JEE Main 2022 (Online) 27th June Evening ShiftQuestion 1
If $m$ and $n$ respectively are the number of local maximum and local minimum points of the function $f(x)=\int_{0}^{x^{2}}\frac{t^{2}-5t+4}{2+e^{t}}dt$, then the ordered pair $(m,n)$ is equal to
Differentiation and its Application
Solution
Concept. Differentiate with the chain rule; the zeros of $f'$ are simple, so each is a sign change. Formulas used. (1) $\dfrac{d}{dx}\int_0^{x^2}h(t)\,dt=2x\,h(x^2)$(2) $t^2-5t+4=(t-1)(t-4)$Step 1 - $f'$. By (1), (2), $$f'(x)=\frac{2x(x^2-1)(x^2-4)}{2+e^{x^2}}.$$ Step 2 - Signs. Zeros at $-2,-1,0,1,2$. From the right: $+$ for $x>2$, then $-,+,-,+,-$ going left. Step 3 - Count. $-\to+$ (minimum) at $-2,0,2$: $n=3$. $+\to-$ (maximum) at $-1,1$: $m=2$. Answer: (B)