Sets, Relation and Function
116917
If \(f(x)=\left(\frac{1}{x}\right)^x\), then the maximum value of \(f(x)\) is:
1 e
2 \((\mathrm{e})^{1 / \mathrm{e}}\)
3 \(\left(\frac{1}{\mathrm{e}}\right)^{\mathrm{e}}\)
4 none of these
Explanation:
B Given,
\(f(x)=\left(\frac{1}{x}\right)^x\)
Then, let \(\mathrm{y}=\left(\frac{1}{\mathrm{x}}\right)^{\mathrm{x}}\)
Taking log both sides, we get -
\(\log y=\log \left(\frac{1}{x}\right)^x\)
\(\log y=x \log \left(\frac{1}{x}\right) \quad\left(\because \log a^m=m \log a\right)\)
Differentiating both side, w.r.t. \(\mathrm{x}\), we get -
\(\frac{1}{y} \frac{d y}{d x}=x \cdot \frac{1}{\frac{1}{x}} \cdot \frac{-1}{x^2}+\log \left(\frac{1}{x}\right) \cdot 1\)
\(\frac{1}{y} \frac{d y}{d x}=x^2 \times\left(-\frac{1}{x^2}\right)+\log \left(\frac{1}{x}\right) \cdot 1\)
\(\frac{1}{y} \frac{d y}{d x}=-1+\log \frac{1}{x}\)
\(\frac{d y}{d x}=y\left(-1+\log \frac{1}{x}\right)\)
\(\frac{d y}{d x}=\left(\frac{1}{x}\right)^x\left(-1+\log \frac{1}{x}\right)\)
We know that, for maximum value, \(\frac{d y}{d x}=0\)
\(0=\left(\frac{1}{x}\right)^x\left(-1+\log \frac{1}{x}\right)\)
\(\log _e \frac{1}{x}=1\)
\(\frac{1}{x}=e^1\)
\(x=\frac{1}{e}\)
Putting the value of \(x=\frac{1}{e}\) in equation (i), we get -
\(\mathrm{f}(\mathrm{x})=\left(\frac{1}{1 / \mathrm{e}}\right)^{\frac{1}{\mathrm{e}}}=(\mathrm{e})^{\frac{1}{\mathrm{e}}}\)