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  • Consider <img alt="f(x)=\int_1^x\left(t+\frac{1}{t}\right) d t" src="https://entrancecorner.oncodecogs.com/gif.latex?f%28x%29%3D%5Cint_1%5Ex%5Cleft%28t&plus;%5Cfrac%7B1%7D%7Bt%7D%5Cright%29%20d

Consider f(x)=\int_1^x\left(t+\frac{1}{t}\right) d tand g(x)=f^{\prime}(x) for x \in\left[\frac{1}{2}, 3\right].
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 \left(\frac{1}{2}, g\left(\frac{1}{2}\right)\right) and (3, g(3)) of the curve, then the coordinates of the point P 

Option: 1

 can't be found out


Option: 2

\left(\frac{7}{4}, \frac{65}{28}\right)


Option: 3

 (1,2)


Option: 4

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


Answers (1)

best_answer

f(x)=\int_1^x\left(t+\frac{1}{t}\right) d t \Rightarrow f^{\prime}(x)=x+\frac{1}{x} \\



  \begin{aligned} \begin{array}{ccc} \therefore & g(x)=x+\frac{1}{x} \text { for } x \in\left[\frac{1}{2}, 3\right] \\ & g\left(\frac{1}{2}\right)=2+\frac{1}{2}=\frac{5}{2}, g(3)=3+\frac{1}{3}=\frac{10}{3}\\ Let &P \equiv(c, g(c)), c \in\left[\frac{1}{2}, 3\right] \end{array} \end{aligned}
\begin{aligned} \begin{array}{ccc} \therefore & g(x)=x+\frac{1}{x} \text { for } x \in\left[\frac{1}{2}, 3\right] \\ & g\left(\frac{1}{2}\right)=2+\frac{1}{2}=\frac{5}{2}, g(3)=3+\frac{1}{3}=\frac{10}{3}\\ Let &P \equiv(c, g(c)), c \in\left[\frac{1}{2}, 3\right] \\ \text{By LMVT,} & g^{\prime}(c)=\frac{g(3)-g\left(\frac{1}{2}\right)}{3-1 / 2} \\ \therefore & 1-\frac{1}{c^2}=\frac{\frac{10}{3}-\frac{5}{2}}{3-\frac{1}{2}} \end{array} \end{aligned}

Posted by

Suraj Bhandari

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