Match List-I with List-II
The function \(f(x) = (x - 1)(x + 1)^2\) has
| List-I | List-II | ||
| (A) | A local maxima at \(x\) = ____ | (I) | \(\displaystyle\frac{1}{3}\) |
| (B) | A local minima at \(x\) = ____ | (II) | 0 |
| (C) | The local minimum value of \(f(x)\) = ____ | (III) | -1 |
| (D) | The local maximum value of \(f(x)\) = ____ | (IV) | \(-\displaystyle\frac{32}{27}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) - (III), (B) - (I), (C) - (IV), (D) - (II)
By the product rule:
\(f'(x) = (x + 1)^2 + 2(x - 1)(x + 1)\)
\(= (x + 1)(x + 1 + 2x - 2)\)
\(= (x + 1)(3x - 1)\)
\(f'(x) = 0 \Rightarrow x = -1\) or \(x = \frac{1}{3}\)
\(f''(x) = (3x - 1) + 3(x + 1) = 6x + 2\)
By the second derivative test:
(A) A local maxima at \(x\) = ____
This matches (III) −1, since \(f''(-1) = -4 \lt 0\).
(B) A local minima at \(x\) = ____
This matches (I) \(\frac{1}{3}\), since \(f''\left(\frac{1}{3}\right) = 4 \gt 0\).
(C) The local minimum value of \(f(x)\) = ____
This matches (IV) \(-\frac{32}{27}\):
\(\displaystyle f\left(\frac{1}{3}\right) = \left(-\frac{2}{3}\right)\left(\frac{4}{3}\right)^2 = -\frac{2}{3}\cdot\frac{16}{9} = -\frac{32}{27}\)
(D) The local maximum value of \(f(x)\) = ____
This matches (II) 0:
\(f(-1) = (-2)(0)^2 = 0\)
So the matching is (A) - (III), (B) - (I), (C) - (IV), (D) - (II).