Circular function graphs and trigonometric equations – IB Maths exam questions with mark scheme

1CoreNon-calculator3 marks

Solve $2\sin x=\sqrt3$ for $0\le x\le2\pi$. Give your answers as exact multiples of $\pi$.

$\sin x=\dfrac{\sqrt3}{2}$M1
$x=\dfrac{\pi}{3}$A1
$x=\dfrac{2\pi}{3}$ (from $\pi-\dfrac\pi3$)A1
Answer: $x=\dfrac\pi3,\ \dfrac{2\pi}3$
Examiner tip: Sine is positive in the first and second quadrants, so there are two solutions in $[0,2\pi]$: do not stop at $\dfrac\pi3$.

2StandardNon-calculator6 marks

Solve the equation $2\cos^2x+\sin x-1=0$ for $0\le x\le2\pi$. Give exact answers in terms of $\pi$.

Uses $\cos^2x=1-\sin^2x$M1
$2\sin^2x-\sin x-1=0$A1
Factorises/solves quadratic in $\sin x$: $(2\sin x+1)(\sin x-1)=0$M1
$x=\dfrac\pi2$ (from $\sin x=1$)A1
$x=\dfrac{7\pi}6$ (from $\sin x=-\tfrac12$)A1
$x=\dfrac{11\pi}6$A1
Answer: $x=\dfrac\pi2,\ \dfrac{7\pi}6,\ \dfrac{11\pi}6$
Examiner tip: Replace $\cos^2x$ so the equation is in one trig function, and never cancel a factor of $\sin x$ away: you lose solutions.

3HardCalculator7 marks

The height, $H$ metres, of the tide in a harbour $t$ hours after midnight is modelled by $H=6+4\sin\left(\dfrac{\pi t}{6}\right)$, for $0\le t\le12$.

(a) Write down the maximum and minimum values of $H$. [2]

(b) Write down the period of the model. [1]

(c) Boats can enter the harbour only when the tide height exceeds $8$ m. Find the times between $t=0$ and $t=12$ when boats can enter, and the total time for which they can enter. [4]

Maximum $10$ mA1
Minimum $2$ mA1
Period $=\dfrac{2\pi}{\pi/6}=12$ hoursA1
Sets $6+4\sin\frac{\pi t}6=8$ (or $>8$), i.e. $\sin\dfrac{\pi t}6=\dfrac12$, or uses intersection of $H$ with $H=8$ on a GDCM1
$t=1$ (from $\frac{\pi t}6=\frac\pi6$)A1
$t=5$ (from $\frac{\pi t}6=\frac{5\pi}6$), so boats can enter for $1A1
Total time $4$ hoursA1
Answer: (a) $10$ m, $2$ m (b) $12$ h (c) between $t=1$ and $t=5$ (4 hours)
Examiner tip: Take care with the inequality: use the graph of $\sin$ to see that the values above $\tfrac12$ lie between the two solutions.

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4CoreNon-calculator5 marks

The diagram shows part of the graph of $y=f(x)$, where $f(x)=p\cos(qx)+r$, for $0\le x\le\pi$ and $p,q>0$. The graph has maximum points at $(0,\,4)$ and $(\pi,\,4)$ and a minimum point at $\left(\dfrac{\pi}{2},\,-2\right)$.

(a) Write down the period of $f$. [1]

(b) Find the value of (i) $r$; (ii) $p$; (iii) $q$. [4]

(a) Period $=\pi$A1
(b)(i) $r=\dfrac{4+(-2)}{2}=1$A1
(ii) $p=\dfrac{4-(-2)}{2}=3$ (do not accept $p=-3$, since $p>0$)A1
(iii) Uses period $=\dfrac{2\pi}{q}$: $\dfrac{2\pi}{q}=\pi$ (follow through their period from (a))M1
$q=2$A1
Answer: (a) $\pi$ (b) $r=1,\ p=3,\ q=2$
Examiner tip: Students often give the amplitude as $6$ (the full height from minimum to maximum) instead of half of it.

5CoreNon-calculator4 marks

Solve the equation $\tan(2x)=-1$ for $0\le x\le\pi$. Give your answers as exact multiples of $\pi$.

Reference angle: $\tan^{-1}1=\dfrac{\pi}{4}$ seen (may be implied by a correct solution)M1
Uses the doubled interval $0\le 2x\le 2\pi$ to find both values of $2x$: $2x=\dfrac{3\pi}{4},\ \dfrac{7\pi}{4}$M1
$x=\dfrac{3\pi}{8}$A1
$x=\dfrac{7\pi}{8}$ and no extra values in the interval (award A0 if any extra solutions, e.g. $-\dfrac{\pi}{8}$, are given)A1
Answer: $x=\dfrac{3\pi}{8},\ \dfrac{7\pi}{8}$
Examiner tip: Change the interval to $0\le2x\le2\pi$ before solving, otherwise the second solution is missed.

6StandardNon-calculator6 marks

Solve the equation $\sin 2x=\sin x$ for $-\pi\le x\le\pi$.

Uses $\sin2x=2\sin x\cos x$: $2\sin x\cos x=\sin x$M1
Rearranges and factorises: $\sin x(2\cos x-1)=0$ (M0 if they divide by $\sin x$)M1
$\sin x=0$ or $\cos x=\dfrac12$A1
$x=-\pi,\ 0,\ \pi$A1
$x=\dfrac{\pi}{3}$A1
$x=-\dfrac{\pi}{3}$ (all five correct and no extras)A1
Answer: $x=-\pi,\ -\dfrac{\pi}{3},\ 0,\ \dfrac{\pi}{3},\ \pi$
Examiner tip: Dividing both sides by $\sin x$ loses the three solutions from $\sin x=0$; always factorise instead.

7StandardCalculator8 marks

A Ferris wheel has a diameter of $36$ m and its centre is $20$ m above the ground. It turns at a constant speed and completes one revolution every $12$ minutes. A passenger boards at the lowest point of the wheel at time $t=0$.

The height, $h$ metres, of the passenger above the ground $t$ minutes later is modelled by $h(t)=d-a\cos(bt)$, where $a,b,d>0$.

(a) Find the value of (i) $a$; (ii) $d$; (iii) $b$. [3]

(b) Find the height of the passenger after $2.5$ minutes. [2]

(c) Find the length of time, during one revolution, for which the passenger is more than $30$ m above the ground. [3]

(a)(i) $a=18$A1
(ii) $d=20$A1
(iii) $b=\dfrac{2\pi}{12}=\dfrac{\pi}{6}$ (accept $0.524$)A1
(b) Substitutes $t=2.5$ into their model: $20-18\cos\left(\dfrac{2.5\pi}{6}\right)$M1
$h=15.3$ m (accept $15.34$)A1
(c) Sets up $20-18\cos\left(\dfrac{\pi t}{6}\right)=30$ or a sketch/GDC intersection with $h=30$M1
$t=4.12$ and $t=7.88$ (accept $4.125$ and $7.875$)A1
Time above $30$ m $=7.875-4.125=3.75$ minutesA1
Answer: (a) $a=18,\ d=20,\ b=\dfrac{\pi}{6}$ (b) $15.3$ m (c) $3.75$ minutes
Examiner tip: Have the GDC in radian mode: $\cos\left(\frac{2.5\pi}{6}\right)$ evaluated in degrees gives a completely wrong height.

8HardNon-calculator7 marks

Let $f(x)=2\sin(3x)+1$, for $0\le x\le\pi$. The diagram shows the graph of $y=f(x)$.

(a) Solve the equation $f(x)=2$, giving your answers as exact multiples of $\pi$. [4]

(b) The equation $f(x)=k$ has exactly two solutions for $0\le x\le\pi$. Find the possible values of $k$. [3]

(a) $\sin(3x)=\dfrac12$A1
Finds at least two correct values of $3x$ in $0\le3x\le3\pi$, e.g. $\dfrac{\pi}{6},\ \dfrac{5\pi}{6}$M1
All four values of $3x$: $\dfrac{\pi}{6},\ \dfrac{5\pi}{6},\ \dfrac{13\pi}{6},\ \dfrac{17\pi}{6}$A1
$x=\dfrac{\pi}{18},\ \dfrac{5\pi}{18},\ \dfrac{13\pi}{18},\ \dfrac{17\pi}{18}$A1
(b) Recognises the horizontal line $y=k$ must cut the graph twice, e.g. considers the maximum $3$ and minimum $-1$ (may be seen on the diagram)M1
$-1A1
$k=3$A1
Answer: (a) $x=\dfrac{\pi}{18},\ \dfrac{5\pi}{18},\ \dfrac{13\pi}{18},\ \dfrac{17\pi}{18}$ (b) $-1
Examiner tip: In (b) check the end values separately: the line $y=1$ also passes through both endpoints of the domain, giving four solutions, not two.

9HardCalculator7 marks

The graph of $y=\sin x$ is transformed onto the graph of $y=g(x)$ by the following sequence of transformations:

a vertical stretch with scale factor $3$;
a horizontal stretch with scale factor $\dfrac12$;
a translation by the vector $\begin{pmatrix}-\frac{\pi}{6}\\ 2\end{pmatrix}$.

(a) Write $g(x)$ in the form $a\sin\big(b(x+c)\big)+d$. [3]

(b) Write down the range of $g$. [1]

(a) $a=3$ and $d=2$A1
$b=2$ (do not accept $b=\frac12$)A1
$c=\dfrac{\pi}{6}$, giving $g(x)=3\sin\left(2\left(x+\dfrac{\pi}{6}\right)\right)+2$ (accept $3\sin\left(2x+\dfrac{\pi}{3}\right)+2$; do not accept $c=-\frac{\pi}{6}$)A1
(b) $-1\le g(x)\le5$ (accept $[-1,5]$)A1
(c) Attempts to solve $3\sin\left(2x+\dfrac{\pi}{3}\right)+2=4$, e.g. GDC intersection of their $g$ with $y=4$, or $\sin\left(2x+\dfrac{\pi}{3}\right)=\dfrac23$M1
$x=0.682$A1
$x=2.98$ (and no other values)A1
Answer: (a) $g(x)=3\sin\left(2\left(x+\dfrac{\pi}{6}\right)\right)+2$ (b) $-1\le g(x)\le5$ (c) $x=0.682,\ 2.98$
Examiner tip: A horizontal stretch with scale factor $\frac12$ multiplies $x$ by $2$ inside the function, and a translation of $-\frac{\pi}{6}$ gives $+\frac{\pi}{6}$ inside the bracket: both signs are commonly reversed.

10StretchNon-calculator9 marks

Consider the equation $2\sin^2x-\cos x=k$, where $k\in\mathbb{R}$.

(a) Show that the equation can be written as $2\cos^2x+\cos x+k-2=0$. [2]

(b) Find the range of values of $k$ for which the equation has at least one solution. [3]

(a) Substitutes $\sin^2x=1-\cos^2x$: $2(1-\cos^2x)-\cos x=k$M1
Expands and rearranges correctly to $2\cos^2x+\cos x+k-2=0$ (AG)A1
(b) Writes $k=2-2c^2-c$ with $c=\cos x$ and recognises that $-1\le c\le1$M1
Finds the maximum value $\dfrac{17}{8}$ (at $c=-\dfrac14$), e.g. by completing the square or vertex $c=-\dfrac{b}{2a}$A1
Finds the minimum value $-1$ (at $c=1$) and states $-1\le k\le\dfrac{17}{8}$A1
(c) $2\cos^2x+\cos x=0$, so $\cos x(2\cos x+1)=0$M1
$\cos x=0$ or $\cos x=-\dfrac12$A1
$x=\dfrac{\pi}{2},\ \dfrac{3\pi}{2}$A1
$x=\dfrac{2\pi}{3},\ \dfrac{4\pi}{3}$A1
Answer: (b) $-1\le k\le\dfrac{17}{8}$ (c) $x=\dfrac{\pi}{2},\ \dfrac{2\pi}{3},\ \dfrac{4\pi}{3},\ \dfrac{3\pi}{2}$
Examiner tip: In (b) the quadratic in $\cos x$ is only valid for $-1\le\cos x\le1$; using the discriminant alone gives $k\le\frac{17}{8}$ but misses the lower bound $-1$.

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