← 2019 Paper 1

UPSC 2019 Maths Optional Paper 1 Q1a — Step-by-Step Solution

10 marks · Section A

Continuity of real functions · Calculus · asked 2× in 13 yrs · Read the full method →

Question

Let $f:\left[0,\dfrac{\pi}{2}\right]\to\mathbb R$ be a continuous function such that $f(x)=\dfrac{\cos^2 x}{4x^2-\pi^2}$, $0\le x<\dfrac{\pi}{2}$. Find the value of $f\!\left(\dfrac{\pi}{2}\right)$.

Technique

$0/0$ indeterminate limit; substitution $x=\frac{\pi}{2}+h$, factor the denominator as a difference of squares, use $\sin h/h\to1$.

Solution

Step 1 — Identify the indeterminacy

As $x\to\dfrac{\pi}{2}^-$, $\cos x\to0$ so $\cos^2 x\to0$, and $4x^2-\pi^2=(2x-\pi)(2x+\pi)\to0$. The quotient is of the form $0/0$. Because $f$ is required to be continuous on the closed interval $\left[0,\frac{\pi}{2}\right]$, the value $f\!\left(\frac{\pi}{2}\right)$ must equal the left-hand limit:

Step 2 — Substitute $x=\dfrac{\pi}{2}+h$

Let $x=\dfrac{\pi}{2}+h$ with $h\to0$. Then $\cos x=\cos\!\left(\dfrac{\pi}{2}+h\right)=-\sin h$, so $\cos^2 x=\sin^2 h$. Also

Hence

Step 3 — Take the limit

Using $\dfrac{\sin h}{h}\to1$ as $h\to0$:

Answer