JEE Main 2019 (Online) 10th April Morning ShiftQuestion 1
Let $f(x)=e^{x}-x$ and $g(x)=x^{2}-x$, $\forall x\in\mathbb R$. Then the set of all $x\in\mathbb R$, where the function $h(x)=(f\circ g)(x)$ is increasing, is :
Differentiation and its Application
Solution
Concept. $h'=f'(g)\,g'=\left(e^{g}-1\right)(2x-1)$; the first factor has the sign of $g=x^2-x$. Formulas used. (1) $f'(u)=e^u-1$, $g'(x)=2x-1$Step 1 - $h'$. By (1), $h'(x)=(2x-1)\left(e^{x^2-x}-1\right)$. Step 2 - Signs. $e^{x^2-x}-1\ge0$ iff $x(x-1)\ge0$, i.e. $x\le0$ or $x\ge1$. $x\le0$: $(-)(+)$, $h'\le0$. $0\le x\le\frac12$: $(-)(-)$, $h'\ge0$. $\frac12\le x\le1$: $(+)(-)$, $h'\le0$. $x\ge1$: $h'\ge0$. Step 3 - Answer. $h$ increases on $\left[0,\frac12\right]\cup[1,\infty)$. Answer: (C)