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Explain solution RD Sharma class 12 chapter Maxima and Minima exercise Multiple choice question, question 28 maths

Answers (1)

Answer: option(c) \frac{-1}{e}

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

Given: f(x)=x \log _ex

Solution:

We have,

f(x)=x \log _ex

\Rightarrow f'(x)= \log _ex+1

For maxima and minima f'(x)=0

\Rightarrow \log _ex+1=0

\Rightarrow \log _ex=-1

\Rightarrow x=e^{-1}

Now,

f''(x)=\frac{1}{x}

f^{\prime \prime}\left(e^{-1}\right)=\frac{1}{e^{-1}}=e>0

So,x=e^{-1}is the local minima.

Minimum value of  f(x)=f\left(e^{-1}\right)=e^{-1} \log _{e}\left(e^{-1}\right)=-e^{-1}=\frac{-1}{e}

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