Write down the value of $a$.

Find the value of $b$.

The outer dome of a large cathedral has the shape of a hemisphere of diameter 32 m, supported by vertical walls of height 17 m. It is also supported by an inner dome which can be modelled by rotating the curve $y=33-0.08x^3$ through 360° about the $y$-axis between $y$ = 0 and $y$ = 33, as indicated in the diagram.

Find the volume of the space between the two domes.

$a$ = 33    A1

[1 mark]

$\frac{1}{\sqrt[3]{0.08}}=2.32$     M1A1

[2 marks]

volume within outer dome

$\frac{2}{3}\pi +{16}^3+\pi \times {16}^2\times 17=22250.85$      M1A1

volume within inner dome

$\pi {\int }_0^{33}{\left(\frac{33-y}{0.08}\right)}^{\frac{2}{3}}\text{d}y=3446.92$       M1A1

volume between = 22 250.85 − 3446.92 = 18 803.93 m³       A1

[5 marks]

Find $g\left(0\right)$.

On the same set of axes draw the graph of $y=g\left(x\right)$, showing any intercepts and asymptotes.

$g\left(0\right)=16$              M1A1

[2 marks]

$y$-asymptote $\left(y=4\right)$            A1

concave up decreasing curve and passing through $(0, 16)$            A1

[2 marks]

This question was not particularly well done. Candidates often failed to apply the transformation of the function correctly and did not understand how to use the algebra of graphical transformations. Others applied the geometry of stretches and translations, often incorrectly. Even if the graph was drawn correctly, some candidates failed to follow the instruction to show the asymptote.

This question was not particularly well done. Candidates often failed to apply the transformation of the function correctly and did not understand how to use the algebra of graphical transformations. Others applied the geometry of stretches and translations, often incorrectly. Even if the graph was drawn correctly, some candidates failed to follow the instruction to show the asymptote.

Find the value of $a$.

Find the value of $b$.

Find the average number of earthquakes in a year with a magnitude of at least $7.2$.

Find $\text{P}(Y>100)$.

${\log }_{10} 100=a-3$        (M1)

$a=5$             A1

[2 marks]

EITHER

$N={10}^{5-M}$        (M1)

$=\frac{{10}^5}{{10}^M}\left(=\frac{100000}{{10}^M}\right)$

OR

$100=\frac{b}{{10}^3}$        (M1)

THEN

$b=100000 \left(={10}^5\right)$             A1

[2 marks]

$N=\frac{{10}^5}{{10}^{7.2}}=0.00631 \left(0.0063095\dots \right)$           A1

Note: Do not accept an answer of ${10}^{-2.2}$.

[1 mark]

METHOD 1

$Y>100\Rightarrow$no earthquakes in the first $100$ years             (M1)

EITHER

let $X$ be the number of earthquakes of at least magnitude $7.2$ in a year

$X~\text{Po}\left(0.0063095\dots \right)$

${\left(\text{P}\left(X=0\right)\right)}^{100}$             (M1)

OR

let $X$ be the number of earthquakes in $100$ years

$X~\text{Po}\left(0.0063095\dots \times 100\right)$             (M1)

$\text{P}\left(X=0\right)$

THEN

$0.532 \left(0.532082\dots \right)$           A1

METHOD 2

$Y>100\Rightarrow$no earthquakes in the first $100$ years             (M1)

let $X$ be the number of earthquakes in $100$ years

since $n$ is large and $p$ is small

$X~\text{B}\left(100, 0.0063095\dots \right)$             (M1)

$\text{P}\left(X=0\right)$

$0.531 \left(0.531019\dots \right)$           A1

[3 marks]

Parts (a), (b), and (c) were accessible to many candidates who earned full marks with the manipulation of logs and indices presenting no problems. Part (d), however, proved to be too difficult for most and very few correct attempts were seen. As in question 9, most candidates relied on calculator notation when using the Poisson distribution. The discipline of defining a random variable in terms of its distribution and parameters helps to conceptualize the problem in terms that aid a better understanding. Most candidates who attempted this question blindly entered values into the Poisson distribution calculator and were unable to earn any marks. There were a couple of correct solutions using a binomial distribution to approximate the given quantity.

The rate, $A$, of a chemical reaction at a fixed temperature is related to the concentration of two compounds, $B$ and $C$, by the equation

$A=k{B}^x{C}^y$, where $x$, $y$, $k\in \mathbb{R}$.

A scientist measures the three variables three times during the reaction and obtains the following values.

Find $x$, $y$ and $k$.

$\text{log}A=x\text{log}B+y\text{log}C+\text{log}k$         (M1)

$\text{log}5.74=x\text{log}2.1+y\text{log}3.4+\text{log}k$

$\text{log}2.88=x\text{log}1.5+y\text{log}2.4+\text{log}k$

$\text{log}0.980=x\text{log}0.8+y\text{log}1.9+\text{log}k$        M1A1

Note: Allow any consistent base, allow numerical equivalents.

attempting to solve their system of equations       (M1)

$x$ = 1.53,  $y$ = 0.505     A1

$k$ = 0.997     A1

[6 marks]

Find the inverse function $f^{-1}$, stating its domain.

Express $g\left(x\right)$ in the form $A+\frac{B}{x-2}$ where A, B are constants.

Sketch the graph of $y=g\left(x\right)$. State the equations of any asymptotes and the coordinates of any intercepts with the axes.

The function $h$ is defined by $h\left(x\right)=\sqrt{x}$, for $x$ ≥ 0.

State the domain and range of $h\circ g$.

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

attempt to make $x$ the subject of $y=\frac{ax+b}{cx+d}$      M1

$y\left(cx+d\right)=ax+b$      A1

$x=\frac{dy-b}{a-cy}$     A1

$f^{-1}\left(x\right)=\frac{dx-b}{a-cx}$     A1

Note: Do not allow $y=$ in place of $f^{-1}\left(x\right)$.

$x\ne \frac{a}{c},\left(x\in \mathbb{R}\right)$     A1

Note: The final A mark is independent.

[5 marks]

$g\left(x\right)=2+\frac{1}{x-2}$     A1A1

[2 marks]

hyperbola shape, with single curves in second and fourth quadrants and third quadrant blank, including vertical asymptote $x=2$     A1

horizontal asymptote $y=2$     A1

intercepts $\left(\frac{3}{2},0\right),\left(0,\frac{3}{2}\right)$     A1

[3 marks]

the domain of $h\circ g$ is $x⩽\frac{3}{2},x>2$     A1A1

the range of $h\circ g$ is $y⩾0,y\ne \sqrt{2}$     A1A1

[4 marks]

Use the data in the second table to find the value of $m$ and the value of $b$ for the regression line, $\ln x=m(\ln d)+b$.

Assuming that the model found in part (a) remains valid, estimate the percentage of trees in stock when $d=25$.

$m=-0.695 \left(-0.695383\dots \right); b=4.63 \left(4.62974\dots \right)$                  A1A1

[2 marks]

$\ln x=-0.695\left(\ln 25\right)+4.63$                  M1

$\ln x=2.39288\dots$                  (A1)

$x=10.9\%$                  A1

[3 marks]

Those candidates who did this question were often successful. There were a number, however, who found an equation of a line through two of the points instead of using their technology to find the equation of the regression line. A common problem was to introduce rounding errors at various stages throughout the problem. Some candidates failed to find the value of $x$ from that of $\ln x$.

Those candidates who did this question were often successful. There were a number, however, who found an equation of a line through two of the points instead of using their technology to find the equation of the regression line. A common problem was to introduce rounding errors at various stages throughout the problem. Some candidates failed to find the value of $x$ from that of $\ln x$.

Write down the range of $f$.

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

Write down the domain and range of $f^{-1}$.

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

$-11⩽f(x)⩽21$     A1A1

Note:     A1 for correct end points, A1 for correct inequalities.

[2 marks]

$f^{-1}(x)=\sqrt[3]{\frac{x-5}{2}}$     (M1)A1

[2 marks]

$-11⩽x⩽21,-2⩽f^{-1}(x)⩽2$     A1A1

[2 marks]

Express $x^2+3x+2$ in the form $(x+h{)}^2+k$.

Factorize $x^2+3x+2$.

Sketch the graph of $f(x)$, indicating on it the equations of the asymptotes, the coordinates of the $y$-intercept and the local maximum.

Show that $\frac{1}{x+1}-\frac{1}{x+2}=\frac{1}{x^2+3x+2}$.

Hence find the value of $p$ if ${\int }_0^{1}f(x)\text{d}x=\ln (p)$.

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

Determine the area of the region enclosed between the graph of $y=f\left(\left|x\right|\right)$, the $x$-axis and the lines with equations $x=-1$ and $x=1$.

$x^2+3x+2={\left(x+\frac{3}{2}\right)}^2-\frac{1}{4}$     A1

[1 mark]

$x^2+3x+2=(x+2)(x+1)$     A1

[1 mark]

A1 for the shape

A1 for the equation $y=0$

A1 for asymptotes $x=-2$ and $x=-1$

A1 for coordinates $\left(-\frac{3}{2},-4\right)$

A1 $y$-intercept $\left(0,\frac{1}{2}\right)$

[5 marks]

$\frac{1}{x+1}-\frac{1}{x+2}=\frac{(x+2)-(x+1)}{(x+1)(x+2)}$     M1

$=\frac{1}{x^2+3x+2}$     AG

[1 mark]

$\underset{0}{\overset{1}{\int }}\frac{1}{x+1}-\frac{1}{x+2}\text{d}x$

$={\left[\ln (x+1)-\ln (x+2)\right]}_0^1$     A1

$=\ln 2-\ln 3-\ln 1+\ln 2$     M1

$=\ln \left(\frac{4}{3}\right)$     M1A1

$\therefore p=\frac{4}{3}$

[4 marks]

symmetry about the $y$-axis     M1

correct shape     A1

Note:     Allow FT from part (b).

[2 marks]

$2{\int }_0^{1}f(x)\text{d}x$     (M1)(A1)

$=2\ln \left(\frac{4}{3}\right)$     A1

Note:     Do not award FT from part (e).

[3 marks]

Write down the $x$-coordinate of the point of inflexion on the graph of $y=f\left(x\right)$.

find the value of $f\left(1\right)$.

find the value of $f\left(4\right)$.

Sketch the curve $y=f\left(x\right)$, 0 ≤ $x$ ≤ 5 indicating clearly the coordinates of the maximum and minimum points and any intercepts with the coordinate axes.

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

3     A1

[1 mark]

attempt to use definite integral of $f^{\prime }\left(x\right)$        (M1)

${\int }_0^1{f}^{\prime }\left(x\right)\text{d}x=0.5$

$f\left(1\right)-f\left(0\right)=0.5$        (A1)

$f\left(1\right)=0.5+3$

= 3.5      A1

[3 marks]

${\int }_1^4{f}^{\prime }\left(x\right)\text{d}x=-2.5$       (A1)

Note: (A1) is for −2.5.

$f\left(4\right)-f\left(1\right)=-2.5$

$f\left(4\right)=3.5-2.5$

= 1      A1

[2 marks]

    A1A1A1

A1 for correct shape over approximately the correct domain
A1 for maximum and minimum (coordinates or horizontal lines from 3.5 and 1 are required),
A1 for $y$-intercept at 3

[3 marks]

Estimate the value of Roger’s laptop after $5$ years.

Find the value of $k$.

Comment on the validity of your answer to part (b).

$\pounds 495\times 0.9^5=\pounds 292 (\pounds 292.292\dots )$        (M1)A1 

[2 marks]

$\pounds 495\times 0.9^k=2200\times 0.{85}^k$        (M1)

$k=26.1 \left(26.0968\dots \right)$        A1 

Note: Award M1A0 for $k-1$ in place of $k$.

[2 marks]

depreciation rates unlikely to be constant (especially over a long time period)        R1

Note: Accept reasonable answers based on the magnitude of $k$ or the fact that “value” depends on factors other than time.

[1 mark]

For $a=-\frac{\pi }{2}$, sketch the graph of $y=g\left(x\right)$. Indicate clearly the maximum and minimum values of the function.

Write down the least value of $a$ such that $g$ has an inverse.

For the value of $a$ found in part (b), write down the domain of $g^{-1}$.

For the value of $a$ found in part (b), find an expression for $g^{-1}\left(x\right)$.

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

concave down and symmetrical over correct domain       A1

indication of maximum and minimum values of the function (correct range)       A1A1

[3 marks]

$a$ = 0      A1

Note: Award A1 for $a$ = 0 only if consistent with their graph.

[1 mark]

$1⩽x⩽5$     A1

Note: Allow FT from their graph.

[1 mark]

$y=4\text{cos}x+1$

$x=4\text{cos}y+1$

$\frac{x-1}{4}=\text{cos}y$      (M1)

$\Rightarrow y=\text{arccos}\left(\frac{x-1}{4}\right)$

$\Rightarrow g^{-1}\left(x\right)=\text{arccos}\left(\frac{x-1}{4}\right)$      A1

[2 marks]

Find an expression for $f^{-1}\left(x\right)$. You are not required to state a domain.

Solve $f\left(x\right)=f^{-1}\left(x\right)$.

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

an attempt to isolate $x$ (or $y$ if switched)         (M1)

${\text{e}}^y=\frac{1}{x-2}$

$x-2={\text{e}}^{-y}$

$x={\text{e}}^{-y}+2$

switching $x$ and $y$ (seen anywhere)          M1

$f^{-1}\left(x\right)={\text{e}}^{-x}+2$          A1

[3 marks]

sketch of $f\left(x\right)$ and $f^{-1}\left(x\right)$          (M1)

$x=2.12 \left(2.12002\dots \right)$          A1

[2 marks]

This question was well answered by most candidates.

Explain why this graph indicates that $P$ is inversely proportional to $d^n$.

Find the equation of the least squares regression line of ${\log }_{10} P$ against ${\log }_{10} d$.

Use your answer to part (b) to write down the value of $n$ to the nearest integer.

Find an expression for $P$ in terms of $d$.

* 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.

a straight line with a negative gradient      A1A1

[2 marks]

$\log P=-2.040\dots \log d-0.12632\dots \approx -2.04 \log d-0.126$     A1A1

Note: A1 for each correct term.

[2 marks]

$n=2$     A1

[1 mark]

$P={10}^{-0.126\dots }{d}^{-2}$          (M1)

$\approx 0.748d^{-2}$          A1

[2 marks]

Find the remainder when $p\left(x\right)$ is divided by $\left(x-2\right)$.

Find the remainder when $p\left(x\right)$ is divided by $\left(x-3\right)$.

Prove that $p\left(x\right)$ has only one real zero.

Write down the transformation that will transform the graph of $y=p\left(x\right)$ onto the graph of $y=8x^3-12x^2+16x-24$.

The random variable $X$ follows a Poisson distribution with a mean of $\lambda$ and $6\text{P}\left(X=3\right)=3\text{P}\left(X=2\right)-2\text{P}\left(X=1\right)+3\text{P}\left(X=0\right)$.

Find the value of $\lambda$.

$p\left(2\right)=8-12+16-24$       (M1)

Note: Award M1 for a valid attempt at remainder theorem or polynomial division.

= −12     A1

remainder = −12

[2 marks]

$p\left(3\right)=27-27+24-24$ = 0      A1 

remainder = 0

[1 mark]

$x=3$ (is a zero)     A1

Note: Can be seen anywhere.

EITHER

factorise to get $\left(x-3\right)\left(x^2+8\right)$      (M1)A1

$x^2+8\ne 0$ (for $x\in \mathbb{R}$) (or equivalent statement)      R1

Note: Award R1 if correct two complex roots are given.

OR

$p^{\prime }\left(x\right)=3x^2-6x+8$    A1

attempting to show $p^{\prime }\left(x\right)\ne 0$       M1

eg discriminant = 36 – 96 < 0, completing the square

no turning points       R1

THEN

only one real zero (as the curve is continuous)      AG

[4 marks]

new graph is $y=p\left(2x\right)$     (M1)

stretch parallel to the $x$-axis (with $x=0$ invariant), scale factor 0.5    A1

Note: Accept “horizontal” instead of “parallel to the $x$-axis”.

[2 marks]

$\frac{6{\lambda }^3{e}^{-\lambda }}{6}=\frac{3{\lambda }^2{e}^{-\lambda }}{2}-2\lambda e^{-\lambda }+3e^{-\lambda }$     M1A1

Note: Allow factorials in the denominator for A1.

$2{\lambda }^3-3{\lambda }^2+4\lambda -6=0$    A1

Note: Accept any correct cubic equation without factorials and $e^{-\lambda }$.

EITHER

$4\left(2{\lambda }^3-3{\lambda }^2+4\lambda -6\right)=8{\lambda }^3-12{\lambda }^2+16\lambda -24=0$       (M1)

$2\lambda =3$      (A1)

OR

$\left(2\lambda -3\right)\left({\lambda }^2+2\right)=0$       (M1)(A1)

THEN

$\lambda$ = 1.5    A1

[6 marks]

Find the equation of the straight line, giving your answer in the form $\text{ln}m=ap+b$, where $a\text{,}b\in \mathbb{R}$.

a formula for $m$ in terms of $p$.

the value of $m$ when $p=0$.

gradient = −1.4              A1

$\text{ln}m-7.3=-1.4\left(p-2.1\right)$       M1

$\text{ln}m=-1.4p+10.24$              A1

[3 marks]

$m=e^{-1.4p+10.24}\left(=28000e^{-1.4p}\right)$            A1

[1 mark]

28000            A1

[1 mark]

Find the range of $f$.

Find an expression for the inverse function$f^{-1}\left(x\right)$. The domain is not required.

Write down the range of $f^{-1}\left(x\right)$.

$\left(f\left(-7\right)=\right) 8$ and $\left(f\left(7\right)=\right) 1$           (A1) 

range is $f\left(x\right)\le 1, f\left(x\right)\ge 8$           A1A1 

Note: Award at most A1A1A0 if strict inequalities are used.

[3 marks]

interchanging $x, y$ at any stage           (A1) 

$y=2-\frac{12}{x+5}$

$\frac{12}{x+5}=2-y$

$\frac{12}{2-y}=x+5$          (A1) 

$\frac{12}{2-y}-5=x$

$\left( f^{-1}\left(x\right)=\right) \frac{12}{2-x}-5 \left(=\frac{2+5x}{2-x}\right)$          A1

[3 marks]

range is $-7\le f^{-1}\left(x\right)\le 7, f^{-1}\left(x\right)\ne -5$          A1

[1 mark]

Describe the transformation by which $f\left(x\right)$ is transformed to $g\left(x\right)$.

State the range of $g$.

Sketch the graphs of $y=f\left(x\right)$ and $y=g\left(x\right)$ on the same axes, clearly stating the points of intersection with any axes.

Find the coordinates of P.

The tangent to $y=f\left(x\right)$ at P passes through the origin (0, 0).

Determine the value of $k$.

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

translation $k$ units to the left (or equivalent)     A1

[1 mark]

range is $\left(g\left(x\right)\in \right)\mathbb{R}$     A1

[1 mark]

correct shape of $y=f\left(x\right)$       A1

their $f\left(x\right)$ translated $k$ units to left (possibly shown by $x=-k$ marked on $x$-axis)       A1

asymptote included and marked as $x=-k$       A1

$f\left(x\right)$ intersects $x$-axis at $x=-1$, $x=1$       A1

$g\left(x\right)$ intersects $x$-axis at $x=-k-1$, $x=-k+1$       A1

$g\left(x\right)$ intersects $y$-axis at $y=\text{ln}k$       A1

Note: Do not penalise candidates if their graphs “cross” as $x\to \pm \mathrm{\infty }$.

Note: Do not award FT marks from the candidate’s part (a) to part (c).

[6 marks]

at P  $\text{ln}\left(x+k\right)=\text{ln}\left(-x\right)$

attempt to solve $x+k=-x$ (or equivalent)       (M1)

$x=-\frac{k}{2}\Rightarrow y=\text{ln}\left(\frac{k}{2}\right)$  (or $y=\text{ln}\left|\frac{k}{2}\right|$)       A1

P$\left(-\frac{k}{2},\text{ln}\frac{k}{2}\right)$  (or P$\left(-\frac{k}{2},\text{ln}\left|\frac{k}{2}\right|\right)$)

[2 marks]

attempt to differentiate $\text{ln}\left(-x\right)$ or $\text{ln}\left|x\right|$       (M1)

$\frac{\text{d}y}{\text{d}x}=\frac{1}{x}$       A1

at P, $\frac{\text{d}y}{\text{d}x}=\frac{-2}{k}$       A1

recognition that tangent passes through origin $\Rightarrow \frac{y}{x}=\frac{\text{d}y}{\text{d}x}$       (M1)

$\frac{\text{ln}\left|\frac{k}{2}\right|}{-\frac{k}{2}}=\frac{-2}{k}$       A1

$\text{ln}\left(\frac{k}{2}\right)=1$       (A1)

$\Rightarrow k=2\text{e}$       A1

Note: For candidates who explicitly differentiate $\text{ln}\left(x\right)$ (rather than $\text{ln}\left(-x\right)$ or $\text{ln}\left|x\right|$, award M0A0A1M1A1A1A1.

[7 marks]

The graph of the function $f\left(x\right)=\ln x$ is translated by $\left(\begin{array}{c}a\\ b\end{array}\right)$ so that it then passes through the points $(0, 1)$ and $({\text{e}}^3, 1+\ln 2)$ .

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

new function is $f\left(x-a\right)+b\left(=\ln \left(x-a\right)+b\right)$            (M1)

$f\left(0\right)=\ln \left(-a\right)+b=1$           A1

$f\left({\text{e}}^3\right)=\ln \left({\text{e}}^3-a\right)+b=1+\ln 2$           A1

$\ln \left(-a\right)=\ln \left({\text{e}}^3-a\right)-\ln 2$            (M1)

$\ln \left(-a\right)=\ln \left(\frac{{\text{e}}^3-a}{2}\right)$

$-a=\frac{{\text{e}}^3-a}{2}$

$-2a={\text{e}}^3-a$

$a=-{\text{e}}^3\left(=-20.0855\dots \right)$           A1

$b=1-\ln {\text{e}}^3=1-3=-2$            (M1)A1

[7 marks]

Show that $\text{ln}\left(T-25\right)=bt+\text{ln}a$.

Find the equation of the regression line of $\text{ln}\left(T-25\right)$ on $t$.

find the value of $a$ and of $b$.

predict the temperature of the metal rod after 3 minutes.

$\text{ln}\left(T-25\right)=\text{ln}\left(a{\text{e}}^{bt}\right)$       M1

$\text{ln}\left(T-25\right)=\text{ln}a+\text{ln}\left({\text{e}}^{bt}\right)$            A1

$\text{ln}\left(T-25\right)=bt+\text{ln}a$            AG

[2 marks]

$\text{ln}\left(T-25\right)=-0.00870t+3.89$      M1A1A1

[3 marks]

$b=-0.00870$       A1

$a=e^{\mathrm{3.89...}}=49.1$     M1A1

[3 marks]

$T\left(180\right)=49.1e^{-0.00870\left(180\right)}+25=35.2$    M1A1

[2 marks]

Consider the function $f\left(x\right)=x^4-6x^2-2x+4$, $x\in \mathbb{R}$.

The graph of $f$ is translated two units to the left to form the function $g\left(x\right)$.

Express $g\left(x\right)$ in the form $a{x}^4+b{x}^3+c{x}^2+dx+e$ where $a$, $b$, $c$, $d$, $e\in \mathbb{Z}$.

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

$g\left(x\right)=f\left(x+2\right)\left(={\left(x+2\right)}^4-6{\left(x+2\right)}^2-2\left(x+2\right)+4\right)$      M1

attempt to expand ${\left(x+2\right)}^4$      M1

${\left(x+2\right)}^4=x^4+4\left(2x^3\right)+6\left(2^2{x}^2\right)+4\left(2^{3}x\right)+2^4$       (A1)

$=x^4+8x^3+24x^2+32x+16$      A1

$g\left(x\right)=x^4+8x^3+24x^2+32x+16-6\left(x^2+4x+4\right)-2x-4+4$

$=x^4+8x^3+18x^2+6x-8$      A1

Note: For correct expansion of $f\left(x-2\right)=x^4-8x^3+18x^2-10x$ award max  M0M1(A1)A0A1.

[5 marks]

Sketch the graph of $y=\frac{1-3x}{x-2}$, showing clearly any asymptotes and stating the coordinates of any points of intersection with the axes.

Hence or otherwise, solve the inequality .

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

correct vertical asymptote     A1

shape including correct horizontal asymptote     A1

$\left(\frac{1}{3},0\right)$     A1

$\left(0,-\frac{1}{2}\right)$     A1

Note:     Accept $x=\frac{1}{3}$ and $y=-\frac{1}{2}$ marked on the axes.

[4 marks]

METHOD 1

$\frac{1-3x}{x-2}=2$     (M1)

$\Rightarrow x=1$    A1

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

Note:     Award this M1 for the line above or a correct sketch identifying a second critical value.

$\Rightarrow x=-3$     A1

solution is     A1

METHOD 2

    (M1)A1

    A1

    (M1)

solution is     A1

METHOD 3

consider     (M1)

Note:     Also allow consideration of “>” or “=” for the awarding of the M mark.

recognition of critical value at $x=1$     A1

consider     (M1)

Note:     Also allow consideration of “>” or “=” for the awarding of the M mark.

recognition of critical value at $x=-3$     A1

solution is     A1

[5 marks]

Sketch the graphs on the same set of axes.

Given that the graphs enclose a region of area 18 square units, find the value of b.

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

graphs sketched correctly (condone missing b)     A1A1

[2 marks]

$\frac{b^2}{2}=18$     (M1)A1

$b=6$     A1

[3 marks]

The graph of $y=f\left(x\right)$ has a local maximum at A. Find the coordinates of A.

Show that there is exactly one point of inflexion, B, on the graph of $y=f\left(x\right)$.

The coordinates of B can be expressed in the form B$\left(2^a,b\times 2^{-3a}\right)$ where a, b$\in \mathbb{Q}$. Find the value of a and the value of b.

Sketch the graph of $y=f\left(x\right)$ showing clearly the position of the points A and B.

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

attempt to differentiate      (M1)

$f^{\prime }\left(x\right)=-3x^{-4}-3x$     A1

Note: Award M1 for using quotient or product rule award A1 if correct derivative seen even in unsimplified form, for example $f^{\prime }\left(x\right)=\frac{-15x^4\times 2x^3-6x^2\left(2-3x^5\right)}{{\left(2x^3\right)}^2}$.

$-\frac{3}{x^4}-3x=0$     M1

$\Rightarrow x^5=-1\Rightarrow x=-1$     A1

$\text{A}\left(-1,-\frac{5}{2}\right)$     A1

[5 marks]

$f^{″}\left(x\right)=0$     M1

$f^{″}\left(x\right)=12x^{-5}-3\left(=0\right)$     A1

Note: Award A1 for correct derivative seen even if not simplified.

$\Rightarrow x=\sqrt[5]{4}\left(=2^{\frac{2}{5}}\right)$     A1

hence (at most) one point of inflexion      R1

Note: This mark is independent of the two A1 marks above. If they have shown or stated their equation has only one solution this mark can be awarded.

$f^{″}\left(x\right)$ changes sign at $x=\sqrt[5]{4}\left(=2^{\frac{2}{5}}\right)$      R1

so exactly one point of inflexion

[5 marks]

$x=\sqrt[5]{4}=2^{\frac{2}{5}}\left(\Rightarrow a=\frac{2}{5}\right)$      A1

$f\left(2^{\frac{2}{5}}\right)=\frac{2-3\times 2^2}{2\times 2^{\frac{6}{5}}}=-5\times 2^{-\frac{6}{5}}\left(\Rightarrow b=-5\right)$     (M1)A1

Note: Award M1 for the substitution of their value for $x$ into $f\left(x\right)$.

[3 marks]

A1A1A1A1

A1 for shape for x < 0
A1 for shape for x > 0
A1 for maximum at A
A1 for POI at B.

Note: Only award last two A1s if A and B are placed in the correct quadrants, allowing for follow through.

[4 marks]

state the value of $a$ and the value of $c$;

find the value of $b$.

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

$a=1$    A1

$c=3$    A1

[2 marks]

use the coordinates of $(1,0)$ on the graph     M1

$f(1)=0\Rightarrow 1+\frac{b}{1-3}=0\Rightarrow b=2$    A1

[2 marks]

Sketch the graphs of $y=\frac{x}{2}+1$ and $y=\left|x-2\right|$ on the following axes.

Solve the equation $\frac{x}{2}+1=\left|x-2\right|$.

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

straight line graph with correct axis intercepts      A1

modulus graph: V shape in upper half plane      A1

modulus graph having correct vertex and y-intercept      A1

[3 marks]

METHOD 1

attempt to solve $\frac{x}{2}+1=x-2$     (M1)

$x=6$      A1

Note: Accept $x=6$ using the graph.

attempt to solve (algebraically) $\frac{x}{2}+1=2-x$     M1

$x=\frac{2}{3}$     A1

[4 marks]

METHOD 2

${\left(\frac{x}{2}+1\right)}^2={\left(x-2\right)}^2$      M1

$\frac{x^2}{4}+x+1=x^2-4x+4$

$0=\frac{3x^2}{4}-5x+3$

$3x^2-20x+12=0$

attempt to factorise (or equivalent)       M1

$\left(3x-2\right)\left(x-6\right)=0$

$x=\frac{2}{3}$     A1

$x=6$      A1

[4 marks]

A function $f$ is of the form $f\left(t\right)=p{\text{e}}^{q \cos \left(rt\right)}, p, q, r\in {\mathrm{ℝ}}^{+}$. Part of the graph of $f$ is shown.

The points $\text{A}$ and $\text{B}$ have coordinates $\text{A}(0, 6.5)$ and $\text{B}(5.2, 0.2)$, and lie on $f$.

The point $\text{A}$ is a local maximum and the point $\text{B}$ is a local minimum.

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

substitute coordinates of $\text{A}$

$f\left(0\right)=p{\text{e}}^{q \cos \left(0\right)}=6.5$

$6.5=p{\text{e}}^q$             (A1)

substitute coordinates of $\text{B}$

$f\left(5.2\right)=p{\text{e}}^{q \cos \left(5.2r\right)}=0.2$

EITHER

$f\text{'}\left(t\right)=-pqr \sin \left(rt\right){\text{e}}^{q \cos \left(rt\right)}$             (M1)

minimum occurs when $-pqr \sin \left(5.2r\right){\text{e}}^{q \cos \left(5.2r\right)}=0$

$\sin \left(rt\right)=0$

$r\times 5.2=\mathrm{\pi }$             (A1)

OR

minimum value occurs when $\cos \left(rt\right)=-1$             (M1)

$r\times 5.2=\mathrm{\pi }$             (A1)

OR

period $=2\times 5.2=10.4$             (A1)

$r=\frac{2\mathrm{\pi }}{10.4}$             (M1)

THEN

$r=\frac{\mathrm{\pi }}{5.2}=0.604152\dots \left(0.604\right)$             A1

$0.2=p {\text{e}}^{-q}$             (A1)

eliminate $p$ or $q$             (M1)

${\text{e}}^{2q}=\frac{6.5}{0.2}$   OR   $0.2=\frac{p^2}{6.5}$

$q=1.74 \left(1.74062\dots \right)$             A1

$p=1.14017\dots \left(1.14\right)$             A1

[8 marks]

This was a challenging question and suitably positioned at the end of the examination. Candidates who attempted it were normally able to substitute points A and B into the given equation. Some were able to determine the first derivative. Only a few candidates were able to earn significant marks for this question.

It is believed that two variables, $v$ and $w$ are related by the equation $v=k{w}^n$, where $k\text{,}n\in \mathbb{R}$.  Experimental values of $v$ and $w$ are obtained. A graph of $\text{ln}v$ against $\text{ln}w$ shows a straight line passing through (−1.7, 4.3) and (7.1, 17.5).

Find the value of $k$ and of $n$. 

$\text{ln}v=n\text{ln}w+\text{ln}k$       M1A1

gradient $=\frac{17.5-4.3}{7.1+1.7}\left(=1.5\right)$       M1

$n=1.5$            A1

$y$-intercept $=1.5\times 1.7+4.3\left(=6.85\right)$       M1

$k=e^{6.85}=944$       M1A1

[7 marks]

Write down a sequence of transformations that will transform the graph of $y=\cos x$ onto the graph of $y=f\left(x\right)$.

Vertical stretch, scale factor $3$        A1