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If L.M.V.T. is applicable for the function $f(x)=x+\frac{1}{x}, x \in[1,3]$, then $c=$
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The correct answer is:
$\sqrt{3}$
(C)
Given $\mathrm{f}(\mathrm{x})=\mathrm{x}+\frac{1}{\mathrm{x}}$ and LMVT holds
$\begin{array}{l}
f^{\prime}(x)=1-\frac{1}{x^{2}} \Rightarrow f^{\prime}(c)=1-\frac{1}{c^{2}} \\
f(1)=1+1=2 \text { and } f(3)=3+\frac{1}{3}=\frac{10}{3}
\end{array}$
$\therefore \quad f^{\prime}(c)=1-\frac{1}{c^{2}}=\frac{\frac{10}{3}-2}{(3-1)} \Rightarrow 1-\frac{1}{c^{2}}=\frac{4}{3(2)}=\frac{2}{3}$
$\therefore \quad \frac{1}{c^{2}}=1-\frac{2}{3}=\frac{1}{3} \Rightarrow c^{2}=3 \Rightarrow c=\pm \sqrt{3}$
Given $\mathrm{f}(\mathrm{x})=\mathrm{x}+\frac{1}{\mathrm{x}}$ and LMVT holds
$\begin{array}{l}
f^{\prime}(x)=1-\frac{1}{x^{2}} \Rightarrow f^{\prime}(c)=1-\frac{1}{c^{2}} \\
f(1)=1+1=2 \text { and } f(3)=3+\frac{1}{3}=\frac{10}{3}
\end{array}$
$\therefore \quad f^{\prime}(c)=1-\frac{1}{c^{2}}=\frac{\frac{10}{3}-2}{(3-1)} \Rightarrow 1-\frac{1}{c^{2}}=\frac{4}{3(2)}=\frac{2}{3}$
$\therefore \quad \frac{1}{c^{2}}=1-\frac{2}{3}=\frac{1}{3} \Rightarrow c^{2}=3 \Rightarrow c=\pm \sqrt{3}$
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