CUET Preparation Today
CUET
-- Mathematics - Section A
Continuity and Differentiability
a
b
c
d
$f(x)=6(2x^4-x^2)$
$⇒f'(x)=6(8x^3-2x)$
$=12x(4x^2-1)$
⇒ Putting $f'(x)=0$
$⇒12x(4x^2-1)=0$
$⇒12x(2x-1)(2x+1)=0$
$⇒x=0,±\frac{1}{2}$
