The interval in which the function
\( f(x)=x^3-3x^2+3x-20 \) is increasing, is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( (-\infty,\infty) \)
Note: The given answer is as per the official NTA Answer Key. However, more than one answer is possible: Option (1), Option (2), Option (3) and Option (4) are all correct, as explained below.
\( f'(x)=3x^2-6x+3 \)
\( =3(x-1)^2 \)
\( f'(x)\ge 0 \) for every x, and it is 0 only at the single point x = 1, so f does not turn back there. Hence f is increasing on the whole real line, and so it is also increasing on every part of the real line.
Option (1) \( (-\infty,\infty) \) — Correct. \( f'(x)=3(x-1)^2\ge 0 \), zero only at x = 1, so f increases on the whole real line, the largest such interval.
Option (2) \( (0,\infty) \) — Should also be correct, though not the answer given by NTA. \( (0,\infty) \) is a part of the real line, so f increases there too.
Option (3) \( (-\infty,1) \) — Should also be correct, though not the answer given by NTA. \( (-\infty,1) \) is a part of the real line, so f increases there too.
Option (4) \( (1,\infty) \) — Should also be correct, though not the answer given by NTA. \( (1,\infty) \) is a part of the real line, so f increases there too.
Advice for the exam: When more than one option is correct, mark the one that answers the question most completely, here the whole real line, \( (-\infty,\infty) \). 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.
NCERT (2026-27 edition):
Increasing and Decreasing Functions
