The function $f(x) = x^4-2x^2$ is increasing on

$(-1, 0) ∪ (1,∞)$

The correct answer is Option (1) → $(-1, 0) ∪ (1,∞)$

$f(x)=x^4-2x^2$

$\frac{df}{dx}=4x^3-4x$

$=4x(x^2-1)$

Critical points $x=-1,0,1$

Sign of $\frac{df}{dx}$

For $\text{ x }<-1$, derivative negative

For $-1<\text{ x }<0$, derivative positive

For $0<\text{ x }<1$, derivative negative

For $\text{ x }>1$, derivative positive

Hence function is increasing on $(-1,0)\cup(1,\infty)$

The correct interval is $(-1,0)\cup(1,\infty)$.