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  • Need solution for RD Sharma maths Class 12 Chapter 15 Tangents and Normals Exercise Very short Answers Question 14 textbook solution.

Need solution for RD Sharma maths Class 12 Chapter 15 Tangents and Normals Exercise Very short Answers Question 14 textbook solution.

Answers (1)

Answer : Required angle =\frac{\pi }{2}

Hint :

  1. If  m_{1}=m_{2}, they are parallel.
  2. If  m_{1} m_{2}=-1, they are perpendicular to each other.

Given :

Given that the curve,

y=e^{-x} \text { and } y=e^{x}

We have to find the angle between the given curves at the point of intersection.

Solution :

Given,

y=e^{-x}                                                                                 ....(i)

y=e^{x}                                                                                    .....(ii)

From (i) and (ii), we get

\begin{aligned} & e^{x}=e^{-x} \\ \Rightarrow & x=0 \end{aligned}

Substituting the value of x in equation (ii), we get

    y=1

So, the point of intersection of the two curves is (0,1)

On differentiating (i) with respect to x, we get

\begin{aligned} &\frac{d y}{d x}=e^{x} \\ &m_{2}=\left(\frac{d y}{d x}\right)_{(0,1)}=1 \end{aligned}

\begin{aligned} & \therefore & & m_{1} m_{2} &=-1 \\ \text { Since } & & m_{1} m_{2} &=-1 \end{aligned},

Hence, they are perpendicular to each other.

Hence, the required angle =\frac{\pi }{2}

 

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