f′(x)=k.x(x+1)(x−1)
=k(x3−x)
Integrating both sides with respect to x, we get
f(x)=k(4x4−2x2)+c
⇒f(0)=c
∵f(x)=f(0)
⇒k4(x4−2x2)+c=c
⇒x2(x2−2)=0
⇒x=0,2,−2
⇒T=0,2,−2
Thus, sum of squares of all the elements of T is (0)2+(2)2+(−2)2=4.
JEE Main 2020 — Mathematics Calculus
Suppose f(x) is a polynomial of degree four having critical points at −1,0,1. If T=x∈R∣f(x)=f(0), then the sum of squares of all the elements of T is :
Held on 3 Sept 2020 · Verified 6 Jul 2026.
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