maximum value of $h$.

minimum value of $h$.

Find the time, in seconds, it takes for the blade $\text{[BC]}$ to make one complete rotation under these conditions.

Calculate the angle, in degrees, that the blade $\text{[BC]}$ turns through in one second.

Write down the amplitude of the function.

Find the period of the function.

Sketch the function $h\left(t\right)$ for $0\le t\le 5$, clearly labelling the coordinates of the maximum and minimum points.

Find the height of $\text{C}$ above the ground when $t=2$.

Find the time, in seconds, that point $\text{C}$ is above a height of $100 \text{m}$, during each complete rotation.

The wind speed increases and the blades rotate faster, but still at a constant rate.

Given that point $\text{C}$ is now higher than $110 \text{m}$ for $1$ second during each complete rotation, find the time for one complete rotation.

maximum $h=130$ metres             A1

[1 mark]

minimum $h=50$ metres             A1

[1 mark]

$\left(60÷12=\right) 5 \text{seconds}$             A1

[1 mark]

$360÷5$            (M1)

Note: Award (M1) for $360$ divided by their time for one revolution.

$=72°$             A1

[2 marks]

(amplitude =)  $40$         A1

[1 mark]

(period $=\frac{360}{72}=$)  $5$         A1

[1 mark]

Maximum point labelled with correct coordinates.         A1

At least one minimum point labelled. Coordinates seen for any minimum points must be correct.         A1

Correct shape with an attempt at symmetry and “concave up" evident as it approaches the minimum points. Graph must be drawn in the given domain.         A1

[3 marks]

$h=90-40 \cos \left(144°\right)$           (M1)

$\left(h=\right) 122 \left(\text{m}\right) \left(122.3606\dots \right)$           A1

[2 marks]

evidence of $h=100$ on graph  OR  $100=90-40 \cos \left(72t\right)$           (M1)

$t$ coordinates $3.55 (3.54892...)$  OR  $1.45 (1.45107...)$ or equivalent           (A1)

Note: Award A1 for either $t$-coordinate seen.

$=2.10$ seconds  $\left(2.09784\dots \right)$           A1

[3 marks]

METHOD 1

$90-40 \cos \left(at°\right)=110$           (M1)

$\cos \left(at°\right)=-0.5$

$at°=120, 240$           (A1)

$1=\frac{240}{a}-\frac{120}{a}$           (M1)

$a=120$           (A1)

period $=\frac{360}{120}=3$ seconds           A1

METHOD 2

attempt at diagram           (M1)

$\cos \alpha =\frac{20}{40}$ (or recognizing special triangle)           (M1)

angle made by $\text{C}$, $2\alpha =120°$           (A1)

one third of a revolution in $1$ second           (M1)

hence one revolution $=3$ seconds           A1

METHOD 3

considering $h\left(t\right)=110$ on original function           (M1)

$t=\frac{5}{3}$  or  $\frac{10}{3}$           (A1)

$\frac{10}{3}-\frac{5}{3}=\frac{5}{3}$           (A1)

Note: Accept $t=1.67$ or equivalent.

so period is $\frac{3}{5}$ of original period           (R1)

so new period is $3$ seconds           A1

[5 marks]

This was perhaps the question with the best responses on the paper. Many candidates got close to full marks on this problem. The issues associated with the question were sometimes due to a lack of understanding of the definitions of amplitude and period. A good number of candidates solved both parts of part (e) suggesting that they had a good understanding of the concept of a function and how it can be applied to mathematical models. Part (f) was also well done by a surprisingly large number of candidates using a variety of approaches. This is evidence that candidates had good problem-solving skills.

This was perhaps the question with the best responses on the paper. Many candidates got close to full marks on this problem. The issues associated with the question were sometimes due to a lack of understanding of the definitions of amplitude and period. A good number of candidates solved both parts of part (e) suggesting that they had a good understanding of the concept of a function and how it can be applied to mathematical models. Part (f) was also well done by a surprisingly large number of candidates using a variety of approaches. This is evidence that candidates had good problem-solving skills.

This was perhaps the question with the best responses on the paper. Many candidates got close to full marks on this problem. The issues associated with the question were sometimes due to a lack of understanding of the definitions of amplitude and period. A good number of candidates solved both parts of part (e) suggesting that they had a good understanding of the concept of a function and how it can be applied to mathematical models. Part (f) was also well done by a surprisingly large number of candidates using a variety of approaches. This is evidence that candidates had good problem-solving skills.

This was perhaps the question with the best responses on the paper. Many candidates got close to full marks on this problem. The issues associated with the question were sometimes due to a lack of understanding of the definitions of amplitude and period. A good number of candidates solved both parts of part (e) suggesting that they had a good understanding of the concept of a function and how it can be applied to mathematical models. Part (f) was also well done by a surprisingly large number of candidates using a variety of approaches. This is evidence that candidates had good problem-solving skills.

This was perhaps the question with the best responses on the paper. Many candidates got close to full marks on this problem. The issues associated with the question were sometimes due to a lack of understanding of the definitions of amplitude and period. A good number of candidates solved both parts of part (e) suggesting that they had a good understanding of the concept of a function and how it can be applied to mathematical models. Part (f) was also well done by a surprisingly large number of candidates using a variety of approaches. This is evidence that candidates had good problem-solving skills.

This was perhaps the question with the best responses on the paper. Many candidates got close to full marks on this problem. The issues associated with the question were sometimes due to a lack of understanding of the definitions of amplitude and period. A good number of candidates solved both parts of part (e) suggesting that they had a good understanding of the concept of a function and how it can be applied to mathematical models. Part (f) was also well done by a surprisingly large number of candidates using a variety of approaches. This is evidence that candidates had good problem-solving skills.

This was perhaps the question with the best responses on the paper. Many candidates got close to full marks on this problem. The issues associated with the question were sometimes due to a lack of understanding of the definitions of amplitude and period. A good number of candidates solved both parts of part (e) suggesting that they had a good understanding of the concept of a function and how it can be applied to mathematical models. Part (f) was also well done by a surprisingly large number of candidates using a variety of approaches. This is evidence that candidates had good problem-solving skills.

This was perhaps the question with the best responses on the paper. Many candidates got close to full marks on this problem. The issues associated with the question were sometimes due to a lack of understanding of the definitions of amplitude and period. A good number of candidates solved both parts of part (e) suggesting that they had a good understanding of the concept of a function and how it can be applied to mathematical models. Part (f) was also well done by a surprisingly large number of candidates using a variety of approaches. This is evidence that candidates had good problem-solving skills.

This was perhaps the question with the best responses on the paper. Many candidates got close to full marks on this problem. The issues associated with the question were sometimes due to a lack of understanding of the definitions of amplitude and period. A good number of candidates solved both parts of part (e) suggesting that they had a good understanding of the concept of a function and how it can be applied to mathematical models. Part (f) was also well done by a surprisingly large number of candidates using a variety of approaches. This is evidence that candidates had good problem-solving skills.

This was perhaps the question with the best responses on the paper. Many candidates got close to full marks on this problem. The issues associated with the question were sometimes due to a lack of understanding of the definitions of amplitude and period. A good number of candidates solved both parts of part (e) suggesting that they had a good understanding of the concept of a function and how it can be applied to mathematical models. Part (f) was also well done by a surprisingly large number of candidates using a variety of approaches. This is evidence that candidates had good problem-solving skills.

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

Hence find the maximum height of the tunnel.

$x=4$.

$x=6$.

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

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

Hence find the cross-sectional area of the tunnel.

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

$\frac{dy}{dx}=-0.3x^2+1.6x$                 A1

[2 marks]

$-0.3x^2+1.6x=0$                 M1

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

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

Note: Award M1 for substituting their zero for $\frac{dy}{dx} \left(5.333\dots \right)$ into $y$.

$7.59 \mathrm{m} \left(7.58519\dots \right)$                 A1

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

[6 marks]

One correct substitution seen             (M1)

$6.4 \text{m}$                 A1

[2 marks]

$7.2 \text{m}$                 A1

[1 mark]

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

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

$=29.6 {\text{m}}^2$                 A1

[3 marks]

$A={\int }_2^8-0.1x^3+0.8x^2 dx$  OR  $A={\int }_2^{8}y dx$                 A1A1

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

[2 marks]

$A=32.4 {\text{m}}^2$                  A2

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

[2 marks]

Show that $\frac{dk}{dT}$ is always positive.

Given that $\underset{T\to \infty }{\lim }k=A$ and $\underset{T\to 0}{\lim }k=0$, sketch the graph of $k$ against $T$.

(i)   the gradient of this line in terms of $c$;

(ii)  the $y$-intercept of this line in terms of $A$.

Find the equation of the regression line for $\ln k$ on $\frac{1}{T}$.

$c$.

It is not required to state units for this value.

$A$.

It is not required to state units for this value.

attempt to use chain rule, including the differentiation of $\frac{1}{T}$          (M1)

$\frac{dk}{dT}=A\times \frac{c}{T^2}\times {\text{e}}^{-\frac{c}{T}}$          A1

this is the product of positive quantities so must be positive          R1

Note: The R1 may be awarded for correct argument from their derivative. R1 is not possible if their derivative is not always positive.

[3 marks]

         A1A1A1

Note: Award A1 for an increasing graph, entirely in first quadrant, becoming concave down for larger values of $T$, A1 for tending towards the origin and A1 for asymptote labelled at $k=A$.

[3 marks]

taking $\ln$ of both sides   OR   substituting $y=\ln x$  and  $x=\frac{1}{T}$           (M1)

$\ln k=\ln A-\frac{c}{T}$  OR  $y=-cx+\ln A$           (A1)

(i)   so gradient is $-c$         A1

(ii)  $y$-intercept is $\ln A$         A1

Note: The implied (M1) and (A1) can only be awarded if both correct answers are seen. Award zero if only one value is correct and no working is seen.

[4 marks]

an attempt to convert data to $\frac{1}{T}$ and $\ln k$           (M1)

e.g. at least one correct row in the following table

line is $\ln k=-13400\times \frac{1}{T}+15.0 \left(=-13383.1\dots \times \frac{1}{T}+15.0107\dots \right)$         A1

[2 marks]

$c=13400 \left(13383.1\dots \right)$         A1

[1 mark]

attempt to rearrange or solve graphically $\ln A=15.0107\dots$          (M1)

$A=3 300 000 \left(3 304 258\dots \right)$         A1

 Note: Accept an $A$ value of $3269017$… from use of $3\mathrm{sf}$ value.

[2 marks]

This question caused significant difficulties for many candidates and many did not even attempt the question. Very few candidates were able to differentiate the expression in part (a) resulting in difficulties for part (b). Responses to parts (c) to (e) illustrated a lack of understanding of linearizing a set of data. Those candidates that were able to do part (d) frequently lost a mark as their answer was given in x and y.

Sketch the graph of $y=f\left(x\right)$ for $-\frac{5\pi }{8}⩽x⩽\frac{\pi }{8}$.

With reference to your graph, explain why $f$ is a function on the given domain.

Explain why $f$ has no inverse on the given domain.

Explain why $f$ is not a function for $-\frac{3\pi }{4}⩽x⩽\frac{\pi }{4}$.

Show that $g\left(t\right)={\left(\frac{1+t}{1-t}\right)}^2$.

Sketch the graph of $y=g\left(t\right)$ for t ≤ 0. Give the coordinates of any intercepts and the equations of any asymptotes.

Find $\alpha$ and β in terms of $k$.

Show that $\alpha$ + β < −2.

     A1A1

A1 for correct concavity, many to one graph, symmetrical about the midpoint of the domain and with two axes intercepts.

Note: Axes intercepts and scales not required.

A1 for correct domain

[2 marks]

for each value of $x$ there is a unique value of $f\left(x\right)$      A1

Note: Accept “passes the vertical line test” or equivalent.

[1 mark]

no inverse because the function fails the horizontal line test or equivalent      R1

Note: No FT if the graph is in degrees (one-to-one).

[1 mark]

the expression is not valid at either of $x=\frac{\pi }{4}\left(\text{or}-\frac{3\pi }{4}\right)$       R1

[1 mark]

METHOD 1

$f\left(x\right)=\frac{\text{tan}\left(x+\frac{\pi }{4}\right)}{\text{tan}\left(\frac{\pi }{4}-x\right)}$     M1

$=\frac{\frac{\text{tan}x+\text{tan}\frac{\pi }{4}}{1-\text{tan}x\text{tan}\frac{\pi }{4}}}{\frac{\text{tan}\frac{\pi }{4}-\text{tan}x}{1+\text{tan}\frac{\pi }{4}\text{tan}x}}$      M1A1

$={\left(\frac{1+t}{1-t}\right)}^2$      AG

METHOD 2

$f\left(x\right)=\text{tan}\left(x+\frac{\pi }{4}\right)\text{tan}\left(\frac{\pi }{2}-\frac{\pi }{4}+x\right)$      (M1)

$=\text{ta}{\text{n}}^2\left(x+\frac{\pi }{4}\right)$     A1

$g\left(t\right)={\left(\frac{\text{tan}x+\text{tan}\frac{\pi }{4}}{1-\text{tan}x\text{tan}\frac{\pi }{4}}\right)}^2$     A1

$={\left(\frac{1+t}{1-t}\right)}^2$      AG

[3 marks]

for t ≤ 0, correct concavity with two axes intercepts and with asymptote $y$ = 1      A1

t intercept at (−1, 0)      A1

$y$ intercept at (0, 1)       A1

[3 marks]

METHOD 1

$\alpha$, β satisfy $\frac{{\left(1+t\right)}^2}{{\left(1-t\right)}^2}=k$     M1

$1+t^2+2t=k\left(1+t^2-2t\right)$     A1

$\left(k-1\right)t^2-2\left(k+1\right)t+\left(k-1\right)=0$     A1

attempt at using quadratic formula      M1

$\alpha$, β $=\frac{k+1\pm 2\sqrt{k}}{k-1}$ or equivalent     A1

METHOD 2

$\alpha$, β satisfy $\frac{1+t}{1-t}=(\pm )\sqrt{k}$      M1

$t+\sqrt{k}t=\sqrt{k}-1$      M1

$t=\frac{\sqrt{k}-1}{\sqrt{k}+1}$ (or equivalent)      A1

$t-\sqrt{k}t=-\left(\sqrt{k}+1\right)$     M1

$t=\frac{\sqrt{k}+1}{\sqrt{k}-1}$ (or equivalent)       A1

so for eg, $\alpha =\frac{\sqrt{k}-1}{\sqrt{k}+1}$, β $=\frac{\sqrt{k}+1}{\sqrt{k}-1}$

[5 marks]

$\alpha$ + β $=2\frac{\left(k+1\right)}{\left(k-1\right)}\left(=-2\frac{\left(1+k\right)}{\left(1-k\right)}\right)$     A1

since $1+k>1-k$     R1

$\alpha$ + β < −2     AG

Note: Accept a valid graphical reasoning.

[2 marks]

Find the equation of the line passing through these two points.

Find the equation of the least squares regression quadratic curve for these four points.

By considering the gradient of this curve when $x=4$, explain why it may not be a good model.

Find the equation of the new model.

Write down an expression for her estimate of the volume as a sum of two integrals.

Find the value of Charlotte’s estimate.

$y=\frac{5}{8}x+\frac{7}{2} \left(y=0.625x+3.5\right)$                  A1A1

Note: Award A1 for $0.625x$, A1 for $3.5$.
Award a maximum of A0A1 if not part of an equation.

[2 marks]

$y=-0.975x^2+9.56x-16.7$                  (M1)A1

$\left(y=-0.974630x^2+9.55919x-16.6569\dots \right)$

[2 marks]

gradient of curve is positive at $x=4$                 R1

Note: Accept a sensible rationale that refers to the gradient.

[1 mark]

METHOD 1

let $y=a{x}^2+bx+c$

differentiating or using $x=\frac{-b}{2a}$                       (M1)

$8a+b=0$

substituting in the coordinates
$7.5^{2}a+7.5b+c=0$                       (A1)
$4^{2}a+4b+c=6$                       (A1)

solve to get
$y=-\frac{24}{49}{x}^2+\frac{192}{49}x-\frac{90}{49}$  OR  $y=-0.490x^2+3.92x-1.84$                       A1

Note: Use of quadratic regression with points using the symmetry of the graph is a valid method.

METHOD 2

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

$0=a{\left(7.5-4\right)}^2+6$                       (M1)

$a=-\frac{24}{49}$                       (A1)

$y=-\frac{24}{49}{\left(x-4\right)}^2+6$  OR  $y=-0.490{\left(x-4\right)}^2+6$                       A1

[4 marks]

$\mathrm{\pi }{\int }_0^4{\left(\frac{5}{8}x+3.5\right)}^{2}dx+\mathrm{\pi }{\int }_4^{7.5}{\left(-\frac{24}{49}{\left(x-4\right)}^2+6\right)}^{2}dx$                       (M1)(M1) (M1)A1

Note: Award (M1)(M1)(M1)A0 if $\mathrm{\pi }$ is omitted but response is otherwise correct. Award (M1) for an integral that indicates volume, (M1) for their part (a) within their volume integral, (M1) for their part (b)(i) within their volume integral, A1 for their correct two integrals with all correct limits.

[4 marks]

$501 {\text{cm}}^3 \left(501.189\dots \right)$                      A1

[1 mark]

Show that the system of coupled first order equations:

$\frac{dx}{dt}=y$

$\frac{dy}{dt}=-2x-3y$

can be written as the given second order differential equation.

Find the eigenvalues of the system of coupled first order equations given in part (a).

Hence find the exact solution of the second order differential equation.

Sketch the graph of $x$ against $t$, labelling the maximum point of the graph with its coordinates.

Use the model to calculate the total amount of time when fishing should be stopped.

Write down one reason, with reference to the context, to support this decision.

differentiating first equation.         M1

$\frac{d^{2}x}{d{t}^2}=\frac{dy}{dt}$

substituting in for $\frac{dy}{dt}$         M1

$=-2x-3y=-2x-3\frac{dx}{dt}$

therefore $\frac{d^{2}x}{d{t}^2}+3\frac{dx}{dt}+2x=0$         AG

Note: The AG line must be seen to award the final M1 mark.

[2 marks]

the relevant matrix is $\left(\begin{array}{cc}0 & 1\\ -2 & -3\end{array}\right)$           (M1)

Note:  $\left(\begin{array}{cc}-3 & -2\\ 1 & 0\end{array}\right)$ is also possible.

(this has characteristic equation) $-\lambda \left(-3-\lambda \right)+2=0$           (A1)

$\lambda =-1, -2$         A1

[3 marks]

EITHER

the general solution is $x=A{\text{e}}^{-t}+B{\text{e}}^{-2t}$             M1

Note: Must have constants, but condone sign error for the M1.

so $\frac{dx}{dt}=-A{\text{e}}^{-t}-2B{\text{e}}^{-2t}$             M1A1

OR

attempt to find eigenvectors           (M1)

respective eigenvectors are $\left(\begin{array}{c}1\\ -1\end{array}\right)$ and $\left(\begin{array}{c}1\\ -2\end{array}\right)$ (or any multiple)

$\left(\begin{array}{c}x\\ y\end{array}\right)=A{\text{e}}^{-t}\left(\begin{array}{c}1\\ -1\end{array}\right)+B{\text{e}}^{-2t}\left(\begin{array}{c}1\\ -2\end{array}\right)$           (M1)A1

THEN

the initial conditions become:

$0=A+B$

$1=-A-2B$             M1

this is solved by $A=1, B=-1$

so the solution is $x={\text{e}}^{-t}-{\text{e}}^{-2t}$            A1

[5 marks]

            A1A1

Note: Award A1 for correct shape (needs to go through origin, have asymptote at $y=0$ and a single maximum; condone $x<0$). Award A1 for correct coordinates of maximum.

[2 marks]

intersecting graph with $y=0.1$         (M1)

so the time fishing is stopped between $2.1830\dots$ and $0.11957\dots$           (A1)

$=2.06 \left(343\dots \right)$  days           A1

[3 marks]

Any reasonable answer. For example:

There are greater downsides to allowing fishing when the levels may be dangerous than preventing fishing when the levels are safe.

The concentration of mercury may not be uniform across the river due to natural variation / randomness.

The situation at the power plant might get worse.

Mercury levels are low in water but still may be high in fish.           R1

Note: Award R1 for a reasonable answer that refers to this specific context (and not a generic response that could apply to any model).

[1 mark]

Many candidates did not get this far, but the attempts at the question that were seen were generally good. The greater difficulties were seen in parts (e) and (f), but this could be a problem with time running out.

Write down the maximum and minimum value of $v$.

Write down two transformations that will transform the graph of $y=v\left(t\right)$ onto the graph of $y=i\left(t\right)$.

Sketch the graph of $y=p\left(t\right)$ for 0 ≤ $t$ ≤ 0.02 , showing clearly the coordinates of the first maximum and the first minimum.

Find the total time in the interval 0 ≤ $t$ ≤ 0.02 for which $p\left(t\right)$ ≥ 3.

Find $p_{av}$(0.007).

With reference to your graph of $y=p\left(t\right)$ explain why $p_{av}\left(T\right)$ > 0 for all $T$ > 0.

Given that $p\left(t\right)$ can be written as $p\left(t\right)=a\text{sin}\left(b\left(t-c\right)\right)+d$ where $a$, $b$, $c$, $d$ > 0, use your graph to find the values of $a$, $b$, $c$ and $d$.

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

3, −3       A1A1 

[2 marks]

stretch parallel to the $y$-axis (with $x$-axis invariant), scale factor $\frac{2}{3}$       A1

translation of $\left(\begin{array}{c}-0.003\\ 0\end{array}\right)$  (shift to the left by 0.003)      A1

Note: Can be done in either order.

[2 marks]

correct shape over correct domain with correct endpoints       A1
first maximum at (0.0035, 4.76)       A1
first minimum at (0.0085, −1.24)       A1

[3 marks]

$p$ ≥ 3 between $t$ = 0.0016762 and 0.0053238 and $t$ = 0.011676 and 0.015324       (M1)(A1)

Note: Award M1A1 for either interval.

= 0.00730       A1

[3 marks]

$p_{av}=\frac{1}{0.007}{\int }_0^{0.007}6\text{sin}\left(100\pi t\right)\text{sin}\left(100\pi \left(t+0.003\right)\right)\text{d}t$     (M1)

= 2.87       A1

[2 marks]

in each cycle the area under the $t$ axis is smaller than area above the $t$ axis      R1

the curve begins with the positive part of the cycle       R1

[2 marks]

$a=\frac{4.76-\left(-1.24\right)}{2}$       (M1)

$a=3.00$       A1

$d=\frac{4.76+\left(-1.24\right)}{2}$

$d=1.76$       A1

$b=\frac{2\pi }{0.01}$

$b=628\left(=200\pi \right)$       A1

$c=0.0035-\frac{0.01}{4}$       (M1)

$c=0.00100$       A1

[6 marks]

Find the largest possible domain $D$ for $f$ to be a function.

Sketch the graph of $y=f(x)$ showing clearly the equations of asymptotes and the coordinates of any intercepts with the axes.

Explain why $f$ is an even function.

Explain why the inverse function $f^{-1}$ does not exist.

Find the inverse function $g^{-1}$ and state its domain.

Find $g^{\prime }(x)$.

Hence, show that there are no solutions to $g^{\prime }(x)=0$;

Hence, show that there are no solutions to $(g^{-1}{)}^{\prime }(x)=0$.

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

$x^2-1>0$     (M1)

or $x>1$     A1

[2 marks]

shape     A1

$x=1$ and $x=-1$     A1

$x$-intercepts     A1

[3 marks]

EITHER

$f$ is symmetrical about the $y$-axis     R1

OR

$f(-x)=f(x)$     R1

[1 mark]

EITHER

$f$ is not one-to-one function     R1

OR

horizontal line cuts twice     R1

Note:     Accept any equivalent correct statement.

[1 mark]

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

${\text{e}}^{2x+2}=y^2-1$     M1

$g^{-1}(x)=\sqrt{{\text{e}}^{2x+2}+1},x\in \mathbb{R}$     A1A1

[4 marks]

$g^{\prime }(x)=\frac{1}{\sqrt{x^2-1}}\times \frac{2x}{2\sqrt{x^2-1}}$     M1A1

$g^{\prime }(x)=\frac{x}{x^2-1}$     A1

[3 marks]

$g^{\prime }(x)=\frac{x}{x^2-1}=0\Rightarrow x=0$     M1

which is not in the domain of $g$ (hence no solutions to $g^{\prime }(x)=0$)     R1

[2 marks]

$(g^{-1}{)}^{\prime }(x)=\frac{{\text{e}}^{2x+2}}{\sqrt{{\text{e}}^{2x+2}+1}}$     M1

as ${\text{e}}^{2x+2}>0\Rightarrow (g^{-1}{)}^{\prime }(x)>0$ so no solutions to $(g^{-1}{)}^{\prime }(x)=0$     R1

Note:     Accept: equation ${\text{e}}^{2x+2}=0$ has no solutions.

[2 marks]

Find the initial speed of the ball.

Find the angle of elevation of the ball as it is launched.

Find the maximum height reached by the ball.

Assuming that the ground is horizontal and the ball is not hit by the arrow, find the $x$ coordinate of the point where the ball lands.

For the path of the ball, find an expression for $y$ in terms of $x$.

Determine the two positions where the path of the arrow intersects the path of the ball.

Determine the time when the arrow should be released to hit the ball before the ball reaches its maximum height.

$\sqrt{{10}^2+8^2}$           (M1)

$=12.8 \left(12.8062\dots , \sqrt{164}\right) \left(\text{m} {\text{s}}^{-1}\right)$          A1

[2 marks]

${\tan }^{-1}\left(\frac{10}{8}\right)$           (M1)

$=0.896$   OR   $51.3$   ($0.896055\dots$   OR   $51.3401\dots °$)           A1

Note: Accept $0.897$ or $51.4$ from use of $\text{arcsin}\left(\frac{10}{12.8}\right)$.

[2 marks]

$y=t\left(10-5t\right)$           (M1)

Note: The M1 might be implied by a correct graph or use of the correct equation.

METHOD 1 – graphical Method

sketch graph           (M1)

Note: The M1 might be implied by correct graph or correct maximum (eg $t=1$).

max occurs when $y=5 \text{m}$           A1

METHOD 2 – calculus

differentiating and equating to zero           (M1)

$\frac{dy}{dt}=10-10t=0$

$t=1$

$y\left(=1\left(10-5\right)\right)=5 \text{m}$           A1

METHOD 3 – symmetry

line of symmetry is $t=1$           (M1)

$y\left(=1\left(10-5\right)\right)=5 \text{m}$           A1

[3 marks]

attempt to solve $t\left(10-5t\right)=0$           (M1)

$t=2$  (or $t=0$)          (A1)

$x \left(=5+8\times 2\right)= 21 \text{m}$           A1

Note: Do not award the final A1 if $x=5$ is also seen.

[3 marks]

METHOD 1

$t=\frac{x-5}{8}$            M1A1

$y=\left(\frac{x-5}{8}\right)\left(10-5\times \frac{x-5}{8}\right)$           A1

METHOD 2

$y=k\left(x-5\right)\left(x-21\right)$           A1

when $x=13, y=5$ so $k=\frac{5}{\left(13-5\right)\left(13-21\right)}=-\frac{5}{64}$            M1A1

$\left(y=-\frac{5}{64}\left(x-5\right)\left(x-21\right)\right)$

METHOD 3

if $y=a{x}^2+bx+c$

 $0=25a+5b+c$
 $5=169a+13b+c$
 $0=441a+21b+c$            M1A1

solving simultaneously, $a=-\frac{5}{64}, b=\frac{130}{64}, c=-\frac{525}{64}$           A1

($y=-\frac{5}{64}{x}^2+\frac{130}{64}x-\frac{525}{64}$)

METHOD 4

use quadratic regression on $(5, 0), (13, 5), (21, 0)$            M1A1

$y=-\frac{5}{64}{x}^2+\frac{130}{64}x-\frac{525}{64}$           A1

Note: Question asks for expression; condone omission of "$y=$".

[3 marks]

trajectory of arrow is $y=x \tan 10+2$             (A1)

intersecting $y=x \tan 10+2$ and their answer to (d)             (M1)

$\left(8.66, 3.53\right) \left(\left(8.65705\dots , 3.52647\dots \right)\right)$           A1

$\left(15.1, 4.66\right) \left(\left(15.0859\dots , 4.66006\dots \right)\right)$           A1

[4 marks]

when $x_{\text{target}}=8.65705\dots , t_{\text{target}}=\frac{8.65705\dots -5}{8}=0.457132\dots \text{s}$             (A1)

attempt to find the distance from point of release to intersection             (M1)

$\sqrt{8.65705{\dots }^2+{\left(3.52647\dots -2\right)}^2} \left(=8.79060\dots \text{m}\right)$

time for arrow to get there is $\frac{8.79060\dots }{60}=0.146510\dots \text{s}$             (A1)

so the arrow should be released when

$t=0.311 \left(\text{s}\right) \left(0.310622\dots \left(\text{s}\right)\right)$           A1 

[4 marks]

This question was found to be the most difficult on the paper. There were a good number of good solutions to parts (a) and part (b), frequently with answers just written down with no working. Part (c) caused some difficulties with confusing variables. The most significant difficulties started with part (d) and became greater to the end of the question. Few candidates were able to work through the final two parts.

Sketch the graphs $y=\text{si}{\text{n}}^{3}x+\text{ln}x$ and $y=1+\text{cos}x$ on the following axes for 0 < $x$ ≤ 9.

Hence solve  in the range 0 < $x$ ≤ 9.

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

       A1A1

Note: Award A1 for each correct curve, showing all local max & mins.

Note: Award A0A0 for the curves drawn in degrees.

[2 marks]

$x$ = 1.35, 4.35, 6.64       (M1)

Note: Award M1 for attempt to find points of intersections between two curves.

0 < $x$ < 1.35      A1

Note: Accept $x$ < 1.35.

4.35 < $x$ < 6.64       A1A1

Note: Award A1 for correct endpoints, A1 for correct inequalities.

Note: Award M1FTA1FTA0FTA0FT for 0 < $x$ < 7.31.

Note: Accept $x$ < 7.31.

[4 marks]

Show that the $x$-coordinate of the minimum point on the curve $y=f(x)$ satisfies the equation $\tan x=2x$.

Determine the values of $x$ for which $f(x)$ is a decreasing function.

Sketch the graph of $y=f(x)$ showing clearly the minimum point and any asymptotic behaviour.

Find the coordinates of the point on the graph of $f$ where the normal to the graph is parallel to the line $y=-x$.

This region is now rotated through $2\pi$ radians about the $x$-axis. Find the volume of revolution.

attempt to use quotient rule or product rule     M1

$f^{\prime }(x)=\frac{\sin x\left(\frac{1}{2}{x}^{-\frac{1}{2}}\right)-\sqrt{x}\cos x}{{\sin }^{2}x}\left(=\frac{1}{2\sqrt{x}\sin x}-\frac{\sqrt{x}\cos x}{{\sin }^{2}x}\right)$     A1A1

Note:     Award A1 for $\frac{1}{2\sqrt{x}\sin x}$ or equivalent and A1 for $-\frac{\sqrt{x}\cos x}{{\sin }^{2}x}$ or equivalent.

setting $f^{\prime }(x)=0$     M1

$\frac{\sin x}{2\sqrt{x}}-\sqrt{x}\cos x=0$

$\frac{\sin x}{2\sqrt{x}}=\sqrt{x}\cos x$ or equivalent     A1

$\tan x=2x$     AG

[5 marks]

$x=1.17$

    A1A1

Note:     Award A1 for and A1 for $x⩽1.17$. Accept .

[2 marks]

concave up curve over correct domain with one minimum point above the $x$-axis.     A1

approaches $x=0$ asymptotically     A1

approaches $x=\pi$ asymptotically     A1

Note:     For the final A1 an asymptote must be seen, and $\pi$ must be seen on the $x$-axis or in an equation.

[3 marks]

$f^{\prime }(x)\left(=\frac{\sin x\left(\frac{1}{2}{x}^{-\frac{1}{2}}\right)-\sqrt{x}\cos x}{{\sin }^{2}x}\right)=1$     (A1)

attempt to solve for $x$     (M1)

$x=1.96$     A1

$y=f(1.96\dots )$

$=1.51$     A1

[4 marks]

$V=\pi {\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{x\text{d}x}{{\sin }^{2}x}$     (M1)(A1)

Note:     M1 is for an integral of the correct squared function (with or without limits and/or $\pi$).

$=2.68(=0.852\pi )$     A1

[3 marks]

Calculate her distance after 8 minutes. Give your answer in km, correct to 3 decimal places.

Find the value of $a$, $b$, $c$ and $d$.

the distance she runs in 20 minutes.

her maximum speed, in ms^–1.

$\frac{2.3\times 8\times 60}{1000}=1.104$    M1A1

[2 marks]

either using a cubic regression or solving a system of 4 equations         M1

$a=-0.00364\text{,}b=0.150\text{,}c=-1.67\text{,}d=6.72$         A1A1A1A1

[5 marks]

$s\left(20\right)=4.21$ km  (Note: Condone $s\left(20\right)=4.2$ km obtained from using rounded values.)      M1A1

[2 marks]

EITHER finding maximum of $\frac{ds}{dt}$ OR solving $\frac{d^{2}s}{d{t}^2}=0$     M1

maximum speed = 0.390… km per minute      A1

maximum speed = 6.51 ms^–1     M1A1

[4 marks]

Sketch the graph of $f$ indicating clearly any intercepts with the axes and the coordinates of any local maximum or minimum points.

State the range of $f$.

Solve the inequality $\left|3x\arccos (x)\right|>1$.

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

correct shape passing through the origin and correct domain     A1

Note: Endpoint coordinates are not required. The domain can be indicated by $-1$ and 1 marked on the axis.

$(0.652,1.68)$    A1

two correct intercepts (coordinates not required)     A1

Note: A graph passing through the origin is sufficient for $(0,0)$.

[3 marks]

$[-9.42,1.68](\text{or }-3\pi ,1.68])$    A1A1

Note: Award A1A0 for open or semi-open intervals with correct endpoints. Award A1A0 for closed intervals with one correct endpoint.

[2 marks]

attempting to solve either $\left|3x\arccos (x)\right|>1$ (or equivalent) or $\left|3x\arccos (x)\right|=1$ (or equivalent) (eg. graphically)     (M1)

$x=-0.189,0.254,0.937$    (A1)

   A1A1

Note: Award A0 for .

[4 marks]

Show that Jorge’s model satisfies the differential equation

$\frac{dN}{dt}=rN$

Show that $r\approx \frac{N\left(n+1\right)-N\left(n\right)}{N\left(n\right)}$

Hence find three approximations for the value of $r$.

Find the equation for Jorge’s model.

Find the sum of the square residuals for Jorge’s model using the values $t=1, 2, 3, 4$.

Comment how well Jorge’s model fits the data.

Give two possible sources of error in the construction of his model.

* 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.

$\frac{dN}{dt}=rA{\text{e}}^{rt}$        (M1)A1

Note: M1 is for an attempt to find $\frac{dN}{dt}$

$=rN$        AG

Note: Accept solution of the differential equation by separating variables

[2 marks]

$N\left(n+1\right)\approx N\left(n\right)+1\times N\text{'}\left(n\right)\Rightarrow N\text{'}\left(n\right)\approx N\left(n+1\right)-N\left(n\right)$        M1

$\Rightarrow rN\left(n\right)\approx N\left(n+1\right)-N\left(n\right)$        M1A1

$\Rightarrow r\approx \frac{N\left(n+1\right)-N\left(n\right)}{N\left(n\right)}$        AG

Note: Do not penalize the use of the $=$ sign.

[3 marks]

Correct method         (M1)

$r\approx \frac{52-40}{40}=0.3$

$r\approx \frac{70-52}{52}=0.346$

$r\approx \frac{98-70}{70}=0.4$        A2

Note: A1 for a single error A0 for two or more errors.

[3 marks]

$r=0.349 \left(0.34871\dots \right)$ or $\frac{68}{195}$        A1

$52=A{\text{e}}^{0.34871\dots \times 2}$        (M1)

$A=25.8887\dots$

$N=25.9{\text{e}}^{0.349t}$        A1

[3 marks]

${\left(36.6904\dots -40\right)}^2+0+{\left(73.6951\dots -70\right)}^2+{\left(104.4435\dots -98\right)}^2$        (M1)

$=66.1 \left(66.126\dots \right)$        A1

[2 marks]

The sum of the square residuals is approximately $10$ times as large as the minimum possible, so Jorge’s model is unlikely to fit the data exactly     R1

[1 mark]

For example

Selecting a single point for the curve to pass through

Approximating the gradient of the curve by the gradient of a chord       R1R1

[2 marks]

Find $\frac{dy}{dx}$.

Hence show that the equation of the tangent to the curve at the point $\left(0.16, 0.4\right)$ is $y=1.25x+0.2$.

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

Find an expression for$h\left(x\right)$.

Find the value of $a$.

Find the value of $b$.

Find the area enclosed by $y=f\left(x\right)$, the $x$-axis and the line $x=0.5$.

Find the area of the shaded region on the diagram.

$y=x^{\frac{1}{2}}$           (M1)

$\frac{dy}{dx}=\frac{1}{2}{x}^{-\frac{1}{2}}$          A1 

[2 marks]

gradient at $x=0.16$ is $\frac{1}{2}\times \frac{1}{\sqrt{0.16}}$          M1

$=1.25$

EITHER

$y-0.4=1.25\left(x-0.16\right)$          M1

OR

$0.4=1.25\left(0.16\right)+b$          M1

Note: Do not allow working backwards from the given answer.

THEN

hence $y=1.25x+0.2$          AG

[2 marks]

$p=0.45, q=0.4125$  (or $0.413$)  (accept " $(0.45, 0.4125)$ ")          A1A1

[2 marks]

$\left(h\left(x\right)=\right) \frac{1}{2}\sqrt{2\left(x-0.2\right)}$          A2

Note: Award A1 if only two correct transformations are seen. 

[2 marks]

$\left(a=\right) 0.28$          A1