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SL Paper 1

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 binomial expansion ${(x + 1)}^{7} = x^{7} + a x^{6} + b x^{5} + 35 x^{4} + \dots + 1$ where $x \neq 0$ and $a, b \in ℤ^{+}$ .

Show that $b = 21$ .

[2]

The third term in the expansion is the mean of the second term and the fourth term in the expansion.

Find the possible values of $x$ .

[5]

Markscheme

EITHER

recognises the required term (or coefficient) in the expansion (M1)

$b x^{5} = {}_{7}C_{2} x^{5} 1^{2}$ OR $b = {}_{7}C_{2}$ OR ${}_{7}C_{5}$

$b = \frac{7!}{2! 5!} (= \frac{7!}{2! (7 - 2)!})$

correct working A1

$\frac{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 1 \times 5 \times 4 \times 3 \times 2 \times 1}$ OR $\frac{7 \times 6}{2!}$ OR $\frac{42}{2}$

lists terms from row $7$ of Pascal’s triangle (M1)

$1, 7, 21, \dots$ A1

THEN

$b = 21$ AG

[2 marks]

$a = 7$ (A1)

correct equation A1

$21 x^{5} = \frac{a x^{6} + 35 x^{4}}{2}$ OR $21 x^{5} = \frac{7 x^{6} + 35 x^{4}}{2}$

correct quadratic equation A1

$7 x^{2} - 42 x + 35 = 0$ OR $x^{2} - 6 x + 5 = 0$ (or equivalent)

valid attempt to solve their quadratic (M1)

$(x - 1) (x - 5) = 0$ OR $x = \frac{6 \pm \sqrt{{(- 6)}^{2} - 4 (1) (5)}}{2 (1)}$

$x = 1, x = 5$ A1

Note: Award final A0 for obtaining $x = 0, x = 1, x = 5$ .

[5 marks]

Examiners report

The majority of candidates answered part (a) correctly, either by using the ${}^{n}C_{r}$ formula or Pascal's Triangle. In part (b) of the question, most candidates were able to correctly find the value of $a = 7$ and set up a correct equation showing the mean of the second and fourth terms. While some struggled to complete the required algebra to solve the equation, the majority of candidates who found a correct quadratic equation were able to solve it correctly. A few candidates included $x = 0$ in their final answer, thus not earning the final mark.

[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]

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 diagram shows the graph of a function $f$ , with domain $- 2 \leqslant x \leqslant 4$ .

N17/5/MATME/SP1/ENG/TZ0/03

The points $( - 2,{\text{ }}0)$ and $(4,{\text{ }}7)$ lie on the graph of $f$ .

On the grid, sketch the graph of ${f^{ - 1}}$ .

Markscheme

N17/5/MATME/SP1/ENG/TZ0/03.c/M A1A1A1 N3

Notes: Award A1 for both end points within circles,

A1 for images of $(2,{\text{ }}3)$ and $(0,{\text{ }}2)$ within circles,

A1 for approximately correct reflection in $y = x$ , concave up then concave down shape (do not accept line segments).

[3 marks]

Examiners report

[N/A]

Olava’s Pizza Company supplies and delivers large cheese pizzas.

The total cost to the customer, $C$ , in Papua New Guinean Kina ( $PGK$ ), is modelled by the function

$C (n) = 34.50 n + 8.50, n \geq 2, n \in ℤ,$

where $n$ , is the number of large cheese pizzas ordered. This total cost includes a fixed cost for delivery.

State, in the context of the question, what the value of $34.50$ represents.

[1]

a.i.

State, in the context of the question, what the value of $8.50$ represents.

[1]

a.ii.

Write down the minimum number of pizzas that can be ordered.

[1]

Kaelani has $450 PGK$ .

Find the maximum number of large cheese pizzas that Kaelani can order from Olava’s Pizza Company.

[3]

Markscheme

the cost of each (large cheese) pizza / a pizza / one pizza / per pizza (A1) (C1)

Note: Award (A0) for “the cost of (large cheese) pizzas”. Do not accept “the minimum cost of a pizza”.

[1 mark]

a.i.

the (fixed) delivery cost (A1) (C1)

[1 mark]

a.ii.

$2$ (A1) (C1)

[1 mark]

$450 = 34.50 n + 8.50$ (M1)

Note: Award (M1) for equating the cost equation to $450$ (may be stated as an inequality).

$12.8 (12.7971 \dots)$ (A1)

$12$ (A1)(ft) (C3)

Note: The final answer must be an integer.
The final (A1)(ft) is awarded for rounding their answer down to a whole number, provided their unrounded answer is seen.

[3 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

The graph of the quadratic function $f(x) = c + bx - {x^2}$ intersects the $x$ -axis at the point ${\text{A}}( - 1,{\text{ }}0)$ and has its vertex at the point ${\text{B}}(3,{\text{ }}16)$ .

N16/5/MATSD/SP1/ENG/TZ0/09

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

[2]

Find the value of $b$ .

[2]

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

[2]

Markscheme

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

$x = 3$ (A1)(A1) (C2)

Note: Award (A1) for $x =$ constant, (A1) for the constant being 3.

The answer must be an equation.

[2 marks]

$\frac{{ - b}}{{2( - 1)}} = 3$ (M1)

Note: Award (M1) for correct substitution into axis of symmetry formula.

$b - 2x = 0$ (M1)

Note: Award (M1) for correctly differentiating and equating to zero.

$c + b( - 1) - {( - 1)^2} = 0$ (or equivalent)

$c + b(3) - {(3)^2} = 16$ (or equivalent) (M1)

Note: Award (M1) for correct substitution of $( - 1,{\text{ }}0)$ and $(3,{\text{ }}16)$ in the original quadratic function.

$(b = ){\text{ }}6$ (A1)(ft) (C2)

Note: Follow through from part (a).

[2 marks]

$( - \infty ,{\text{ 16]}}$ OR $] - \infty ,{\text{ }}16]$ (A1)(A1)

Note: Award (A1) for two correct interval endpoints, (A1) for left endpoint excluded and right endpoint included.

[2 marks]

Examiners report

[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]

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]

A function $f$ is defined by $f (x) = \frac{2 x - 1}{x + 1}$ , where $x \in ℝ, x \neq - 1$ .

The graph of $y = f (x)$ has a vertical asymptote and a horizontal asymptote.

Write down the equation of the vertical asymptote.

[1]

a.i.

Write down the equation of the horizontal asymptote.

[1]

a.ii.

On the set of axes below, sketch the graph of $y = f (x)$ .

On your sketch, clearly indicate the asymptotes and the position of any points of intersection with the axes.

[3]

Hence, solve the inequality $0 < \frac{2 x - 1}{x + 1} < 2$ .

[1]

Markscheme

$x = - 1$ A1

[1 mark]

a.i.

$y = 2$ A1

[1 mark]

a.ii.

rational function shape with two branches in opposite quadrants, with two correctly positioned asymptotes and asymptotic behaviour shown A1

Note: The equations of the asymptotes are not required on the graph provided there is a clear indication of asymptotic behaviour at $x = - 1$ and $y = 2$ (or at their FT asymptotes from part (a)).

axes intercepts clearly shown at $x = \frac{1}{2}$ and $y = - 1$ A1A1

[3 marks]

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

Note: Accept correct alternative correct notation, such as $(\frac{1}{2}, \infty)$ and $] \frac{1}{2}, \infty [$ .

[1 mark]

Examiners report

It is pleasing to note that many candidates were familiar with the shape of the graph of a rational function of the form $f (x) = \frac{a x + b}{c x + d}$ , and a large number of them were able to sketch an appropriate graph. Part (c) was a struggle for the majority of candidates, with only a few answering correctly. Despite the word "hence" and the single mark available in this part, most candidates who attempted part (c) did so by trying to solve the inequality algebraically, rather than seeing the connection to the values in their graph.

a.i.

[N/A]

a.ii.

[N/A]

Consider the function $f (x) = - 2 (x - 1) (x + 3)$ , for $x \in ℝ$ . The following diagram shows part of the graph of $f$ .

For the graph of $f$

find the $x$ -coordinates of the $x$ -intercepts.

[2]

a.i.

find the coordinates of the vertex.

[3]

a.ii.

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

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

[2]

Markscheme

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

$x = 1, x = - 3$ (accept $(1, 0), (- 3, 0)$ ) A1

[2 marks]

a.i.

METHOD 1

$x = - 1$ A1

substituting their $x$ -coordinate into $f$ (M1)

$y = 8$ A1

$(- 1, 8)$

METHOD 2

attempt to complete the square (M1)

$- 2 ({(x + 1)}^{2} - 4)$ (M1)

$x = - 1, y = 8$ A1A1

$(- 1, 8)$

[3 marks]

a.ii.

$h = - 1$ A1

$k = 8$ A1

[2 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

Consider the function $f (x) = a^{x}$ where $x, a \in ℝ$ and $x > 0, a > 1$ .

The graph of $f$ contains the point $(\frac{2}{3}, 4)$ .

Consider the arithmetic sequence $\log_{8} 27, \log_{8} p, \log_{8} q, \log_{8} 125,$ where $p > 1$ and $q > 1$ .

Show that $a = 8$ .

[2]

Write down an expression for $f^{- 1} (x)$ .

[1]

Find the value of $f^{- 1} (\sqrt{32})$ .

[3]

Show that $27, p, q$ and $125$ are four consecutive terms in a geometric sequence.

[4]

d.i.

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

[5]

d.ii.

Markscheme

$f (\frac{2}{3}) = 4$ OR $a^{\frac{2}{3}} = 4$ (M1)

$a = 4^{\frac{3}{2}}$ OR $a = {(2^{2})}^{\frac{3}{2}}$ OR $a^{2} = 64$ OR $\sqrt[3]{a} = 2$ A1

$a = 8$ AG

[2 marks]

$f^{- 1} (x) = \log_{8} x$ A1

Note: Accept $f^{- 1} (x) = \log_{a} x$ .
Accept any equivalent expression for $f^{- 1}$ e.g. $f^{- 1} (x) = \frac{\ln x}{\ln 8}$ .

[1 mark]

correct substitution (A1)

$\log_{8} \sqrt{32}$ OR $8^{x} = 32^{\frac{1}{2}}$

correct working involving log/index law (A1)

$\frac{1}{2} \log_{8} 32$ OR $\frac{5}{2} \log_{8} 2$ OR $\log_{8} 2 = \frac{1}{3}$ OR $\log_{2} 2^{\frac{5}{2}}$ OR $\log_{2} 8 = 3$ OR $\frac{\ln 2^{\frac{5}{2}}}{\ln 2^{3}}$ OR $2^{3 x} = 2^{\frac{5}{2}}$

$f^{- 1} (\sqrt{32}) = \frac{5}{6}$ A1

[3 marks]

METHOD 1

equating a pair of differences (M1)

$u_{2} - u_{1} = u_{4} - u_{3} (= u_{3} - u_{2})$

$\log_{8} p - \log_{8} 27 = \log_{8} 125 - \log_{8} q$

$\log_{8} 125 - \log_{8} q = \log_{8} q - \log_{8} p$

$\log_{8} (\frac{p}{27}) = \log_{8} (\frac{125}{q}), \log_{8} (\frac{125}{q}) = \log_{8} (\frac{q}{p})$ A1A1

$\frac{p}{27} = \frac{125}{q}$ and $\frac{125}{q} = \frac{q}{p}$ A1

$27, p, q$ and $125$ are in geometric sequence AG

Note: If candidate assumes the sequence is geometric, award no marks for part (i). If $r = \frac{5}{3}$ has been found, this will be awarded marks in part (ii).

METHOD 2

expressing a pair of consecutive terms, in terms of $d$ (M1)

$p = 8^{d} \times 27$ and $q = 8^{2 d} \times 27$ OR $q = 8^{2 d} \times 27$ and $125 = 8^{3 d} \times 27$

two correct pairs of consecutive terms, in terms of $d$ A1

$\frac{8^{d} \times 27}{27} = \frac{8^{2 d} \times 27}{8^{d} \times 27} = \frac{8^{3 d} \times 27}{8^{2 d} \times 27}$ (must include 3 ratios) A1

all simplify to $8^{d}$ A1

$27, p, q$ and $125$ are in geometric sequence AG

[4 marks]

d.i.

METHOD 1 (geometric, finding $r$ )

$u_{4} = u_{1} r^{3}$ OR $125 = 27 {(r)}^{3}$ (M1)

$r = \frac{5}{3}$ (seen anywhere) A1

$p = 27 r$ OR $\frac{125}{q} = \frac{5}{3}$ (M1)

$p = 45, q = 75$ A1A1

METHOD 2 (arithmetic)

$u_{4} = u_{1} + 3 d$ OR $\log_{8} 125 = \log_{8} 27 + 3 d$ (M1)

$d = \log_{8} (\frac{5}{3})$ (seen anywhere) A1

$\log_{8} p = \log_{8} 27 + \log_{8} (\frac{5}{3})$ OR $\log_{8} q = \log_{8} 27 + 2 \log_{8} (\frac{5}{3})$ (M1)

$p = 45, q = 75$ A1A1

METHOD 3 (geometric using proportion)

recognizing proportion (M1)

$p q = 125 \times 27$ OR $q^{2} = 125 p$ OR $p^{2} = 27 q$

two correct proportion equations A1

attempt to eliminate either $p$ or $q$ (M1)

$q^{2} = 125 \times \frac{125 \times 27}{q}$ OR $p^{2} = 27 \times \frac{125 \times 27}{p}$

$p = 45, q = 75$ A1A1

[5 marks]

d.ii.

Examiners report

[N/A]

d.i.

[N/A]

d.ii.

The diagram shows the graph of the quadratic function $f (x) = a x^{2} + b x + c$ , with vertex $(- 2, 10)$ .

The equation $f (x) = k$ has two solutions. One of these solutions is $x = 2$ .

Write down the other solution of $f (x) = k$ .

[2]

Complete the table below placing a tick (✔) to show whether the unknown parameters $a$ and $b$ are positive, zero or negative. The row for $c$ has been completed as an example.

[2]

State the values of $x$ for which $f (x)$ is decreasing.

[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.

$(x =) (- 2) - 4$ OR $(x =) (- 2) - (2 - (- 2))$ (M1)

Note: Award (M1) for correct calculation of the left symmetrical point.

$(x =) - 6$ (A1) (C2)

[2 marks]

(A1)(A1) (C2)

Note: Award (A1) for each correct row.

[2 marks]

$x > - 2$ OR $x \geq - 2$ (A1)(A1) (C2)

Note: Award (A1) for $- 2$ seen as part of an inequality, (A1) for completely correct notation. Award (A1)(A1) for correct equivalent statement in words, for example “decreasing when $x$ is greater than negative $2$ ”.

[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]

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]

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]

The following table shows the probability distribution of a discrete random variable $X$ where $x = 1, 2, 3, 4$ .

Find the value of $k$ , justifying your answer.

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.

uses $\sum P (X = x) (= 1)$ (M1)

$k^{2} + (7 k + 2) + (- 2 k) + (3 k^{2}) (= 1)$

$4 k^{2} + 5 k + 1 (= 0)$ A1

EITHER

attempts to factorize their quadratic M1

$(k + 1) (4 k + 1) = 0$

attempts use of the quadratic formula on their equation M1

$k = \frac{- 5 \pm \sqrt{5^{2} - 4 (4) (1)}}{8} (= \frac{- 5 \pm 3}{8})$

THEN

$k = - 1, - \frac{1}{4}$ A1

rejects $k = - 1$ as this value leads to invalid probabilities, for example, $P (X = 2) = - 5 < 0$ R1

so $k = - \frac{1}{4}$ A1

Note: Award R0A1 if $k = - \frac{1}{4}$ is stated without a valid reason given for rejecting $k = - 1$ .

[6 marks]

Examiners report

[N/A]

Let $f (x) = m x^{2} - 2 m x$ , where $x \in ℝ$ and $m \in ℝ$ . The line $y = m x - 9$ meets the graph of $f$ at exactly one point.

The function $f$ can be expressed in the form $f (x) = 4 (x - p) (x - q)$ , where $p, q \in ℝ$ .

The function $f$ can also be expressed in the form $f (x) = 4 {(x - h)}^{2} + k$ , where $h, k \in ℝ$ .

Show that $m = 4$ .

[6]

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

[2]

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

[3]

Hence find the values of $x$ where the graph of $f$ is both negative and increasing.

[3]

Markscheme

METHOD 1 (discriminant)

$m x^{2} - 2 m x = m x - 9$ (M1)

$m x^{2} - 3 m x + 9 = 0$

recognizing $Δ = 0$ (seen anywhere) M1

$Δ = {(- 3 m)}^{2} - 4 (m) (9)$ (do not accept only in quadratic formula for $x$ ) A1

valid approach to solve quadratic for $m$ (M1)

$9 m (m - 4) = 0$ OR $m = \frac{36 \pm \sqrt{36^{2} - 4 \times 9 \times 0}}{2 \times 9}$

both solutions $m = 0, 4$ A1

$m \neq 0$ with a valid reason R1

the two graphs would not intersect OR $0 \neq - 9$

$m = 4$ AG

METHOD 2 (equating slopes)

$m x^{2} - 2 m x = m x - 9$ (seen anywhere) (M1)

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

equating slopes, $f' (x) = m$ (seen anywhere) M1

$2 m x - 2 m = m$

$x = \frac{3}{2}$ A1

substituting their $x$ value (M1)

${(\frac{3}{2})}^{2} m - 2 m \times \frac{3}{2} = m \times \frac{3}{2} - 9$

$\frac{9}{4} m - \frac{12}{4} m = \frac{6}{4} m - 9$ A1

$\frac{- 9 m}{4} = - 9$

$m = 4$ AG

METHOD 3 (using $\frac{- b}{2 a}$ )

$m x^{2} - 2 m x = m x - 9$ (M1)

$m x^{2} - 3 m x + 9 = 0$

attempt to find $x$ -coord of vertex using $\frac{- b}{2 a}$ (M1)

$\frac{- (- 3 m)}{2 m}$ A1

$x = \frac{3}{2}$ A1

substituting their $x$ value (M1)

${(\frac{3}{2})}^{2} m - 3 m \times \frac{3}{2} + 9 = 0$

$\frac{9}{4} m - \frac{9}{2} m + 9 = 0$ A1

$- 9 m = - 36$

$m = 4$ AG

[6 marks]

$4 x (x - 2)$ (A1)

$p = 0$ and $q = 2$ OR $p = 2$ and $q = 0$ A1

[2 marks]

attempt to use valid approach (M1)

$\frac{0 + 2}{2}, \frac{- (- 8)}{2 \times 4}, f (1), 8 x - 8 = 0$ OR $4 (x^{2} - 2 x + 1 - 1) (= 4 {(x - 1)}^{2} - 4)$

$h = 1, k = - 4$ A1A1

[3 marks]

EITHER

recognition $x = h$ to $2$ (may be seen on sketch) (M1)

recognition that $f (x) < 0$ and $f' (x) > 0$ (M1)

THEN

$1 < x < 2$ A1A1

Note: Award A1 for two correct values, A1 for correct inequality signs.

[3 marks]

Examiners report

[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.

Jean-Pierre jumps out of an airplane that is flying at constant altitude. Before opening his parachute, he goes through a period of freefall.

Jean-Pierre’s vertical speed during the time of freefall, $S$ , in ${m s}^{- 1}$ , is modelled by the following function.

$S (t) = K - 60 (1 . 2^{- t}), t \geq 0$

where $t$ , is the number of seconds after he jumps out of the airplane, and $K$ is a constant. A sketch of Jean-Pierre’s vertical speed against time is shown below.

Jean-Pierre’s initial vertical speed is $0 {m s}^{- 1}$ .

Find the value of $K$ .

[2]

In the context of the model, state what the horizontal asymptote represents.

[1]

Find Jean-Pierre’s vertical speed after $10$ seconds. Give your answer in $km h^{- 1}$ .

[3]

Markscheme

$0 = K - 60 (1 . 2^{0})$ (M1)

Note: Award (M1) for correctly substituted function equated to zero.

$(K =) 60$ (A1) (C2)

[2 marks]

the (vertical) speed that Jean-Pierre is approaching (as $t$ increases) (A1) (C1)
OR
the limit of the (vertical) speed of Jean-Pierre (A1) (C1)

Note: Accept “maximum speed” or “terminal speed”.

[1 mark]

$(S =) 60 - 60 (1 . 2^{- 10})$ (M1)

Note: Award (M1) for correctly substituted function.

$(S =) 50.3096 \dots ({m s}^{- 1})$ (A1)(ft)

Note: Follow through from part (a).

$181 ({km h}^{- 1}) (181.114 \dots ({km h}^{- 1}))$ (A1)(ft) (C3)

Note: Award the final (A1)(ft) for correct conversion of their speed to $km h^{- 1}$ .

[3 marks]

Examiners report

[N/A]

Let $f(x) = 1 + {{\text{e}}^{ - x}}$ and $g(x) = 2x + b$ , for $x \in \mathbb{R}$ , where $b$ is a constant.

Find $(g \circ f)(x)$ .

[2]

Given that $\mathop {\lim }\limits_{x \to + \infty } (g \circ f)(x) = - 3$ , find the value of $b$ .

[4]

Markscheme

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

attempt to form composite (M1)

eg $\,\,\,\,\,$ $g(1 + {{\text{e}}^{ - x}})$

correct function A1 N2

eg $\,\,\,\,\,$ $(g \circ f)(x) = 2 + b + 2{{\text{e}}^{ - x}},{\text{ }}2(1 + {{\text{e}}^{ - x}}) + b$

[2 marks]

evidence of $\mathop {\lim }\limits_{x \to \infty } (2 + b + 2{{\text{e}}^{ - x}}) = 2 + b + \mathop {\lim }\limits_{x \to \infty } (2{{\text{e}}^{ - x}})$ (M1)

eg $\,\,\,\,\,$ $2 + b + 2{{\text{e}}^{ - \infty }}$ , graph with horizontal asymptote when $x \to \infty$

Note: Award M0 if candidate clearly has incorrect limit, such as $x \to 0,{\text{ }}{{\text{e}}^\infty },{\text{ }}2{{\text{e}}^0}$ .

evidence that ${{\text{e}}^{ - x}} \to 0$ (seen anywhere) (A1)

eg $\,\,\,\,\,$ $\mathop {\lim }\limits_{x \to \infty } ({{\text{e}}^{ - x}}) = 0,{\text{ }}1 + {{\text{e}}^{ - x}} \to 1,{\text{ }}2(1) + b = - 3,{\text{ }}{{\text{e}}^{{\text{large negative number}}}} \to 0$ , graph of $y = {{\text{e}}^{ - x}}$ or

$y = 2{{\text{e}}^{ - x}}$ with asymptote $y = 0$ , graph of composite function with asymptote $y = - 3$

correct working (A1)

eg $\,\,\,\,\,$ $2 + b = - 3$

$b = - 5$ A1 N2

[4 marks]

Examiners report

[N/A]

The following diagram shows the graph of $y = - 1 - \sqrt{x + 3}$ for $x \geq - 3$ .

A function $f$ is defined by $f (x) = - 1 - \sqrt{x + 3}$ for $x \geq - 3$ .

Describe a sequence of transformations that transforms the graph of $y = \sqrt{x}$ for $x \geq 0$ to the graph of $y = - 1 - \sqrt{x + 3}$ for $x \geq - 3$ .

[3]

State the range of $f$ .

[1]

Find an expression for $f^{- 1} (x)$ , stating its domain.

[5]

Find the coordinates of the point(s) where the graphs of $y = f (x)$ and $y = f^{- 1} (x)$ intersect.

[5]

Markscheme

for example,

a reflection in the $x$ -axis (in the line $y = 0$ ) A1

a horizontal translation (shift) $3$ units to the left A1

a vertical translation (shift) down by $1$ unit A1

Note: Award A1 for each correct transformation applied in a correct position in the sequence. Do not accept use of the “move” for a translation.

Note: Award A1A1A1 for a correct alternative sequence of transformations. For example,

a vertical translation (shift) down by $1$ unit, followed by a horizontal translation (shift) $3$ units to the left and then a reflection in the line $y = - 1$ .

[3 marks]

range is $f (x) \leq - 1$ A1

Note: Correct alternative notations include $] - \infty, - 1]$ , $(- \infty, - 1]$ or $y \leq - 1$ .

[1 mark]

$- 1 - \sqrt{y + 3} = x$ M1

Note: Award M1 for interchanging $x$ and $y$ (can be done at a later stage).

$\sqrt{y + 3} = - x - 1 (= - (x + 1))$ A1

$y + 3 = {(x + 1)}^{2}$ A1

so $f^{- 1} (x) = {(x + 1)}^{2} - 3 (f^{- 1} (x) = x^{2} + 2 x - 2)$ A1

domain is $x \leq - 1$ A1

Note: Correct alternative notations include $] - \infty, - 1]$ or $(- \infty, - 1]$ .

[5 marks]

the point of intersection lies on the line $y = x$

EITHER

${(x + 1)}^{2} - 3 = x$ M1

attempts to solve their quadratic equation M1

for example, $(x + 2) (x - 1) = 0$ or $x = \frac{- 1 \pm \sqrt{1^{2} - 4 (1) (- 2)}}{2} (x = \frac{- 1 \pm 3}{2})$

$- 1 - \sqrt{x + 3} = x$ M1

${(- 1 - \sqrt{x + 3})}^{2} = x^{2} \Rightarrow 2 \sqrt{x + 3} + x + 4 = x^{2}$

substitutes $2 \sqrt{x + 3} = - 2 (x + 1)$ to obtain $- 2 (x + 1) + x + 4 = x^{2}$

attempts to solve their quadratic equation M1

for example, $(x + 2) (x - 1) = 0$ or $x = \frac{- 1 \pm \sqrt{1^{2} - 4 (1) (- 2)}}{2} (x = \frac{- 1 \pm 3}{2})$

THEN

$x = - 2, 1$ A1

as $x \leq - 1$ , the only solution is $x = - 2$ R1

so the coordinates of the point of intersection are $(- 2, - 2)$ A1

Note: Award R0A1 if $(- 2, - 2)$ is stated without a valid reason given for rejecting $(1, 1)$ .

[5 marks]

Examiners report

[N/A]

Line $L$ intersects the $x$ -axis at point A and the $y$ -axis at point B, as shown on the diagram.

M17/5/MATSD/SP1/ENG/TZ2/04

The length of line segment OB is three times the length of line segment OA, where O is the origin.

Point ${\text{(2, 6)}}$ lies on $L$ .

Find the equation of $L$ in the form $y = mx + c$ .

[2]

Find the $x$ -coordinate of point A.

[2]

Markscheme

$6 = - 3(2) + c$ $\,\,\,$ OR $\,\,\,$ $(y - 6) = - 3(x - 2)$ (M1)

Note: Award (M1) for substitution of their gradient from part (a) into a correct equation with the coordinates $(2,{\text{ }}6)$ correctly substituted.

$y = - 3x + 12$ (A1)(ft) (C2)

Notes: Award (A1)(ft) for their correct equation. Follow through from part (a).

If no method seen, award (A1)(A0) for $y = - 3x$ .

Award (A1)(A0) for $- 3x + 12$ .

[2 marks]

$0 = - 3x + 12$ (M1)

Note: Award (M1) for substitution of $y = 0$ in their equation from part (b).

$(x = ){\text{ }}4$ (A1)(ft) (C2)

Notes: Follow through from their equation from part (b). Do not follow through if no method seen. Do not award the final (A1) if the value of $x$ is negative or zero.

[2 marks]

Examiners report

[N/A]

Consider the functions $f\left( x \right) = {x^4} - 2$ and $g\left( x \right) = {x^3} - 4{x^2} + 2x + 6$

The functions intersect at points P and Q. Part of the graph of $y = f\left( x \right)$ and part of the graph of $y = g\left( x \right)$ are shown on the diagram.

Find the range of f.

[2]

Write down the x-coordinate of P and the x-coordinate of Q.

[2]

Write down the values of x for which $f\left( x \right) > g\left( x \right)$ .

[2]

Markscheme

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

$\left[ { - 2,\,\,\infty } \right[{\text{ or }}\left[ { - 2,\,\,\infty } \right)$ OR $f\left( x \right) \geqslant - 2{\text{ or }}y \geqslant - 2$ OR $- 2 \leqslant f\left( x \right) < \infty$ (A1)(A1) (C2)

Note: Award (A1) for −2 and (A1) for completely correct mathematical notation, including weak inequalities. Accept $f \geqslant - 2$ .

[2 marks]

–1 and 1.52 (1.51839…) (A1)(A1) (C2)

Note: Award (A1) for −1 and (A1) for 1.52 (1.51839).

[2 marks]

$x < - 1,\,\,\,x > 1.52$ OR $\left( { - \infty ,\,\, - 1} \right) \cup \left( {1.52,\,\,\infty } \right)$ . (A1)(ft)(A1)(ft) (C2)

Note: Award (A1)(ft) for both critical values in inequality or range statements such as $x < - 1,\,\,\left( { - \infty ,\,\, - 1} \right),\,\,x > 1.52\,{\text{ or }}\left( {1.52,\,\,\infty } \right)$ .

Award the second (A1)(ft) for correct strict inequality statements used with their critical values. If an incorrect use of strict and weak inequalities has already been penalized in (a), condone weak inequalities for this second mark and award (A1)(ft).

[2 marks]

Examiners report

[N/A]

Let $g\left( x \right) = {x^2} + bx + 11$ . The point $\left( { - 1{\text{, }}8} \right)$ lies on the graph of $g$ .

Find the value of $b$ .

[3]

The graph of $f\left( x \right) = {x^2}$ is transformed to obtain the graph of $g$ .

Describe this transformation.

[4]

Markscheme

valid attempt to substitute coordinates (M1)

eg $g\left( { - 1} \right) = 8$

correct substitution (A1)

eg ${\left( { - 1} \right)^2} + b\left( { - 1} \right) + 11 = 8$ , $1 - b + 11 = 8$

$b = 4$ A1 N2

[3 marks]

valid attempt to solve (M1)

eg $\left( {{x^2} + 4x + 4} \right) + 7$ , $h = \frac{{ - 4}}{2}$ , $k = g\left( { - 2} \right)$

correct working A1

eg ${\left( {x + 2} \right)^2} + 7$ , $h = - 2$ , $k = 7$

translation or shift (do not accept move) of vector $\left( {\begin{array}{*{20}{c}} { - 2} \\ 7 \end{array}} \right)$ (accept left by $2$ and up by $7$ ) A1A1 N2

[4 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]

The following table shows values of $f (x)$ and $g (x)$ for different values of $x$ .

Both $f$ and $g$ are one-to-one functions.

Find $g (0)$ .

[1]

Find $(f \circ g) (0)$ .

[2]

Find the value of $x$ such that $f (x) = 0$ .

[2]

Markscheme

$g (0) = - 2$ A1

[1 mark]

evidence of using composite function (M1)

$f (g (0))$ OR $f (- 2)$

$(f \circ g) (0) = 8$ A1

[2 marks]

$x = 3$ A2

[2 marks]

Examiners report

This question was completed successfully by most of the candidates. In part (b) of the question, a few candidates did not recognize the notation for a composite function and instead incorrectly thought they were supposed to multiply values for $f (0)$ and $g (0)$ .

[N/A]

The function $f$ is defined by $f (x) = \frac{2 x + 4}{3 - x}$ , where $x \in ℝ, x \neq 3$ .

Write down the equation of

Find the coordinates where the graph of $f$ crosses

the vertical asymptote of the graph of $f$ .

[1]

a.i.

the horizontal asymptote of the graph of $f$ .

[1]

a.ii.

the $x$ -axis.

[1]

b.i.

the $y$ -axis.

[1]

b.ii.

Sketch the graph of $f$ on the axes below.

[1]

Markscheme

$x = 3$ A1

[1 mark]

a.i.

$y = - 2$ A1

[1 mark]

a.ii.

$(- 2, 0)$ (accept $x = - 2$ ) A1

[1 mark]

b.i.

$(0, \frac{4}{3})$ (accept $y = \frac{4}{3}$ and $f (0) = \frac{4}{3}$ ) A1

[1 mark]

b.ii.

Note: Award A1 for completely correct shape: two branches in correct quadrants with asymptotic behaviour.

[1 mark]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

Consider the series $\ln x + p \ln x + \frac{1}{3} \ln x + \dots$ , where $x \in ℝ, x > 1$ and $p \in ℝ, p \neq 0$ .

Consider the case where the series is geometric.

Now consider the case where the series is arithmetic with common difference $d$ .

Show that $p = \pm \frac{1}{\sqrt{3}}$ .

[2]

a.i.

Given that $p > 0$ and $S_{\infty} = 3 + \sqrt{3}$ , find the value of $x$ .

[3]

a.ii.

Show that $p = \frac{2}{3}$ .

[3]

b.i.

Write down $d$ in the form $k \ln x$ , where $k \in ℚ$ .

[1]

b.ii.

The sum of the first $n$ terms of the series is $- 3 \ln x$ .

Find the value of $n$ .

[6]

b.iii.

Markscheme

EITHER

attempt to use a ratio from consecutive terms M1

$\frac{p \ln x}{\ln x} = \frac{\frac{1}{3} \ln x}{p \ln x}$ OR $\frac{1}{3} \ln x = (\ln x) r^{2}$ OR $p \ln x = \ln x (\frac{1}{3 p})$

Note: Candidates may use $\ln x^{1} + \ln x^{p} + \ln x^{\frac{1}{3}} \dots$ and consider the powers of $x$ in geometric sequence

Award M1 for $\frac{p}{1} = \frac{\frac{1}{3}}{p}$ .

$r = p$ and $r^{2} = \frac{1}{3}$ M1

THEN

$p^{2} = \frac{1}{3}$ OR $r = \pm \frac{1}{\sqrt{3}}$ A1

$p = \pm \frac{1}{\sqrt{3}}$ AG

Note: Award M0A0 for $r^{2} = \frac{1}{3}$ or $p^{2} = \frac{1}{3}$ with no other working seen.

[2 marks]

a.i.

$\frac{\ln x}{1 - \frac{1}{\sqrt{3}}} (= 3 + \sqrt{3})$ (A1)

$\ln x = 3 - \frac{3}{\sqrt{3}} + \sqrt{3} - \frac{\sqrt{3}}{\sqrt{3}}$ OR $\ln x = 3 - \sqrt{3} + \sqrt{3} - 1 (\Rightarrow \ln x = 2)$ A1

$x = e^{2}$ A1

[3 marks]

a.ii.

METHOD 1

attempt to find a difference from consecutive terms or from $u_{2}$ M1

correct equation A1

$p \ln x - \ln x = \frac{1}{3} \ln x - p \ln x$ OR $\frac{1}{3} \ln x = \ln x + 2 (p \ln x - \ln x)$

Note: Candidates may use $\ln x^{1} + \ln x^{p} + \ln x^{\frac{1}{3}} + \dots$ and consider the powers of $x$ in arithmetic sequence.

Award M1A1 for $p - 1 = \frac{1}{3} - p$

$2 p \ln x = \frac{4}{3} \ln x (\Rightarrow 2 p = \frac{4}{3})$ A1

$p = \frac{2}{3}$ AG

METHOD 2

attempt to use arithmetic mean $u_{2} = \frac{u_{1} + u_{3}}{2}$ M1

$p \ln x = \frac{\ln x + \frac{1}{3} \ln x}{2}$ A1

$2 p \ln x = \frac{4}{3} \ln x (\Rightarrow 2 p = \frac{4}{3})$ A1

$p = \frac{2}{3}$ AG

METHOD 3

attempt to find difference using $u_{3}$ M1

$\frac{1}{3} \ln x = \ln x + 2 d (\Rightarrow d = - \frac{1}{3} \ln x)$

$u_{2} = \ln x + \frac{1}{2} (\frac{1}{3} \ln x - \ln x)$ OR $p \ln x - \ln x = - \frac{1}{3} \ln x$ A1

$p \ln x = \frac{2}{3} \ln x$ A1

$p = \frac{2}{3}$ AG

[3 marks]

b.i.

$d = - \frac{1}{3} \ln x$ A1

[1 mark]

b.ii.

METHOD 1

$S_{n} = \frac{n}{2} [2 \ln x + (n - 1) \times (- \frac{1}{3} \ln x)]$

attempt to substitute into $S_{n}$ and equate to $- 3 \ln x$ (M1)

$\frac{n}{2} [2 \ln x + (n - 1) \times (- \frac{1}{3} \ln x)] = - 3 \ln x$

correct working with $S_{n}$ (seen anywhere) (A1)

$\frac{n}{2} [2 \ln x - \frac{n}{3} \ln x + \frac{1}{3} \ln x]$ OR $n \ln x - \frac{n (n - 1)}{6} \ln x$ OR $\frac{n}{2} (\ln x + (\frac{4 - n}{3}) \ln x)$

correct equation without $\ln x$ A1

$\frac{n}{2} (\frac{7}{3} - \frac{n}{3}) = - 3$ OR $n - \frac{n (n - 1)}{6} = - 3$ or equivalent

Note: Award as above if the series $1 + p + \frac{1}{3} + \dots$ is considered leading to $\frac{n}{2} (\frac{7}{3} - \frac{n}{3}) = - 3$ .

attempt to form a quadratic $= 0$ (M1)

$n^{2} - 7 n - 18 = 0$

attempt to solve their quadratic (M1)

$(n - 9) (n + 2) = 0$

$n = 9$ A1

METHOD 2

listing the first $7$ terms of the sequence (A1)

$\ln x + \frac{2}{3} \ln x + \frac{1}{3} \ln x + 0 - \frac{1}{3} \ln x - \frac{2}{3} \ln x - \ln x + \dots$

recognizing first $7$ terms sum to $0$ M1

$8$ ^th term is $- \frac{4}{3} \ln x$ (A1)

$9$ ^th term is $- \frac{5}{3} \ln x$ (A1)

sum of $8$ ^th and $9$ ^th term $= - 3 \ln x$ (A1)

$n = 9$ A1

[6 marks]

b.iii.

Examiners report

Many candidates were able to identify the key relationship between consecutive terms for both geometric and arithmetic sequences. Substitution into the infinity sum formula was good with solving involving the natural logarithm done quite well. The complexity of the equation formed using 𝑆𝑛 was a stumbling block for some candidates. Those who factored out and cancelled the ln𝑥 expression were typically successful in solving the resulting quadratic.

a.i.

[N/A]

a.ii.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

b.iii.

Let $f (x) = - x^{2} + 4 x + 5$ and $g (x) = - f (x) + k$ .

Find the values of $k$ so that $g (x) = 0$ has no real roots.

Markscheme

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

METHOD 1 – (discriminant)

correct expression for $g$ (A1)

eg $- (- x^{2} + 4 x + 5) + k, x^{2} - 4 x - 5 + k = 0$

evidence of discriminant (M1)

eg $b^{2} - 4 a c, Δ$

correct substitution into discriminant of $g$ (A1)

eg ${(- 4)}^{2} - 4 (1) (- 5 + k), 16 - 4 (k - 5)$

recognizing discriminant is negative (M1)

eg $Δ < 0, {(- 4)}^{2} - 4 (1) (- 5 + k) < 0, 16 < 4 (k - 5), 16 - 4 (- 1) (5) < 0$

correct working (must be correct inequality) (A1)

eg $- 4 k < - 36, k - 5 > 4, 16 + 20 - 4 k < 0$

$k > 9$ A1 N3

METHOD 2 – (transformation of vertex of $f$ )

valid approach for finding $f (x)$ vertex (M1)

eg $- \frac{b}{2 a} = 2, f' (x) = 0$

correct vertex of $f (x)$ (A1)

eg $(2, 9)$

correct vertex of $- f (x)$ (A1)

eg $(2, - 9)$

correct vertex of $g (x)$ (A1)

eg $(\begin{matrix} 2 \\ - 9 \end{matrix}) + (\begin{matrix} 0 \\ k \end{matrix}), (2, - 9 + k)$

recognizing when vertex is above $x$ -axis (M1)

eg $- 9 + k > 0$ , sketch

$k > 9$ A1 N3

METHOD 3 – (transformation of $f$ )

recognizing vertical reflection of $f (x)$ (M1)

eg $- f (x), x^{2} - 4 x - 5$ , sketch

correct expression for $g (x)$ (A1)

eg $x^{2} - 4 x - 5 + k$

valid approach for finding vertex of $g (x)$ (M1)

eg $- \frac{b}{2 a} = 2, g' (x) = 0$

correct $y$ coordinate of vertex of $g (x)$ (A1)

eg $y = - 9 + k, (2, - 9 + k)$

recognizing when vertex is above $x$ -axis (M1)

eg $- 9 + k > 0$ , sketch

$k > 9$ A1 N3

[6 marks]

Examiners report

[N/A]

The functions $f$ and $g$ are defined such that $f\left( x \right) = \frac{{x + 3}}{4}$ and $g\left( x \right) = 8x + 5$ .

Show that $\left( {g \circ f} \right)\left( x \right) = 2x + 11$ .

[2]

Given that ${\left( {g \circ f} \right)^{ - 1}}\left( a \right) = 4$ , find the value of $a$ .

[3]

Markscheme

attempt to form composition M1

correct substitution $g\left( {\frac{{x + 3}}{4}} \right) = 8\left( {\frac{{x + 3}}{4}} \right) + 5$ A1

$\left( {g \circ f} \right)\left( x \right) = 2x + 11$ AG

[2 marks]

attempt to substitute 4 (seen anywhere) (M1)

correct equation $a = 2 \times 4 + 11$ (A1)

$a$ = 19 A1

[3 marks]

Examiners report

[N/A]

Let $f (x) = a \log_{3} (x - 4)$ , for $x > 4$ , where $a > 0$ .

Point $A (13, 7)$ lies on the graph of $f$ .

Find the value of $a$ .

[3]

The $x$ -intercept of the graph of $f$ is $(5, 0)$ .

On the following grid, sketch the graph of $f$ .

[3]

Markscheme

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

attempt to substitute coordinates (in any order) into $f$ (M1)

eg $a \log_{3} (13 - 4) = 7, a \log_{3} (7 - 4) = 13, a \log 9 = 7$

finding $\log_{3} 9 = 2$ (seen anywhere) (A1)

eg $\log_{3} 9 = 2, 2 a = 7$

$a = \frac{7}{2}$ A1 N2

[3 marks]

A1A1A1 N3

Note: Award A1 for correct shape of logarithmic function (must be increasing and concave down).
Only if the shape is correct, award the following:
A1 for being asymptotic to $x = 4$
A1 for curve including both points in circles.

[3 marks]

Examiners report

[N/A]

Consider the graph of the function $f (x) = x + \frac{12}{x^{2}}, x \neq 0$ .

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

[2]

a.i.

Write down the coordinates of the local minimum point.

[2]

a.ii.

Consider the function $g (x) = 3 - x$ .

Solve $f (x) = g (x)$ .

[2]

Markscheme

$0 = x + \frac{12}{x^{2}}$ (M1)

Note: Award (M1) for equating the function to zero.

$(x =) - 2.29 (- 2.28942 \dots)$ (A1) (C2)

Note: Award (C1) for a correct $x$ -value given as part of a coordinate pair or alongside an explicitly stated $y$ -value.

[2 marks]

a.i.

$(2.88, 4.33)$ $((2.88449 \dots, 4.32674 \dots))$ (A1)(A1) (C2)

Note: Accept $x = 2.88, y = 4.33$ .

[2 marks]

a.ii.

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

Note: Award (M1) for equating the functions or for a sketch of the two functions.

$(x =) - 1.43 (- 1.43080 \dots)$ (A1) (C2)

Note: Do not award the final (A1) if the answer is seen as part of a coordinate pair or a $y$ -value is explicitly stated, unless already penalized in part (a).

[2 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

The points ${\text{A}}$ and ${\text{B}}$ have position vectors $\left( {\begin{array}{*{20}{c}} { - 2} \\ 4 \\ { - 4} \end{array}} \right)$ and $\left( {\begin{array}{*{20}{c}} 6 \\ 8 \\ 0 \end{array}} \right)$ respectively.

Point ${\text{C}}$ has position vector $\left( {\begin{array}{*{20}{c}} { - 1} \\ k \\ 0 \end{array}} \right)$ . Let ${\text{O}}$ be the origin.

Find, in terms of $k$ ,

$\overrightarrow {{\text{OA}}} \bullet \overrightarrow {{\text{OC}}}$ .

[2]

a.i.

$\overrightarrow {{\text{OB}}} \bullet \overrightarrow {{\text{OC}}}$ .

[1]

a.ii.

Given that ${\text{A}}\widehat {\text{O}}{\text{C}} = {\text{B}}\widehat {\text{O}}{\text{C}}$ , show that $k = 7$ .

[8]

Calculate the area of triangle ${\text{AOC}}$ .

[6]

Markscheme

correct substitution into either $\overrightarrow {{\text{OA}}} \bullet \overrightarrow {{\text{OC}}}$ or into $\overrightarrow {{\text{OB}}} \bullet \overrightarrow {{\text{OC}}}$ (in (ii)) (A1)

eg $- 2 \times \left( { - 1} \right) + 4 \times k$ , $6 \times \left( { - 1} \right) + 8 \times k$

correct expression A1 N1

eg $2 + 4k$ , $4k + 2$

[2 marks]

a.i.

correct expression A1 N1

eg $8k - 6$ , $- 6 + 8k$

[1 mark]

a.ii.

finding magnitudes (seen anywhere) A1A1

eg $\sqrt {{{\left( { - 2} \right)}^2} + {{\left( 4 \right)}^2} + {{\left( { - 4} \right)}^2}} \,\,\left( { = 6} \right)$ , $\sqrt {{{\left( 6 \right)}^2} + {{\left( 8 \right)}^2} + {0^2}} \,\,\left( { = 10} \right)$

correct substitution of their values into formula for angle ${\text{AOC}}$ (A1)

eg ${\text{cos}}\,\theta = \frac{{2 + 4k}}{{\sqrt {{{\left( { - 2} \right)}^2} + {{\left( 4 \right)}^2} + {{\left( { - 4} \right)}^2}} \left| {\overrightarrow {{\text{OC}}} } \right|}}$

correct substitution of their values into formula for angle ${\text{BOC}}$ (A1)

eg ${\text{cos}}\,\theta = \frac{{8k - 6}}{{\sqrt {{{\left( 6 \right)}^2} + {{\left( 8 \right)}^2} + {0^2}} \left| {\overrightarrow {{\text{OC}}} } \right|}}$

recognizing that ${\text{cos}}\,{\text{A}}\widehat {\text{O}}{\text{C}} = {\text{cos}}\,{\text{B}}\widehat {\text{O}}{\text{C}}$ (seen anywhere) (M1)

eg $\frac{{2 + 4k}}{{\left| {\overrightarrow {{\text{OC}}} } \right|\sqrt {{{\left( { - 2} \right)}^2} + {{\left( 4 \right)}^2} + {{\left( { - 4} \right)}^2}} }} = \frac{{8k - 6}}{{\left| {\overrightarrow {{\text{OC}}} } \right|\sqrt {{6^2} + {{\left( 8 \right)}^2} + {0^2}} }}$ , $\frac{{2 + 4k}}{{6\sqrt {1 + {k^2}} }} = \frac{{8k - 6}}{{10\sqrt {1 + {k^2}} }}$

correct working (without radicals) (A2)

eg $10\left( {2 + 4k} \right) = 6\left( {8k - 6} \right)$ , $11{k^2} - 79k + 14 = 0$

correct working clearly leading to the required answer A1

eg $20 + 36 = 48k$ , $56 = 8k$ , $k = 7$ and $k = \frac{2}{{11}}$ , $\left( {k - 7} \right)\left( {11k - 2} \right) = 0$

$k = 7$ AG N0

[8 marks]

finding magnitude of $\overrightarrow {{\text{OC}}}$ (seen anywhere) A1

eg $\sqrt {{{\left( { - 1} \right)}^2} + {7^2} + {0^2}}$ , $\sqrt {50}$

valid attempt to find ${\text{cos}}\,\theta$ (M1)

eg ${\text{cos}}\,\theta = \frac{{2 + 28}}{{6\sqrt {{{\left( { - 1} \right)}^2} + {7^2} + {0^2}} }}$ , ${\text{cos}}\,\theta = \frac{{56 - 6}}{{10\sqrt {{{\left( { - 1} \right)}^2} + {7^2} + {0^2}} }}$ , ${\left( {\sqrt {26} } \right)^2} = {6^2} + {\left( {\sqrt {50} } \right)^2} - 2\left( 6 \right)\sqrt {50} \,{\text{cos}}\,\theta$

finding ${\text{cos}}\,\theta$ A1

eg ${\text{cos}}\,\theta = \frac{5}{{\sqrt {50} }}\,\,\,\left( { = \frac{1}{{\sqrt 2 }}} \right)$

valid approach to find ${\text{sin}}\,\theta$ (seen anywhere) (M1)

eg $\theta = \frac{\pi }{4}$ , ${\text{sin}}\,\theta = {\text{cos}}\,\theta$ , ${\text{sin}}\,\theta = \sqrt {1 - \frac{{25}}{{50}}}$ , ${\text{sin}}\,\theta = \sqrt {1 - {\text{co}}{{\text{s}}^2}\,\theta }$ , ${\text{sin}}\,\theta = \frac{{\sqrt 2 }}{2}$

correct substitution of their values into $\frac{1}{2}ab\,{\text{sin}}\,{\text{C}}$ (A1)

eg $\frac{1}{2} \times 6 \times \sqrt {50}$ , $\frac{1}{2} \times 6 \times \sqrt {50} \times \frac{5}{{\sqrt {50} }}$

area is $15$ A1 N3

[6 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

The graph of $y = f (x)$ for $- 4 \leq x \leq 6$ is shown in the following diagram.

Write down the value of $f (2)$ .

[1]

a.i.

Write down the value of $(f \circ f) (2)$ .

[1]

a.ii.

Let $g (x) = \frac{1}{2} f (x) + 1$ for $- 4 \leq x \leq 6$ . On the axes above, sketch the graph of $g$ .

[3]

Markscheme

$f (2) = 6$ A1

[1 mark]

a.i.

$(f \circ f) (2) = - 2$ A1

[1 mark]

a.ii.

M1A1A1

Note: Award M1 for an attempt to apply any vertical stretch or vertical translation, A1 for a correct horizontal line segment between $- 4$ and $0$ (located roughly at $y = 3$ ),
A1 for a correct concave down parabola including max point at $(2, 4)$ and for correct end points at $(0, 3)$ and $(6, 0)$ (within circles). Points do not need to be labelled.

[3 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

Consider the points $A (- 2, 20)$ , $B (4, 6)$ and $C (- 14, 12)$ . The line $L$ passes through the point $A$ and is perpendicular to $[BC]$ .

Find the equation of $L$ .

[3]

The line $L$ passes through the point $(k, 2)$ .

Find the value of $k$ .

[2]

Markscheme

$m_{BC} = \frac{12 - 6}{- 14 - 4} (= - \frac{1}{3})$ (A1)

finding $m_{L} = \frac{- 1}{m_{BC}}$ using their $m_{BC}$ (M1)

$m_{L} = 3$

$y - 20 = 3 (x + 2), y = 3 x + 26$ A1

Note: Do not accept $L = 3 x + 26$

[3 marks]

substituting $(k, 2)$ into their $L$ (M1)

$2 - 20 = 3 (k + 2)$ OR $2 = 3 k + 26$

$k = - 8$ A1

[2 marks]

Examiners report

Finding the gradient of a line was well understood and many candidates also correctly found the perpendicular slope. Even with an error in their part (a), follow through marks in part (b) allowed many candidates to earn full marks for finding k despite their incorrect equation resulting in arithmetic of greater complexity.

[N/A]

The functions $f$ and $g$ are defined for $x \in ℝ$ by $f (x) = x - 2$ and $g (x) = a x + b$ , where $a, b \in ℝ$ .

Given that $(f \circ g) (2) = - 3$ and $(g \circ f) (1) = 5$ , find the value of $a$ and the value of $b$ .

Markscheme

$(f \circ g) (x) = a x + b - 2$ (M1)

$(f \circ g) (2) = - 3 \Rightarrow 2 a + b - 2 = - 3 (2 a + b = - 1)$ A1

$(g \circ f) (x) = a (x - 2) + b$ (M1)

$(g \circ f) (1) = 5 \Rightarrow - a + b = 5$ A1

a valid attempt to solve their two linear equations for $a$ and $b$ M1

so $a = - 2$ and $b = 3$ A1

[6 marks]

Examiners report

[N/A]

Consider the functions $f (x) = \sqrt{3} \sin x + \cos x$ where $0 \leq x \leq π$ and $g (x) = 2 x$ where $x \in ℝ$ .

Find $(f \circ g) (x)$ .

[2]

Solve the equation $(f \circ g) (x) = 2 \cos 2 x$ where $0 \leq x \leq π$ .

[5]

Markscheme

$(f \circ g) (x) = f (2 x)$ (A1)

$f (2 x) = \sqrt{3} \sin 2 x + \cos 2 x$ A1

[2 marks]

$\sqrt{3} \sin 2 x + \cos 2 x = 2 \cos 2 x$

$\sqrt{3} \sin 2 x = \cos 2 x$

recognising to use $\tan$ or $cot$ M1

$\tan 2 x = \frac{1}{\sqrt{3}}$ OR $cot 2 x = \sqrt{3}$ (values may be seen in right triangle) (A1)

$(arctan (\frac{1}{\sqrt{3}}) =) \frac{π}{6}$ (seen anywhere) (accept degrees) (A1)

$2 x = \frac{π}{6}, \frac{7 π}{6}$

$x = \frac{π}{12}, \frac{7 π}{12}$ A1A1

Note: Do not award the final A1 if any additional solutions are seen.
Award A1A0 for correct answers in degrees.
Award A0A0 for correct answers in degrees with additional values.

[5 marks]

Examiners report

Determining the composite function was very well done. In part (b) very few candidates showed any recognition that tan (or cot) were required to solve this trigonometric equation. Many saw the 2x and simply employed one of the double angle rules but could not then progress to an answer.

[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]

The function $f$ is of the form $f(x) = ax + b + \frac{c}{x}$ , where $a$ , $b$ and $c$ are positive integers.

Part of the graph of $y = f(x)$ is shown on the axes below. The graph of the function has its local maximum at $( - 2,{\text{ }} - 2)$ and its local minimum at $(2,{\text{ }}6)$ .

M17/5/MATSD/SP1/ENG/TZ1/12

Draw the line $y = - 6$ on the axes.

[1]

b.i.

Write down the number of solutions to $f(x) = - 6$ .

[1]

b.ii.

Find the range of values of $k$ for which $f(x) = k$ has no solution.

[2]

Markscheme

M17/5/MATSD/SP1/ENG/TZ1/21.b.i/M (A1) (C1)

Note: The command term “Draw” states: “A ruler (straight edge) should be used for straight lines”; do not accept a freehand $y = - 6$ line.

[1 mark]

b.i.

2 (A1)(ft) (C1)

Note: Follow through from part (b)(i).

[1 mark]

b.ii.

$- 2 < k < 6$ (A1)(A1) (C2)

Note: Award (A1) for both end points correct and (A1) for correct strict inequalities.

Award at most (A1)(A0) if the stated variable is different from $k$ or $y$ for example $- 2 < x < 6$ is (A1)(A0).

[2 marks]

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

Consider the vectors a = $\left( {\begin{array}{*{20}{c}} 3 \\ {2p} \end{array}} \right)$ and b = $\left( {\begin{array}{*{20}{c}} {p + 1} \\ 8 \end{array}} \right)$ .

Find the possible values of p for which a and b are parallel.

Markscheme

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

METHOD 1 (eliminating k)

recognizing parallel vectors are multiples of each other (M1)

eg a = kb, $\left( {\begin{array}{*{20}{c}} 3 \\ {2p} \end{array}} \right)$ = k $\left( {\begin{array}{*{20}{c}} {p + 1} \\ 8 \end{array}} \right)$ , $\frac{{p + 1}}{3} = \frac{8}{{2p}}$ , 3k = p + 1 and 2kp = 8

correct working (must be quadratic) (A1)

eg 2p² + 2p = 24, p² + p – 12, $3 = \frac{{{p^2} + p}}{4}$

valid attempt to solve their quadratic equation (M1)

eg factorizing, formula, completing the square

evidence of correct working (A1)

eg (p + 4)(p – 3), $x = \frac{{ - 2 \pm \sqrt {4 - 4\left( 2 \right)\left( { - 24} \right)} }}{4}$

p = –4, p = 3 A1A1 N4

METHOD 2 (solving for k)

recognizing parallel vectors are multiples of each other (M1)

eg a = kb, $\left( {\begin{array}{*{20}{c}} 3 \\ {2p} \end{array}} \right)$ = k $\left( {\begin{array}{*{20}{c}} {p + 1} \\ 8 \end{array}} \right)$ , 3k = p + 1 and 2kp = 8

correct working (must be quadratic) (A1)

eg 3k² – k = 4, 3k² – k – 4, 4k² = 3 – k

one correct value for k (A1)

eg k = –1, k = $\frac{4}{3}$ , k = $\frac{3}{4}$

substituting their value(s) of k (M1)

eg $\left( {\begin{array}{*{20}{c}} 3 \\ {2p} \end{array}} \right) = \frac{3}{4}\left( {\begin{array}{*{20}{c}} {p + 1} \\ 8 \end{array}} \right)$ , $3\left( {\frac{4}{3}} \right) = p + 1$ and $2\left( {\frac{4}{3}} \right)p = 8$ , $\left( { - 1} \right)\left( {\begin{array}{*{20}{c}} 3 \\ {2p} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {p + 1} \\ 8 \end{array}} \right)$

p = –4, p = 3 A1A1 N4

METHOD 3 (working with angles and cosine formula)

recognizing angle between parallel vectors is 0 and/or 180° M1

eg cos θ = ±1, $a \bullet b = \left| a \right|\left| b \right|$

correct substitution of scalar product and magnitudes into equation (A1)

eg $\frac{{3\left( {p + 1} \right) + 2p\left( 8 \right)}}{{\sqrt {{3^2} + {{\left( {2p} \right)}^2}} \sqrt {{{\left( {p + 1} \right)}^2} + {8^2}} }} = \pm 1$ , $19p + 3 = \sqrt {4{p^2} + 9} \sqrt {{p^2} + 2p + 65}$

correct working (must include both ± ) (A1)

eg $3\left( {p + 1} \right) + 2p\left( 8 \right) = \pm \sqrt {{3^2} + {{\left( {2p} \right)}^2}} \sqrt {{{\left( {p + 1} \right)}^2} + {8^2}}$ , $19p + 3 = \pm \sqrt {4{p^2} + 9} \sqrt {{p^2} + 2p + 65}$

correct quartic equation (A1)

eg $361\,{p^2} + 114p + 9 = 4{p^4} + 8{p^3} + 269{p^2} + 18p + 585$ , $4{p^4} + 8{p^3} - 92{p^2} - 96p + 576 = 0$ , ${p^4} + 2{p^3} - 23{p^2} - 24p + 144 = 0$ , ${\left( {p + 4} \right)^2}{\left( {p - 3} \right)^2} = 0$

p = –4, p = 3 A2 N4

[6 marks]

Examiners report

[N/A]

Let $f(x) = {x^2} - 4x + 5$ .

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

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

(ii) Find the value of $k$ .

Markscheme

(i) $h = 2$ A1 N1

(ii) METHOD 1

valid attempt to find $k$ (M1)

eg $\,\,\,\,\,$ $f(2)$

correct substitution into their function (A1)

eg $\,\,\,\,\,$ ${(2)^2} - 4(2) + 5$

$k = 1$ A1 N2

METHOD 2

valid attempt to complete the square (M1)

eg $\,\,\,\,\,$ ${x^2} - 4x + 4$

correct working (A1)

eg $\,\,\,\,\,$ $({x^2} - 4x + 4) - 4 + 5,{\text{ }}{(x - 2)^2} + 1$

$k = 1$ A1 N2

[4 marks]

Examiners report

[N/A]

Consider the function $f$ , with derivative $f'\left( x \right) = 2{x^2} + 5kx + 3{k^2} + 2$ where $x{\text{, }}k \in \mathbb{R}$ .

Show that the discriminant of $f'\left( x \right)$ is ${k^2} - 16$ .

[2]

Given that $f$ is an increasing function, find all possible values of $k$ .

[4]

Markscheme

correct substitution into ${b^2} - 4ac$ (A1)

eg ${\left( {5k} \right)^2} - 4\left( 2 \right)\left( {3{k^2} + 2} \right)$ , ${\left( {5k} \right)^2} - 8\left( {3{k^2} + 2} \right)$

correct expansion of each term A1

eg $25{k^2} - 24{k^2} - 16$ , $25{k^2} - \left( {24{k^2} + 16} \right)$

${k^2} - 16$ AG N0

[2 marks]

valid approach M1

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

recognizing discriminant $< 0$ or $\leqslant 0$ M1

eg $D < 0$ , ${k^2} - 16 \leqslant 0$ , ${k^2} < 16$

two correct values for $k$ /endpoints (even if inequalities are incorrect) (A1)

eg $k = \pm 4$ , $k < - 4$ and $k > 4$ , $\left| k \right| < 4$

correct interval A1 N2

eg $- 4 < k < 4$ , $- 4 \leqslant k \leqslant 4$

Note: Candidates may work with an equation, then write the intervals with inequalities at the end. If inequalities are not seen until the candidate’s final correct answer, M0M0A1A1 may be awarded.
If candidate is working with incorrect inequalitie(s) at the beginning, then gets the correct final answer, award M0M0A1A0 or M1M0A1A0 or M0M1A1A0 in line with the markscheme.

[4 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]