Question 3
The derivative of the function $f(x)$ is given by $f'(x)=\sin(2x)$. It is known that $f\left(\dfrac{\pi}{2}\right)=4$.
Determine $f(x)$.
Not quite. $-\cos(2x)+3$ does give $f\left(\tfrac{\pi}{2}\right)=4$, but its derivative is $2\sin(2x)$. Integrating $\sin(2x)$ divides by 2, so you need $-\tfrac12\cos(2x)$. In 2024, 10.6% of Queensland students chose (A).
Not quite. $\cos(2x)+5$ passes through $\left(\tfrac{\pi}{2},4\right)$, but its derivative is $-2\sin(2x)$. The antiderivative of $\sin$ is $-\cos$, and the 2 inside means you divide by 2. In 2024, 3.7% of Queensland students chose (B).
That’s it. Step through the hints below to see why. In 2024, 74.2% of Queensland students chose (C).
Not quite. The size is right but the sign is wrong. $\tfrac12\cos(2x)$ differentiates to $-\sin(2x)$. The antiderivative of $\sin(2x)$ is $-\tfrac12\cos(2x)$. In 2024, 10.8% of Queensland students chose (D).
Option (C)’s gradient equals $\sin(2x)$ at every $x$, so the orange tangent lies on the dashed blue line. It also passes through $\left(\tfrac{\pi}{2},4\right)$, so it’s the answer.
Here $\sin(2x)=0$, so every option has a flat tangent and the gradients agree. Move the point to see them split.
Option () passes through $\left(\tfrac{\pi}{2},4\right)$ too, but here its gradient is and you need . The orange tangent and the dashed blue line don’t match.
To undo the derivative, integrate $\sin(2x)$. The antiderivative of $\sin$ is $-\cos$, and the 2 inside means you divide by 2.
$\displaystyle f(x)=\int\sin(2x)\,dx$$\displaystyle =-\tfrac12\cos(2x)+c$
Substitute the known point to find $c$. Remember $\cos(\pi)=-1$.
$\displaystyle f\left(\tfrac{\pi}{2}\right)=-\tfrac12\cos(\pi)+c$$\displaystyle =\tfrac12+c$
$$\tfrac12+c=4\ \Rightarrow\ c=3.5$$
Every option gives $f\left(\tfrac{\pi}{2}\right)=4$, so the point alone can’t decide. Differentiate each one instead.
(A) gives $2\sin(2x)$, (B) gives $-2\sin(2x)$ and (D) gives $-\sin(2x)$. Only (C) gives $\sin(2x)$.
$$f(x)=-\tfrac12\cos(2x)+c$$
$$\tfrac12+c=4\ \Rightarrow\ c=3.5$$
$$f(x)=-\tfrac12\cos(2x)+3.5$$
Check by differentiating: $-\tfrac12\times(-\sin(2x))\times2=\sin(2x)$.
How Queensland students answered in 2024
From the QCAA Mathematical Methods subject report for 2024. 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 2024 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2024, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
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