Match List-I with List-II
| List-I (Function) | List-II (Minimum value) | ||
| (A) | \(x^2 + 4x + 7,\ x \in R\) | (I) | \(-13\) |
| (B) | \(x^4 - 8x^2 + 3,\ x \in R\) | (II) | 2 |
| (C) | \(2x^3 - 6x,\ x \in [0, 2]\) | (III) | 3 |
| (D) | \((x - 1)^2 + (x - 3)^2,\ x \in R\) | (IV) | \(-4\) |
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)
In each case the minimum is found from the critical points (and, on a closed interval, the end points).
(A) \(x^2 + 4x + 7,\ x \in R\)
This matches (III) 3.
\(f'(x) = 2x + 4 = 0\), so \(x = -2\)
\(f(-2) = 4 - 8 + 7 = 3\)
(B) \(x^4 - 8x^2 + 3,\ x \in R\)
This matches (I) −13.
\(f'(x) = 4x^3 - 16x = 4x(x - 2)(x + 2)\)
The critical points are 0 and ±2; f(0) = 3 is a local maximum, and
\(f(\pm 2) = 16 - 32 + 3 = -13\)
(C) \(2x^3 - 6x,\ x \in [0, 2]\)
This matches (IV) −4.
\(f'(x) = 6x^2 - 6 = 0\), so \(x = 1\) in [0, 2]
\(f(0) = 0,\ f(1) = -4,\ f(2) = 4\)
(D) \((x - 1)^2 + (x - 3)^2,\ x \in R\)
This matches (II) 2.
\(f'(x) = 2(x - 1) + 2(x - 3) = 4x - 8 = 0\), so \(x = 2\)
\(f(2) = 1 + 1 = 2\)