The number of points of the local maximum value of the function \( f(x)=\ 3x^4+4x^3-12x^2+2 \) is,
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1
By the first derivative test, a point of local maximum is where \( f'(x) \) changes sign from + to −.
\( f'(x)=12x^3+12x^2-24x \)
\( =12x(x^2+x-2) \)
\( =12x(x+2)(x-1) \)
The critical points are x = −2, 0 and 1. Signs of \( f'(x) \):
For \( x\lt -2 \): \( f'(x)\lt 0 \)
For \( -2\lt x\lt 0 \): \( f'(x)\gt 0 \)
For \( 0\lt x\lt 1 \): \( f'(x)\lt 0 \)
For \( x\gt 1 \): \( f'(x)\gt 0 \)
\( f'(x) \) changes from + to − only at x = 0, so there is exactly one point of local maximum. At x = −2 and x = 1 it changes from − to +, so these are points of local minimum.
Why the other options are incorrect
Option (1) 2 — Incorrect. Two is the number of points of local minimum (x = −2 and x = 1), not maximum.
Option (3) \( -2 \) — Incorrect. x = −2 is a point of local minimum, and a number of points cannot be negative.
Option (4) 0 — Incorrect. \( f'(x) \) changes from + to − at x = 0, so there is one point of local maximum.
NCERT (2026-27 edition):
Maxima and Minima
