JEE Main 2020 (Online) 3rd September Evening ShiftQuestion 1
Suppose $f(x)$ is a polynomial of degree four, having critical points at $-1,0,1$. If $T=\{x\in\mathbf{R}\mid f(x)=f(0)\}$, then the sum of squares of all the elements of $T$ is :
Differentiation and its Application
Solution
Concept. With critical points $-1,0,1$, the derivative is $4a\,x(x^2-1)$; integrate and solve $f(x)=f(0)$. Formulas used. (1) $f'=4a\,x(x-1)(x+1)=4a(x^3-x)$ $\Rightarrow$ $f=ax^4-2ax^2+c$Step 1 - Solve $f(x)=f(0)$. By (1), $ax^4-2ax^2=0$, i.e. $ax^2(x^2-2)=0$. Step 2 - Set $T$. $T=\{0,\sqrt2,-\sqrt2\}$. Step 3 - Sum of squares. $0+2+2=4$. Answer: (A)