Verified Solution • 2025–26 Session moderate

QJEEM20Maca62: JEE Main 2020 Mathematics - Class 12 JEEM20Maca62 Solution: Suppose f(x) is a polynomial of degree four, having critical…

Suppose f(x) is a polynomial of degree four, having critical points at –1, 0, 1. If
T = {x ∈\in R | f(x) = f(0)}, then the sum of squares of all the elements of T is :

Multiple Choice Options

  • A 6
  • B 2
  • C 8
  • D 4

Step-by-Step Verified Solution

Critical points = −- 1, 0, 1.

∴\therefore f'(x) = a(x −- 1)(x + 1)x

∴\therefore f(x) = a (x44−x22)+C\left( {{{{x^4}} \over 4} - {{{x^2}} \over 2}} \right) + C

∵\because f(x) = f(0)

−a(x44−x22)+C=C- a\left( {{{{x^4}} \over 4} - {{{x^2}} \over 2}} \right) + C = C

ax24(x2−2)=0a{{{x^2}} \over 4}\left( {{x^2} - 2} \right) = 0

∴\therefore x = 0, 2\sqrt 2 , −2- \sqrt 2

∴\therefore T = {0,2,−2}\left\{ {0,\sqrt 2 , - \sqrt 2 } \right\}

Sum of square of elements of T=02+(2)2+(−2)2=4T = {0^2} + {\left( {\sqrt 2 } \right)^2} + {\left( { - \sqrt 2 } \right)^2} = 4

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