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.

The answer, in its entirety:

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.

I’d ask “who taught you this stuff?”, but sadly I already know the answer.

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One Response to Exam howlers

  1. Pingback: On the banks of denial: the Higher Maths “crocodile” question | New-cleckit dominie

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