Q111 / 200 · WBJEE 2013 · Mathematics
Let \(\exp (x)\) denote exponential function ex. If \(f(x)=\exp \left(x^{\frac{1}{x}}\right), x>0\) then the minimum value of \(f\) in the interval \([2,5]\) is
- A\(\exp \left(e^{\frac{1}{c}}\right)\)
- B\(\exp \left(2^{\frac{1}{2}}\right)\)
- Correct answerC\(\exp \left(5^{\frac{1}{5}}\right)\)
- D\(\exp \left(3^{\frac{1}{3}}\right)\)
Solution
Given that,
$$
f(x)=e^{(x)^{\frac{1}{x}}, x>0}
$$
Taking log on both sides, we get
$$
\log +(x)=(x)^{\frac{1}{x}}=g(x)
$$
Here, $g(x)=x^{\frac{1}{x}}$
$\Rightarrow \quad \log g(x)=\frac{1}{x} \log x$
On differentiating w.r.t. $x$, we get
$$
\begin{aligned}
\frac{1}{g(x)} \cdot g^{\prime}(x) &=\frac{x \cdot \frac{1}{x}-\log x}{x^{2}} \\
&=\left(\frac{1-\log x}{2}\right)
\end{aligned}
$$
$\Rightarrow \quad g^{\prime}(x)=x^{\left(\frac{1}{x}-2\right)}(1-\log x)$
For maximum or minimum of $g(x)$ put
$$
g^{\prime}(x)=0
$$
$\Rightarrow \quad x^{\left(\frac{1}{x}-2\right)}(1-\log x)=0$
$$
\begin{array}{lr}
\Rightarrow & \log x=1=\log e \\
\Rightarrow & x=e
\end{array}
$$
and $\left.\quad g^{\prime \prime}(x)\right|_{x=e}>0$
So, $g(x)$ is minimum at $x=e$ $\therefore g(x)$ increases in $(0, e)$ and decreases in $(\theta, \infty)$ it will be minimum at either 2 or 5 $\therefore 2^{\frac{1}{2}}>5^{\frac{1}{5}} \Rightarrow$ Minimum value of $f(x)=e^{(5)^{\frac{1}{5}}}$
Hence, the correct answer is option: 'C'.