KCET · Maths · Application of Derivatives
The set of real values of \(x\) for which \(f(x)=\frac{x}{\log x}\) is increasing, is
- A \(\{x: x \geq e\}\)
- B empty
- C \(\{\mathrm{x}: \mathrm{x} < \mathrm{e}\}\)
- D \(\{1\}\)
Answer & Solution
Correct Answer
(A) \(\{x: x \geq e\}\)
Step-by-step Solution
Detailed explanation
\[
\begin{aligned}
f(x)=& \frac{x}{\log x} \\
f^{\prime}(x) &=\frac{\log x \cdot 1-x \cdot \frac{1}{x}}{(\log x)^{2}}=\frac{(\log x-1)}{(\log x)^{2}}
\end{aligned}
\]
We know that,
\(\mathrm{f}(\mathrm{x})\) is increasing (strictly)
When \(\mathrm{f}^{\prime}(\mathrm{x})>0\)
\(\Rightarrow \quad \frac{(\log x-1)}{(\log x)^{2}}>0\)
\(\Rightarrow \quad(\log x-1)>0\)
\(\Rightarrow \quad \log x>1\)
\(\Rightarrow \quad \log _{\mathrm{e}} \mathrm{x}>\log _{\mathrm{e}} \mathrm{e}\)
\(\Rightarrow \quad \quad \mathrm{x}>\mathrm{e}\)
Hence, \(\mathrm{x}: \mathrm{x} \geq \mathrm{e}\)
\begin{aligned}
f(x)=& \frac{x}{\log x} \\
f^{\prime}(x) &=\frac{\log x \cdot 1-x \cdot \frac{1}{x}}{(\log x)^{2}}=\frac{(\log x-1)}{(\log x)^{2}}
\end{aligned}
\]
We know that,
\(\mathrm{f}(\mathrm{x})\) is increasing (strictly)
When \(\mathrm{f}^{\prime}(\mathrm{x})>0\)
\(\Rightarrow \quad \frac{(\log x-1)}{(\log x)^{2}}>0\)
\(\Rightarrow \quad(\log x-1)>0\)
\(\Rightarrow \quad \log x>1\)
\(\Rightarrow \quad \log _{\mathrm{e}} \mathrm{x}>\log _{\mathrm{e}} \mathrm{e}\)
\(\Rightarrow \quad \quad \mathrm{x}>\mathrm{e}\)
Hence, \(\mathrm{x}: \mathrm{x} \geq \mathrm{e}\)
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