# The tangent to the curve $y = e^{2x}$ at the point (0, 1) meets x-axis at: A. (0, 1) B. $\left ( -\frac{1}{2},0 \right )$ C. (2, 0) D. (0, 2)

Given the equation of the curve is

$y = e^{2x}$

Differentiating on both sides with respect to x, we get

$\frac{d(y)}{d x}=\frac{d\left(e^{2 x}\right)}{d x}$
Applying the exponential rule of differentiation, we get
$\\\Rightarrow \frac{d y}{d x}=e^{2 x} \frac{d(2 x)}{d x} \\ \Rightarrow \frac{d y}{d x}=2 e^{2 x} \ldots . .(i)$
As it is given the curve has tangent at (0,1), so the curve passes through the point (0,1), so above equation at (0,1), becomes
$\\\Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(0.1)}=2 \mathrm{e}^{2 \mathrm{x}} \\ \Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(0,1)}=2 \mathrm{e}^{2(0)} \\ \Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(0,1)}=2$

So, the slope of the tangent to the curve at point (0,1) is 2

Hence the equation of the tangent is given by

$\\y-1=2(x-0) \\\Rightarrow y-1=2x \\\Rightarrow y=2x+1$

It is given that the tangent to the curve $y=e\textsuperscript{2x}$  at the point (0,1) meet x-axis i.e., y=0

So the equation on tangent becomes,

$\\ { \Rightarrow y=2x+1}\\ { \Rightarrow 0=2x+1}\\ { \Rightarrow 2x=-1}\\$

$\Rightarrow \mathrm{x}=-\frac{1}{2}$
Hence, the required point is $\left(-\frac{1}{2}, 0\right)$
Therefore, the tangent to the curve $y=e^{2 x}$ at the point (0,1) meets x -axis at $\left(-\frac{1}{2}, 0\right)$
So, the correct option is option B.