The function \( f(x)=2x^3-3x^2-36x+7 \) is :
(A) Increasing in \( (-\infty,-2) \)
(B) Decreasing in \( (-2,3) \)
(C) Neither increasing nor decreasing in \( (-2,3) \)
(D) Increasing in \( (3,\infty) \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A), (B) and (D) only
Differentiating,
\( \displaystyle f'(x)=6x^2-6x-36 \)
\( \displaystyle =6(x+2)(x-3) \)
The critical points x = −2 and x = 3 split the line into three intervals. The sign of f′(x) on each:
\( \displaystyle x\lt -2:\ (x+2)\lt 0,\ (x-3)\lt 0,\ f'(x)\gt 0 \)
\( \displaystyle -2\lt x\lt 3:\ (x+2)\gt 0,\ (x-3)\lt 0,\ f'(x)\lt 0 \)
\( \displaystyle x\gt 3:\ (x+2)\gt 0,\ (x-3)\gt 0,\ f'(x)\gt 0 \)
So f′(x) > 0 on \( (-\infty,-2) \) and on \( (3,\infty) \), and f′(x) < 0 on \( (-2,3) \).
Checking the statements
(A) Increasing in \( (-\infty,-2) \) — Correct. f′(x) > 0 on \( (-\infty,-2) \).
(B) Decreasing in \( (-2,3) \) — Correct. f′(x) < 0 on \( (-2,3) \).
(C) Neither increasing nor decreasing in \( (-2,3) \) — Incorrect. f′(x) < 0 on the whole of \( (-2,3) \), so f is decreasing there.
(D) Increasing in \( (3,\infty) \) — Correct. f′(x) > 0 on \( (3,\infty) \).
NCERT (2026-27 edition):
Increasing and Decreasing Functions
