#### Please solve RD Sharma class 12 chapter 16 Increasing and Decreasing Function Excercise Fill in the blanks Question 10: Maths Textbook Solution.

Answer: The largest interval in which $f(x)$ is increasing is

Hint: Function is increasing if $f'(x)>0$ .

Given: $f(x)=x^{1/x}$

Explanation: We have,

$f(x)=x^{1/x}$

Taking logarithm  on both sides,

$\log f(x)=\frac{1}{x}\log x$                                             …(i)

Differentiate (i) with respect to x

$\frac{1}{f(x)}f'(x)=\frac{1}{x}\cdot \frac{1}{x}+\log x\cdot \left ( -\frac{1}{x^{2}} \right )$

$\therefore f'(x)=\left ( \frac{1}{x^{2}}-\frac{1}{x^{2}}\log x \right )\cdot x^{1/x}$                               $\left [ \because f(x)=x^{\frac{1}{x}} \right ]$

$f'(x)=\frac{1}{x^{2}}(1-\log x)x^{\frac{1}{x}}$

$\because f(x)$  is strictly increasing,

$f'(x)>0$

$\Rightarrow \frac{1}{x^{2}}(1-\log x)x^{1/x}>0$

Now, if  $1-\log x>0$

$\Rightarrow \log x<1$

$\Rightarrow x

$\Rightarrow1

So, the largest interval in which $f(x)$  in increasing in $(1,e )$