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explain solution RD Sharma class 12 chapter Mean Value Theoram exercise 14.2 question 9 maths

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Answer:  \left(\pm \sqrt{\frac{13}{3}} \cdot\left(\frac{13}{3}\right)^{\frac{1}{2}}+1\right)

Hint: You must know the slope of the tangent.

Given:y=x^{3}+1 ;(1,2),(3,28)


\begin{aligned} &y=x^{3}+1 \\ &\frac{d y}{d x}=3 x^{2} \end{aligned}

Slope of tangent = 3x2

Slope of line joining (1, 2) and (3, 28) =\frac{28-2}{3-1}

                                                                    \begin{aligned} &=\frac{26}{2} \\ &=13 \end{aligned}

The tangent is parallel to the line:

Slope is equal to 3x^{2} = 13

\begin{aligned} &x^{2}=\frac{13}{3} \\ &x=\pm \sqrt{\frac{13}{3}} \end{aligned}

\Rightarrow \quad y=x^{3}+1

             \begin{aligned} &=\left(\sqrt{\frac{13}{3}}\right)^{3}+1 \\ &=\left(\frac{13}{3}\right)^{\frac{3}{2}}+1 \end{aligned}

Points are \left(\pm \sqrt{\frac{13}{3}} ,\left (\frac{13}{3}\right)^{\frac{1}{2}}+1\right)

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