Q138 / 200 · Manipal (MET) 2014 · Mathematics
The minimum value of $\frac{x}{\log x}$ is
- Correct answerA$e$
- B$\frac{1}{e}$
- C$e^2$
- D$e^3$
Solution
Let
$\begin{aligned}
f(x) & =\frac{x}{\log x} \\
f^{\prime}(x) & =\frac{\log x-1}{(\log x)^2}
\end{aligned}$
Pur $f^{\prime}(x)=0$, for maxima or minima, we get $\log x-1=0$ $\Rightarrow \quad x=e$ Now, $f^{\prime \prime}(x)=\frac{(\log x)^2 \cdot \frac{1}{x}-(\log x-1) \cdot \frac{2 \log x}{x}}{(\log x)^4}$ $\Rightarrow \quad f^{\prime \prime}(e)=\frac{\frac{1}{e}-0}{1}=\frac{1}{e}\gt0$ $\therefore$ Function is minimum at $x=e$. Hence, minimum value of $f(x)$ at $x=e$ is $f(e)=e .$
Pur $f^{\prime}(x)=0$, for maxima or minima, we get $\log x-1=0$ $\Rightarrow \quad x=e$ Now, $f^{\prime \prime}(x)=\frac{(\log x)^2 \cdot \frac{1}{x}-(\log x-1) \cdot \frac{2 \log x}{x}}{(\log x)^4}$ $\Rightarrow \quad f^{\prime \prime}(e)=\frac{\frac{1}{e}-0}{1}=\frac{1}{e}\gt0$ $\therefore$ Function is minimum at $x=e$. Hence, minimum value of $f(x)$ at $x=e$ is $f(e)=e .$
Hence, the correct answer is option: 'A'.