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  • Explain solution RD Sharma class 12 chapter Maxima and Minima exercise Multiple choice question, question 22 maths

Explain solution RD Sharma class 12 chapter Maxima and Minima exercise Multiple choice question, question 22 maths

Answers (1)

Answer: option(a) \frac{4}{3}

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

Given: f(x)=\frac{1}{4x^2+2x+1}

Solution:

We have,

Maximum Value of \frac{1}{4x^2+2x+1} = Minimum Value of 4x^2+2x+1

f(x)=4x^2+2x+1               

\Rightarrow f'(x) =8x+2           

For maxima and minima f'(x)=0

\Rightarrow 8x+2=0

\Rightarrow x=\frac{-2}{8}

\Rightarrow x=\frac{-1}{4}

Now,

f''(x)=8

f''(1)=8>0

So,x=\frac{-1}{4} is a local minima.

Thus, \frac{1}{4x^2+2x+1} is maximum at x=\frac{-1}{4}

Maximum Value of 

\frac{1}{4 x^{2}+2 x+1}=\frac{1}{4\left(\frac{-1}{4}\right)^{2}+2\left(\frac{-1}{4}\right)+1}=\frac{1}{4\left(\frac{1}{16}\right)+2\left(\frac{-1}{4}\right)+1}=\frac{1}{\frac{1}{4}+\frac{-1}{2}+1}=\frac{1}{\frac{-1}{4}+1}=\frac{1}{\frac{3}{4}}=\frac{4}{3}

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