#### Please solve RD Sharma class 12 Chapter Maxima and Minima exercise Multiple choice question, question 2 maths textbook solution.

Answer: option (b) $ab\geq \frac{c^2}{4}$

Hint: For local maxima or minima, we must have f'(x) =0.

Given: $ax+\frac{b}{x}\geq c$

Solution:

We have,

$ax+\frac{b}{x}\geq c$

Minimum value of $ax+\frac{b}{x}= c$

Now,

$f(x)=ax+\frac{b}{x}$

$f'(x)=a-\frac{b}{x^2}$

$f'(x)=0$

$a-\frac{b}{x^2}=0$

$ax^2-b=0$

$x^2=\frac{b}{a}$

$x=\pm \sqrt{\frac{b}{a}}$

$f''(x)=\frac{2b}{x^3}$

Taking $x=\sqrt{\frac{x}{b}}$

\begin{aligned} &f^{\prime \prime}\left(\sqrt{\frac{b}{a}}\right)=\frac{2 b}{\left(\sqrt{\frac{b}{a}}\right)^{3}} \\ &f^{\prime \prime}\left(\sqrt{\frac{b}{a}}\right)=\frac{2 b(a)^{\frac{3}{2}}}{(b)^{\frac{3}{2}}}>0 \end{aligned}

So, $x=\sqrt{\frac{x}{b}}$ is a point of local minima.

\begin{aligned} &f\left(\frac{\sqrt{b}}{\sqrt{a}}\right)=a\left(\frac{\sqrt{b}}{\sqrt{a}}\right)+\frac{b}{\left(\frac{\sqrt{b}}{\sqrt{a}}\right)} \geq c \\ &f\left(\frac{\sqrt{b}}{\sqrt{a}}\right)=\sqrt{a b}+\sqrt{a b} \geq c \end{aligned}

$2\sqrt{ab}\geq c$

$\sqrt{ab}\geq \frac{c}{2}$

$ab\geq \frac{c^2}{4}$

Option (b) is correct.