Q93 / 180 · KCET 2016 · Mathematics
The maximum value of \( \left(\frac{1}{x}\right)^{x} \) is
- A\( \mathrm{e} \)
- B$e^{e}$
- Correct answerC\( e^{\frac{1}{e}} \)
- D\( \left(\frac{1}{e}\right)^{e} \)
Solution
Let \( y=\left(\frac{1}{x}\right)^{x} \)
Taking log on both the sides, we get
\[ \begin{array}{l} \log y=x \log \left(\frac{1}{x}\right)=-x \log x \\ \log y=-x \log x \rightarrow(1) \end{array} \]
Now, differentiating Eq. (1) with respect to \( x \), we get
\[ \begin{array}{l} \frac{1}{y} \frac{d y}{d x}=-\left(x \cdot \frac{1}{x}+\log x\right) \\ \Rightarrow \frac{1}{y} \frac{d y}{d x}=-(1+\log x) \\ \Rightarrow \frac{d y}{d x}=-y(1+\log x) \rightarrow(2) \\ \Rightarrow \frac{d y}{d x}=-\left(\frac{1}{x}\right)^{x}(1+\log x) \end{array} \]
\[ \begin{array}{l} \text { Now, } \frac{d y}{d x}=0 \\ -y(1+\log x)=0 \\ \Rightarrow \log x=-1 \Rightarrow x=e^{-1} \end{array} \]
Now, differentiating Eq. (2) with respect to \( x \), we get
\[ \begin{array}{l} \frac{d^{2} y}{d x^{2}}=-(1+\log x) \frac{d y}{d x}-\frac{y}{x} \\ \frac{d^{2} y}{d x^{2}}=y(1+\log x)^{2}-\frac{y}{x} \\ \frac{d^{2} y}{d x^{2}}=\left(\frac{1}{x}\right)^{x}(1+\log x)^{2}-\left(\frac{1}{x}\right)^{x} \frac{1}{x} \\ \text { At } x=e^{-1}, \text { we get } \end{array} \]
Taking log on both the sides, we get
\[ \begin{array}{l} \log y=x \log \left(\frac{1}{x}\right)=-x \log x \\ \log y=-x \log x \rightarrow(1) \end{array} \]
Now, differentiating Eq. (1) with respect to \( x \), we get
\[ \begin{array}{l} \frac{1}{y} \frac{d y}{d x}=-\left(x \cdot \frac{1}{x}+\log x\right) \\ \Rightarrow \frac{1}{y} \frac{d y}{d x}=-(1+\log x) \\ \Rightarrow \frac{d y}{d x}=-y(1+\log x) \rightarrow(2) \\ \Rightarrow \frac{d y}{d x}=-\left(\frac{1}{x}\right)^{x}(1+\log x) \end{array} \]
\[ \begin{array}{l} \text { Now, } \frac{d y}{d x}=0 \\ -y(1+\log x)=0 \\ \Rightarrow \log x=-1 \Rightarrow x=e^{-1} \end{array} \]
Now, differentiating Eq. (2) with respect to \( x \), we get
\[ \begin{array}{l} \frac{d^{2} y}{d x^{2}}=-(1+\log x) \frac{d y}{d x}-\frac{y}{x} \\ \frac{d^{2} y}{d x^{2}}=y(1+\log x)^{2}-\frac{y}{x} \\ \frac{d^{2} y}{d x^{2}}=\left(\frac{1}{x}\right)^{x}(1+\log x)^{2}-\left(\frac{1}{x}\right)^{x} \frac{1}{x} \\ \text { At } x=e^{-1}, \text { we get } \end{array} \]
Hence, the correct answer is option: 'C'.