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HL Paper 3

This question will diagonalize a matrix and apply this to the transformation of a curve.

Let the matrix $M = \left( {\begin{array}{*{20}{c}} {\tfrac{5}{2}}&{\tfrac{1}{2}} \\ {\tfrac{1}{2}}&{\tfrac{5}{2}} \end{array}} \right)$ .

Let $\left( {\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} {\tfrac{1}{{\sqrt 2 }}}&{\tfrac{1}{{\sqrt 2 }}} \end{array}} \\ {\begin{array}{*{20}{c}} {\tfrac{{ - 1}}{{\sqrt 2 }}}&{\tfrac{1}{{\sqrt 2 }}} \end{array}} \end{array}} \right) = {{\mathbf{R}}^{ - 1}}$ .

Let ${\mathbf{R}}\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) = \left( {\begin{array}{*{20}{c}} X \\ Y \end{array}} \right)$ .

Let $\left( {\begin{array}{*{20}{c}} {\tfrac{1}{{\sqrt 3 }}}&0 \\ 0&{\tfrac{1}{{\sqrt 2 }}} \end{array}} \right)\left( {\begin{array}{*{20}{c}} X \\ Y \end{array}} \right) = \left( {\begin{array}{*{20}{c}} u \\ v \end{array}} \right)$ .

Hence state the geometrical shape represented by

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

[8]

Show that the matrix equation $\left( {x\,\,\,y} \right){\mathbf{M}}\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) = \left( 6 \right)$ is equivalent to the Cartesian equation $\frac{5}{2}{x^2} + xy + \frac{5}{2}{y^2} = 6$ .

[2]

Show that $\left( {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}} \\ {\frac{{ - 1}}{{\sqrt 2 }}} \end{array}} \right)$ and $\left( {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}} \\ {\frac{1}{{\sqrt 2 }}} \end{array}} \right)$ are unit eigenvectors and that they correspond to different eigenvalues.

[2]

c.i.

Hence, show that $M\left( {\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}} \end{array}} \\ {\begin{array}{*{20}{c}} {\frac{{ - 1}}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}} \end{array}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}} \end{array}} \\ {\begin{array}{*{20}{c}} {\frac{{ - 1}}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}} \end{array}} \end{array}} \right)\left( {\begin{array}{*{20}{c}} 2&0 \\ 0&3 \end{array}} \right)$ .

[1]

c.ii.

Find matrix R.

[2]

d.i.

Show that ${\mathbf{M}} = {{\mathbf{R}}^{ - 1}}\left( {\begin{array}{*{20}{c}} 2&0 \\ 0&3 \end{array}} \right){\mathbf{R}}$ .

[1]

d.ii.

Verify that $\left( {\begin{array}{*{20}{c}} X&Y \end{array}} \right) = \left( {\begin{array}{*{20}{c}} x&y \end{array}} \right){{\mathbf{R}}^{ - 1}}$ .

[3]

e.i.

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

[2]

e.ii.

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

[2]

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

[2]

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

[2]

h.i.

the curve in $x$ and $y$ in part (b).

[1]

h.ii.

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

[2]

Markscheme

$\left| {\begin{array}{*{20}{c}} {\frac{5}{2} - \lambda }&{\frac{1}{2}} \\ {\frac{1}{2}}&{\frac{5}{2} - \lambda } \end{array}} \right| = 0 \Rightarrow {\left( {\frac{5}{2} - \lambda } \right)^2} - {\left( {\frac{1}{2}} \right)^2} = 0 \Rightarrow \frac{5}{2} - \lambda = \pm \frac{1}{2} \Rightarrow \lambda = 2\,\,{\text{or}}\,\,{\text{3}}$ M1M1A1A1

$\lambda = 2$ $\left( {\begin{array}{*{20}{c}} {\frac{1}{2}}&{\frac{1}{2}} \\ {\frac{1}{2}}&{\frac{1}{2}} \end{array}} \right)\left( {\begin{array}{*{20}{c}} p \\ q \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0 \\ 0 \end{array}} \right) \Rightarrow q = - p$ eigenvalues are of the form $t\left( {\begin{array}{*{20}{c}} 1 \\ { - 1} \end{array}} \right)$ M1A1

$\lambda = 3$ $\left( {\begin{array}{*{20}{c}} { - \frac{1}{2}}&{\frac{1}{2}} \\ {\frac{1}{2}}&{ - \frac{1}{2}} \end{array}} \right)\left( {\begin{array}{*{20}{c}} p \\ q \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0 \\ 0 \end{array}} \right) \Rightarrow q = p$ eigenvalues are of the form $t\left( {\begin{array}{*{20}{c}} 1 \\ {1} \end{array}} \right)$ M1A1

[8 marks]

$\left( {x\,\,\,y} \right)\left( {\begin{array}{*{20}{c}} {\frac{5}{2}}&{\frac{1}{2}} \\ {\frac{1}{2}}&{\frac{5}{2}} \end{array}} \right)\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) = \left( 6 \right) \Rightarrow \left( {\begin{array}{*{20}{c}} {\frac{5}{2}x + \frac{1}{2}y}&{\frac{1}{2}x + \frac{5}{2}y} \end{array}} \right)\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) = \left( 6 \right)$ 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]

$\left( {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}} \\ {\frac{{ - 1}}{{\sqrt 2 }}} \end{array}} \right) = \frac{1}{{\sqrt 2 }}\left( {\begin{array}{*{20}{c}} 1 \\ { - 1} \end{array}} \right)$ corresponding to $\lambda = 2$ , $\left( {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}} \\ {\frac{1}{{\sqrt 2 }}} \end{array}} \right) = \frac{1}{{\sqrt 2 }}\left( {\begin{array}{*{20}{c}} 1 \\ 1 \end{array}} \right)$ corresponding to $\lambda = 3$ R1R1

[2 marks]

c.i.

$M\left( {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}} \\ {\frac{{ - 1}}{{\sqrt 2 }}} \end{array}} \right) = 2\left( {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}} \\ {\frac{{ - 1}}{{\sqrt 2 }}} \end{array}} \right)\,\,{\text{and}}\,\,M\left( {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}} \\ {\frac{1}{{\sqrt 2 }}} \end{array}} \right) = 3\left( {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}} \\ {\frac{1}{{\sqrt 2 }}} \end{array}} \right) \Rightarrow M\left( {\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}} \end{array}} \\ {\begin{array}{*{20}{c}} {\frac{{ - 1}}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}} \end{array}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}} \end{array}} \\ {\begin{array}{*{20}{c}} {\frac{{ - 1}}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}} \end{array}} \end{array}} \right)\left( {\begin{array}{*{20}{c}} 2&0 \\ 0&3 \end{array}} \right)$ A1AG

[1 mark]

c.ii.

Determinant is 1. ${\mathbf{R}} = \left( {\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} {\tfrac{1}{{\sqrt 2 }}}&{\tfrac{{ - 1}}{{\sqrt 2 }}} \end{array}} \\ {\begin{array}{*{20}{c}} {\tfrac{1}{{\sqrt 2 }}}&{\tfrac{1}{{\sqrt 2 }}} \end{array}} \end{array}} \right)$ M1A1

[2 marks]

d.i.

${\mathbf{M}}{{\mathbf{R}}^{ - 1}} = {{\mathbf{R}}^{ - 1}}\left( {\begin{array}{*{20}{c}} 2&0 \\ 0&3 \end{array}} \right)$ so post multiplying by ${\mathbf{R}}$ gives ${\mathbf{M}} = {{\mathbf{R}}^{ - 1}}\left( {\begin{array}{*{20}{c}} 2&0 \\ 0&3 \end{array}} \right){\mathbf{R}}$ M1AG

[1 mark]

d.ii.

$\left( {\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} {\tfrac{1}{{\sqrt 2 }}}&{\tfrac{{ - 1}}{{\sqrt 2 }}} \end{array}} \\ {\begin{array}{*{20}{c}} {\tfrac{1}{{\sqrt 2 }}}&{\tfrac{1}{{\sqrt 2 }}} \end{array}} \end{array}} \right)\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) = \left( {\begin{array}{*{20}{c}} X \\ Y \end{array}} \right) \Rightarrow \left( {\begin{array}{*{20}{c}} {\tfrac{1}{{\sqrt 2 }}x - \tfrac{1}{{\sqrt 2 }}y} \\ {\tfrac{1}{{\sqrt 2 }}x + \tfrac{1}{{\sqrt 2 }}y} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} X \\ Y \end{array}} \right) \Rightarrow \left( {\begin{array}{*{20}{c}} X&Y \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {\tfrac{1}{{\sqrt 2 }}x - \tfrac{1}{{\sqrt 2 }}y}&{\tfrac{1}{{\sqrt 2 }}x + \tfrac{1}{{\sqrt 2 }}y} \end{array}} \right)$ M1A1

and $\left( {\begin{array}{*{20}{c}} x&y \end{array}} \right)\left( {\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} {\tfrac{1}{{\sqrt 2 }}}&{\tfrac{1}{{\sqrt 2 }}} \end{array}} \\ {\begin{array}{*{20}{c}} {\tfrac{{ - 1}}{{\sqrt 2 }}}&{\tfrac{1}{{\sqrt 2 }}} \end{array}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {\tfrac{1}{{\sqrt 2 }}x - \tfrac{1}{{\sqrt 2 }}y}&{\tfrac{1}{{\sqrt 2 }}x + \tfrac{1}{{\sqrt 2 }}y} \end{array}} \right)$ completing the proof A1AG

[3 marks]

e.i.

$\left( {\begin{array}{*{20}{c}} x&y \end{array}} \right){\mathbf{M}}\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) = \left( 6 \right) \Rightarrow \left( {\begin{array}{*{20}{c}} x&y \end{array}} \right){{\mathbf{R}}^{ - 1}}\left( {\begin{array}{*{20}{c}} 2&0 \\ 0&3 \end{array}} \right){\mathbf{R}}\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) = \left( 6 \right) \Rightarrow \left( {\begin{array}{*{20}{c}} X&Y \end{array}} \right)\left( {\begin{array}{*{20}{c}} 2&0 \\ 0&3 \end{array}} \right)\left( {\begin{array}{*{20}{c}} X \\ Y \end{array}} \right) = \left( 6 \right)$

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

[2 marks]

e.ii.

$\frac{X}{{\sqrt 3 }} = u{\text{,}}\,\,\frac{Y}{{\sqrt 2 }} = v \Rightarrow {u^2} + {v^2} = 1$ , 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]

h.i.

an ellipse A1

[1 mark]

h.ii.

$y = x$ , $y = - x$ A1A1

[2 marks]

Examiners report

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

d.i.

[N/A]

d.ii.

[N/A]

e.i.

[N/A]

e.ii.

[N/A]

h.i.

[N/A]

h.ii.

[N/A]

A graphic designer, Ben, wants to create an animation in which a sequence of squares is created by a composition of successive enlargements and translations and then rotated about the origin and reduced in size.

Ben outlines his plan with the following storyboards.

The first four frames of the animation are shown below in greater detail.

The sides of each successive square are one half the size of the adjacent larger square. Let the sequence of squares be $U_{0}, U_{1}, U_{2}, \dots$

The first square, $U_{0}$ , has sides of length $4 cm$ .

Ben decides the animation will continue as long as the width of the square is greater than the width of one pixel.

Ben decides to generate the squares using the transformation

$(\begin{matrix} x_{n} \\ y_{n} \end{matrix}) = A_{n} (\begin{matrix} x_{0} \\ y_{0} \end{matrix}) + b_{n}$

where $A_{n}$ is a $2 \times 2$ matrix that represents an enlargement, $b_{n}$ is a $2 \times 1$ column vector that represents a translation, $(x_{0}, y_{0})$ is a point in $U_{0}$ and $(x_{n}, y_{n})$ is its image in $U_{n}$ .

By considering the case where $(x_{0}, y_{0})$ is $(0, 0)$ ,

Once the image of squares has been produced, Ben wants to continue the animation by rotating the image counter clockwise about the origin and having it reduce in size during the rotation.

Let $E_{θ}$ be the enlargement matrix used when the original sequence of squares has been rotated through $θ$ degrees.

Ben decides the enlargement scale factor, $s$ , should be a linear function of the angle, $θ$ , and after a rotation of $360 °$ the sequence of squares should be half of its original length.

Find an expression for the width of $U_{n}$ in centimetres.

[2]

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

[3]

Write down $A_{1}$ .

[1]

c.i.

Write down $A_{n}$ , in terms of $n$ .

[1]

c.ii.

state the coordinates, $(x_{1}, y_{1})$ , of its image in $U_{1}$ .

[1]

d.i.

hence find $b_{1}$ .

[2]

d.ii.

show that $b_{n} = (\begin{matrix} 8 (1 - 2^{- n}) \\ 8 (1 - 2^{- n}) \end{matrix})$ .

[3]

d.iii.

Hence or otherwise, find the coordinates of the top left-hand corner in $U_{7}$ .

[3]

Find, $s$ , in the form $s (θ) = m θ + c$ .

[4]

f.i.

Write down $E_{θ}$ .

[1]

f.ii.

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

[4]

f.iii.

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

[1]

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.

[3]

Markscheme

* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.

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

[2 marks]

$\frac{4}{2^{n}} > 0.025$ (A1)

$2^{n} < 160$

$n \leq 7$ (A1)

Note: Accept equations in place of inequalities.

Hence there are $8$ squares A1

[3 marks]

$(\begin{matrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{matrix})$ A1

[1 mark]

c.i.

$A_{n} = (\begin{matrix} \frac{1}{2^{n}} & 0 \\ 0 & \frac{1}{2^{n}} \end{matrix})$ A1

[1 mark]

c.ii.

$(4, 4)$ A1

[1 mark]

d.i.

$A_{1} (\begin{matrix} 0 \\ 0 \end{matrix}) + b_{1} = (\begin{matrix} 4 \\ 4 \end{matrix})$ (M1)

$b_{1} = (\begin{matrix} 4 \\ 4 \end{matrix})$ A1

[2 marks]

d.ii.

Recognise the geometric series $b_{n} = (\begin{matrix} 4 + 2 + 1 + \dots \\ 4 + 2 + 1 + \dots \end{matrix})$ M1

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

$(\begin{matrix} 8 (1 - \frac{1}{2^{n}}) \\ 8 (1 - \frac{1}{2^{n}}) \end{matrix})$ AG

[3 marks]

d.iii.

$(\begin{matrix} \frac{1}{2^{7}} & 0 \\ 0 & \frac{1}{2^{7}} \end{matrix}) (\begin{matrix} 0 \\ 4 \end{matrix}) + (\begin{matrix} 8 (1 - \frac{1}{2^{7}}) \\ 8 (1 - \frac{1}{2^{7}}) \end{matrix})$ M1A1

$(7.9375, 7.96875)$ A1

[3 marks]

$s (θ) = m θ + c$

$s (0) = 1, c = 1$ M1A1

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

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

$s (θ) = - \frac{θ}{720} + 1$

[4 marks]

f.i.

$E_{θ} = (\begin{matrix} - \frac{θ}{720} + 1 & 0 \\ 0 & - \frac{θ}{720} + 1 \end{matrix})$ A1

[1 mark]

f.ii.

$(\begin{matrix} - \frac{135}{720} + 1 & 0 \\ 0 & - \frac{135}{720} + 1 \end{matrix}) (\begin{matrix} \cos 135 ° & - \sin 135 ° \\ \sin 135 ° & \cos 135 ° \end{matrix}) (\begin{matrix} 1 \\ 1 \end{matrix})$ M1A1A1

$(- 1.15, 0)$ A1

[4 marks]

f.iii.

$θ = 720 °$ 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]

Examiners report

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

d.i.

[N/A]

d.ii.

[N/A]

d.iii.

[N/A]

f.i.

[N/A]

f.ii.

[N/A]

f.iii.

[N/A]

A suitable site for the landing of a spacecraft on the planet Mars is identified at a point, $A$ . The shortest time from sunrise to sunset at point $A$ must be found.

Radians should be used throughout this question. All values given in the question should be treated as exact.

Mars completes a full orbit of the Sun in $669$ Martian days, which is one Martian year.

On day $t$ , where $t \in ℤ$ , the length of time, in hours, from the start of the Martian day until sunrise at point $A$ can be modelled by a function, $R (t)$ , where

$R (t) = a \sin (b t) + c, t \in ℝ$ .

The graph of $R$ is shown for one Martian year.

Mars completes a full rotation on its axis in $24$ hours and $40$ minutes.

The time of sunrise on Mars depends on the angle, $δ$ , at which it tilts towards the Sun. During a Martian year, $δ$ varies from $- 0.440$ to $0.440$ radians.

The angle, $ω$ , through which Mars rotates on its axis from the start of a Martian day to the moment of sunrise, at point $A$ , is given by $\cos ω = 0.839 \tan δ$ , $0 \leq ω \leq π$ .

Use your answers to parts (b) and (c) to find

Let $S (t)$ be the length of time, in hours, from the start of the Martian day until sunset at point $A$ on day $t$ . $S (t)$ can be modelled by the function

$S (t) = 1.5 \sin (0.00939 t + 2.83) + 18.65$ .

The length of time between sunrise and sunset at point $A$ , $L (t)$ , can be modelled by the function

$L (t) = 1.5 \sin (0.00939 t + 2.83) - 1.6 \sin (0.00939 t) + d$ .

Let $f (t) = 1.5 \sin (0.00939 t + 2.83) - 1.6 \sin (0.00939 t)$ and hence $L (t) = f (t) + d$ .

$f (t)$ can be written in the form $Im (z_{1} - z_{2})$ , where $z_{1}$ and $z_{2}$ are complex functions of $t$ .

Show that $b \approx 0.00939$ .

[2]

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

[3]

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

[3]

c.i.

Find the minimum value of $ω$ .

[1]

c.ii.

the maximum value of $R (t)$ .

[2]

d.i.

the minimum value of $R (t)$ .

[1]

d.ii.

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

[2]

Find the value of $c$ .

[2]

Find the value of $d$ .

[2]

Write down $z_{1}$ and $z_{2}$ in exponential form, with a constant modulus.

[3]

h.i.

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

[4]

h.ii.

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

[2]

h.iii.

Markscheme

recognition that period $= 669$ (M1)

$b = \frac{2 π}{669}$ 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 $\frac{2}{3}, 0.666 \dots, 0 . \bar{6}$ or $0.667$ .

$\frac{2 π}{24 \frac{2}{3}}$ (M1)

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

$= 0.255$ radians $(0.254723 \dots, \frac{3 π}{37}, 14.5945 \dots °)$ A1

[3 marks]

substitution of either value of $δ$ into equation (M1)

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

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

$\cos ω = 0.839 \tan (- 0.440)$ A1

$ω = 1.97684 \dots$

$\approx 1.98$ AG

Note: For substitution of $1.98$ award M0A0.

[3 marks]

c.i.

$δ = 0.440$

$ω = 1.16 (1.16474 \dots)$ A1

[1 mark]

c.ii.

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

$= 7.76 hours (7.76075 \dots)$ A1

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

[2 marks]

d.i.

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

$= 4.57 hours (4.57258 \dots)$ A1

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

[1 mark]

d.ii.

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

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

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

THEN

$= 6.17 (6.16666 \dots)$ 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)

$= 12.5 (12.4833 \dots)$ A1

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

[2 marks]

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

$z_{1} = 1.5 e^{(0.00939 t + 2.83) i}, z_{2} = 1.6 e^{(0.00939 t) i}$ A1A1

[3 marks]

h.i.

EITHER

$z_{1} - z_{2} = 1.5 e^{(0.00939 t + 2.83) i} - 1.6 e^{(0.00939 t) i}$

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

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

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

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

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

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

[4 marks]

h.ii.

shortest time between sunrise and sunset

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

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

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

[2 marks]

h.iii.

Examiners report

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

d.i.

[N/A]

d.ii.

[N/A]

h.i.

[N/A]

h.ii.

[N/A]

h.iii.