Question 7
Determine $f(x)$, given $f'(x)=6x^2+\dfrac{1}{x^2}+\dfrac{1}{x}$ and $f(1)=5$.
Not quite. $\ln(x)$ is right, but $\dfrac{3}{x^3}$ differentiates to $-\dfrac{9}{x^4}$, not $\dfrac{1}{x^2}$. Write $\dfrac{1}{x^2}$ as $x^{-2}$ and it integrates to $-x^{-1}=-\dfrac{1}{x}$. Option (A) also gives $f(1)=4$, not 5. In 2021, 11.1% of Queensland students chose (A).
That’s it. Step through the hints below to see why. In 2021, 73.7% of Queensland students chose (B).
Not quite. $-\dfrac{1}{x}$ is right, and (C) does give $f(1)=5$. But $\dfrac{2}{x^2}$ differentiates to $-\dfrac{4}{x^3}$, not $\dfrac{1}{x}$. The antiderivative of $\dfrac{1}{x}$ is $\ln(x)$. In 2021, 8.6% of Queensland students chose (C).
Not quite. (D) gives $f(1)=5$, but $\dfrac{3}{x^3}$ and $\dfrac{2}{x^2}$ differentiate to $-\dfrac{9}{x^4}$ and $-\dfrac{4}{x^3}$. Neither is in $f'(x)$. You need $-\dfrac{1}{x}$ and $\ln(x)$. In 2021, 5.9% of Queensland students chose (D).
Option (A) misses the hollow point: it gives $f(1)=4$. Its gradient is wrong too. Here it is and you need .
Option (A) misses the hollow point: it gives $f(1)=4$. Here the two lines nearly line up, but they are not equal: its gradient is and you need . Move the point left and the lines split.
Option (B)’s gradient equals $f'(x)$ at every $x$, so the orange tangent lies on the dashed blue line. It also passes through $(1,5)$, so it’s the answer.
Very close, but not equal. Here the two lines nearly line up, but option ()’s gradient is and you need . Move the point left and they split.
Option () passes through $(1,5)$ too, but here its gradient is and you need . The orange tangent and the dashed blue line don’t match.
Write each term as a power of $x$ where you can, then integrate term by term. The power rule fails for $x^{-1}$, so $\dfrac{1}{x}$ integrates to $\ln(x)$.
$\displaystyle f(x)$$\displaystyle =\int\left(6x^2+x^{-2}+x^{-1}\right)dx$$\displaystyle =2x^3-x^{-1}+\ln(x)+c$
So $f(x)=2x^3-\dfrac{1}{x}+\ln(x)+c$.
Substitute $f(1)=5$ to find $c$. Remember $\ln(1)=0$.
$\displaystyle f(1)=2-1+0+c$$\displaystyle =1+c$
$$1+c=5\ \Rightarrow\ c=4$$
Three options, (B), (C) and (D), give $f(1)=5$, so the point alone can’t decide. Check the terms instead.
Only (B) has $-\dfrac{1}{x}$ and $\ln(x)$. Differentiating it gives back $6x^2+\dfrac{1}{x^2}+\dfrac{1}{x}$.
$\displaystyle f(x)=2x^3-\frac{1}{x}+\ln(x)+c$
$\displaystyle f(1)=2-1+0+c=5$$\displaystyle \Rightarrow\ c=4$
$\displaystyle f(x)=2x^3-\frac{1}{x}+\ln(x)+4$
Check by differentiating: $6x^2+\dfrac{1}{x^2}+\dfrac{1}{x}$.
How Queensland students answered in 2021
From the QCAA Mathematical Methods 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 Mathematical Methods 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.
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