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Given a function f(x)=ax+bxf(x) = ax + \frac{b}{x}; a>0,b>0a > 0, b > 0, then

Match List-I with List-II

List-IList-II
(A) f(x)f(x) has local maxima at xx is equal to(I) ba\sqrt{\frac{b}{a}}
(B) f(x)f(x) has local minima at xx is equal to(II) 2ab2\sqrt{ab}
(C) Local Maxima of f(x)f(x)(III) −ba-\sqrt{\frac{b}{a}}
(D) Local Minima of f(x)f(x)(IV) −2ab-2\sqrt{ab}

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 2

f′(x)=a−bx2=0f'(x)=a-\frac{b}{x^2}=0 gives x=±b/ax=\pm\sqrt{b/a}, and f′′(x)=2bx3f''(x)=\frac{2b}{x^3}. At x=−b/ax=-\sqrt{b/a}, f′′<0f''<0, so it is a local maximum with value −2ab-2\sqrt{ab}. At x=b/ax=\sqrt{b/a}, f′′>0f''>0, so it is a local minimum with value 2ab2\sqrt{ab}. Hence (A)-(III), (B)-(I), (C)-(IV), (D)-(II).

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