Given a function \( f(x)=ax+\displaystyle\frac{b}{x};\ a\gt0,\ b\gt0 \), then
Match List-I with List-II
| List-I | List-II | ||
| (A) | \( f(x) \) has local maxima at \( x \) is equal to | (I) | \( \sqrt{\displaystyle\frac{b}{a}} \) |
| (B) | \( f(x) \) has local minima at \( x \) is equal to | (II) | \( 2\sqrt{ab} \) |
| (C) | Local Maxima of \( f(x) \) | (III) | \( -\sqrt{\displaystyle\frac{b}{a}} \) |
| (D) | Local Minima of \( f(x) \) | (IV) | \( -2\sqrt{ab} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) - (III), (B) - (I), (C) - (IV), (D) - (II)
At a local extremum, \( f'(x)=0 \):
\( \displaystyle f'(x)=a-\frac{b}{x^{2}}=0 \)
\( \displaystyle x=\pm\sqrt{\frac{b}{a}} \)
\( \displaystyle f''(x)=\frac{2b}{x^{3}} \)
Since b > 0, f″ > 0 at \( x=\sqrt{b/a} \) (minimum) and f″ < 0 at \( x=-\sqrt{b/a} \) (maximum).
(A) \( f(x) \) has local maxima at \( x \) is equal to
This matches (III) \( -\sqrt{b/a} \), where f″ < 0.
(B) \( f(x) \) has local minima at \( x \) is equal to
This matches (I) \( \sqrt{b/a} \), where f″ > 0.
(C) Local Maxima of \( f(x) \)
This matches (IV) \( -2\sqrt{ab} \):
\( \displaystyle f\left(-\sqrt{\frac{b}{a}}\right)=-\sqrt{ab}-\sqrt{ab}=-2\sqrt{ab} \)
(D) Local Minima of \( f(x) \)
This matches (II) \( 2\sqrt{ab} \):
\( \displaystyle f\left(\sqrt{\frac{b}{a}}\right)=\sqrt{ab}+\sqrt{ab}=2\sqrt{ab} \)
So the matching is (A) - (III), (B) - (I), (C) - (IV), (D) - (II).
Tip: A local maximum value can be smaller than a local minimum value; “local” compares only nearby points. Here the maximum value is \( -2\sqrt{ab} \) and the minimum value is \( 2\sqrt{ab} \).