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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise Multiple Choice Question Question 11 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise Multiple Choice Question Question 11 Maths Textbook Solution.

Answers (1)

Answer:

            \left ( b \right )x+y-1=x-y-2

Hint:

         Use slope of the tangent m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}

Given:

            y=x^{2}-3x+2

Solution:

y=x^{2}-3x+2

Let the tangent meet the x-axis at point \left ( x,o \right )

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

The tangent passes through point \left ( x,o \right )

\begin{aligned} &0=x^{2}-3 x+2 \\ &(x-2)(x-1)=0 \\ &\Rightarrow x=2, x=1 \end{aligned}

Case 1

When x=2

Slope of tangent \frac{dy}{dx}|\left ( _{2,0} \right )=2\left ( 2 \right )-3=4-3=1

\therefore \left ( x_{1},y_{1} \right )=\left ( 2,0 \right )

Equation of tangent

\begin{aligned} &y-y_{1}=m\left(x-x_{1}\right) \\ &y-0=1(x-2) \\ &x-y-2=0 \end{aligned}

Case 2

When x=1

Slope of tangent

\frac{dy}{dx}|\left ( 2,0 \right )=2-3=-1

\left ( x_{1}-y_{1} \right )=\left ( 1,0 \right )

Equation of tangent

\begin{aligned} &y-y_{1}=m\left(x-x_{1}\right) \\ &y-0=-1(x-1) \\ &\Rightarrow x+y-1=0 \end{aligned}

                

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