Question

Let exp(x) denote exponential function ex. If $f\left(x\right)=exp\left(x^{\frac{1}{x}}\right),x>0$ then the minimum value of f in the interval [2, 5] is

• A. $exp\left(e^{\frac{1}{e}}\right)$
• B. $exp\left(2^{\frac{1}{2}}\right)$
• C. $exp\left(5^{\frac{1}{5}}\right)$
• D. $exp\left(3^{\frac{1}{3}}\right)$

Answer 1

Answer: (C)

Given that,
$f(x)=e^{(x{)}^{\frac{1}{x}},x>0}$
Taking log on both sides, we get
$\log +(x)=(x{)}^{\frac{1}{x}}=g(x)(\text{ say })\dots (i)$
Here,$g(x)=x^{\frac{1}{x}}$
$\Rightarrow \log g(x)=\frac{1}{x}\log x$
On differentiating w.r.t. $x$, we get
$\cdot \frac{1}{g(x)}\cdot g^{\prime }(x)=\frac{x\cdot \frac{1}{x}-\log x}{x^2}$
$=\left(\frac{1-\log x}{x^2}\right)$
$\Rightarrow g^{\prime }(x)=x^{\left(\frac{1}{x}-2\right)}(1-\log \bar{x})$
For maximum or minimum of $g(x)$ put $g^{\prime }(x)=0$
$\Rightarrow x^{\left(\frac{1}{x}-2\right)}(1-\log x)=0$
$\Rightarrow \log x=1=\log e$
$\Rightarrow x=e$
and ${g^{\prime \prime }(x)|}_{x=e}>0$
So, $g(x)$ is minimum at $x=e$
$\therefore g(x)$ increases in $(0,e)$ and decreases in $(e,\infty )$ it will be minimum at either 2 or 5
$\therefore 2^{\frac{1}{2}}>5^{\frac{1}{5}}$
$\Rightarrow$ Minimum value of $f(x)=e^{(5{)}^{\frac{1}{5}}}$