Match List-I with List-II
| List-I | List-II | ||
| (A) | Minimum value of \(x\,e^x\) | (I) | 2 |
| (B) | Maximum value of \(8x - x^2\) | (II) | −16 |
| (C) | Maximum value of \(3x - x^3\), x > 0 | (III) | \(-\displaystyle\frac{1}{e}\) |
| (D) | Minimum value of \(x^4 - 8x^2\) | (IV) | 16 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
(A) Minimum value of \(x\,e^x\)
This matches (III) \(-\frac{1}{e}\). By the product rule:
\(f'(x) = e^x + x e^x = e^x(x + 1)\)
f′ is negative for x < −1 and positive for x > −1, so the minimum is at x = −1:
\( \displaystyle f(-1) = -e^{-1} = -\frac{1}{e} \)
(B) Maximum value of \(8x - x^2\)
This matches (IV) 16.
\(f'(x) = 8 - 2x = 0 \ \Rightarrow \ x = 4\)
\(f''(x) = -2 \lt 0\)
\(f(4) = 32 - 16 = 16\)
(C) Maximum value of \(3x - x^3\), x > 0
This matches (I) 2.
\(f'(x) = 3 - 3x^2 = 0 \ \Rightarrow \ x = 1\)
\(f''(1) = -6 \lt 0\)
\(f(1) = 3 - 1 = 2\)
(D) Minimum value of \(x^4 - 8x^2\)
This matches (II) −16.
\(f'(x) = 4x^3 - 16x = 4x(x - 2)(x + 2)\)
The critical points are x = 0 and x = ±2. The values are f(0) = 0 and
\(f(\pm 2) = 16 - 32 = -16\)
so the minimum value is −16 (x = 0 gives a local maximum).