Match List-I with List-II
| List-I (Function) | List-II (Local maximum value) | ||
| (A) | \(x^{3} - 3x\) | (I) | 16 |
| (B) | \(6x - x^{2} - 5\) | (II) | 1 |
| (C) | \(12x - x^{3}\) | (III) | 2 |
| (D) | \(\displaystyle\frac{2x}{1 + x^{2}}\) | (IV) | 4 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
For each function the critical points come from f′(x) = 0, and the point where f′ changes from + to − (or f″ < 0) gives the local maximum.
(A) \(x^{3} - 3x\)
This matches (III) 2.
\( f' = 3x^{2} - 3 = 0,\ x = \pm 1;\ f'' = 6x,\ f''(-1) = -6 \lt 0 \)
\( f(-1) = -1 + 3 = 2 \)
(B) \(6x - x^{2} - 5\)
This matches (IV) 4.
\( f' = 6 - 2x = 0,\ x = 3;\ f'' = -2 \lt 0 \)
\( f(3) = 18 - 9 - 5 = 4 \)
(C) \(12x - x^{3}\)
This matches (I) 16.
\( f' = 12 - 3x^{2} = 0,\ x = \pm 2;\ f'' = -6x,\ f''(2) = -12 \lt 0 \)
\( f(2) = 24 - 8 = 16 \)
(D) \(\displaystyle\frac{2x}{1 + x^{2}}\)
This matches (II) 1.
\( \displaystyle f' = \frac{2(1 - x^{2})}{(1 + x^{2})^{2}} = 0,\ x = \pm 1 \)
f′ changes from + to − at x = 1, and \( f(1) = \frac{2}{2} = 1 \)