3,$ gradient of $L_{2} =$ gradient of $L_{1}$ , $D (4, 3), 2p+4,$ one correct shift

correct working (A1)

eg area of triangle $DEF = 2 k, CD = 3, DF = 2 p, CG = 2 p, E (4, \frac{2 k}{p} + 3), F (2 p + 4, 3), Q (p + 4, \frac{k}{p} + 3),$

gradient of $L_{2} = - \frac{k}{p^{2}}, g' (x) = - \frac{k}{{(x - 4)}^{2}},$ area of rectangle $CDFG = 2 k$

valid approach (M1)

eg $\frac{ED \times DF}{2} = CD \times DF, 2 p \cdot 3 = 2 k, ED = 2 CD, \int_{4}^{2 p + 4} L_{2} d x = 4 k$

correct working (A1)

eg $ED = 6, E (4, 9), k = 3 p, gradient = \frac{3 - (\frac{2 k}{p} + 3)}{(2 p + 4) - 4}, \frac{- 6}{(\frac{2 k}{3})}, - \frac{9}{k}$

correct expression for gradient (in terms of $p$ ) (A1)

eg $\frac{- 6}{2 p}, \frac{9 - 3}{4 - (2 p + 4)}, - \frac{3 p}{p^{2}}, \frac{3 - (\frac{2 (3 p)}{p} + 3)}{(2 p + 4) - 4}, - \frac{9}{3 p}$

gradient of $L_{2}$ is $- \frac{3}{p} (= - 3 p^{- 1})$ A1 N3

[6 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

The following diagram shows the graph of $y = 4 - x^{2}$ , $0 \leq x \leq 2$ and rectangle $ORST$ . The rectangle has a vertex at the origin $O$ , a vertex on the $y$ -axis at the point $R (0, y)$ , a vertex on the $x$ -axis at the point $T (x, 0)$ and a vertex at point $S (x, y)$ on the graph.

Let $P$ represent the perimeter of rectangle $ORST$ .

Let $A$ represent the area of rectangle $ORST$ .

Show that $P = - 2 x^{2} + 2 x + 8$ .

[2]

Find the dimensions of rectangle $ORST$ that has maximum perimeter and determine the value of the maximum perimeter.

[6]

Find an expression for $A$ in terms of $x$ .

[2]

Find the dimensions of rectangle $ORST$ that has maximum area.

[5]

Determine the maximum area of rectangle $ORST$ .

[1]

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.

$P = 2 x + 2 y$ (A1)

$= 2 x + 2 (4 - x^{2})$ A1

so $P = - 2 x^{2} + 2 x + 8$ AG

[2 marks]

METHOD 1

EITHER

uses the axis of symmetry of a quadratic (M1)

$x = - \frac{2}{2 (- 2)}$

forms $\frac{d P}{d x} = 0$ (M1)

$- 4 x + 2 = 0$

THEN

$x = \frac{1}{2}$ A1

substitutes their value of $x$ into $y = 4 - x^{2}$ (M1)

$y = 4 - {(\frac{1}{2})}^{2}$

$y = \frac{15}{4}$ A1

so the dimensions of rectangle $ORST$ of maximum perimeter are $\frac{1}{2}$ by $\frac{15}{4}$

EITHER

substitutes their value of $x$ into $P = - 2 x^{2} + 2 x + 8$ (M1)

$P = - 2 {(\frac{1}{2})}^{2} + 2 (\frac{1}{2}) + 8$

substitutes their values of $x$ and $y$ into $P = 2 x + 2 y$ (M1)

$P = 2 (\frac{1}{2}) + 2 (\frac{15}{4})$

$P = \frac{17}{2}$ A1

so the maximum perimeter is $\frac{17}{2}$

METHOD 2

attempts to complete the square M1

$P = - 2 {(x - \frac{1}{2})}^{2} + \frac{17}{2}$ A1

$x = \frac{1}{2}$ A1

substitutes their value of $x$ into $y = 4 - x^{2}$ (M1)

$y = 4 - {(\frac{1}{2})}^{2}$

$y = \frac{15}{4}$ A1

so the dimensions of rectangle $ORST$ of maximum perimeter are $\frac{1}{2}$ by $\frac{15}{4}$

$P = \frac{17}{2}$ A1

so the maximum perimeter is $\frac{17}{2}$

[6 marks]

substitutes $y = 4 - x^{2}$ into $A = x y$ (M1)

$A = x (4 - x^{2}) (= 4 x - x^{3})$ A1

[2 marks]

$\frac{d A}{d x} = 4 - 3 x^{2}$ A1

attempts to solve their $\frac{d A}{d x} = 0$ for $x$ (M1)

$4 - 3 x^{2} = 0$

$\Rightarrow x = \frac{2}{\sqrt{3}} (= \frac{2 \sqrt{3}}{3}) (x > 0)$ A1

substitutes their (positive) value of $x$ into $y = 4 - x^{2}$ (M1)

$y = 4 - {(\frac{2}{\sqrt{3}})}^{2}$

$y = \frac{8}{3}$ A1

[5 marks]

$A = \frac{16}{3 \sqrt{3}} (= \frac{16 \sqrt{3}}{9})$ A1

[1 mark]

Examiners report

[N/A]

Let $f (x) = \sqrt{12 - 2 x}, x \leq a$ . The following diagram shows part of the graph of $f$ .

The shaded region is enclosed by the graph of $f$ , the $x$ -axis and the $y$ -axis.

The graph of $f$ intersects the $x$ -axis at the point $(a, 0)$ .

Find the value of $a$ .

[2]

Find the volume of the solid formed when the shaded region is revolved $360 °$ about the $x$ -axis.

[5]

Markscheme

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

recognize $f (x) = 0$ (M1)

eg $\sqrt{12 - 2 x} = 0, 2 x = 12$

$a = 6$ (accept $x = 6$ , $(6, 0)$ ) A1 N2

[2 marks]

attempt to substitute either their limits or the function into volume formula (must involve $f^{2}$ ) (M1)

eg $\int_{0}^{6} f^{2} d x, π \int {(\sqrt{12 - 2 x})}^{2}, π \int_{0}^{6} 12 - 2 x d x$

correct integration of each term A1 A1

eg $12 x - x^{2}, 12 x - x^{2} + c, {[12 x - x^{2}]}_{0}^{6}$

substituting limits into their integrated function and subtracting (in any order) (M1)

eg $π (12 (6) - {(6)}^{2}) - π (0), 72 π - 36 π, (12 (6) - {(6)}^{2}) - (0)$

Note: Award M0 if candidate has substituted into $f, f^{2}$ or $f'$ .

volume $= 36 π$ A1 N2

[5 marks]

Examiners report

[N/A]

The graph of a function $f$ passes through the point $(\ln 4, 20)$ .

Given that $f' (x) = 6 e^{2 x}$ , find $f (x)$ .

Markscheme

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

evidence of integration (M1)

eg $\int f' (x) d x, \int 6 e^{2 x}$

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

eg $\frac{1}{2} \times 6 e^{2 x}, 3 e^{2 x} + c$

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

eg $3 e^{2 \times \ln 4} + c = 20$

Note: Award M0 if candidate has substituted into $f'$ or $f''$ .

correct application of $\log (a^{b}) = b \log a$ rule (seen anywhere) (A1)

eg $2 \ln 4 = \ln 16, e^{\ln 16}, \ln 4^{2}$

correct application of $e^{\ln a} = a$ rule (seen anywhere) (A1)

eg $e^{\ln 16} = 16, {(e^{\ln 4})}^{2} = 4^{2}$

correct working (A1)

eg $3 \times 16 + c = 20, 3 \times {(4)}^{2} + c = 20, c = - 28$

$f (x) = 3 e^{2 x} - 28$ A1 N4

[7 marks]

Examiners report

[N/A]

Consider the functions $f (x) = - {(x - h)}^{2} + 2 k$ and $g (x) = e^{x - 2} + k$ where $h, k \in ℝ$ .

The graphs of $f$ and $g$ have a common tangent at $x = 3$ .

Find $f' (x)$ .

[1]

Show that $h = \frac{e + 6}{2}$ .

[3]

Hence, show that $k = e + \frac{e^{2}}{4}$ .

[3]

Markscheme

$f' (x) = - 2 (x - h)$ A1

[1 mark]

$g' (x) = e^{x - 2}$ OR $g' (3) = e^{3 - 2}$ (may be seen anywhere) A1

Note: The derivative of $g$ must be explicitly seen, either in terms of $x$ or $3$ .

recognizing $f' (3) = g' (3)$ (M1)

$- 2 (3 - h) = e^{3 - 2} (= e)$

$- 6 + 2 h = e$ OR $3 - h = - \frac{e}{2}$ A1

Note: The final A1 is dependent on one of the previous marks being awarded.

$h = \frac{e + 6}{2}$ AG

[3 marks]

$f (3) = g (3)$ (M1)

$- {(3 - h)}^{2} + 2 k = e^{3 - 2} + k$

correct equation in $k$

EITHER

$- {(3 - \frac{e + 6}{2})}^{2} + 2 k = e^{3 - 2} + k$ A1

$k = e + {(\frac{6 - e - 6}{2})}^{2} (= e + {(\frac{- e}{2})}^{2})$ A1

$k = e + {(3 - \frac{e + 6}{2})}^{2}$ A1

$k = e + 9 - 3 e - 18 + \frac{e^{2} + 12 e + 36}{4}$ A1

THEN

$k = e + \frac{e^{2}}{4}$ AG

[3 marks]

Examiners report

[N/A]

Let $y = \frac{\ln x}{x^{4}}$ for $x > 0$ .

Consider the function defined by $f (x) \frac{\ln x}{x^{4}}$ for $x > 0$ and its graph $y = f (x)$ .

Show that $\frac{d y}{d x} = \frac{1 - 4 \ln x}{x^{5}}$ .

[3]

The graph of $f$ has a horizontal tangent at point $P$ . Find the coordinates of $P$ .

[5]

Given that $f'' (x) = \frac{20 \ln x - 9}{x^{6}}$ , show that $P$ is a local maximum point.

[3]

Solve $f (x) > 0$ for $x > 0$ .

[2]

Sketch the graph of $f$ , showing clearly the value of the $x$ -intercept and the approximate position of point $P$ .

[3]

Markscheme

attempt to use quotient or product rule (M1)

$\frac{d y}{d x} = \frac{x^{4} (\frac{1}{x}) - (\ln x) (4 x^{3})}{{(x^{4})}^{2}}$ OR $(\ln x) (- 4 x^{- 5}) + (x^{- 4}) (\frac{1}{x})$ A1

correct working A1

$= \frac{x^{3} (1 - 4 \ln x)}{x^{8}}$ OR cancelling $x^{3}$ OR $\frac{- 4 \ln x}{x^{5}} + \frac{1}{x^{5}}$

$= \frac{1 - 4 \ln x}{x^{5}}$ AG

[3 marks]

$f' (x) = \frac{d y}{d x} = 0$ (M1)

$\frac{1 - 4 \ln x}{x^{5}} = 0$

$\ln x = \frac{1}{4}$ (A1)

$x = e^{\frac{1}{4}}$ A1

substitution of their $x$ to find $y$ (M1)

$y = \frac{\ln e^{\frac{1}{4}}}{{(e^{\frac{1}{4}})}^{4}}$

$= \frac{1}{4 e} (= \frac{1}{4} e^{- 1})$ A1

$P (e^{\frac{1}{4}}, \frac{1}{4 e})$

[5 marks]

$f'' (e^{\frac{1}{4}}) = \frac{20 \ln e^{\frac{1}{4}} - 9}{{(e^{\frac{1}{4}})}^{6}}$ (M1)

$= \frac{5 - 9}{e^{1.5}} (= - \frac{4}{e^{1.5}})$ A1

which is negative R1

hence $P$ is a local maximum AG

Note: The R1 is dependent on the previous A1 being awarded.

[3 marks]

$\ln x > 0$ (A1)

$x > 1$ A1

[2 marks]

A1A1A1

Note: Award A1 for one $x$ -intercept only, located at $1$

A1 for local maximum, $P$ , in approximately correct position
A1 for curve approaching $x$ -axis as $x \to \infty$ (including change in concavity).

[3 marks]

Examiners report

[N/A]

Let $f’(x) = \frac{{3{x^2}}}{{{{({x^3} + 1)}^5}}}$ . Given that $f(0) = 1$ , find $f(x)$ .

Markscheme

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

valid approach (M1)

eg $\,\,\,\,\,$ $\int {f'{\text{d}}x,{\text{ }}\int {\frac{{3{x^2}}}{{{{({x^3} + 1)}^5}}}{\text{d}}x} }$

correct integration by substitution/inspection A2

eg $\,\,\,\,\,$ $f(x) = - \frac{1}{4}{({x^3} + 1)^{ - 4}} + c,{\text{ }}\frac{{ - 1}}{{4{{({x^3} + 1)}^4}}}$

correct substitution into their integrated function (must include $c$ ) M1

eg $\,\,\,\,\,$ $1 = \frac{{ - 1}}{{4{{({0^3} + 1)}^4}}} + c,{\text{ }} - \frac{1}{4} + c = 1$

Note: Award M0 if candidates substitute into $f^{'}$ or $f^{″}$ .

$c = \frac{5}{4}$ (A1)

$f(x) = - \frac{1}{4}{({x^3} + 1)^{ - 4}} + \frac{5}{4}{\text{ }}\left( { = \frac{{ - 1}}{{4{{({x^3} + 1)}^4}}} + \frac{5}{4},{\text{ }}\frac{{5{{({x^3} + 1)}^4} - 1}}{{4{{({x^3} + 1)}^4}}}} \right)$ A1 N4

[6 marks]

Examiners report

[N/A]

A particle P starts from point O and moves along a straight line. The graph of its velocity, $v$ ms⁻¹ after $t$ seconds, for 0 ≤ $t$ ≤ 6 , is shown in the following diagram.

The graph of $v$ has $t$ -intercepts when $t$ = 0, 2 and 4.

The function $s\left( t \right)$ represents the displacement of P from O after $t$ seconds.

It is known that P travels a distance of 15 metres in the first 2 seconds. It is also known that $s\left( 2 \right) = s\left( 5 \right)$ and $\int_2^4 {v\,{\text{d}}t} = 9$ .

Find the value of $s\left( 4 \right) - s\left( 2 \right)$ .

[2]

Find the total distance travelled in the first 5 seconds.

[5]

Markscheme

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

recognizing relationship between $v$ and $s$ (M1)

eg $\int {v = s}$ , $s' = v$

$s\left( 4 \right) - s\left( 2 \right) = 9$ A1 N2

[2 marks]

correctly interpreting distance travelled in first 2 seconds (seen anywhere, including part (a) or the area of 15 indicated on diagram) (A1)

eg $\int_0^2 {v = 15}$ , $s\left( 2 \right) = 15$

valid approach to find total distance travelled (M1)

eg sum of 3 areas, $\int_0^4 {v + } \int_4^5 v$ , shaded areas in diagram between 0 and 5

Note: Award M0 if only $\int_0^5 {\left| v \right|}$ is seen.

correct working towards finding distance travelled between 2 and 5 (seen anywhere including within total area expression or on diagram) (A1)

eg $\int_2^4 {v - } \int_4^5 v$ , $\int_2^4 {v = } \int_4^5 {\left| v \right|}$ , $\int_4^5 {v\,{\text{d}}t} = - 9$ , $s\left( 4 \right) - s\left( 2 \right) - \left[ {s\left( 5 \right) - s\left( 4 \right)} \right]$ ,

equal areas

correct working using $s\left( 5 \right) = s\left( 2 \right)$ (A1)

eg $15 + 9 - \left( { - 9} \right)$ , $15 + 2\left[ {s\left( 4 \right) - s\left( 2 \right)} \right]$ , $15 + 2\left( 9 \right)$ , $2 \times s\left( 4 \right) - s\left( 2 \right)$ , $48 - 15$

total distance travelled = 33 (m) A1 N2

[5 marks]

Examiners report

[N/A]

Let $f'(x) = {\sin ^3}(2x)\cos (2x)$ . Find $f(x)$ , given that $f\left( {\frac{\pi }{4}} \right) = 1$ .

Markscheme

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

evidence of integration (M1)

eg $\,\,\,\,\,$ $\int {f'(x){\text{d}}x}$

correct integration (accept missing $C$ ) (A2)

eg $\,\,\,\,\,$ $\frac{1}{2} \times \frac{{{{\sin }^4}(2x)}}{4},{\text{ }}\frac{1}{8}{\sin ^4}(2x) + C$

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

eg $\,\,\,\,\,$ $1 = \frac{1}{8}{\sin ^4}\left( {\frac{\pi }{2}} \right) + C$

Note: Award M0 if they substitute into the original or differentiated function.

recognizing $\sin \left( {\frac{\pi }{2}} \right) = 1$ (A1)

eg $\,\,\,\,\,$ $1 = \frac{1}{8}{(1)^4} + C$

$C = \frac{7}{8}$ (A1)

$f(x) = \frac{1}{8}{\sin ^4}(2x) + \frac{7}{8}$ A1 N5

[7 marks]

Examiners report

[N/A]

A quadratic function $f$ is given by $f(x) = a{x^2} + bx + c$ . The points $(0,{\text{ }}5)$ and $( - 4,{\text{ }}5)$ lie on the graph of $y = f(x)$ .

The $y$ -coordinate of the minimum of the graph is 3.

Find the equation of the axis of symmetry of the graph of $y = f(x)$ .

[2]

Write down the value of $c$ .

[1]

Find the value of $a$ and of $b$ .

[3]

Markscheme

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

$x = - 2$ (A1)(A1) (C2)

Note: Award (A1) for $x =$ (a constant) and (A1) for $- 2$ .

[2 marks]

$(c = ){\text{ }}5$ (A1) (C1)

[1 mark]

$- \frac{b}{{2a}} = - 2$

$a{( - 2)^2} - 2b + 5 = 3$ or equivalent

$a{( - 4)^2} - 4b + 5 = 5$ or equivalent

$2a( - 2) + b = 0$ or equivalent (M1)

Note: Award (M1) for two of the above equations.

$a = 0.5$ (A1)(ft)

$b = 2$ (A1)(ft) (C3)

Note: Award at most (M1)(A1)(ft)(A0) if the answers are reversed.

Follow through from parts (a) and (b).

[3 marks]

Examiners report

[N/A]

Let $f\left( x \right) = \frac{{6 - 2x}}{{\sqrt {16 + 6x - {x^2}} }}$ . The following diagram shows part of the graph of $f$ .

The region R is enclosed by the graph of $f$ , the $x$ -axis, and the $y$ -axis. Find the area of R.

Markscheme

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

METHOD 1 (limits in terms of $x$ )

valid approach to find $x$ -intercept (M1)

eg $f\left( x \right) = 0$ , $\frac{{6 - 2x}}{{\sqrt {16 + 6x - {x^2}} }} = 0$ , $6 - 2x = 0$

$x$ -intercept is 3 (A1)

valid approach using substitution or inspection (M1)

eg $u = 16 + 6x - {x^2}$ , $\int_0^3 {\frac{{6 - 2x}}{{\sqrt u }}} {\text{d}}x$ , ${\text{d}}u = 6 - 2x$ , $\int {\frac{1}{{\sqrt u }}}$ , $2{u^{\frac{1}{2}}}$ ,

$u = \sqrt {16 + 6x - {x^2}}$ , $\frac{{{\text{d}}u}}{{{\text{d}}x}} = \left( {6 - 2x} \right)\frac{1}{2}{\left( {16 + 6x - {x^2}} \right)^{ - \frac{1}{2}}}$ , $\int 2 \,{\text{d}}u$ , $2u$

$\int {f\left( x \right)} \,{\text{d}}x = 2\sqrt {16 + 6x - {x^2}}$ (A2)

substituting both of their limits into their integrated function and subtracting (M1)

eg $2\sqrt {16 + 6\left( 3 \right) - {3^2}} - 2\sqrt {16 + 6{{\left( 0 \right)}^2} - {0^2}}$ , $2\sqrt {16 + 18 - 9} - 2\sqrt {16}$

Note: Award M0 if they substitute into original or differentiated function. Do not accept only “– 0” as evidence of substituting lower limit.

correct working (A1)

eg $2\sqrt {25} - 2\sqrt {16}$ , $10 - 8$

area = 2 A1 N2

METHOD 2 (limits in terms of $u$ )

valid approach to find $x$ -intercept (M1)

eg $f\left( x \right) = 0$ , $\frac{{6 - 2x}}{{\sqrt {16 + 6x - {x^2}} }} = 0$ , $6 - 2x = 0$

$x$ -intercept is 3 (A1)

valid approach using substitution or inspection (M1)

eg $u = 16 + 6x - {x^2}$ , $\int_0^3 {\frac{{6 - 2x}}{{\sqrt u }}} {\text{d}}x$ , ${\text{d}}u = 6 - 2x$ , $\int {\frac{1}{{\sqrt u }}}$ ,

$u = \sqrt {16 + 6x - {x^2}}$ , $\frac{{{\text{d}}u}}{{{\text{d}}x}} = \left( {6 - 2x} \right)\frac{1}{2}{\left( {16 + 6x - {x^2}} \right)^{ - \frac{1}{2}}}$ , $\int 2 \,{\text{d}}u$

correct integration (A2)

eg $\int {\frac{1}{{\sqrt u }}} \,{\text{d}}u = 2{u^{\frac{1}{2}}}$ , $\int 2 \,{\text{d}}u = 2u$

both correct limits for $u$ (A1)

eg $u$ = 16 and $u$ = 25, $\int_{16}^{25} {\frac{1}{{\sqrt u }}{\text{d}}u}$ , $\left[ {2{u^{\frac{1}{2}}}} \right]_{16}^{25}$ , $u$ = 4 and $u$ = 5, $\int_4^5 2 \,{\text{d}}u$ , $\left[ {2u} \right]_4^5$

substituting both of their limits for $u$ (do not accept 0 and 3) into their integrated function and subtracting (M1)

eg $2\sqrt {25} - 2\sqrt {16}$ , $10 - 8$

Note: Award M0 if they substitute into original or differentiated function, or if they have not attempted to find limits for $u$ .

area = 2 A1 N2

[8 marks]

Examiners report

[N/A]

Consider the graph of the function $f (x) = x^{2} - \frac{k}{x}$ .

The equation of the tangent to the graph of $y = f (x)$ at $x = - 2$ is $2 y = 4 - 5 x$ .

Write down $f' (x)$ .

[3]

Write down the gradient of this tangent.

[1]

Find the value of $k$ .

[2]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure. It appeared in a paper that permitted the use of a calculator, and so might not be suitable for all forms of practice.

$2 x + \frac{k}{x^{2}}$ (A1)(A1)(A1) (C3)

Note: Award (A1) for $2 x$ , (A1) for $+ k$ , and (A1) for $x^{- 2}$ or $\frac{1}{x^{2}}$ .
Award at most (A1)(A1)(A0) if additional terms are seen.

[3 marks]

$- 2.5 (\frac{- 5}{2})$ (A1) (C1)

[1 mark]

$- 2.5 = 2 \times (- 2) + \frac{k}{{(- 2)}^{2}}$ (M1)

Note: Award (M1) for equating their gradient from part (b) to their substituted derivative from part (a).

$(k =) 6$ (A1)(ft) (C2)

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

[2 marks]

Examiners report

[N/A]

Consider the curve with equation $y = (2 x - 1) e^{k x}$ , where $x \in ℝ$ and $k \in ℚ$ .

The tangent to the curve at the point where $x = 1$ is parallel to the line $y = 5 e^{k} x$ .

Find the value of $k$ .

Markscheme

evidence of using product rule (M1)

$\frac{d y}{d x} = (2 x - 1) \times (k e^{k x}) + 2 \times e^{k x} (= e^{k x} (2 k x - k + 2))$ A1

correct working for one of (seen anywhere) A1

$\frac{d y}{d x}$ at $x = 1 \Rightarrow k e^{k} + 2 e^{k}$

slope of tangent is $5 e^{k}$

their $\frac{d y}{d x}$ at $x = 1$ equals the slope of $y = 5 e^{k} x (= 5 e^{k})$ (seen anywhere) (M1)

$k e^{k} + 2 e^{k} = 5 e^{k}$

$k = 3$ A1

[5 marks]

Examiners report

The product rule was well recognised and used with 𝑥=1 properly substituted into this expression. Although the majority of the candidates tried equating the derivative to the slope of the tangent line, the slope of the tangent line was not correctly identified; many candidates incorrectly substituted 𝑥=1 into the tangent equation, thus finding the y-coordinate instead of the slope.

Let $f\left( x \right) = \frac{{{\text{ln}}\,5x}}{{kx}}$ where $x > 0$ , $k \in {\mathbb{R}^ + }$ .

The graph of $f$ has exactly one maximum point P.

The second derivative of $f$ is given by $f''\left( x \right) = \frac{{2\,{\text{ln}}\,5x - 3}}{{k{x^3}}}$ . The graph of $f$ has exactly one point of inflexion Q.

Show that $f'\left( x \right) = \frac{{1 - {\text{ln}}\,5x}}{{k{x^2}}}$ .

[3]

Find the x-coordinate of P.

[3]

Show that the x-coordinate of Q is $\frac{{\text{1}}}{5}{{\text{e}}^{\frac{3}{2}}}$ .

[3]

The region R is enclosed by the graph of $f$ , the x-axis, and the vertical lines through the maximum point P and the point of inflexion Q.

Given that the area of R is 3, find the value of $k$ .

[7]

Markscheme

attempt to use quotient rule (M1)

correct substitution into quotient rule

$f'\left( x \right) = \frac{{5kx\left( {\frac{1}{{5x}}} \right) - k\,{\text{ln}}\,5x}}{{{{\left( {kx} \right)}^2}}}$ (or equivalent) A1

$= \frac{{k - k\,{\text{ln}}\,5x}}{{{k^2}{x^2}}}$ , $\left( {k \in {\mathbb{R}^ + }} \right)$ A1

$= \frac{{1 - {\text{ln}}\,5x}}{{k{x^2}}}$ AG

[3 marks]

$f'\left( x \right) = 0$ M1

$\frac{{1 - {\text{ln}}\,5x}}{{k{x^2}}} = 0$

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

$x = \frac{{\text{e}}}{5}$ A1

[3 marks]

$f''\left( x \right) = 0$ M1

$\frac{{2\,{\text{ln}}\,5x - 3}}{{k{x^3}}} = 0$

${\text{ln}}\,5x = \frac{3}{2}$ A1

$5x = {{\text{e}}^{\frac{3}{2}}}$ A1

so the point of inflexion occurs at $x = \frac{{\text{1}}}{5}{{\text{e}}^{\frac{3}{2}}}$ AG

[3 marks]

attempt to integrate (M1)

$u = {\text{ln}}\,5x \Rightarrow \frac{{{\text{d}}u}}{{{\text{d}}x}} = \frac{1}{x}$

$\int {\frac{{{\text{ln}}\,5x}}{{kx}}} {\text{d}}x = \frac{1}{k}\int {u\,} {\text{d}}u$ (A1)

EITHER

= $\frac{{{u^2}}}{{2k}}$ A1

so $\frac{1}{k}\int\limits_1^{\frac{3}{2}} {u\,{\text{d}}u = } \left[ {\frac{{{u^2}}}{{2k}}} \right]_1^{\frac{3}{2}}$ A1

$= \frac{{{{\left( {{\text{ln}}\,5x} \right)}^2}}}{{2k}}$ A1

so $\int\limits_{\frac{{\text{e}}}{5}}^{\frac{{\text{1}}}{5}{{\text{e}}^{\frac{3}{2}}}} {\frac{{{\text{ln}}\,5x}}{{kx}}} {\text{d}}x = \left[ {\frac{{{{\left( {{\text{ln}}\,5x} \right)}^2}}}{{2k}}} \right]_{\frac{{\text{e}}}{5}}^{\frac{{\text{1}}}{5}{{\text{e}}^{\frac{3}{2}}}}$ A1

THEN

$= \frac{1}{{2k}}\left( {\frac{9}{4} - 1} \right)$

$= \frac{5}{{8k}}$ A1

setting their expression for area equal to 3 M1

$\frac{5}{{8k}} = 3$

$k = \frac{5}{{24}}$ A1

[7 marks]

Examiners report

[N/A]

Find $\int {x{{\text{e}}^{{x^2} - 1}}{\text{d}}x}$ .

[4]

Find $f(x)$ , given that $f’(x) = x{{\text{e}}^{{x^2} - 1}}$ and $f( - 1) = 3$ .

[3]

Markscheme

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

valid approach to set up integration by substitution/inspection (M1)

eg $\,\,\,\,\,$ $u = {x^2} - 1,{\text{ d}}u = 2x,{\text{ }}\int {2x{{\text{e}}^{{x^2} - 1}}{\text{d}}x}$

correct expression (A1)

eg $\,\,\,\,\,$ $\frac{1}{2}\int {2x{{\text{e}}^{{x^2} - 1}}{\text{d}}x,{\text{ }}\frac{1}{2}\int {{{\text{e}}^u}{\text{d}}u} }$

$\frac{1}{2}{{\text{e}}^{{x^2} - 1}} + c$ A2 N4

Notes: Award A1 if missing “ $+ c$ ”.

[4 marks]

substituting $x = - 1$ into their answer from (a) (M1)

eg $\,\,\,\,\,$ $\frac{1}{2}{{\text{e}}^0},{\text{ }}\frac{1}{2}{{\text{e}}^{1 - 1}} = 3$

correct working (A1)

eg $\,\,\,\,\,$ $\frac{1}{2} + c = 3,{\text{ }}c = 2.5$

$f(x) = \frac{1}{2}{{\text{e}}^{{x^2} - 1}} + 2.5$ A1 N2

[3 marks]

Examiners report

[N/A]

A cylinder with radius $r$ and height $h$ is shown in the following diagram.

The sum of $r$ and $h$ for this cylinder is 12 cm.

Write down an equation for the area, $A$ , of the curved surface in terms of $r$ .

[2]

Find $\frac{{{\text{d}}A}}{{{\text{d}}r}}$ .

[2]

Find the value of $r$ when the area of the curved surface is maximized.

[2]

Markscheme

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

$A = 2\pi r\left( {12 - r} \right)$ OR $A = 24\pi r - 2\pi {r^2}$ (A1)(M1) (C2)

Note: Award (A1) for $r + h = 12$ or $h = 12 - r$ seen. Award (M1) for correctly substituting into curved surface area of a cylinder. Accept $A = 2\pi r\left( {12 - r} \right)$ OR $A = 24\pi r - 2\pi {r^2}$ .

[2 marks]

$24\pi - 4\pi r$ (A1)(ft)(A1)(ft) (C2)

Note: Award (A1)(ft) for $24\pi$ and (A1)(ft) for $- 4\pi r$ . Follow through from part (a). Award at most (A1)(ft)(A0) if additional terms are seen.

[2 marks]

$24\pi - 4\pi r = 0$ (M1)

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

6 (cm) (A1)(ft) (C2)

Note: Follow through from part (b).

[2 marks]

Examiners report

[N/A]

A quadratic function $f$ can be written in the form $f(x) = a(x - p)(x - 3)$ . The graph of $f$ has axis of symmetry $x = 2.5$ and $y$ -intercept at $(0,{\text{ }} - 6)$

Find the value of $p$ .

[3]

Find the value of $a$ .

[3]

The line $y = kx - 5$ is a tangent to the curve of $f$ . Find the values of $k$ .

[8]

Markscheme

METHOD 1 (using x-intercept)

determining that 3 is an $x$ -intercept (M1)

eg $\,\,\,\,\,$ $x - 3 = 0$ , M17/5/MATME/SP1/ENG/TZ1/09.a/M

valid approach (M1)

eg $\,\,\,\,\,$ $3 - 2.5,{\text{ }}\frac{{p + 3}}{2} = 2.5$

$p = 2$ A1 N2

METHOD 2 (expanding f (x))

correct expansion (accept absence of $a$ ) (A1)

eg $\,\,\,\,\,$ $a{x^2} - a(3 + p)x + 3ap,{\text{ }}{x^2} - (3 + p)x + 3p$

valid approach involving equation of axis of symmetry (M1)

eg $\,\,\,\,\,$ $\frac{{ - b}}{{2a}} = 2.5,{\text{ }}\frac{{a(3 + p)}}{{2a}} = \frac{5}{2},{\text{ }}\frac{{3 + p}}{2} = \frac{5}{2}$

$p = 2$ A1 N2

METHOD 3 (using derivative)

correct derivative (accept absence of $a$ ) (A1)

eg $\,\,\,\,\,$ $a(2x - 3 - p),{\text{ }}2x - 3 - p$

valid approach (M1)

eg $\,\,\,\,\,$ $f’(2.5) = 0$

$p = 2$ A1 N2

[3 marks]

attempt to substitute $(0,{\text{ }} - 6)$ (M1)

eg $\,\,\,\,\,$ $- 6 = a(0 - 2)(0 - 3),{\text{ }}0 = a( - 8)( - 9),{\text{ }}a{(0)^2} - 5a(0) + 6a = - 6$

correct working (A1)

eg $\,\,\,\,\,$ $- 6 = 6a$

$a = - 1$ A1 N2

[3 marks]

METHOD 1 (using discriminant)

recognizing tangent intersects curve once (M1)

recognizing one solution when discriminant = 0 M1

attempt to set up equation (M1)

eg $\,\,\,\,\,$ $g = f,{\text{ }}kx - 5 = - {x^2} + 5x - 6$

rearranging their equation to equal zero (M1)

eg $\,\,\,\,\,$ ${x^2} - 5x + kx + 1 = 0$

correct discriminant (if seen explicitly, not just in quadratic formula) A1

eg $\,\,\,\,\,$ ${(k - 5)^2} - 4,{\text{ }}25 - 10k + {k^2} - 4$

correct working (A1)

eg $\,\,\,\,\,$ $k - 5 = \pm 2,{\text{ }}(k - 3)(k - 7) = 0,{\text{ }}\frac{{10 \pm \sqrt {100 - 4 \times 21} }}{2}$

$k = 3,{\text{ }}7$ A1A1 N0

METHOD 2 (using derivatives)

attempt to set up equation (M1)

eg $\,\,\,\,\,$ $g = f,{\text{ }}kx - 5 = - {x^2} + 5x - 6$

recognizing derivative/slope are equal (M1)

eg $\,\,\,\,\,$ $f’ = {m_T},{\text{ }}f' = k$

correct derivative of $f$ (A1)

eg $\,\,\,\,\,$ $- 2x + 5$

attempt to set up equation in terms of either $x$ or $k$ M1

eg $\,\,\,\,\,$ $( - 2x + 5)x - 5 = - {x^2} + 5x - 6,{\text{ }}k\left( {\frac{{5 - k}}{2}} \right) - 5 = - {\left( {\frac{{5 - k}}{2}} \right)^2} + 5\left( {\frac{{5 - k}}{2}} \right) - 6$

rearranging their equation to equal zero (M1)

eg $\,\,\,\,\,$ ${x^2} - 1 = 0,{\text{ }}{k^2} - 10k + 21 = 0$

correct working (A1)

eg $\,\,\,\,\,$ $x = \pm 1,{\text{ }}(k - 3)(k - 7) = 0,{\text{ }}\frac{{10 \pm \sqrt {100 - 4 \times 21} }}{2}$

$k = 3,{\text{ }}7$ A1A1 N0

[8 marks]

Examiners report

[N/A]

Consider the function $f$ defined by $f (x) = \ln (x^{2} - 16)$ for $x > 4$ .

The following diagram shows part of the graph of $f$ which crosses the $x$ -axis at point $A$ , with coordinates $(a, 0)$ . The line $L$ is the tangent to the graph of $f$ at the point $B$ .

Find the exact value of $a$ .

[3]

Given that the gradient of $L$ is $\frac{1}{3}$ , find the $x$ -coordinate of $B$ .

[6]

Markscheme

$\ln (x^{2} - 16) = 0$ (M1)

$e^{0} = x^{2} - 16 (= 1)$

$x^{2} = 17$ OR $x = \pm \sqrt{17}$ (A1)

$a = \sqrt{17}$ A1

[3 marks]

attempt to differentiate (must include $2 x$ and/or $\frac{1}{x^{2} - 16}$ ) (M1)

$f' (x) = \frac{2 x}{x^{2} - 16}$ A1

setting their derivative $= \frac{1}{3}$ M1

$\frac{2 x}{x^{2} - 16} = \frac{1}{3}$

$x^{2} - 16 = 6 x$ OR $x^{2} - 6 x - 16 = 0$ (or equivalent) A1

valid attempt to solve their quadratic (M1)

$x = 8$ A1

Note: Award A0 if the candidate’s final answer includes additional solutions (such as $x = - 2, 8$ ).

[6 marks]

Examiners report

[N/A]

Consider f(x), g(x) and h(x), for x∈ $\mathbb{R}$ where h(x) = $\left( {f \circ g} \right)$ (x).

Given that g(3) = 7 , g′ (3) = 4 and f ′ (7) = −5 , find the gradient of the normal to the curve of h at x = 3.

Markscheme

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

recognizing the need to find h′ (M1)

recognizing the need to find h′ (3) (seen anywhere) (M1)

evidence of choosing chain rule (M1)

eg $\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{\text{d}}y}}{{{\text{d}}u}} \times \frac{{{\text{d}}u}}{{{\text{d}}x}},\,\,f'\left( {g\left( 3 \right)} \right) \times g'\left( 3 \right),\,\,f'\left( g \right) \times g'$

correct working (A1)

eg $f'\left( 7 \right) \times 4,\,\, - 5 \times 4$

$h'\left( 3 \right) = - 20$ (A1)

evidence of taking their negative reciprocal for normal (M1)

eg $- \frac{1}{{h'\left( 3 \right)}},\,\,{m_1}{m_2} = - 1$

gradient of normal is $\frac{1}{{20}}$ A1 N4

[7 marks]

Examiners report

[N/A]

Particle A travels in a straight line such that its displacement, $s$ metres, from a fixed origin after $t$ seconds is given by $s (t) = 8 t - t^{2}$ , for $0 \leq t \leq 10$ , as shown in the following diagram.

Particle A starts at the origin and passes through the origin again when $t = p$ .

Particle A changes direction when $t = q$ .

The total distance travelled by particle A is given by $d$ .

Find the value of $p$ .

[2]

Find the value of $q$ .

[2]

b.i.

Find the displacement of particle A from the origin when $t = q$ .

[2]

b.ii.

Find the distance of particle A from the origin when $t = 10$ .

[2]

Find the value of $d$ .

[2]

A second particle, particle B, travels along the same straight line such that its velocity is given by $v (t) = 14 - 2 t$ , for $t \geq 0$ .

When $t = k$ , the distance travelled by particle B is equal to $d$ .

Find the value of $k$ .

[4]

Markscheme

setting $s (t) = 0$ (M1)

$8 t - t^{2} = 0$

$t (8 - t) = 0$

$p = 8$ (accept $t = 8, (8, 0)$ ) A1

Note: Award A0 if the candidate’s final answer includes additional solutions (such as $p = 0, 8$ ).

[2 marks]

recognition that when particle changes direction $v = 0$ OR local maximum on graph of $s$ OR vertex of parabola (M1)

$q = 4$ (accept $t = 4$ ) A1

[2 marks]

b.i.

substituting their value of $q$ into $s (t)$ OR integrating $v (t)$ from $t = 0$ to $t = 4$ (M1)

$displacement = 16 (m)$ A1

[2 marks]

b.ii.

$s (10) = - 20$ OR $distance = |s (t)|$ OR integrating $v (t)$ from $t = 0$ to $t = 10$ (M1)

$distance = 20 (m)$ A1

[2 marks]

$16$ forward $+ 36$ backward OR $16 + 16 + 20$ OR $\int_{0}^{10} |v (t)| d t$ (M1)

$d = 52 (m)$ A1

[2 marks]

METHOD 1

graphical method with triangles on $v (t)$ graph M1

$49 + (\frac{x (2 x)}{2})$ (A1)

$49 + x^{2} = 52, x = \sqrt{3}$ (A1)

$k = 7 + \sqrt{3}$ A1

METHOD 2

recognition that distance $= \int |v (t)| d t$ M1

$\int_{0}^{7} (14 - 2 t) d t + \int_{7}^{k} (2 t - 14) d t$

${[14 t - t^{2}]}_{0}^{7} + {[t^{2} - 14 t]}_{7}^{k}$ (A1)

$14 (7) - 7^{2} + ((k^{2} - 14 k) - (7^{2} - 14 (7))) = 52$ (A1)

$k = 7 + \sqrt{3}$ A1

[4 marks]

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

Maria owns a cheese factory. The amount of cheese, in kilograms, Maria sells in one week, $Q$ , is given by

$Q = 882 - 45p$ ,

where $p$ is the price of a kilogram of cheese in euros (EUR).

Maria earns $(p - 6.80){\text{ EUR}}$ for each kilogram of cheese sold.

To calculate her weekly profit $W$ , in EUR, Maria multiplies the amount of cheese she sells by the amount she earns per kilogram.

Write down how many kilograms of cheese Maria sells in one week if the price of a kilogram of cheese is 8 EUR.

[1]

Find how much Maria earns in one week, from selling cheese, if the price of a kilogram of cheese is 8 EUR.

[2]

Write down an expression for $W$ in terms of $p$ .

[1]

Find the price, $p$ , that will give Maria the highest weekly profit.

[2]

Markscheme

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

522 (kg) (A1) (C1)

[1 mark]

$522(8 - 6.80)$ or equivalent (M1)

Note: Award (M1) for multiplying their answer to part (a) by $(8 - 6.80)$ .

626 (EUR) (626.40) (A1)(ft) (C2)

Note: Follow through from part (a).

[2 marks]

$(W = ){\text{ }}(882 - 45p)(p - 6.80)$ (A1)

$(W = ) - 45{p^2} + 1188p - 5997.6$ (A1) (C1)

[1 mark]

sketch of $W$ with some indication of the maximum (M1)

$- 90p + 1188 = 0$ (M1)

Note: Award (M1) for equating the correct derivative of their part (c) to zero.

$(p = ){\text{ }}\frac{{ - 1188}}{{2 \times ( - 45)}}$ (M1)

Note: Award (M1) for correct substitution into the formula for axis of symmetry.

$(p = ){\text{ }}13.2{\text{ (EUR)}}$ (A1)(ft) (C2)

Note: Follow through from their part (c), if the value of $p$ is such that $6.80 < p < 19.6$ .

[2 marks]

Examiners report

[N/A]

The point A has coordinates (4 , −8) and the point B has coordinates (−2 , 4).

The point D has coordinates (−3 , 1).

Write down the coordinates of C, the midpoint of line segment AB.

[2]

Find the gradient of the line DC.

[2]

Find the equation of the line DC. Write your answer in the form ax + by + d = 0 where a , b and d are integers.

[2]

Markscheme

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

(1, −2) (A1)(A1) (C2)
Note: Award (A1) for 1 and (A1) for −2, seen as a coordinate pair.

Accept x = 1, y = −2. Award (A1)(A0) if x and y coordinates are reversed.

[2 marks]

$\frac{{1 - \left( { - 2} \right)}}{{ - 3 - 1}}$ (M1)

Note: Award (M1) for correct substitution, of their part (a), into gradient formula.

$= - \frac{3}{4}\,\,\,\left( { - 0.75} \right)$ (A1)(ft) (C2)

Note: Follow through from part (a).

[2 marks]

$y - 1 = - \frac{3}{4}\left( {x + 3} \right)$ OR $y + 2 = - \frac{3}{4}\left( {x - 1} \right)$ OR $y = - \frac{3}{4}x - \frac{5}{4}$ (M1)

Note: Award (M1) for correct substitution of their part (b) and a given point.

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

Note: Award (M1) for correct substitution of their part (b) and a given point.

$3x + 4y + 5 = 0$ (accept any integer multiple, including negative multiples) (A1)(ft) (C2)

Note: Follow through from parts (a) and (b). Where the gradient in part (b) is found to be $\frac{5}{0}$ , award at most (M1)(A0) for either $x = - 3$ or $x + 3 = 0$ .

[2 marks]

Examiners report

[N/A]

A potter sells $x$ vases per month.

His monthly profit in Australian dollars (AUD) can be modelled by

$P\left( x \right) = - \frac{1}{5}{x^3} + 7{x^2} - 120{\text{,}}\,\,x \geqslant 0.$

Find the value of $P$ if no vases are sold.

[1]

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

[2]

Hence, find the number of vases that will maximize the profit.

[3]

Markscheme

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

−120 (AUD) (A1) (C1)

[1 mark]

$- \frac{3}{5}{x^2} + 14x$ (A1)(A1) (C2)

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

[2 marks]

$- \frac{3}{5}{x^2} + 14x = 0$ (M1)

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

sketch of their derivative (approximately correct shape) with $x$ -intercept seen (M1)

$23\frac{1}{3}\,\,\,\left( {23.3{\text{,}}\,\,23.3333 \ldots {\text{,}}\,\,\frac{{70}}{3}} \right)$ (A1)(ft)

Note: Award (C2) for $23\frac{1}{3}\,\,\,\left( {23.3{\text{,}}\,\,23.3333 \ldots {\text{,}}\,\,\frac{{70}}{3}} \right)$ seen without working.

23 (A1)(ft) (C3)

Note: Follow through from part (b).

[3 marks]

Examiners report

[N/A]

Let $f(x) = 15 - {x^2}$ , for $x \in \mathbb{R}$ . The following diagram shows part of the graph of $f$ and the rectangle OABC, where A is on the negative $x$ -axis, B is on the graph of $f$ , and C is on the $y$ -axis.

N17/5/MATME/SP1/ENG/TZ0/06

Find the $x$ -coordinate of A that gives the maximum area of OABC.

Markscheme

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

attempt to find the area of OABC (M1)

eg $\,\,\,\,\,$ ${\text{OA}} \times {\text{OC, }}x \times f(x),{\text{ }}f(x) \times ( - x)$

correct expression for area in one variable (A1)

eg $\,\,\,\,\,$ ${\text{area}} = x(15 - {x^2}),{\text{ }}15x - {x^3},{\text{ }}{x^3} - 15x$

valid approach to find maximum area (seen anywhere) (M1)

eg $\,\,\,\,\,$ $A’(x) = 0$

correct derivative A1

eg $\,\,\,\,\,$ $15 - 3{x^2},{\text{ }}(15 - {x^2}) + x( - 2x) = 0,{\text{ }} - 15 + 3{x^2}$

correct working (A1)

eg $\,\,\,\,\,$ $15 = 3{x^2},{\text{ }}{x^2} = 5,{\text{ }}x = \sqrt 5$

$x = - \sqrt 5 {\text{ }}\left( {{\text{accept A}}\left( { - \sqrt 5 ,{\text{ }}0} \right)} \right)$ A2 N3

[7 marks]

Examiners report

[N/A]

The derivative of a function $f$ is given by $f'(x) = 2{{\text{e}}^{ - 3x}}$ . The graph of $f$ passes through $\left( {\frac{1}{3}{\text{,}}\,\,5} \right)$ .

Find $f(x)$ .

Markscheme

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

recognizing to integrate (M1)

eg $\int {f'}$ , $\int {2{{\text{e}}^{ - 3x}}{\text{d}}x}$ , ${\text{d}}u = - 3$

correct integral (do not penalize for missing + $C$ ) (A2)

eg $- \frac{2}{3}{{\text{e}}^{ - 3x}} + C$

substituting $\left( {\frac{1}{3}{\text{,}}\,\,5} \right)$ (in any order) into their integrated expression (must have + $C$ ) M1

eg $- \frac{2}{3}{{\text{e}}^{ - 3\left( {1/3} \right)}} + C = 5$

Note: Award M0 if they substitute into original or differentiated function.

$f(x) = - \frac{2}{3}{{\text{e}}^{ - 3x}} + 5 + \frac{2}{3}{{\text{e}}^{ - 1}}$ (or any equivalent form, eg $- \frac{2}{3}{{\text{e}}^{ - 3x}} + 5 - \frac{2}{{ - 3{\text{e}}}}$ ) A1 N4

[5 marks]

Examiners report

[N/A]

The diagram shows part of the graph of a function $y = f(x)$ . The graph passes through point ${\text{A}}(1,{\text{ }}3)$ .

M17/5/MATSD/SP1/ENG/TZ2/13

The tangent to the graph of $y = f(x)$ at A has equation $y = - 2x + 5$ . Let $N$ be the normal to the graph of $y = f(x)$ at A.

Write down the value of $f(1)$ .

[1]

Find the equation of $N$ . Give your answer in the form $ax + by + d = 0$ where $a$ , $b$ , $d \in \mathbb{Z}$ .

[3]

Draw the line $N$ on the diagram above.

[2]

Markscheme

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

3 (A1) (C1)

Notes: Accept $y = 3$

[1 mark]

$3 = 0.5(1) + c$ $\,\,\,$ OR $\,\,\,$ $y - 3 = 0.5(x - 1)$ (A1)(A1)

Note: Award (A1) for correct gradient, (A1) for correct substitution of ${\text{A}}(1,{\text{ }}3)$ in the equation of line.

$x - 2y + 5 = 0$ or any integer multiple (A1)(ft) (C3)

Note: Award (A1)(ft) for their equation correctly rearranged in the indicated form.

The candidate’s answer must be an equation for this mark.

[3 marks]

M17/5/MATSD/SP1/ENG/TZ2/13.c/M (M1)(A1)(ft) (C2)

Note: Award M1) for a straight line, with positive gradient, passing through $(1,{\text{ }}3)$ , (A1)(ft) for line (or extension of their line) passing approximately through 2.5 or their intercept with the $y$ -axis.

[2 marks]

Examiners report

[N/A]

Consider the function $f$ defined by $f (x) = 6 + 6 \cos x$ , for $0 \leq x \leq 4 π$ .

The following diagram shows the graph of $y = f (x)$ .

The graph of $f$ touches the $x$ -axis at points $A$ and $B$ , as shown. The shaded region is enclosed by the graph of $y = f (x)$ and the $x$ -axis, between the points $A$ and $B$ .

The right cone in the following diagram has a total surface area of $12 π$ , equal to the shaded area in the previous diagram.

The cone has a base radius of $2$ , height $h$ , and slant height $l$ .

Find the $x$ -coordinates of $A$ and $B$ .

[3]

Show that the area of the shaded region is $12 π$ .

[5]

Find the value of $l$ .

[3]

Hence, find the volume of the cone.

[4]

Markscheme

$6 + 6 \cos x = 0$ (or setting their $f' (x) = 0$ ) (M1)

$\cos x = - 1$ (or $\sin x = 0$ )

$x = π, x = 3 π$ A1A1

[3 marks]

attempt to integrate $\int_{π}^{3 π} (6 + 6 \cos x) d x$ (M1)

$= {[6 x + 6 \sin x]}_{π}^{3 π}$ A1A1

substitute their limits into their integrated expression and subtract (M1)

$= (18 π + 6 \sin 3 π) - (6 π + 6 \sin π)$

$= (6 (3 π) + 0) - (6 π + 0) (= 18 π - 6 π)$ A1

area $= 12 π$ AG

[5 marks]

attempt to substitute into formula for surface area (including base) (M1)

$π (2^{2}) + π (2) (l) = 12 π$ (A1)

$4 π + 2 π l = 12 π$

$2 π l = 8 π$

$l = 4$ A1

[3 marks]

valid attempt to find the height of the cone (M1)

e.g. $2^{2} + h^{2} = {(their l)}^{2}$

$h = \sqrt{12} (= 2 \sqrt{3})$ (A1)

attempt to use $V = \frac{1}{3} π r^{2} h$ with their values substituted M1

$(\frac{1}{3} π (2^{2}) (\sqrt{12}))$

$volume = \frac{4 π \sqrt{12}}{3} (= \frac{8 π \sqrt{3}}{3} = \frac{8 π}{\sqrt{3}})$ A1

[4 marks]

Examiners report

[N/A]

The following diagram shows a ball attached to the end of a spring, which is suspended from a ceiling.

The height, $h$ metres, of the ball above the ground at time $t$ seconds after being released can be modelled by the function $h (t) = 0.4 \cos (π t) + 1.8$ where $t \geq 0$ .

Find the height of the ball above the ground when it is released.

[2]

Find the minimum height of the ball above the ground.

[2]

Show that the ball takes $2$ seconds to return to its initial height above the ground for the first time.

[2]

For the first 2 seconds of its motion, determine the amount of time that the ball is less than $1.8 + 0.2 \sqrt{2}$ metres above the ground.

[5]

Find the rate of change of the ball’s height above the ground when $t = \frac{1}{3}$ . Give your answer in the form $p π \sqrt{q} {ms}^{- 1}$ where $p \in ℚ$ and $q \in ℤ^{+}$ .

[4]

Markscheme

attempts to find $h (0)$ (M1)

$h (0) = 0.4 \cos (0) + 1.8 (= 2.2)$

$2.2$ (m) (above the ground) A1

[2 marks]

EITHER

uses the minimum value of $\cos (π t)$ which is $- 1$ M1

$0.4 (- 1) + 1.8$ (m)

the amplitude of motion is $0.4$ (m) and the mean position is $1.8$ (m) M1

finds $h' (t) = - 0.4 π \sin (π t)$ , attempts to solve $h' (t) = 0$ for $t$ and determines that the minimum height above the ground occurs at $t = 1, 3, \dots$ M1

$0.4 (- 1) + 1.8$ (m)

THEN

$1.4$ (m) (above the ground) A1

[2 marks]

EITHER

the ball is released from its maximum height and returns there a period later R1

the period is $\frac{2 π}{π} (= 2) (s)$ A1

attempts to solve $h (t) = 2.2$ for $t$ M1

$\cos (π t) = 1$

$t = 0, 2, \dots$ A1

THEN

so it takes $2$ seconds for the ball to return to its initial position for the first time AG

[2 marks]

$0.4 \cos (π t) + 1.8 = 1.8 + 0.2 \sqrt{2}$ (M1)

$0.4 \cos (π t) = 0.2 \sqrt{2}$

$\cos (π t) = \frac{\sqrt{2}}{2}$ A1

$π t = \frac{π}{4}, \frac{7 π}{4}$ (A1)

Note: Accept extra correct positive solutions for $π t$ .

$t = \frac{1}{4}, \frac{7}{4} (0 \leq t \leq 2)$ A1

Note: Do not award A1 if solutions outside $0 \leq t \leq 2$ are also stated.

the ball is less than $1.8 + 0.2 \sqrt{2}$ metres above the ground for $\frac{7}{4} - \frac{1}{4}$ (s)

$1.5$ (s) A1

[5 marks]

EITHER

attempts to find $h' (t)$ (M1)

recognizes that $h' (t)$ is required (M1)

THEN

$h' (t) = - 0.4 π \sin (π t)$ A1

attempts to evaluate their $h' (\frac{1}{3})$ (M1)

$h' (\frac{1}{3}) = - 0.4 π \sin \frac{π}{3}$

$= 0.2 π \sqrt{3} ({ms}^{- 1})$ A1

Note: Accept equivalent correct answer forms where $p \in ℚ$ . For example, $- \frac{1}{5} π \sqrt{3}$ .

[4 marks]

Examiners report

[N/A]

Given that $\frac{d y}{d x} = \cos (x - \frac{π}{4})$ and $y = 2$ when $x = \frac{3 π}{4}$ , find $y$ in terms of $x$ .

Markscheme

METHOD 1

recognition that $y = \int \cos (x - \frac{π}{4}) d x$ (M1)

$y = \sin (x - \frac{π}{4}) (+ c)$ (A1)

substitute both $x$ and $y$ values into their integrated expression including $c$ (M1)

$2 = \sin \frac{π}{2} + c$

$c = 1$

$y = \sin (x - \frac{π}{4}) + 1$ A1

METHOD 2

$\int_{2}^{y} d y = \int_{\frac{3 π}{4}}^{x} \cos (x - \frac{π}{4}) d x$ (M1)(A1)

$y - 2 = \sin (x - \frac{π}{4}) - \sin \frac{π}{2}$ A1

$y = \sin (x - \frac{π}{4}) + 1$ A1

[4 marks]

Examiners report

[N/A]

The derivative of a function $f$ is given by $f' (x) = 3 \sqrt{x}$ .

Given that $f (1) = 3$ , find the value of $f (4)$ .

Markscheme

correct working (A1)

eg −5 + (8 − 1)(3)

u₈ = 16 A1 N2

[2 marks]

METHOD 1

$f (x) = \int 3 \sqrt{x} d x$ (A1)

attempts to integrate (M1)

$f (x) = 2 x^{\frac{3}{2}} + C (= 2 x \sqrt{x} + C)$ A1

uses $f (1) = 3$ to obtain $3 = 2 {(1)}^{\frac{3}{2}} + C$ and so $C = 1$ M1

substitutes $x = 4$ into their expression for $f (x)$ (M1)

so $f (4) = 17$ A1

METHOD 2

$\int_{1}^{4} f' (x) d x = \int_{1}^{4} 3 \sqrt{x} d x$ (A1)

attempts to integrate both sides (M1)

${[f (x)]}_{1}^{4} = {[2 x^{\frac{3}{2}}]}_{1}^{4}$ A1

$f (4) - f (1) = 16 - 2$ M1

uses $f (1) = 3$ to find their value of $f (4)$ (M1)

$f (4) - 3 = 16 - 2$

so $f (4) = 17$ A1

[6 marks]

Examiners report

[N/A]

Let $f'\left( x \right) = \frac{{8x}}{{\sqrt {2{x^2} + 1} }}$ . Given that $f\left( 0 \right) = 5$ , find $f\left( x \right)$ .

Markscheme

attempt to integrate (M1)

$u = 2{x^2} + 1 \Rightarrow \frac{{{\text{d}}u}}{{{\text{d}}x}} = 4x$

$\int {\frac{{8x}}{{\sqrt {2{x^2} + 1} }}} {\text{d}}x = \int {\frac{2}{{\sqrt u }}} {\text{d}}u$ (A1)

EITHER

$= 4\sqrt u \left( { + C} \right)$ A1

$= 4\sqrt {2{x^2} + 1} \left( { + C} \right)$ A1

THEN

correct substitution into their integrated function (must have C) (M1)

$5 = 4 + C \Rightarrow C = 1$

$f\left( x \right) = 4\sqrt {2{x^2} + 1} + 1$ A1

[5 marks]

Examiners report

[N/A]

Consider the curve y = 5x³− 3x.

The curve has a tangent at the point P(−1, −2).

Find the equation of this tangent. Give your answer in the form y = mx + c.

Markscheme

(y − (−2)) = 12 (x − (−1)) (M1)

−2 = 12(−1) + c (M1)

Note: Award (M1) for point and their gradient substituted into the equation of a line.