JEE PYQ: Application Of Derivatives Question 31
Question 31 - 2020 (03 Sep Shift 1)
Suppose $f(x)$ is a polynomial of degree four, having critical points at $-1, 0, 1$. If $T = {x \in \mathbb{R} \mid f(x) = f(0)}$, then the sum of squares of all the elements of $T$ is:
(1) $4$
(2) $6$
(3) $2$
(4) $8$
Show Answer
Answer: (1)
Solution
$f’(x) = kx(x+1)(x-1) = k(x^3-x)$. $f(x) = k\left(\frac{x^4}{4} - \frac{x^2}{2}\right) + C$. $f(x) = f(0)$ gives $x^2\left(\frac{x^2}{2} - 1\right) = 0$, so $x = 0, \pm\sqrt{2}$. Sum of squares $= 0 + 2 + 2 = 4$.
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