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  • Please Solve R.D.Sharma Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 3 Sub Question 2 Maths Textbook Solution.

Please Solve R.D.Sharma Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 3 Sub Question 2 Maths Textbook Solution.

Answers (1)

Answer: Equation of tangent 2 x-y+1=0

              Equation  of normal x+2 y-7=0

HINTS:

 Putting x=1 in the given equation and differentiate it.

GIVEN:

y=x^{4}-6 x^{3}+13 x^{2}-10 x+5 \text { at } x=1

Solution:

y=1-6+13-10+5=3

So,\left(x_{1}, y_{1}\right)=(1,3)

Now,

\begin{aligned} &y=x^{4}-6 x^{3}+13 x^{2}-10 x+5 \\ &\frac{d y}{d x}=4 x^{3}-18 x^{2}+26 x-10 \end{aligned}

Slope of tangent ,

\begin{aligned} m &=\left(\frac{d y}{d x}\right)_{(1,3)} \\ &=4-18+26-10 \\ &=2 \end{aligned}

Equation of tangent is,

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

Equation of Normal  is,

\begin{aligned} &y-y_{1}=-\frac{1}{m}\left(x-x_{1}\right) \\ &\Rightarrow y-3=-\frac{1}{2}(x-1) \\ &\Rightarrow 2 y-6=-x+1 \\ &\Rightarrow x+2 y-7=0 \end{aligned}

 

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