Find the eigenvalues for $M$. For each eigenvalue find the set of associated eigenvectors.

Show that the matrix equation $\left(xy\right)\mathbf{M}\left(\begin{array}{c}x\\ y\end{array}\right)=\left(6\right)$ is equivalent to the Cartesian equation $\frac{5}{2}{x}^2+xy+\frac{5}{2}{y}^2=6$.

Show that $\left(\begin{array}{c}\frac{1}{\sqrt{2}}\\ \frac{-1}{\sqrt{2}}\end{array}\right)$ and $\left(\begin{array}{c}\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}\end{array}\right)$ are unit eigenvectors and that they correspond to different eigenvalues.

Hence, show that .

Find matrix R.

Show that .

Verify that .

Hence, find the Cartesian equation satisfied by $X$ and $Y$.

Find the Cartesian equation satisfied by $u$ and $v$ and state the geometric shape that this curve represents.

State geometrically what transformation the matrix $\mathbf{R}$ represents.

the curve in $X$ and $Y$ in part (e) (ii), giving a reason.

the curve in $x$ and $y$ in part (b).

Write down the equations of two lines of symmetry for the curve in $x$ and $y$ in part (b).

       M1M1A1A1

$\lambda =2$           eigenvalues are of the form $t\left(\begin{array}{c}1\\ -1\end{array}\right)$       M1A1

$\lambda =3$          eigenvalues are of the form $t\left(\begin{array}{c}1\\ 1\end{array}\right)$       M1A1

[8 marks]

      M1A1

$\Rightarrow \left(\frac{5}{2}{x}^2+\frac{1}{2}xy+\frac{1}{2}xy+\frac{5}{2}{y}^2\right)=\left(6\right)\Rightarrow \frac{5}{2}{x}^2+xy+\frac{5}{2}{y}^2=6.$       AG

[2 marks]

$\left(\begin{array}{c}\frac{1}{\sqrt{2}}\\ \frac{-1}{\sqrt{2}}\end{array}\right)=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\ -1\end{array}\right)$ corresponding to $\lambda =2$,     $\left(\begin{array}{c}\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}\end{array}\right)=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\ 1\end{array}\right)$ corresponding to $\lambda =3$      R1R1

[2 marks]

      A1AG

[1 mark]

Determinant is 1.           M1A1

[2 marks]

 so post multiplying by $\mathbf{R}$ gives        M1AG

[1 mark]

         M1A1

and  completing the proof       A1AG

[3 marks]

$\Rightarrow \left(2X^2+3Y^2\right)=\left(6\right)\Rightarrow \frac{X^2}{3}+\frac{Y^2}{2}=1$       M1A1

[2 marks]

$\frac{X}{\sqrt{3}}=u\text{,}\frac{Y}{\sqrt{2}}=v\Rightarrow u^2+v^2=1$, a circle (centre at the origin radius of 1)     A1A1

[2 marks]

A rotation about the origin through an angle of 45° anticlockwise.    A1A1

[2 marks]

an ellipse, since the matrix represents a vertical and a horizontal stretch    R1A1

[2 marks]

an ellipse      A1

[1 mark]

$y=x$, $y=-x$      A1A1

[2 marks]

Find an expression for the width of $U_n$ in centimetres.

Given the width of a pixel is approximately $0.025 \text{cm}$, find the number of squares in the final image.

Write down ${\mathit{A}}_1$.

Write down ${\mathit{A}}_n$, in terms of $n$.

state the coordinates, $\left(x_1, y_1\right)$, of its image in $U_1$.

hence find ${\mathit{b}}_1$.

show that ${\mathit{b}}_n=\left(\begin{array}{c}8\left(1-2^{-n}\right)\\ 8\left(1-2^{-n}\right)\end{array}\right)$.

Hence or otherwise, find the coordinates of the top left-hand corner in $U_7$.

Find, $s$, in the form $s\left(\theta \right)=m\theta +c$.

Write down $E_{\theta }$.

Hence find the image of $(1, 1)$ after it is rotated $135°$ and enlarged.

Find the value of $\theta$ at which the enlargement scale factor equals zero.

After the enlargement scale factor equals zero, Ben continues to rotate the image for another two revolutions.

Describe the animation for these two revolutions, stating the final position of the sequence of squares.

* 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.

$4\left(\frac{1}{2^n}\right)$       M1A1

[2 marks]

$\frac{4}{2^n}>0.025$        (A1)

$2^n<160$

$n\le 7$        (A1)

Note: Accept equations in place of inequalities.  

Hence there are $8$ squares        A1

[3 marks]

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

[1 mark]

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

[1 mark]

$\left(4, 4\right)$        A1

[1 mark]

${\mathit{A}}_1\left(\begin{array}{c}0\\ 0\end{array}\right)+{\mathit{b}}_1=\left(\begin{array}{c}4\\ 4\end{array}\right)$        (M1)

 ${\mathit{b}}_1=\left(\begin{array}{c}4\\ 4\end{array}\right)$        A1

[2 marks]

Recognise the geometric series ${\mathit{b}}_n=\left(\begin{array}{c}4+2+1+\dots \\ 4+2+1+\dots \end{array}\right)$        M1

Each component is equal to $\frac{4\left(1-{\displaystyle \frac{1}{2^n}}\right)}{{\displaystyle \frac{1}{2}}} \left(=8\left(1-\frac{{\displaystyle 1}}{{\displaystyle 2^n}}\right)\right)$        M1A1

$\left(\begin{array}{c}8\left(1-\frac{1}{2^n}\right)\\ 8\left(1-\frac{1}{2^n}\right)\end{array}\right)$        AG

[3 marks]

$\left(\begin{array}{cc}\frac{1}{2^7}& 0\\ 0& \frac{1}{2^7}\end{array}\right)\left(\begin{array}{c}0\\ 4\end{array}\right)+\left(\begin{array}{c}8\left(1-\frac{1}{2^7}\right)\\ 8\left(1-\frac{1}{2^7}\right)\end{array}\right)$        M1A1

$\left(7.9375, 7.96875\right)$        A1

[3 marks]

$s\left(\theta \right)=m\theta +c$

$s\left(0\right)=1, c=1$        M1A1

$s\left(360\right)=\frac{1}{2}$        A1

$\frac{1}{2}=360m+1\Rightarrow m=-\frac{1}{720}$        A1

$s\left(\theta \right)=-\frac{\theta }{720}+1$

[4 marks]

$E_{\theta }=\left(\begin{array}{cc}-\frac{\theta }{720}+1& 0\\ 0& -\frac{\theta }{720}+1\end{array}\right)$        A1

[1 mark]

$\left(\begin{array}{cc}-\frac{135}{720}+1& 0\\ 0& -\frac{135}{720}+1\end{array}\right)\left(\begin{array}{cc}\cos 135°& -\sin 135°\\ \sin 135°& \cos 135°\end{array}\right)\left(\begin{array}{c}1\\ 1\end{array}\right)$        M1A1A1

 $\left(-1.15, 0\right)$        A1

[4 marks]

$\theta =720°$      A1

[1 mark]

The image will expand from zero (accept equivalent answers)

It will rotate counter clockwise

The design will (re)appear in the opposite (third) quadrant         A1A1

Note: Accept any two of the above

Its final position will be in the opposite (third) quadrant or $180˚$ from its original position or equivalent statement.         A1

[3 marks]

Show that $b\approx 0.00939$.

Find the angle through which Mars rotates on its axis each hour.

Show that the maximum value of $\omega =1.98$, correct to three significant figures.

Find the minimum value of $\omega$.

the maximum value of $R\left(t\right)$.

the minimum value of $R\left(t\right)$.

Hence show that $a=1.6$, correct to two significant figures.

Find the value of $c$.

Find the value of $d$.

Write down $z_1$ and $z_2$ in exponential form, with a constant modulus.

Hence or otherwise find an equation for $L$ in the form $L\left(t\right)=p \sin (qt+r)+d$, where $p, q, r, d\in \mathrm{ℝ}$.

Find, in hours, the shortest time from sunrise to sunset at point $\text{A}$ that is predicted by this model.

recognition that period $=669$                  (M1)

$b=\frac{2\mathrm{\pi }}{669}$  OR  $b=0.00939190\dots$                A1

Note: Award A1 for a correct expression leading to the given value or for a correct value of $b$ to 4 sf or greater accuracy.

$b\approx 0.00939$             AG

[2 marks]

length of day$=24\frac{2}{3}$ hours                (A1)

Note: Award A1 for $\frac{2}{3}, 0.666\dots , 0.\overline{6}$ or $0.667$.

$\frac{2\mathrm{\pi }}{24{\displaystyle \frac{2}{3}}}$               (M1)

Note: Accept $\left(\frac{360}{24{\displaystyle \frac{2}{3}}}\right)$.

$=0.255$ radians $\left(0.254723\dots , \frac{3\mathrm{\pi }}{37}, 14.5945\dots °\right)$               A1

[3 marks]

substitution of either value of $\delta$ into equation              (M1)

correct use of arccos to find a value for $\omega$              (M1)

Note: Both (M1) lines may be seen in either part (c)(i) or part (c)(ii).

$\cos \omega =0.839 \tan \left(-0.440\right)$               A1

$\omega =1.97684\dots$

$\approx 1.98$               AG

Note: For substitution of $1.98$ award M0A0.

[3 marks]

$\delta =0.440$

$\omega =1.16 (1.16474\dots )$           A1

[1 mark]

$R_{\text{max}}=\frac{1.97684\dots }{0.25472\dots }$           (M1)

$=7.76 \text{hours} (7.76075\dots )$           A1

Note: Accept $7.70$ from use of $1.98$.

[2 marks]

$R_{\text{min}}=\frac{1.16474\dots }{0.25472\dots }$

$=4.57 \text{hours} (4.57258\dots )$           A1

Note: Accept $4.55$ and $4.56$ from use of rounded values.

[1 mark]

$a=\frac{7.76075\dots -4.57258\dots }{2}$           M1

$\approx 1.59408\dots$           A1

Note: Award M1 for substituting their values into a correct expression. Award A1 for a correct value of $a$ from their expression which has at least 3 significant figures and rounds correctly to $1.6$.

$\approx 1.6$ (correct to $2$ sf)          AG

[2 marks]

EITHER

$c=\frac{7.76075\dots +4.57258\dots }{2} \left(=\frac{12.333\dots }{2}\right)$           (M1)

OR

$c=4.57258\dots +1.59408\dots$  or  $c=7.76075\dots -1.59408\dots$

THEN

$=6.17 \left(6.16666\dots \right)$           A1

Note: Accept $6.16$ from use of rounded values. Follow through on their answers to part (d) and $1.6$.


[2 marks]

$d=18.65-6.16666\dots$           (M1)

$=12.5 \left(12.4833\dots \right)$           A1

Note: Follow through for $18.65$ minus their answer to part (f).


[2 marks]

at least one expression in the form $r{\text{e}}^{g\left(t\right)\text{i}}$           (M1)

$z_1=1.5{\text{e}}^{\left(0.00939t+2.83\right)\text{i}}\mathrm{,} {\mathrm{z}}_2=1.6{\text{e}}^{\left(0.00939t\right)\text{i}}$           A1A1

[3 marks]

EITHER

$z_1-z_2=1.5{\text{e}}^{\left(0.00939t+2.83\right)\text{i}}-1.6{\text{e}}^{\left(0.00939t\right)\text{i}}$

$={\text{e}}^{0.00939t\text{i}}\left(1.5{\text{e}}^{2.83\text{i}}-1.6\right)$               (M1)

$={\text{e}}^{0.00939t\text{i}}\left(3.06249\dots {\text{e}}^{2.99086\dots \text{i}}\right)$            (A1)(A1)

OR

graph of $L$ or $f$

$p=3.06249...$            (A1)

$r=-0.150729...$  OR  $r=2.99086...$            (M1)(A1)

Note: The $p$ and $r$ variables (or equivalent) must be seen.


THEN

$L\left(t\right)=3.06 \sin (0.00939t+2.99)+12.5$                 A1

$\left(L\left(t\right)=3.06248\dots \sin (0.00939t+2.99086\dots )+12.4833\dots \right)$

Note: Accept equivalent forms, e.g. $L\left(t\right)=3.06 \sin (0.00939t-0.151)+12.5$.
Follow through on their answer to part (g) replacing $12.5$.

 
[4 marks]

shortest time between sunrise and sunset

$12.4833\dots -3.06249\dots$               (M1)

$=9.42$ hours  $\left(9.420843\dots \right)$                 A1

Note: Accept $9.44$ from use of $3$ sf values.

[2 marks]