JEE Main 2020 (Online) 9th January Evening ShiftQuestion 1
Let a function $f:[0,5]\to\mathbf{R}$ be continuous, $f(1)=3$ and $F$ be defined as : $F(x)=\int_1^xt^2g(t)\,dt$, where $g(t)=\int_1^tf(u)\,du$. Then for the function $F$, the point $x=1$ is :
Differentiation and its Application
Solution
Concept. $F'(1)=0$ because $g(1)=0$; the sign of $F''(1)$ decides the type. Formulas used. (1) $F'(x)=x^2g(x)$, $\ g'(x)=f(x)$Step 1 - Critical point. $g(1)=\int_1^1=0$, so by (1) $F'(1)=0$. Step 2 - $F''(1)$. $F''(x)=2xg(x)+x^2f(x)$, so $F''(1)=0+f(1)=3>0$. Step 3 - Conclude. Local minimum at $x=1$. Answer: (C)