JEE Main 2026 (Online) 24th January Evening ShiftQuestion 1
Consider the following three statements for the function $f:(0,\infty)\to\mathbf{R}$ defined by $f(x)=|\log_ex|-|x-1|$ :
(I) $f$ is differentiable at all $x>0$.
(II) $f$ is increasing in $(0,1)$.
(III) $f$ is decreasing in $(1,\infty)$.
Then,
Differentiation and its Application
Solution
Concept. On $(0,1)$ and $(1,\infty)$ both moduli open with fixed signs; compare slopes at $x=1$. Formulas used. (1) $\ln x<0$ for $x<1$ and $\ln x>0$ for $x>1$Step 1 - Pieces. By (1), $$0<x<1:\ f=-\ln x+x-1,\qquad x>1:\ f=\ln x-x+1.$$ Step 2 - (I) at $x=1$. $f'(1^-)=-1+1=0$ and $f'(1^+)=1-1=0$. Equal, so $f$ is differentiable everywhere. (I) is true. Step 3 - (II). On $(0,1)$, $f'=1-\frac1x<0$, so $f$ is decreasing. (II) is false. Step 4 - (III). On $(1,\infty)$, $f'=\frac1x-1<0$, so $f$ is decreasing. (III) is true. Answer: (A)