← 2017 Paper 2

UPSC 2017 Maths Optional Paper 2 Q1c — Step-by-Step Solution

10 marks · Section A

Properties of Continuous Functions on Compact Sets · Real Analysis · Read the full method →

Question

Find the supremum and the infimum of $\dfrac{x}{\sin x}$ on the interval $\left(0,\dfrac{\pi}{2}\right]$.

Technique

Reduce to monotonicity via the sign of $g'(x)$; the key sub-lemma $\sin x>x\cos x$ on $(0,\pi/2]$ comes from $h(x)=\sin x-x\cos x$ with $h(0)=0,\ h'(x)=x\sin x>0$.

Solution

Let $g(x)=\dfrac{x}{\sin x}$ on $(0,\pi/2]$. We show $g$ is strictly increasing, so its values fill the interval between the (limiting) value at $0^+$ and the value at $\pi/2$.

Step 1 — Monotonicity: $g$ is strictly increasing

Differentiate:

Let $h(x)=\sin x-x\cos x$. Then $h(0)=0$ and

On $(0,\pi/2]$ we have $x>0$ and $\sin x>0$, so $h'(x)>0$; with $h(0)=0$ this gives $h(x)>0$ on $(0,\pi/2]$. Since $\sin^2x>0$, we conclude $g'(x)>0$ throughout, so $g$ is strictly increasing on $(0,\pi/2]$.

Step 2 — Infimum (value at the open end $0^+$)

As $g$ is increasing, its infimum is the limit at the left endpoint:

using the standard limit $\lim_{x\to0}\frac{\sin x}{x}=1$. This value is not attained, because $x=0\notin(0,\pi/2]$; for every $x$ in the interval $g(x)>1$. So $\inf g=1$ (an infimum, not a minimum).

Step 3 — Supremum (value at the closed end $\pi/2$)

As $g$ is increasing, its supremum is the value at the right endpoint $x=\pi/2$, which is in the interval:

This is attained, so it is in fact a maximum.

Conclusion

Answer