Match List-I with List-II
| List-I (Function on R) | List-II (Number of points of local extremum) | ||
| (A) | \(x^{3} + 3x^{2} + 3x\) | (I) | 2 |
| (B) | \(x^{4} - 32x\) | (II) | 0 |
| (C) | \(x^{3} - 6x^{2}\) | (III) | 3 |
| (D) | \(x^{4} - 8x^{2}\) | (IV) | 1 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
A point of local extremum is a critical point where f′(x) changes sign. A critical point without a change of sign is not counted.
(A) \(x^{3} + 3x^{2} + 3x\)
This matches (II) 0.
\( f'(x) = 3x^{2} + 6x + 3 = 3(x + 1)^{2} \ge 0 \)
f′ is zero at x = −1 but does not change sign, so there is no extremum
(B) \(x^{4} - 32x\)
This matches (IV) 1.
\( f'(x) = 4x^{3} - 32 = 4(x^{3} - 8) \), zero only at x = 2
\( f''(2) = 12 \times 4 = 48 \gt 0 \): one local minimum
(C) \(x^{3} - 6x^{2}\)
This matches (I) 2.
\( f'(x) = 3x^{2} - 12x = 3x(x - 4) \), zero at x = 0 and x = 4
\( f''(x) = 6x - 12 \): f″(0) = −12 (maximum), f″(4) = 12 (minimum)
(D) \(x^{4} - 8x^{2}\)
This matches (III) 3.
\( f'(x) = 4x^{3} - 16x = 4x(x - 2)(x + 2) \), zero at x = 0, 2, −2
\( f''(x) = 12x^{2} - 16 \): f″(0) = −16 (maximum), f″(±2) = 32 (minima)