## Exam howlers

I’ve just emerged from the cave of slightly grubby paper that I inhabit during the marking season. As ever, most of the idiocies with which I’ve been vexing my spirit are sad rather than amusing — or even vaguely sweet, like the student who admitted that “these answers don’t agree because I made up my own maths to do the question”. There’s a small crop, though, of specimens for any future researcher on the subject of what our students do instead of learning.

Let’s start with a simple one: an intricate calculation to determine $A$, defined as the area of a particular finite region, concluded that

$A = \displaystyle\int_{-\infty}^{\infty}\cos(x)-\sin(x)\mathrm{d}x = -4.$

This was accompanied by a perfectly acceptable sketch of the cos and sin functions: why the alarm bells didn’t ring is a mystery to me.

Sometimes, of course, the alarm bells ring a little too readily:

$\dfrac{0}{1}$ is impossible.

Sometimes you can see what’s going on… roughly, at least. Two examples:

$\displaystyle\int 1\mathrm{d}t = \dfrac{1^2}{2} = \dfrac{1}{2}.$

and

$|\mathbf{v}|^2 = -4 \implies \mathbf{v} = -2i$. The velocity is -2 in the x-direction.

Our imaginary friend $\sqrt{-1}$ featured in a more mystifying specimen as well, which concluded

So $f^{-1}(x)$ is defined when $x < i$, for all values of $i$.

This year’s champion, however, is in a class of its own when it comes to getting from one line to the next by vague association.

Q: Given the function $f(x) = \sqrt{1-x^2}$ defined on the interval $[0,1]$, find the inverse function $f^{-1}(x)$ and explain why an inverse function cannot be found if $f(x)$ is defined on its natural domain.

$f(x) = \sqrt{1-x^2}.$

$f^{-1}(x) = \dfrac{1}{\sqrt{1-x^2}}.$

$f^{-1}(x) = \sin^{-1}(x).$

This is already an inverse function and you can’t have the inverse of an inverse.