JEE Main 2021 (Online) 4th August Morning ShiftQuestion 1
The number of points in $(0,2\pi)$, where the function $f:\mathbf{R}\to\mathbf{R}$ given by $f(x)=|\sin x|+|\cos x|$ is NOT differentiable, is $\_\_\_\_\_\_$ .
Differentiation and its Application
Answer$3$
Solution
Concept. $|\sin x|$ has corners where $\sin x=0$, $|\cos x|$ where $\cos x=0$; at such points the other term is smooth, so the corner stays. Formulas used. (1) $|u|$ has a corner at a simple zero of $u$ (slope jumps from $-u'$ to $u'$)Step 1 - Candidates in $(0,2\pi)$. $\cos x=0$: $\frac\pi2,\frac{3\pi}2$. $\sin x=0$: $\pi$. Step 2 - Check. At each, by (1) one term has a slope jump of size $2$, and the other term is smooth there (it is $\pm\sin x$ or $\pm\cos x$ near the point). So $f$ is not differentiable. Step 3 - Count. $3$. Answer: 3