Match List-I with List-II
| List-I (Function) | List-II (Point of local minimum) | ||
| (A) | \(2x^2 - 12x + 5\) | (I) | \(x = 0\) |
| (B) | \(2x^3 - 24x\) | (II) | \(x = 1\) |
| (C) | \(e^{x} - x\) | (III) | \(x = 2\) |
| (D) | \(x^4 - 4x\) | (IV) | \(x = 3\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
In each case the local minimum is at a critical point where the second derivative is positive.
(A) \(2x^2 - 12x + 5\)
This matches (IV) x = 3.
\(f'(x) = 4x - 12 = 0\), so \(x = 3\)
\(f''(x) = 4 \gt 0\)
(B) \(2x^3 - 24x\)
This matches (III) x = 2.
\(f'(x) = 6x^2 - 24 = 0\), so \(x = \pm 2\)
\(f''(2) = 24 \gt 0,\ f''(-2) = -24 \lt 0\)
So x = 2 is the local minimum (x = −2 is a local maximum).
(C) \(e^{x} - x\)
This matches (I) x = 0.
\(f'(x) = e^{x} - 1 = 0\), so \(x = 0\)
\(f''(0) = e^{0} = 1 \gt 0\)
(D) \(x^4 - 4x\)
This matches (II) x = 1.
\(f'(x) = 4x^3 - 4 = 0\), so \(x = 1\)
\(f''(1) = 12 \gt 0\)