For the function \( f(x)=3x^4+4x^3-12x^2+12 \), then
Match List-I with List-II
| List-I | List-II | ||
| (A) | \( x=0 \) | (I) | Local minimum value of \( f(x) \) |
| (B) | \( x=-2 \) | (II) | Local maximum value of \( f(x) \) |
| (C) | \( f(0) \) | (III) | Point of local minima |
| (D) | \( f(1) \) | (IV) | Point of local maxima |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) - (IV), (B) - (III), (C) - (II), (D) - (I)
The critical points come from f′(x) = 0, and the second derivative test classifies them.
\( \displaystyle f'(x)=12x^3+12x^2-24x \)
\( \displaystyle =12x(x+2)(x-1) \)
\( \displaystyle f'(x)=0\ \Rightarrow\ x=0,\ -2,\ 1 \)
\( \displaystyle f''(x)=36x^2+24x-24 \)
\( \displaystyle f''(0)=-24\lt 0 \)
\( \displaystyle f''(-2)=144-48-24=72\gt 0 \)
\( \displaystyle f''(1)=36+24-24=36\gt 0 \)
So x = 0 is a point of local maxima, and x = −2 and x = 1 are points of local minima.
(A) \( x=0 \)
This matches (IV) Point of local maxima, since f″(0) < 0.
(B) \( x=-2 \)
This matches (III) Point of local minima, since f″(−2) > 0.
(C) \( f(0) \)
This matches (II) Local maximum value of f(x).
\( \displaystyle f(0)=12 \)
(D) \( f(1) \)
This matches (I) Local minimum value of f(x), since x = 1 is a point of local minima.
\( \displaystyle f(1)=3+4-12+12=7 \)
So the matching is (A)-(IV), (B)-(III), (C)-(II), (D)-(I).
Tip: Keep "point of local maxima" (a value of x) apart from "local maximum value" (the value of f there); items like x = 0 and f(0) are chosen to test exactly this difference.
NCERT (2026-27 edition):
Maxima and Minima
