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need solution for RD Sharma maths class 12 chapter Mean Value Theoram exercise 14.2 question 8

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Answer:  \left(\pm \sqrt{\frac{7}{3}}, \mp \frac{2}{3} \sqrt{\frac{7}{3}}\right)

Hint: You must know the slope of the tangent.

Given:y=x^{3}-3 x ;(1,-2),(2,2)


\begin{aligned} &y=x^{3}-3 x \\ &\frac{d y}{d x}=3 x^{2}-3 \end{aligned}

Slope of tangent =3 x^{2}-3

Slope of line joining (1, -2) and (2, 2) =\frac{2-(-2)}{2-1}

                                                      \begin{aligned} &=\frac{2+2}{1} \\ &=4 \end{aligned}

The tangent is parallel to the line: y=x^{3}-x

Slope is equal to 3 x^{2}-3=4

\begin{gathered} 3 x^{2}=7 \\ x^{2}=\frac{7}{3} \\ x=\pm \sqrt{\frac{7}{3}} \\ y=x^{3}-x \end{gathered}

    \begin{aligned} &=\left(\sqrt{\frac{7}{3}}\right)^{3}+3 \sqrt{\frac{7}{3}} \\ &=\sqrt{\frac{7}{3}}\left(\frac{7}{3}-3\right) \\ &=\sqrt{\frac{7}{3}}\left(\frac{7-9}{3}\right) \\ &=\frac{2}{3} \sqrt{\frac{7}{3}} \end{aligned}

    Points are \left(\pm \sqrt{\frac{7}{3}}, \mp \frac{2}{3} \sqrt{\frac{7}{3}}\right)




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