Question 2
Determine the area of the shaded region between the graphs of the functions $y=\dfrac13\sec\left(\dfrac{x}{3}\right)$ and $y=2\cos\left(\dfrac{x}{2}\right)$, as shown.
Not quite. 5.29 is the area between $x=-2$ and $x=2$: the integral comes to $5.287$. On the sketch the curves seem to cross near $x=\pm2$, but the sketch is not to scale. Solving the equation puts the crossings at $x\approx\pm2.617$. In 2021, 6.0% of Queensland students chose (A).
Not quite. 5.51 is the area between $x=-2.2$ and $x=2.2$: the integral comes to $5.511$. Those bounds stop before the curves meet at $x\approx\pm2.617$, so the ends of the region are missing. Find the crossings with your GDC’s solve or intersect tool. In 2021, 15.2% of Queensland students chose (B).
Not quite. 5.65 is the area between $x=-2.4$ and $x=2.4$: the integral comes to $5.652$. That is closer, but the curves meet at $x\approx\pm2.617$, so thin slivers at each end are still missing. In 2021, 8.0% of Queensland students chose (C).
That’s it. Step through the hints below to see why. In 2021, 69.9% of Queensland students chose (D).
Drawn to scale, unlike the question’s sketch. The orange dots are where the curves cross. Move the bounds and watch the integral of $f(x)-g(x)$ between them.
At the bounds the black curve is still above the blue one, so the curves haven’t met yet. The two ends of the region are left out, and the integral is too small.
These bounds are where the curves cross. The shading covers the whole enclosed region, and the integral is as large as it can get.
Past the crossings the blue curve is on top, so $f(x)-g(x)$ is negative there. The grey strips are subtracted, and the integral falls again.
The region is enclosed between the curves, so its ends are where the curves cross. The question’s sketch is not to scale, so don’t read the crossings off it. Solve on your GDC in radians. Many GDCs have no sec key, so enter it as $1\div\left(3\cos(x/3)\right)$.
$$2\cos\left(\frac{x}{2}\right)=\frac13\sec\left(\frac{x}{3}\right)$$
$$x\approx\pm2.617$$
Between the crossings, $y=2\cos\left(\frac{x}{2}\right)$ is on top. The area is the integral of top minus bottom.
$$\begin{aligned}A=\int_{-2.617}^{2.617}\Big(&2\cos\left(\frac{x}{2}\right)\\&-\frac13\sec\left(\frac{x}{3}\right)\Big)\,dx\end{aligned}$$
Store the crossing in your GDC at full accuracy and use the stored value as the bound.
Evaluate the integral on your GDC. Both curves are symmetric about the $y$-axis, so doubling the integral from $0$ to $2.617$ gives the same value and is a good check.
$$A\approx5.706$$
$$A\approx5.71\text{ units}^2$$
Solve $2\cos\left(\frac{x}{2}\right)=\frac13\sec\left(\frac{x}{3}\right)$ on a GDC in radians. The curves cross at $x\approx\pm2.617$. Between them, $y=2\cos\left(\frac{x}{2}\right)$ is on top.
$$\begin{aligned}A=\int_{-2.617}^{2.617}\Big(&2\cos\left(\frac{x}{2}\right)\\&-\frac13\sec\left(\frac{x}{3}\right)\Big)\,dx\end{aligned}$$
$$\approx5.706$$
The area is about $5.71$ units2.
How Queensland students answered in 2021
From the QCAA Specialist Mathematics subject report for 2021. The percentages can add to less than 100 because some students didn’t answer.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2021 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2021, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
Spot a mistake on this page? Report it