← 2019 Paper 1

UPSC 2019 Maths Optional Paper 1 Q2a — Step-by-Step Solution

15 marks · Section A

Differentiability · Calculus · asked 3× in 14 yrs · Read the full method →

Question

Is $f(x)=|\cos x|+|\sin x|$ differentiable at $x=\dfrac{\pi}{2}$? If yes, then find its derivative at $x=\dfrac{\pi}{2}$. If no, then give a proof of it.

Technique

Piecewise removal of $|\cdot|$ around the point; compute one-sided derivatives; differentiability $\iff$ they agree.

Solution

Step 1 — Resolve the absolute values near $x=\pi/2$

Near $x=\dfrac{\pi}{2}$, $\sin x>0$, so $|\sin x|=\sin x$ on a neighbourhood. But $\cos x$ changes sign at $\dfrac{\pi}{2}$:

Therefore, on a punctured neighbourhood of $\dfrac{\pi}{2}$,

Note $f\!\left(\frac{\pi}{2}\right)=0+1=1$, and $f$ is continuous there (both branches $\to1$).

Step 2 — Left-hand derivative

For $x<\dfrac{\pi}{2}$, $f'(x)=-\sin x+\cos x$, so

Step 3 — Right-hand derivative

For $x>\dfrac{\pi}{2}$, $f'(x)=\sin x+\cos x$, so

(Rigorously via difference quotients with $x=\frac\pi2+h$: $f\big(\frac\pi2+h\big)=|\!-\!\sin h|+|\cos h|=|\sin h|+\cos h$ for small $h$, and $f(\frac\pi2)=1$, so $\dfrac{f(\frac\pi2+h)-1}{h}=\dfrac{|\sin h|+\cos h-1}{h}$. As $h\to0^+$ this $\to\dfrac{\sin h}{h}+\dfrac{\cos h-1}{h}\to1+0=1$; as $h\to0^-$, $|\sin h|=-\sin h$ gives $\to-1$.)

Step 4 — Conclusion

The one-sided derivatives differ:

Hence $f$ is not differentiable at $x=\dfrac{\pi}{2}$ (there is a corner).

Answer