Mathematical thinking and STEP

In my last blog I highlighted how establishing patterns in Maths is often the route to discovering theorems and solving complex looking problems.

STEP questions often challenge the student to recognise an approach established in an early part of a question and apply it to a more complex problem posed in later parts of the same question. This is demonstrated by the following STEP II question and solution.

In this question, you may assume without proof that any function f for which f'(x) \geq 0 is increasing, that is f(x_2) \geq f(x_1) if x_2 \geq x_1

(i)   (a) Let f(x) = \sin x -x\cos x. Show that f(x) is increasing for 0\leq x \leq\dfrac{1}{2} \pi and deduce that f(x) \geq 0 for 0 \leq x \leq \dfrac{1}{2} \pi.

f'(x) = \cos x - \cos x +x\sin x = x\sin x \geq0 for 0\leq x\leq\dfrac{1}{2}\pi

Also f(0)=0 \Rightarrow f(x)\geq0 for 0\leq x\leq \dfrac{1}{2}\pi

(b) Given that \dfrac{d}{dx}(\arcsin x) \geq 1 for 0\leq x < 1, show that

    \[\arcsin x \geq x\qquad  (0\leq x<1).\]

Let f(x)= \arcsin x-x then f'(x)=\dfrac{d}{dx}(\arcsin x)-1 \geq 0 \qquad (0\leq x<1)

f(0)=0 \Rightarrow f(x)=\arcsin x-x\geq 0 \qquad \Rightarrow \arcsin x \geq x \qquad (0\leq x<1)

(c) Let g(x)=x\csc x for 0<x<\dfrac{1}{2}\pi. Show that g is increasing and deduce that

    \[(\arcsin x)x^{-1} \geq x\csc x \qquad (0<x<1)\]

g'(x)=\csc x - x\csc x\cot x = \dfrac{\sin x - x\cos x}{\sin ^2 x} \geq 0 \quad (0<x<\dfrac{1}{2}\pi )\quad from (a).

(Note that the re-arrangement of g'(x)allows the use of (a) to prove the result).

It is now tempting to rewrite the result from (b) as (\arcsin x)x^{-1} \geq 1 to prove the desired inequality. However this would require showing that x\csc x\leq 1 \quad (0<x<1) which is not true! So we need to look further!

This is essentially the crux of the whole question. Experience suggests that we look at results demonstrated previously to give  clues as to how best to proceed, but our first attempt to use (b) has proved unsuccessful.

However, the moment of inspiration arrives when we spot that g(\arcsin x) yields the left hand side of the inequality.

Since g(x) is increasing

    \[\arcsin x\geq x \qquad (0\leq x<1)\]

    \[\Rightarrow g(\arcsin x)\geq g(x)\qquad (0<x<1)\]

    \[\Rightarrow (\arcsin x)x^{-1}\geq x\csc x\qquad (0<x<1)\]

(ii) Given that \dfrac{d}{dx}(\arctan x)\leq 1 for x\geq 0, show that

    \[(\tan x)(\arctan x)\geq x^2 \qquad (0<x<\dfrac{1}{2}\pi )\]

Again we now look carefully at previous results. We know that the given inequality yields \arctan x \leq x so we look for an increasing function h(x) such that h(\arctan x)\leq h(x) will yield the desired result. At first this does not seem obvious, but a simple re-arrangement of the inequality as follows provides the solution.

    \[\dfrac{x}{\arctan x} \leq \dfrac{\tan x}{x}\]

So letting h(x)=\dfrac{\tan x}{x} it is merely required to show that h(x) is increasing.

    \[h'(x)=\dfrac{x\sec ^2x - \tan x}{x^2} \]

and 

    \[\dfrac{d}{dx}(x\sec ^2x - \tan x) = 2x\sec ^2x\tan x \geq 0 \quad (0<x<\dfrac{1}{2}\pi )\]

The rest of the proof follows trivially.

I hope this has provided some insight into developing Mathematical thinking and how many STEP questions are designed to test the ability to build on previously established results.

After all, that is how the whole of Mathematical theory has been developed.