MHT CET · Maths · Application of Derivatives
The value of C for which Mean value Theorem holds for the function \(\mathrm{f}(x)=\log _{\mathrm{e}} x\) on the interval \([1,3]\) is
- A \(\log _3 \mathrm{e}\)
- B \(\log _{\mathrm{e}} 3\)
- C \(\frac{1}{2} \log _{\mathrm{e}} 3\)
- D \(2 \log _3 \mathrm{e}\)
Answer & Solution
Correct Answer
(D) \(2 \log _3 \mathrm{e}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
\mathrm{f}(x) & =\log _{\mathrm{e}} x \\
\mathrm{f}(1) & =\log _{\mathrm{e}} 1=0 \\
\mathrm{f}(3) & =\log _{\mathrm{e}} 3 \text { and } \mathrm{f}^{\prime}(x)=\frac{1}{x}
\end{aligned}\)
By Lagrange's mean value theorem,
\(\begin{aligned}
& f^{\prime}(c)=\frac{f(3)-f(1)}{3-1} \\
& \Rightarrow \frac{1}{c}=\frac{\log _e 3-0}{2} \Rightarrow c=\frac{2}{\log _e 3} \Rightarrow \mathrm{c}=2 \log _3 e
\end{aligned}\)
\mathrm{f}(x) & =\log _{\mathrm{e}} x \\
\mathrm{f}(1) & =\log _{\mathrm{e}} 1=0 \\
\mathrm{f}(3) & =\log _{\mathrm{e}} 3 \text { and } \mathrm{f}^{\prime}(x)=\frac{1}{x}
\end{aligned}\)
By Lagrange's mean value theorem,
\(\begin{aligned}
& f^{\prime}(c)=\frac{f(3)-f(1)}{3-1} \\
& \Rightarrow \frac{1}{c}=\frac{\log _e 3-0}{2} \Rightarrow c=\frac{2}{\log _e 3} \Rightarrow \mathrm{c}=2 \log _3 e
\end{aligned}\)
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