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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise Multiple Choice Questions Question 6 Maths Textbook Solution.

Provide Solution For  R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise Multiple Choice Questions Question 6 Maths Textbook Solution.

Answers (1)

Answer:

                \left ( b \right )\left ( 1,0 \right )

Hint:

         Use differentiation

Given:

            y=x^{2}-3x+2

             y=x

Solution:

Differentiate both the side w.r.t x we get

\frac{dy}{dx}=1

Let \left ( x,y \right )be the required point

It is given that point lies on the curve

\begin{aligned} &y=x_{1}^{2}-3 x_{1}+2 \\ &y=x^{2}-3 x+2 \end{aligned}

Differentiating both the sides w.r.t x, we get

\frac{dy}{dx}=2x-3

Slope of the tangent=\left ( \frac{dy}{dx} \right )\left ( _{x_{1},y_{1}} \right )=2x_{1}-3

The tangent is perpendicular to the line

Slope of the tangent =-1/slope of the line=\frac{-1}{1}=-1

Now,

\begin{aligned} &2 x_{1}-3=-1 \\ &2 x_{1}=2 \\ &x_{1}=1 \\ &y_{1}=x_{1}^{2}-3 x_{1}+2 \\ &y_{1}=(1)^{2}-3(1)+2 \\ &y_{1}=1-3+2 \\ &y_{1}= \\ &\left(x_{1}, y_{1}\right)=(1,0) \end{aligned}

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