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Please Solve RD Sharma Class 12 Chapter15 Tangents and Normals Exercise Fill in the blanks Question 9 Maths Textbook Solution.

Answers (1)


Point  =\left(\frac{-1}{2}, 0\right)


  1. Using equation of tangent is \left(y-y_{1}\right)=\frac{d y}{d x}\left(x-x_{1}\right)
  2. Tangent line meets the  x-axis. i.e. y= 0


Given equation of curve,

                y=e^{2 x}

To find:

We have to find the point where curve at  \left ( 0,1 \right ) cut x-axis.


The given equation of curve,

                y=e^{2 x}                                                                                                                                                … (i)

Differentiating equation (i), we get

\begin{aligned} &\Rightarrow \quad \frac{d y}{d x}=2 \cdot e^{2 x} \\\\ &\Rightarrow \quad\left(\frac{d y}{d x}\right)_{(0.1)}=2 \cdot e^{2(0)}=2 \end{aligned}

Slope of tangent to the curve =2

So equation of tangent is


                \Rightarrow \quad y=2 x+1

Above tangent line cuts the x-  axis, where  y= 0

\therefore \quad x=\frac{-1}{2}

Hence the required point is  \left ( \frac{-1}{2} ,0\right )

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