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

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

Answers (1)

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.

Posted by

infoexpert24

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