The minimum value of f(x) = sin^4 x + cos^4 x , x ∈ R , is

The correct answer is Option (2) → \(\displaystyle\frac{1}{2}\)

Using \(\sin^{2} x + \cos^{2} x = 1\):

    \( f(x) = (\sin^{2} x + \cos^{2} x)^{2} - 2\sin^{2} x\cos^{2} x \)

    \( \displaystyle = 1 - \frac{1}{2}\sin^{2} 2x \)

Differentiating:

    \( f'(x) = -\sin 2x\cos 2x \times 2 = -\sin 4x \)

The critical points have \(\sin 4x = 0\), where \(\sin^{2} 2x\) is 0 or 1. The least value comes where \(\sin^{2} 2x = 1\) (for example \(x = \frac{\pi}{4}\)):

    \( \displaystyle f\left(\frac{\pi}{4}\right) = 1 - \frac{1}{2} = \frac{1}{2} \)