The function f(x)=sin^4(x)+cos^4(x) , on interval ≤ft(0,(π)/2) , has

The correct answer is Option (2) → A local minima in the interval \( \left(0,\displaystyle\frac{\pi}{2}\right) \)

Note: The given answer is as per the official NTA Answer Key. However, more than one answer is possible: Option (2) and Option (4) are both correct, as explained below.

Since \( \sin^{2}x+\cos^{2}x=1 \),

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

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

Differentiating,

    \( \displaystyle f'(x)=-\frac{1}{2}\cdot2\sin2x\cos2x\cdot2 \)

    \( \displaystyle =-\sin4x \)

In \( \left(0,\frac{\pi}{2}\right) \), \( f'(x)=0 \) only at \( x=\frac{\pi}{4} \). The sign of f′ on either side is

    \( \displaystyle 0\lt x\lt\frac{\pi}{4}:\ f'\lt0 \)

    \( \displaystyle \frac{\pi}{4}\lt x\lt\frac{\pi}{2}:\ f'\gt0 \)

Since f′ changes from negative to positive, f has a local minimum at \( x=\frac{\pi}{4} \), and no local maximum in the interval.

Differentiating again,

    \( \displaystyle f''(x)=-4\cos4x \)

This is 0 at \( x=\frac{\pi}{8} \) and \( x=\frac{3\pi}{8} \), and its sign changes at each of them:

    \( \displaystyle 0\lt x\lt\frac{\pi}{8}:\ f''\lt0 \)

    \( \displaystyle \frac{\pi}{8}\lt x\lt\frac{3\pi}{8}:\ f''\gt0 \)

    \( \displaystyle \frac{3\pi}{8}\lt x\lt\frac{\pi}{2}:\ f''\lt0 \)

So \( x=\frac{\pi}{8} \) and \( x=\frac{3\pi}{8} \) are points of inflexion of f in the interval.

Option (1) A local maxima in the interval \( \left(0,\displaystyle\frac{\pi}{2}\right) \) — Incorrect. The only point where the slope is zero is \( x=\frac{\pi}{4} \), and there the function stops falling and starts rising, which is a minimum, not a maximum.

Option (2) A local minima in the interval \( \left(0,\displaystyle\frac{\pi}{2}\right) \) — Correct. The function falls up to \( x=\frac{\pi}{4} \) and rises after it, so it has a local minimum there.

Option (3) Neither local maxima nor local minima in the interval \( \left(0,\displaystyle\frac{\pi}{2}\right) \) — Incorrect. There is a local minimum at \( x=\frac{\pi}{4} \).

Option (4) A point of inflexion in \( \left(0,\displaystyle\frac{\pi}{2}\right) \) — Should also be correct, though not the answer given by NTA. At \( x=\frac{\pi}{8} \) the curve changes from bending downward to bending upward, and at \( x=\frac{3\pi}{8} \) it changes back, because \( f''(x)=-4\cos4x \) changes sign at both points. Such points are points of inflexion, and both lie inside the interval.

Advice for the exam: When more than one option is correct, mark the one that the question is mainly testing, here the local minimum at \( x=\frac{\pi}{4} \), since three of the four options are about local maxima and minima. If NTA’s provisional answer key marks a different option, challenge it during the answer-key challenge window, giving proper reasoning. NTA does revise its key when a challenge is well justified, so a sound challenge can raise your marks.