Consider f(x) = intlimits_0^x {left( {t + fra | vedclass

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(D) Given $f(x) = \int\limits_0^x {\left( {t + \frac{1}{t}} \right)\,dt}$.By the Fundamental Theorem of Calculus,$f'(x) = x + \frac{1}{x}$.Thus,$g(x)

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

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(D) Given $f(x) = \int\limits_0^x {\left( {t + \frac{1}{t}} \right)\,dt}$.By the Fundamental Theorem of Calculus,$f'(x) = x + \frac{1}{x}$.Thus,$g(x) = x + \frac{1}{x}$ for $x \in \left[ {\frac{1}{2}, 3} \right]$.We have $g\left( {\frac{1}{2}} \right) = \frac{1}{2} + 2 = \frac{5}{2}$ and $g(3) = 3 + \frac{1}{3} = \frac{10}{3}$.Let $P = (c, g(c))$ be a point on the curve where the tangent is parallel to the chord joining $\left( {\frac{1}{2}, \frac{5}{2}} \right)$ and $\left( {3, \frac{10}{3}} \right)$.By the Mean Value Theorem $(LMVT)$,there exists $c \in \left( {\frac{1}{2}, 3} \right)$ such that $g'(c) = \frac{g(3) - g(1/2)}{3 - 1/2}$.Since $g'(x) = 1 - \frac{1}{x^2}$,we have $1 - \frac{1}{c^2} = \frac{10/3 - 5/2}{3 - 1/2} = \frac{(20-15)/6}{5/2} = \frac{5/6}{5/2} = \frac{1}{3}$.Therefore,$1 - \frac{1}{c^2} = \frac{1}{3} \implies \frac{1}{c^2} = \frac{2}{3} \implies c^2 = \frac{3}{2} \implies c = \sqrt{\frac{3}{2}}$.The $y$-coordinate is $g(c) = \sqrt{\frac{3}{2}} + \frac{1}{\sqrt{3/2}} = \sqrt{\frac{3}{2}} + \sqrt{\frac{2}{3}} = \frac{3 + 2}{\sqrt{6}} = \frac{5}{\sqrt{6}}$.Thus,the point $P$ is $\left( {\sqrt {\frac{3}{2}}, \frac{5}{\sqrt 6 }} \right)$.

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