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  • Provide solution for RD Sharma maths class12 Chapter Maxima and Minima exercise Multiple choice question, question 9.

Provide solution for RD Sharma maths class12 Chapter Maxima and Minima exercise Multiple choice question, question 9.

Answers (1)

Answer: option (b) \frac{1}{2}

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

Given: f(x)=\frac{1}{x}+\frac{1}{y}

Solution:

Let the two non-zero numbers be x and y.

x + y =8

y = 8 -x                                                                                                         ...(i)

We have,

f(x)=\frac{1}{x}+\frac{1}{y}

f(x)=\frac{1}{x}+\frac{1}{(8-x)}                                                  …(ii)                   From Equation (i)                 

On differentiating w.r.t x

f'(x)=\frac{-1}{x^2}+\frac{1}{(8-x)^2}
for maxima and minimaf'(x)=0

\Rightarrow \frac{-1}{x^2}+\frac{1}{(8-x)^2}=0

\Rightarrow \frac{-(8-x)^2+x^2}{x^2(8-x)^2}=0

\Rightarrow -64-x^2+16x+x^2=0

\Rightarrow 16x=64

\Rightarrow x=4

Now, again differentiating (ii) w.r.t  x

\begin{aligned} &f^{\prime \prime}(x)=\frac{2}{x^{3}}-\frac{2}{(8-x)^{3}} \\ &f^{\prime \prime}(4)=\frac{2}{4^{3}}-\frac{2}{(8-4)^{3}} \\ &f^{\prime \prime}(4)=\frac{2}{64}-\frac{2}{64}=0 \end{aligned}

Minimum Value =\frac{1}{4}+\frac{1}{4}=\frac{2}{4}=\frac{1}{2}

Posted by

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