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provide solution for RD Sharma maths class 12 chapter Mean Value Theoram exercise 14.2 question 7

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Answer:  \left(\frac{7}{2}, \frac{1}{4}\right)

Hint: You must know the value of tangent parabola.

Given: y=(x-3)^{2} parallel to chord: (3,0)(4,1)

Solution:

y=(x-3)^{2}

Since tangent is parallel to chord joining (3, 0) and (4, 1), so, we get:

2(x-3)=\frac{1-0}{4-3}

\begin{aligned} &2 x-6=\frac{1}{1} \\ &2 x=7 \\ &x=\frac{7}{2} \end{aligned}

\begin{aligned} \text {When} \; \; \; \; \; &x=\frac{7}{2}, \text { then } &y=\left(\frac{7}{2}-3\right)^{2} \end{aligned}

\begin{aligned} &=\left(\frac{7-6}{2}\right)^{2} \\ &=\left(\frac{1}{2}\right)^{2} \\ &=\frac{1}{4} \end{aligned}

Hence, the point is \left(\frac{7}{2}, \frac{1}{4}\right)

 

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