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Date	May 2021	Marks available	2	Reference code	21M.3.AHL.TZ1.1
Level	Additional Higher Level	Paper	Paper 3	Time zone	Time zone 1
Command term	Find	Question number	1	Adapted from	N/A

Question

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 $12.5$ .