Suppose that $u_1$ is the first term of a geometric series with common ratio $r$.

Prove, by mathematical induction, that the sum of the first $n$ terms, $s_n$ is given by

$s_n=\frac{u_1\left(1-r^n\right)}{1-r}$, where $n\in {\mathbb{Z}}^{+}$.

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

$n=1\Rightarrow s_1=u_1$, so true for $n=1$              R1

assume true for $n=k$, ie. $s_k=\frac{u_1\left(1-r^k\right)}{1-r}$              M1

Note: Award M0 for statements such as “let $n=k$”.

Note: Subsequent marks after the first M1 are independent of this mark and can be awarded.

$s_{k+1}=s_k+u_1{r}^k$              M1

$s_{k+1}=\frac{u_1\left(1-r^k\right)}{1-r}+u_1{r}^k$         A1

$s_{k+1}=\frac{u_1\left(1-r^k\right)}{1-r}+\frac{u_1{r}^k\left(1-r\right)}{1-r}$

$s_{k+1}=\frac{u_1-u_1{r}^k+u_1{r}^k-r{u}_1{r}^k}{1-r}$         A1

$s_{k+1}=\frac{u_1\left(1-r^{k+1}\right)}{1-r}$         A1

true for $n=1$ and if true for $n=k$ then true for $n=k+1$, the statement is true for any positive integer (or equivalent).        R1

Note: Award the final R1 mark provided at least four of the previous marks are gained.

[7 marks]

Write down the value of $b$.

What does $b$ represent in this context?

Find the eigenvalues of $\mathit{M}$.

Find the eigenvectors of $\mathit{M}$.

when $n=5$.

in the long term.

$0.02$         A1

[1 mark]

the probability of mutating from ‘not normal state’ to ‘normal state’         A1

Note: The A1 can only be awarded if it is clear that transformation is from the mutated state.

[1 mark]

$\text{det}\left(\begin{array}{cc}0.94-\lambda & 0.02\\ 0.06 & 0.98-\lambda \end{array}\right)=0$         (M1)

Note: Award M1 for an attempt to find eigenvalues. Any indication that $\text{det}\left(\mathit{M}-\lambda \mathit{I}\right)=0$ has been used is sufficient for the (M1).

$\left(0.94-\lambda \right)\left(0.98-\lambda \right)-0.0012=0$  OR  ${\lambda }^2-1.92\lambda +0.92=0$         (A1)

$\lambda =1, 0.92 \left(\frac{23}{25}\right)$         A1

[3 marks]

$\left(\begin{array}{cc}0.94 & 0.02\\ 0.06 & 0.98\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}x\\ y\end{array}\right)$  OR  $\left(\begin{array}{cc}0.94 & 0.02\\ 0.06 & 0.98\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=0.92\left(\begin{array}{c}x\\ y\end{array}\right)$         (M1)

Note: Award M1 can be awarded for attempting to find either eigenvector.

$0.02y-0.06x=0$  OR  $0.02y+0.02x=0$

$\left(\begin{array}{c}1\\ 3\end{array}\right)$  and  $\left(\begin{array}{c}1\\ -1\end{array}\right)$         A1A1

Note: Accept any multiple of the given eigenvectors.

[3 marks]

${\left(\begin{array}{cc}0.94 & 0.02\\ 0.06 & 0.98\end{array}\right)}^5\left(\begin{array}{c}1\\ 0\end{array}\right)$  OR  $\left(\begin{array}{cc}0.744 & 0.0852\\ 0.256 & 0.915\end{array}\right)\left(\begin{array}{c}1\\ 0\end{array}\right)$         (M1)

Note: Condone omission of the initial state vector for the M1.

$0.744 \left(0.744311\dots \right)$            A1

[2 marks]

$\left(\begin{array}{c}0.25\\ 0.75\end{array}\right)$         (A1)

Note: Award A1 for $\left(\begin{array}{c}0.25\\ 0.75\end{array}\right)$  OR  $\left(\begin{array}{cc}0.25 & 0.25\\ 0.75 & 0.75\end{array}\right)$ seen.

$0.25$          A1

[2 marks]

There was some difficulty in interpreting the meaning of the values in the transition matrix, but most candidates did well with the rest of the question. In part (d) there was frequently evidence of a correct method, but a failure to identify the correct probabilities. It was surprising to see a significant number of candidates diagonalizing the matrix in part (d) and this often led to errors. Clearly this was not necessary.

By considering the image of $(0, 0)$, find $\mathit{q}$.

By considering the image of $(1, 0)$ and $(0, 1)$, show that

$\mathit{P}=\left(\begin{array}{cc}\frac{\sqrt{3}}{4} & \frac{1}{4}\\ -\frac{1}{4} & \frac{\sqrt{3}}{4}\end{array}\right)$.

Write down the matrix $\mathit{S}$.

Use $\mathit{P}=\mathit{R}\mathit{S}$ to find the matrix $\mathit{R}$.

Hence find the angle and direction of the rotation represented by $\mathit{R}$.

Write down an equation satisfied by $\left(\begin{array}{c}a\\ b\end{array}\right)$.

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

$\mathit{P}\left(\begin{array}{c}0\\ 0\end{array}\right)+\mathit{q}=\left(\begin{array}{c}0\\ 1\end{array}\right)$        (M1)

$\mathit{q}=\left(\begin{array}{c}0\\ 1\end{array}\right)$          A1

[2 marks]

EITHER

$\mathit{P}\left(\begin{array}{c}1\\ 0\end{array}\right)+\left(\begin{array}{c}0\\ 1\end{array}\right)=\left(\begin{array}{c}\frac{\sqrt{3}}{4}\\ \frac{3}{4}\end{array}\right)$          M1

hence $\mathit{P}\left(\begin{array}{c}1\\ 0\end{array}\right)=\left(\begin{array}{c}\frac{\sqrt{3}}{4}\\ -\frac{1}{4}\end{array}\right)$          A1

$\mathit{P}\left(\begin{array}{c}0\\ 1\end{array}\right)+\left(\begin{array}{c}0\\ 1\end{array}\right)=\left(\begin{array}{c}\frac{1}{4}\\ 1+\frac{\sqrt{3}}{4}\end{array}\right)$          M1

hence $\mathit{P}\left(\begin{array}{c}0\\ 1\end{array}\right)=\left(\begin{array}{c}\frac{1}{4}\\ \frac{\sqrt{3}}{4}\end{array}\right)$          A1

OR

$\left(\begin{array}{cc}a & b\\ c & d\end{array}\right)\left(\begin{array}{c}1\\ 0\end{array}\right)+\left(\begin{array}{c}0\\ 1\end{array}\right)=\left(\begin{array}{c}\frac{\sqrt{3}}{4}\\ \frac{3}{4}\end{array}\right)$          M1

hence $\left(\begin{array}{cc}a & b\\ c & d\end{array}\right)\left(\begin{array}{c}1\\ 0\end{array}\right)=\left(\begin{array}{c}\frac{\sqrt{3}}{4}\\ -\frac{1}{4}\end{array}\right)$          A1

$\left(\begin{array}{c}a\\ c\end{array}\right)=\left(\begin{array}{c}\frac{\sqrt{3}}{4}\\ -\frac{1}{4}\end{array}\right)$

$\left(\begin{array}{cc}a & b\\ c & d\end{array}\right)\left(\begin{array}{c}0\\ 1\end{array}\right)+\left(\begin{array}{c}0\\ 1\end{array}\right)=\left(\begin{array}{c}\frac{1}{4}\\ 1+\frac{\sqrt{3}}{4}\end{array}\right)$          M1

$\left(\begin{array}{cc}a & b\\ c & d\end{array}\right)\left(\begin{array}{c}0\\ 1\end{array}\right)=\left(\begin{array}{c}\frac{1}{4}\\ \frac{\sqrt{3}}{4}\end{array}\right)$          A1

$\left(\begin{array}{c}b\\ d\end{array}\right)=\left(\begin{array}{c}\frac{1}{4}\\ \frac{\sqrt{3}}{4}\end{array}\right)$

THEN

$\Rightarrow \mathit{P}=\left(\begin{array}{cc}\frac{\sqrt{3}}{4} & \frac{1}{4}\\ -\frac{1}{4} & \frac{\sqrt{3}}{4}\end{array}\right)$          AG

[4 marks]

$\left(\begin{array}{cc}\frac{1}{2} & 0\\ 0 & \frac{1}{2}\end{array}\right)$          A1

[1 mark]

EITHER

${\mathit{S}}^{-1}=\left(\begin{array}{cc}2 & 0\\ 0 & 2\end{array}\right)$         (A1)

$\mathit{R}=\mathit{P}{\mathit{S}}^{-1}$         (M1)

Note: The M1 is for an attempt at rearranging the matrix equation. Award even if the order of the product is reversed.

$\mathit{R}=\left(\begin{array}{cc}\frac{\sqrt{3}}{4} & \frac{1}{4}\\ -\frac{1}{4} & \frac{\sqrt{3}}{4}\end{array}\right)\left(\begin{array}{cc}2 & 0\\ 0 & 2\end{array}\right)$         (A1)

OR

$\left(\begin{array}{cc}\frac{\sqrt{3}}{4} & \frac{1}{4}\\ -\frac{1}{4} & \frac{\sqrt{3}}{4}\end{array}\right)=\mathit{R}\left(\begin{array}{cc}0.5 & 0\\ 0 & 0.5\end{array}\right)$

let $\mathit{R}=\left(\begin{array}{cc}a & b\\ c & d\end{array}\right)$

attempt to solve a system of equations         M1

$\frac{\sqrt{3}}{4}=0.5a, \frac{1}{4}=0.5b$

$-\frac{1}{4}=0.5c, \frac{\sqrt{3}}{4}=0.5d$          A2

Note: Award A1 for two correct equations, A2 for all four equations correct.

THEN

$\mathit{R}=\left(\begin{array}{cc}\frac{\sqrt{3}}{2} & \frac{1}{2}\\ -\frac{1}{2} & \frac{\sqrt{3}}{2}\end{array}\right)$  OR  $\left(\begin{array}{cc}0.866 & 0.5\\ -0.5 & 0.866\end{array}\right)$  OR  $\left(\left(\begin{array}{cc}0.866025\dots & 0.5\\ -0.5 & 0.866025\dots \end{array}\right)\right)$          A1

Note: The correct answer can be obtained from reversing the matrices, so do not award if incorrect product seen. If the given answer is obtained from the product $\mathit{R}={\mathit{S}}^{-1}\mathit{P}$, award (A1)(M1)(A0)A0.

[4 marks]

clockwise         A1

arccosine or arcsine of value in matrix seen         (M1)

$30°$         A1

Note: Both A1 marks are dependent on the answer to part (c)(i) and should only be awarded for a valid rotation matrix.

[3 marks]

METHOD 1

$\left(\begin{array}{c}a\\ b\end{array}\right)=\mathit{P}\left(\begin{array}{c}a\\ b\end{array}\right)+\mathit{q}$         A1

METHOD 2

$\left(\begin{array}{c}x\text{'}\\ y\text{'}\end{array}\right)=\mathit{P}\left(\begin{array}{c}x-a\\ y-b\end{array}\right)+\left(\begin{array}{c}a\\ b\end{array}\right)$         A1

Note: Accept substitution of $x$ and $y$ (and $x\text{'}$ and $y\text{'}$) with particular points given in the question.

[1 mark]

METHOD 1

solving $\left(\begin{array}{c}a\\ b\end{array}\right)=\mathit{P}\left(\begin{array}{c}a\\ b\end{array}\right)+\mathit{q}$ using simultaneous equations or $\mathit{a}={\left(\mathit{I}-\mathit{P}\right)}^{-1}\mathit{q}$         (M1)

$a=0.651 \left(0.651084\dots \right), b=1.48 \left(1.47662\dots \right)$        A1A1

$\left(a=\frac{5+2\sqrt{3}}{13}, b=\frac{14+3\sqrt{3}}{13}\right)$

METHOD 2

$\left(\begin{array}{c}0\\ 1\end{array}\right)=\mathit{P}\left(\begin{array}{c}0-a\\ 0-b\end{array}\right)+\left(\begin{array}{c}a\\ b\end{array}\right)$         (M1)

Note: This line, with any of the points substituted, may be seen in part (d)(i) and if so the M1 can be awarded there.

$\left(\begin{array}{c}0\\ 1\end{array}\right)=\left(\mathit{I}-\mathit{P}\right)\left(\begin{array}{c}a\\ b\end{array}\right)$

$a=0.651084\dots , b=1.47662\dots$        A1A1

$\left(a=\frac{5+2\sqrt{3}}{13}, b=\frac{14+3\sqrt{3}}{13}\right)$

[3 marks]

Part (i) proved to be straightforward for most candidates. A common error in part (ii) was for candidates to begin with the matrix P and to show it successfully transformed the points to their images. This received no marks. For a ‘show that’ question it is expected that the work moves to rather than from the given answer.

(b), (c) These two parts dealt generally with more familiar aspects of matrix transformations and were well done.

(b), (c) These two parts dealt generally with more familiar aspects of matrix transformations and were well done.

The trick of recognizing that $(a,b)$ was invariant was generally not seen and as such the question could not be successfully answered.

Use the substitution $y=\frac{dx}{dt}$ to show that this equation can be written as

$\left(\begin{array}{c}\frac{dx}{dt}\\ \frac{dy}{dt}\end{array}\right)=\left(\begin{array}{cc}0 & 1\\ -6 & -5\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)$.

Find the eigenvalues for the matrix $\left(\begin{array}{cc}0 & 1\\ -6 & -5\end{array}\right)$.

Hence state the long-term velocity of the particle.

Use the substitution $y=\frac{dx}{dt}$ to write the differential equation as a system of coupled, first order differential equations.

Use Euler’s method with a step length of $0.1$ to find the displacement of the particle when $t=1$.

Find the long-term velocity of the particle.

$y=\frac{dx}{dt}\Rightarrow \frac{dy}{dt}+5\frac{dx}{dt}+6x=0$   OR   $\frac{dy}{dt}+5y+6x=0$         M1

Note: Award M1 for substituting $\frac{dy}{dt}$ for $\frac{d^{2}x}{d{t}^2}$.

$\left(\begin{array}{c}\frac{dx}{dt}\\ \frac{dy}{dt}\end{array}\right)=\left(\begin{array}{cc}0 & 1\\ -6 & -5\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)$        AG

[1 mark]

$\text{det}\left(\begin{array}{cc}-\lambda & 1\\ -6 & -5-\lambda \end{array}\right)=0$         (M1)

Note: Award M1 for an attempt to find eigenvalues. Any indication that $\text{det}\left(\mathit{M}-\lambda \mathit{I}\right)=0$ has been used is sufficient for the (M1).

$-\lambda \left(-5-\lambda \right)+6=0$   OR   ${\lambda }^2+5\lambda +6=0$         (A1)

$\lambda =-2, -3$        A1

[3 marks]

(on a phase portrait the particle approaches $(0, 0)$ as $t$ increases so long term velocity ($y$) is)

$0$        A1

Note: Only award A1 for $0$ if both eigenvalues in part (a)(ii) are negative. If at least one is positive accept an answer of ‘no limit’ or ‘infinity’, or in the case of one positive and one negative also accept ‘no limit or $\mathit{0}$ (depending on initial conditions)’.

[1 mark]

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

$\frac{d^{2}x}{d{t}^2}=\frac{dy}{dt}$         (A1)

$\frac{dy}{dt}+5y+6x=3t+4$        A1

[2 marks]

recognition that $h=0.1$ in any recurrence formula           (M1)

$\left(t_{n+1}=t_n+0.1\right)$

$x_{n+1}=x_n+0.1y_n$           (A1)

$y_{n+1}=y_n+0.1\left(3t_n+4-5y_n-6x_n\right)$           (A1)

(when $t=1$,) $x=0.64402\dots \approx 0.644 \text{m}$        A2

[5 marks]

recognizing that $y$ is the velocity

$0.5 {\text{m\hspace{0.05em}s}}^{-1}$         A1 

[2 marks]

It was clear that second order differential equations had not been covered by many schools. Fortunately, many were able to successfully answer part (ii) as this was independent of the other two parts. For part (iii) it was expected that candidates would know that two negative eigenvalues mean the system tends to the origin and so the long-term velocity is 0. Some candidates tried to solve the system. It should be noted that when the command term is ‘state’ then no further working out is expected to be seen.

Forming a coupled system from a second order differential equation and solving it using Euler’s method is a technique included in the course guide. Candidates who had learned this technique were successful in this question.

the eigenvalues.

the eigenvectors.

Hence write down the general solution of the system of equations.

Sketch the phase portrait for this system, for $x, y>0$.

On your sketch show

• the equation of the line defined by the eigenvector in the first quadrant
• at least two trajectories either side of this line using arrows on those trajectories to represent the change in populations as t increases

Write down a condition on the size of the initial population of $\text{Y}$ if it is to avoid its population reducing to zero.

Find the value of $t$ at which $x=0$.

Find the population of $\text{Y}$ at this value of $t$. Give your answer to the nearest $10$ herbivores.

* 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.

$\left|\begin{array}{cc}0.3-\lambda & -0.1\\ -0.2& 0.4-\lambda \end{array}\right|=0$        (M1)(A1)

 $\lambda =0.5$ and $0.2$        A1

[3 marks]

Attempt to solve either

$\left(\begin{array}{cc}-0.2& -0.1\\ -0.2& -0.1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)$  or  $\left(\begin{array}{cc}0.1& -0.1\\ -0.2& 0.2\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)$

or equivalent        (M1)

$\left(\begin{array}{c}1\\ -2\end{array}\right)$ or $\left(\begin{array}{c}1\\ 1\end{array}\right)$        A1A1

Note: accept equivalent forms

[3 marks]

$\left(\begin{array}{c}x\\ y\end{array}\right)=A{\text{e}}^{0.5t}\left(\begin{array}{c}1\\ -2\end{array}\right)+B{\text{e}}^{0.2t}\left(\begin{array}{c}1\\ 1\end{array}\right)$      A1

[1 mark]

        A1A1A1

Note: A1 for $y=x$ correctly labelled, A1 for at least two trajectories above $y=x$ and A1 for at least two trajectories below $y=x$, including arrows.

[3 marks]

$y>2000$        A1

[1 mark]

$\left(\begin{array}{c}x\\ y\end{array}\right)=A{\mathit{\text{e}}}^{0.5t}\left(\begin{array}{c}1\\ -2\end{array}\right)+B{\mathit{\text{e}}}^{0.2t}\left(\begin{array}{c}1\\ 1\end{array}\right)$

At $t=0$  $2000=A+B, 2900=-2A+B$         M1A1

Note: Award M1 for the substitution of $2000$ and $2900$

Hence $A=-300, B=2300$        A1A1

$0=-300{\text{e}}^{0.5t}+2300{\text{e}}^{0.2t}$       M1

$t=6.79 \left(6.7896\dots \right)$ (years)        A1

[6 marks]

$y=600{\text{e}}^{0.5\times 6.79}+2300{\text{e}}^{0.2\times 6.79}$       (M1)

$=26827.9\dots$

$=26830$  (to the nearest $10$ animals)         A1

[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.

Find the amount Phil would owe the bank after 20 years. Give your answer to the nearest dollar.

Show that the total value of Phil’s savings after 20 years is $\frac{({1.02}^{20}-1)P}{(1.02-1)}$.

Given that Phil’s aim is to own the house after 20 years, find the value for $P$ to the nearest dollar.

David wishes to withdraw $5000 at the end of each year for a period of $n$ years. Show that an expression for the minimum value of $Q$ is

$\frac{5000}{1.028}+\frac{5000}{{1.028}^2}+\dots +\frac{5000}{{1.028}^n}$.

Hence or otherwise, find the minimum value of $Q$ that would permit David to withdraw annual amounts of $5000 indefinitely. Give your answer to the nearest dollar.

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

$150000\times {1.035}^{20}$     (M1)(A1)

$=\mathrm{\$}298468$     A1

Note:     Only accept answers to the nearest dollar. Accept $298469.

[3 marks]

attempt to look for a pattern by considering 1 year, 2 years etc     (M1)

recognising a geometric series with first term $P$ and common ratio 1.02     (M1)

EITHER

$P+1.02P+\dots +{1.02}^{19}P\left(=P(1+1.02+\dots +{1.02}^{19})\right)$     A1

OR

explicitly identify $u_1=P,r=1.02$ and $n=20$ (may be seen as $S_{20}$).     A1

THEN

$s_{20}=\frac{({1.02}^{20}-1)P}{(1.02-1)}$     AG

[3 marks]

$24.297\dots P=298468$     (M1)(A1)

$P=12284$     A1

Note:     Accept answers which round to 12284.

[3 marks]

METHOD 1

$Q({1.028}^n)=5000(1+1.028+{1.028}^2+{1.028}^3+\dots +{1.028}^{n-1})$     M1A1

$Q=\frac{5000\left(1+1.028+{1.028}^2+{1.028}^3+...+{1.028}^{n-1}\right)}{{1.028}^n}$    A1

$=\frac{5000}{1.028}+\frac{5000}{{1.028}^2}+\dots +\frac{5000}{{1.028}^n}$     AG

METHOD 2

the initial value of the first withdrawal is $\frac{5000}{1.028}$     A1

the initial value of the second withdrawal is $\frac{5000}{{1.028}^2}$     R1

the investment required for these two withdrawals is $\frac{5000}{1.028}+\frac{5000}{{1.028}^2}$     R1

$Q=\frac{5000}{1.028}+\frac{5000}{{1.028}^2}+\dots +\frac{5000}{{1.028}^n}$     AG

[3 Marks]

sum to infinity is $\frac{\frac{5000}{1.028}}{1-\frac{1}{1.028}}$     (M1)(A1)

$=178571.428\dots$

so minimum amount is $178572     A1

Note:     Accept answers which round to $178571 or $178572.

[3 Marks]

Given that $y=\dot{x}$, show that $\dot{y}=-1.25x-3y$.

Write down the matrix $\mathit{A}$.

Find the eigenvalues of matrix $\mathit{A}$.

Find the eigenvectors of matrix $\mathit{A}$.

Given that when $t=0$ the shock absorber is displaced $8 \text{cm}$ and its velocity is zero, find an expression for $x$ in terms of $t$.

$y=\dot{x}\Rightarrow \dot{y}=\ddot{x}$           A1

$\dot{y}+3\left(y\right)+1.25x=0$           R1

Note: If no explicit reference is made to $\dot{y}=\ddot{x}$, or equivalent, award A0R1 if second line is seen. If $\frac{dy}{dx}$ used instead of $\frac{dy}{dt}$, award A0R0.

$\dot{y}=-3y-1.25x$           AG

[2 marks]

$\mathit{A}=\left(\begin{array}{cc}0& 1\\ -1.25& -3\end{array}\right)$           A1

[1 mark]

$\left|\begin{array}{cc}-\lambda & 1\\ -1.25& -3-\lambda \end{array}\right|=0$           (M1)

$\lambda \left(\lambda +3\right)+1.25=0$           (A1)

$\lambda =-2.5 ; \lambda =-0.5$           A1

[3 marks]

$\left(\begin{array}{cc}2.5& 1\\ -1.25& -0.5\end{array}\right)\left(\begin{array}{c}a\\ b\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)$           (M1)

$2.5a+b=0$

$v_1=\left(\begin{array}{c}-2\\ 5\end{array}\right)$           A1

$\left(\begin{array}{cc}0.5& 1\\ -1.25& -2.5\end{array}\right)\left(\begin{array}{c}a\\ b\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)$

$0.5a+b=0$

$v_2=\left(\begin{array}{c}-2\\ 1\end{array}\right)$           A1

Note: Award M1 for a valid attempt to find either eigenvector. Accept equivalent forms of the eigenvectors.
Do not award FT for eigenvectors that do not satisfy both rows of the matrix.

[3 marks]

$\left(\begin{array}{c}x\\ y\end{array}\right)=A{\text{e}}^{-2.5t}\left(\begin{array}{c}-2\\ 5\end{array}\right)+B{\text{e}}^{-0.5t}\left(\begin{array}{c}-2\\ 1\end{array}\right)$           M1A1

$t=0 \Rightarrow x=8, \dot{x}=y=0$           (M1)

$-2A-2B=8$

$5A+B=0$           (M1)

$A=1 ; B=-5$           A1

$x=-2{\text{e}}^{-2.5t}+10{\text{e}}^{-0.5t}$           A1

Note: Do not award the final A1 if the answer is given in the form $\left(\begin{array}{c}x\\ y\end{array}\right)=A{\text{e}}^{-2.5t}\left(\begin{array}{c}-2\\ 5\end{array}\right)+B{\text{e}}^{-0.5t}\left(\begin{array}{c}-2\\ 1\end{array}\right)$.

[6 marks]

There were many good attempts at this problem, although simple errors often complicated things. In part (a) an explicit statement of the relationship between the second derivative of $x$ and the first derivative of $y$ was often  issing. Then in part (b) there seemed to be confusion about the matrix, with the correct values often placed in the wrong row or column of the matrix. Despite these errors, candidates made good attempts at finding eigenvalues and eigenvectors. It is to be noted that an error in solving the quadratic equation to find the eigenvectors means that follow-through marks are unlikely to be awarded since the eigenvectors are not reasonable answers and will not be consistent with the eigenvalues. Candidates need to take real care at this point of a question in part (c)(i). A significant number of candidates did not write down the final answer correctly, leaving their final answer in vector form, rather than “$x=$ ….” as asked for in the question.

There were many good attempts at this problem, although simple errors often complicated things. In part (a) an explicit statement of the relationship between the second derivative of $x$ and the first derivative of $y$ was often  issing. Then in part (b) there seemed to be confusion about the matrix, with the correct values often placed in the wrong row or column of the matrix. Despite these errors, candidates made good attempts at finding eigenvalues and eigenvectors. It is to be noted that an error in solving the quadratic equation to find the eigenvectors means that follow-through marks are unlikely to be awarded since the eigenvectors are not reasonable answers and will not be consistent with the eigenvalues. Candidates need to take real care at this point of a question in part (c)(i). A significant number of candidates did not write down the final answer correctly, leaving their final answer in vector form, rather than “$x=$ ….” as asked for in the question.

There were many good attempts at this problem, although simple errors often complicated things. In part (a) an explicit statement of the relationship between the second derivative of $x$ and the first derivative of $y$ was often  issing. Then in part (b) there seemed to be confusion about the matrix, with the correct values often placed in the wrong row or column of the matrix. Despite these errors, candidates made good attempts at finding eigenvalues and eigenvectors. It is to be noted that an error in solving the quadratic equation to find the eigenvectors means that follow-through marks are unlikely to be awarded since the eigenvectors are not reasonable answers and will not be consistent with the eigenvalues. Candidates need to take real care at this point of a question in part (c)(i). A significant number of candidates did not write down the final answer correctly, leaving their final answer in vector form, rather than “$x=$ ….” as asked for in the question.

There were many good attempts at this problem, although simple errors often complicated things. In part (a) an explicit statement of the relationship between the second derivative of $x$ and the first derivative of $y$ was often  issing. Then in part (b) there seemed to be confusion about the matrix, with the correct values often placed in the wrong row or column of the matrix. Despite these errors, candidates made good attempts at finding eigenvalues and eigenvectors. It is to be noted that an error in solving the quadratic equation to find the eigenvectors means that follow-through marks are unlikely to be awarded since the eigenvectors are not reasonable answers and will not be consistent with the eigenvalues. Candidates need to take real care at this point of a question in part (c)(i). A significant number of candidates did not write down the final answer correctly, leaving their final answer in vector form, rather than “$x=$ ….” as asked for in the question.

There were many good attempts at this problem, although simple errors often complicated things. In part (a) an explicit statement of the relationship between the second derivative of $x$ and the first derivative of $y$ was often  issing. Then in part (b) there seemed to be confusion about the matrix, with the correct values often placed in the wrong row or column of the matrix. Despite these errors, candidates made good attempts at finding eigenvalues and eigenvectors. It is to be noted that an error in solving the quadratic equation to find the eigenvectors means that follow-through marks are unlikely to be awarded since the eigenvectors are not reasonable answers and will not be consistent with the eigenvalues. Candidates need to take real care at this point of a question in part (c)(i). A significant number of candidates did not write down the final answer correctly, leaving their final answer in vector form, rather than “$x=$ ….” as asked for in the question.

Write down each of the transformations in matrix form, clearly stating which matrix represents each transformation.

Find a single matrix $\mathit{P}$ that defines a transformation that represents the overall change in position.

Find ${\mathit{P}}^2$.

Hence state what the value of ${\mathit{P}}^2$ indicates for the possible movement of the drone.

Three drones are initially positioned at the points $\text{A}$, $\text{B}$ and $\text{C}$. After performing the movements listed above, the drones are positioned at points $\text{A}\prime$, $\text{B}\prime$ and $\text{C}\prime$ respectively.

Show that the area of triangle $\text{ABC}$ is equal to the area of triangle $\text{A}\prime \text{B}\prime \text{C}\prime$ .

Find a single transformation that is equivalent to the three transformations represented by matrix $\mathit{P}$.

Note: For clarity, exact answers are used throughout this markscheme. However it is perfectly acceptable for candidates to write decimal values $\left(\text{e.g.} \frac{\sqrt{3}}{2}=0.866\right)$.

rotation anticlockwise $\frac{\mathrm{\pi }}{6}$ is $\left(\begin{array}{cc}0.866& -0.5\\ 0.5& 0.866\end{array}\right)$  OR  $\left(\begin{array}{cc}\frac{\sqrt{3}}{2}& -\frac{1}{2}\\ \frac{1}{2}& \frac{\sqrt{3}}{2}\end{array}\right)$           (M1)A1

reflection in $y=\frac{x}{\sqrt{3}}$

$\tan \theta =\frac{1}{\sqrt{3}}$           (M1)

$\Rightarrow 2\theta =\frac{\mathrm{\pi }}{3}$           (A1)

matrix is $\left(\begin{array}{cc}0.5& 0.866\\ 0.866& -0.5\end{array}\right)$  OR  $\left(\begin{array}{cc}\frac{1}{2}& \frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}& -\frac{1}{2}\end{array}\right)$            A1

rotation clockwise $\frac{\mathrm{\pi }}{3}$ is $\left(\begin{array}{cc}0.5& 0.866\\ -0.866& 0.5\end{array}\right)$  OR  $\left(\begin{array}{cc}\frac{1}{2}& \frac{\sqrt{3}}{2}\\ -\frac{\sqrt{3}}{2}& \frac{1}{2}\end{array}\right)$            A1

[6 marks]

Note: For clarity, exact answers are used throughout this markscheme. However it is perfectly acceptable for candidates to write decimal values $\left(\text{e.g.} \frac{\sqrt{3}}{2}=0.866\right)$.

an attempt to multiply three matrices           (M1)

$\mathit{P}=\left(\begin{array}{cc}\frac{1}{2}& \frac{\sqrt{3}}{2}\\ -\frac{\sqrt{3}}{2}& \frac{1}{2}\end{array}\right)\left(\begin{array}{cc}\frac{1}{2}& \frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}& -\frac{1}{2}\end{array}\right)\left(\begin{array}{cc}\frac{\sqrt{3}}{2}& -\frac{1}{2}\\ \frac{1}{2}& \frac{\sqrt{3}}{2}\end{array}\right)$           (A1)

$\mathit{P}=\left(\begin{array}{cc}\frac{\sqrt{3}}{2}& -\frac{1}{2}\\ -\frac{1}{2}& -\frac{\sqrt{3}}{2}\end{array}\right)$  OR  $\left(\begin{array}{cc}0.866& -0.5\\ -0.5& -0.866\end{array}\right)$            A1

[3 marks]

Note: For clarity, exact answers are used throughout this markscheme. However it is perfectly acceptable for candidates to write decimal values $\left(\text{e.g.} \frac{\sqrt{3}}{2}=0.866\right)$.

$\left({\mathit{P}}^2=\left(\begin{array}{cc}\frac{\sqrt{3}}{2}& -\frac{1}{2}\\ -\frac{1}{2}& -\frac{\sqrt{3}}{2}\end{array}\right)\left(\begin{array}{cc}\frac{\sqrt{3}}{2}& -\frac{1}{2}\\ -\frac{1}{2}& -\frac{\sqrt{3}}{2}\end{array}\right)=\right) \left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$            A1

Note: Do not award A1 if final answer not resolved into the identity matrix $I$.

[1 mark]

Note: For clarity, exact answers are used throughout this markscheme. However it is perfectly acceptable for candidates to write decimal values $\left(\text{e.g.} \frac{\sqrt{3}}{2}=0.866\right)$.

if the overall movement of the drone is repeated          A1

the drone would return to its original position          A1

[2 marks]

Note: For clarity, exact answers are used throughout this markscheme. However it is perfectly acceptable for candidates to write decimal values $\left(\text{e.g.} \frac{\sqrt{3}}{2}=0.866\right)$.

METHOD 1

$\left|\text{det} \mathit{P}\right|=\left|\left(-\frac{3}{4}\right)-\left(\frac{1}{4}\right)\right|=1$            A1

area of triangle $\text{ABC}=$ area of triangle $\text{A}\prime \text{B}\prime \text{C}\prime$ $\times \left|\text{det} \mathit{P}\right|$            R1

area of triangle $\text{ABC}=$ area of triangle $\text{A}\prime \text{B}\prime \text{C}\prime$            AG

Note: Award at most A1R0 for responses that omit modulus sign.

METHOD 2

statement of fact that rotation leaves area unchanged            R1

statement of fact that reflection leaves area unchanged            R1

area of triangle $\text{ABC}=$ area of triangle $\text{A}\prime \text{B}\prime \text{C}\prime$            AG

[2 marks]

Note: For clarity, exact answers are used throughout this markscheme. However it is perfectly acceptable for candidates to write decimal values $\left(\text{e.g.} \frac{\sqrt{3}}{2}=0.866\right)$.

attempt to find angles associated with values of elements in matrix $\mathit{P}$            (M1)

$\left(\begin{array}{cc}\frac{\sqrt{3}}{2}& -\frac{1}{2}\\ -\frac{1}{2}& -\frac{\sqrt{3}}{2}\end{array}\right)=\left(\begin{array}{cc}\cos \left(-\frac{\mathrm{\pi }}{6}\right)& \sin \left(-\frac{\mathrm{\pi }}{6}\right)\\ \sin \left(-\frac{\mathrm{\pi }}{6}\right)& -\cos \left(-\frac{\mathrm{\pi }}{6}\right)\end{array}\right)$

reflection (in $y=\left(\tan \theta \right)x$)            (M1)

where $2\theta =-\frac{\mathrm{\pi }}{6}$            A1

reflection in $y=\tan \left(-\frac{\mathrm{\pi }}{12}\right)x \left(=-0.268x\right)$            A1

[4 marks]

There were many good attempts at this problem. In most cases these good attempts were undermined by a key lack of understanding. Whilst candidates were able to find the correct matrices in part (a)(i), they then invariably went onto multiply the matrices in the wrong order in part (a)(ii). Whilst follow through marks were readily available after this, the incorrect matrix for $\mathit{P}$ then caused issues in part (c). If these candidates had multiplied correctly, it seems that many of them could have gained close to full marks on this question. At the same time there was a lack of precision in the description of the transformation in part (c). As a general point, it would also help candidates if they resolved the trig ratios in the matrices; writing $0.5$ or $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ rather than, for example $\cos \left(\frac{\mathrm{\pi }}{3}\right)$. Finally, there were many attempts in part (c) that suggested candidates had a good knowledge and understanding of the concepts of matrices and affine transformations.

There were many good attempts at this problem. In most cases these good attempts were undermined by a key lack of understanding. Whilst candidates were able to find the correct matrices in part (a)(i), they then invariably went onto multiply the matrices in the wrong order in part (a)(ii). Whilst follow through marks were readily available after this, the incorrect matrix for $\mathit{P}$ then caused issues in part (c). If these candidates had multiplied correctly, it seems that many of them could have gained close to full marks on this question. At the same time there was a lack of precision in the description of the transformation in part (c). As a general point, it would also help candidates if they resolved the trig ratios in the matrices; writing $0.5$ or $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ rather than, for example $\cos \left(\frac{\mathrm{\pi }}{3}\right)$. Finally, there were many attempts in part (c) that suggested candidates had a good knowledge and understanding of the concepts of matrices and affine transformations.

There were many good attempts at this problem. In most cases these good attempts were undermined by a key lack of understanding. Whilst candidates were able to find the correct matrices in part (a)(i), they then invariably went onto multiply the matrices in the wrong order in part (a)(ii). Whilst follow through marks were readily available after this, the incorrect matrix for $\mathit{P}$ then caused issues in part (c). If these candidates had multiplied correctly, it seems that many of them could have gained close to full marks on this question. At the same time there was a lack of precision in the description of the transformation in part (c). As a general point, it would also help candidates if they resolved the trig ratios in the matrices; writing $0.5$ or $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ rather than, for example $\cos \left(\frac{\mathrm{\pi }}{3}\right)$. Finally, there were many attempts in part (c) that suggested candidates had a good knowledge and understanding of the concepts of matrices and affine transformations.

There were many good attempts at this problem. In most cases these good attempts were undermined by a key lack of understanding. Whilst candidates were able to find the correct matrices in part (a)(i), they then invariably went onto multiply the matrices in the wrong order in part (a)(ii). Whilst follow through marks were readily available after this, the incorrect matrix for $\mathit{P}$ then caused issues in part (c). If these candidates had multiplied correctly, it seems that many of them could have gained close to full marks on this question. At the same time there was a lack of precision in the description of the transformation in part (c). As a general point, it would also help candidates if they resolved the trig ratios in the matrices; writing $0.5$ or $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ rather than, for example $\cos \left(\frac{\mathrm{\pi }}{3}\right)$. Finally, there were many attempts in part (c) that suggested candidates had a good knowledge and understanding of the concepts of matrices and affine transformations.

There were many good attempts at this problem. In most cases these good attempts were undermined by a key lack of understanding. Whilst candidates were able to find the correct matrices in part (a)(i), they then invariably went onto multiply the matrices in the wrong order in part (a)(ii). Whilst follow through marks were readily available after this, the incorrect matrix for $\mathit{P}$ then caused issues in part (c). If these candidates had multiplied correctly, it seems that many of them could have gained close to full marks on this question. At the same time there was a lack of precision in the description of the transformation in part (c). As a general point, it would also help candidates if they resolved the trig ratios in the matrices; writing $0.5$ or $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ rather than, for example $\cos \left(\frac{\mathrm{\pi }}{3}\right)$. Finally, there were many attempts in part (c) that suggested candidates had a good knowledge and understanding of the concepts of matrices and affine transformations.

There were many good attempts at this problem. In most cases these good attempts were undermined by a key lack of understanding. Whilst candidates were able to find the correct matrices in part (a)(i), they then invariably went onto multiply the matrices in the wrong order in part (a)(ii). Whilst follow through marks were readily available after this, the incorrect matrix for $\mathit{P}$ then caused issues in part (c). If these candidates had multiplied correctly, it seems that many of them could have gained close to full marks on this question. At the same time there was a lack of precision in the description of the transformation in part (c). As a general point, it would also help candidates if they resolved the trig ratios in the matrices; writing $0.5$ or $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ rather than, for example $\cos \left(\frac{\mathrm{\pi }}{3}\right)$. Finally, there were many attempts in part (c) that suggested candidates had a good knowledge and understanding of the concepts of matrices and affine transformations.

Write down the value of $q$.

Show that $27m+9n+3p=18$.

Write down the other two linear equations in $m$, $n$ and $p$.

Write down these three equations as a matrix equation.

Solve this matrix equation.

The function $f$ can also be written $f(x)=x\left(x-1\right)\left(rx-s\right)$ where $r$ and $s$ are integers. Find $r$ and $s$.

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

$q$ = 0     A1  N1

[1 mark]

Attempting to substitute (3, 18)          (M1)

$m{3}^3+n{3}^2+p3=18$       A1

$27m+9n+3p=18$        AG  N0

[2 marks]

$m$ + $n$ + $p$ = 0      A1    N1

−$m$ + $n$ − $p$ = −10      A1    N1

[2 marks]

Evidence of attempting to set up a matrix equation                (M1)

Correct matrix equation representing the given equations          A2   N3

eg 

[3 marks]

$\left(\begin{array}{c}2\\ -5\\ 3\end{array}\right)$               A1A1A1    N3

[3 marks]

Factorizing       (M1)

eg $f(x)=x\left(2x^2-5x+3\right)$,  $f(x)=\left(x^2-x\right)\left(rx-s\right)$ 

$r=2$  $s=3$      (accept $f(x)=x\left(x-1\right)\left(2x-3\right)$)              A1A1    N3

[3 marks]

Show that $abc=8$.

Show that one of the real roots is equal to 1.

Given that $q=8d^2$, find the other two real roots.

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

recognition of the other root $=-d\text{i}$       (A1)

$\text{lo}{\text{g}}_{2}a+\text{lo}{\text{g}}_{2}b+\text{lo}{\text{g}}_{2}c+d\text{i}-d\text{i}=3$        M1A1

Note: Award M1 for sum of the roots, A1 for 3. Award A0M1A0 for just $\text{lo}{\text{g}}_{2}a+\text{lo}{\text{g}}_{2}b+\text{lo}{\text{g}}_{2}c=3$.

$\text{lo}{\text{g}}_{2}abc=3$       (M1)

$\Rightarrow abc=2^3$       A1

$abc=8$       AG

[5 marks]

METHOD 1

let the geometric series be $u_1$, $u_{1}r$, $u_1{r}^2$

${\left(u_{1}r\right)}^3=8$      M1

$u_{1}r=2$       A1

hence one of the roots is $\text{lo}{\text{g}}_{2}2=1$      R1

METHOD 2

$\frac{b}{a}=\frac{c}{b}$

$b^2=ac\Rightarrow b^3=abc=8$      M1

$b=2$       A1

hence one of the roots is $\text{lo}{\text{g}}_{2}2=1$      R1

[3 marks]

METHOD 1

product of the roots is $r_1\times r_2\times 1\times d\text{i}\times -d\text{i}=-8d^2$       (M1)(A1)

$r_1\times r_2=-8$       A1

sum of the roots is $r_1+r_2+1+d\text{i}+-d\text{i}=3$       (M1)(A1)

$r_1+r_2=2$       A1

solving simultaneously       (M1)

$r_1=-2$, $r_2=4$       A1A1

METHOD 2

product of the roots $\text{lo}{\text{g}}_{2}a\times \text{lo}{\text{g}}_{2}b\times \text{lo}{\text{g}}_{2}c\times d\text{i}\times -d\text{i}=-8d^2$       M1A1

$\text{lo}{\text{g}}_{2}a\times \text{lo}{\text{g}}_{2}b\times \text{lo}{\text{g}}_{2}c=-8$       A1

EITHER

$a$, $b$, $c$ can be written as $\frac{2}{r}$, $2$, $2r$       M1

$\left(\text{lo}{\text{g}}_2\frac{2}{r}\right)\left(\text{lo}{\text{g}}_{2}2\right)\left(\text{lo}{\text{g}}_{2}2r\right)=-8$

attempt to solve       M1

$\left(1-\text{lo}{\text{g}}_{2}r\right)\left(1+\text{lo}{\text{g}}_{2}r\right)=-8$

$\text{lo}{\text{g}}_{2}r=\pm 3$

$r=\frac{1}{8}\text{,}8$       A1A1

OR

$a$, $b$, $c$ can be written as $a$, $2$, $\frac{4}{a}$      M1

$\left(\text{lo}{\text{g}}_{2}a\right)\left(\text{lo}{\text{g}}_{2}2\right)\left(\text{lo}{\text{g}}_2\frac{4}{a}\right)=-8$

attempt to solve       M1

$a=\frac{1}{4}\text{,}16$       A1A1

THEN

$a$ and $c$ are $\frac{1}{4}\text{,}16$       (A1)

roots are −2, 4       A1

[9 marks]

Plot the position of $z$ on an Argand Diagram.

Express $z$ in the form $z=a{\text{e}}^{\text{i}b}$, where $a, b\in \mathrm{ℝ}$, giving the exact value of $a$ and the exact value of $b$.

Find $w_1+w_2$ in the form ${\text{e}}^{\text{i}x}\left(c+\text{i}d\right)$.

Hence find $\text{Re}\left(w_1+w_2\right)$ in the form $A \cos \left(x-a\right)$, where $A>0$ and $0<a\le \frac{\mathrm{\pi }}{2}$.

Find the maximum value of $I_{\text{T}}$.

Find the phase shift of $I_{\text{T}}$.

         A1

[1 mark]

$z=\sqrt{2}{\text{e}}^{\frac{\text{i}\mathrm{\pi }}{4}}$          A1A1

Note: Accept an argument of $\frac{7\mathrm{\pi }}{4}$. Do NOT accept answers that are not exact.

[2 marks]

$w_1+w_2={\text{e}}^{\text{i}x}+{\text{e}}^{\text{i}\left(x-\frac{\mathrm{\pi }}{2}\right)}$

                $={\text{e}}^{\text{i}x}\left(1+{\text{e}}^{-\frac{\text{i}\mathrm{\pi }}{2}}\right)$          (M1)

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

[2 marks]

$w_1+w_2={\text{e}}^{\text{i}x}\times \sqrt{2}{\text{e}}^{-\frac{\text{i}\mathrm{\pi }}{4}}$           M1