The interval on which the function \(f(x) = x^4 -\displaystyle\frac{x^3}{3}\) is strictly decreasing, is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\left(-\infty,\frac{1}{4}\right)\)
Note: The given answer is as per the official NTA Answer Key. However, more than one answer is possible: Option (3) and Option (4) are both correct, as explained below.
\(f'(x) = 4x^3 - x^2 = x^2(4x - 1)\)
\(x^2 \ge 0\), and it is 0 only at x = 0. Also \(4x - 1 \lt 0\) for \(x \lt \frac{1}{4}\) and \(4x - 1 \gt 0\) for \(x \gt \frac{1}{4}\).
So \(f'(x) \lt 0\) on \(\left(-\infty, \frac{1}{4}\right)\) except at the single point x = 0, which does not stop f from being strictly decreasing there; and \(f'(x) \gt 0\) on \(\left(\frac{1}{4}, \infty\right)\), where f is strictly increasing.
Option (1) \((4, \infty)\) — Incorrect. \(f' \gt 0\) there, so f is increasing.
Option (2) \(\left(\frac{1}{4}, \infty\right)\) — Incorrect. This is the interval where f is strictly increasing.
Option (3) \(\left(-\infty,\frac{1}{4}\right)\) — Correct. f is strictly decreasing on the whole of this interval, the largest such interval.
Option (4) \(\left(0,\frac{1}{4}\right)\) — Should also be correct, though not the answer given by NTA. f is strictly decreasing on the whole of \(\left(-\infty, \frac{1}{4}\right)\). The interval \(\left(0, \frac{1}{4}\right)\) is just a part of it, so f is strictly decreasing there too.
Advice for the exam: When more than one option is correct, mark the one that answers the question most completely, here the largest interval on which f is strictly decreasing. 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.