Match List-I with List-II
| List-I | List-II | ||
| (A) | Greatest product of two positive numbers whose sum is 12 | (I) | 4 |
| (B) | Least value of \(x + \displaystyle\frac{4}{x}\), \(x \gt 0\) | (II) | 25 |
| (C) | Least value of \(2x^{2} - 12x + 43\), \(x \in R\) | (III) | 8 |
| (D) | Least value of \(x^{2} + \displaystyle\frac{16}{x^{2}}\), \(x \ne 0\) | (IV) | 36 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
In each case the derivative is set equal to zero and the sign of the second derivative confirms the optimum.
(A) Greatest product of two positive numbers whose sum is 12
This matches (IV) 36.
\( P = x(12 - x),\ P' = 12 - 2x = 0,\ x = 6 \)
\( P'' = -2 \lt 0,\ P = 6 \times 6 = 36 \)
(B) Least value of \(x + \displaystyle\frac{4}{x}\), \(x \gt 0\)
This matches (I) 4.
\( \displaystyle f' = 1 - \frac{4}{x^{2}} = 0,\ x = 2 \)
\( \displaystyle f'' = \frac{8}{x^{3}} \gt 0,\ f(2) = 2 + 2 = 4 \)
(C) Least value of \(2x^{2} - 12x + 43\), \(x \in R\)
This matches (II) 25.
\( g' = 4x - 12 = 0,\ x = 3 \)
\( g'' = 4 \gt 0,\ g(3) = 18 - 36 + 43 = 25 \)
(D) Least value of \(x^{2} + \displaystyle\frac{16}{x^{2}}\), \(x \ne 0\)
This matches (III) 8.
\( \displaystyle h' = 2x - \frac{32}{x^{3}} = 0,\ x^{4} = 16,\ x = \pm 2 \)
\( \displaystyle h'' = 2 + \frac{96}{x^{4}} \gt 0,\ h(\pm 2) = 4 + 4 = 8 \)