The function f(x) = 6(2x^4 - x^2) is strictly increasing in

The correct answer is Option (2) → \(\left(-\displaystyle\frac{1}{2}, 0\right) \cup \left(\displaystyle\frac{1}{2}, \infty\right)\)

A function is strictly increasing where \(f'(x) \gt 0\).

    \(f'(x) = 6(8x^3 - 2x)\)
    \(= 12x(4x^2 - 1)\)
    \(= 12x(2x - 1)(2x + 1)\)

\(f'(x) = 0\) at \(x = -\frac{1}{2}, 0, \frac{1}{2}\). These points split the line into four intervals. Sign of f′(x) in each:

    \(x \lt -\frac{1}{2}\): three negative factors, \(f'(x) \lt 0\)
    \(-\frac{1}{2} \lt x \lt 0\): two negative factors, \(f'(x) \gt 0\)
    \(0 \lt x \lt \frac{1}{2}\): one negative factor, \(f'(x) \lt 0\)
    \(x \gt \frac{1}{2}\): no negative factor, \(f'(x) \gt 0\)

So f is strictly increasing in \(\left(-\frac{1}{2}, 0\right) \cup \left(\frac{1}{2}, \infty\right)\).