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$f (x) = \int_{1}^{x} (t + \frac{1}{t}) d t$ and $g (x) = f^{'} (x)$ for $x \in [\frac{1}{2}, 3]$ If P is a point on the curve $y = g (x)$ such that the tangent to this curve at P is parallel to a chord joining the points $(\frac{1}{2}, g (\frac{1}{2}))$ and $(3, g (3))$ of the curve then the co ordinates of the point P are

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$(1, 2)$

$(\sqrt{\frac{3}{2}} \frac{5}{\sqrt{6}})$

can’t be find out

$(\frac{7}{4}, \frac{65}{28})$

answer is D.

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Detailed Solution

$f (x) = \int_{1}^{x} (t + \frac{1}{t}) d t \Rightarrow f^{1} (x) = x + \frac{1}{x}$

$g (x) = f^{'} (x) = x + \frac{1}{x}$

for $x \in [\frac{1}{2}, 3]$

$g (\frac{1}{2}) = 2 + \frac{1}{2} = \frac{5}{2}$

$g (3) = 3 + \frac{1}{3} = \frac{10}{3}$

Let $p \in (c, g (c)) c \in [\frac{1}{2}, 3]$

By LMVT $g^{'} (c) = \frac{g (3) - g (\frac{1}{2})}{3 - \frac{1}{2}}$

$1 - \frac{1}{c^{2}} = \frac{\frac{10}{3} - \frac{5}{2}}{3 - \frac{1}{2}} \Rightarrow c^{2} = \frac{3}{2} \Rightarrow c = \sqrt{\frac{3}{2}}$

$g (c) = \sqrt{\frac{3}{2}} + \frac{1}{\sqrt{\frac{3}{2}}} = \sqrt{\frac{3}{2}} + \sqrt{\frac{2}{3}} = \frac{5}{\sqrt{6}}$

$p \in (\sqrt{\frac{3}{2}}, \frac{5}{\sqrt{6}})$

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