MHT CET · Maths · Application of Derivatives
If \(\mathrm{f}(x)=\frac{\log x}{x}(x\gt0)\), then it is increasing in
- A \((0, \mathrm{e})\)
- B \((\mathrm{e}, \infty)\)
- C \((0, \infty)\)
- D \((-\infty, \infty)\)
Answer & Solution
Correct Answer
(A) \((0, \mathrm{e})\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& \mathrm{f}(x)=\frac{\log x}{x} \\
& \therefore \quad \mathrm{f}^{\prime}(x)=\frac{1}{x^2}-\frac{\log x}{x^2}=\frac{1-\log x}{x^2}
\end{aligned}\)
For \(\mathrm{f}(x)\) to be increasing, \(\mathrm{f}^{\prime}(x) \gt 0\)
\(\Rightarrow 1-\log x \gt 0 \Rightarrow 1 \gt \log x \Rightarrow \mathrm{e} \gt x\)
\(\therefore \quad \mathrm{f}(x)\) is increasing in the interval \((0, \mathrm{e})\).
& \mathrm{f}(x)=\frac{\log x}{x} \\
& \therefore \quad \mathrm{f}^{\prime}(x)=\frac{1}{x^2}-\frac{\log x}{x^2}=\frac{1-\log x}{x^2}
\end{aligned}\)
For \(\mathrm{f}(x)\) to be increasing, \(\mathrm{f}^{\prime}(x) \gt 0\)
\(\Rightarrow 1-\log x \gt 0 \Rightarrow 1 \gt \log x \Rightarrow \mathrm{e} \gt x\)
\(\therefore \quad \mathrm{f}(x)\) is increasing in the interval \((0, \mathrm{e})\).
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