JEE Advanced 2026 Paper 1Question 1
Consider the function $f:(0,\infty)\to(-\infty,\infty)$ given by $$f(x)=\sqrt{x}\,\log_e(x)-x+1.$$ Then which one of the following statements is TRUE ?
Differentiation and its Application
Solution
Concept. Put $u=\sqrt x$; the numerator of $f'$ becomes $2(\ln u+1-u)$, which is never positive. Formulas used. (1) $\ln u\le u-1$ for $u>0$, with equality only at $u=1$Step 1 - $f'$. $$f'(x)=\frac{\ln x}{2\sqrt x}+\frac{1}{\sqrt x}-1=\frac{\ln x+2-2\sqrt x}{2\sqrt x}.$$ Step 2 - Sign. With $u=\sqrt x$, $\ln x=2\ln u$ and the numerator is $2(\ln u+1-u)$. By (1) this is $\le0$, and $0$ only at $x=1$. Step 3 - Conclude. $f'\le0$ everywhere with a single zero, so $f$ is strictly decreasing on $(0,\infty)$: no local maximum and no local minimum. So (B), (C) are false and (D) is true. Step 4 - Option (A). On $(0,1)$ we have $f''>0$ (e.g. $f''(0.5)\approx0.49$), so $f'$ is increasing there, not decreasing. Answer: (D)