Explanation

(b)
Given curve is $y = e^{2x}$
On differentiating w.r.t. $x$, we get
$\frac{dy}{dx} = 2e^{2x}\ \ \Rightarrow \ \ \left( \frac{dy}{dx} \right)_{(0,\ 1)} = 2e^{0} = 2$
Equation of tangent at (0, 1) with slope 2 is
$y - 1 = 2(x - 0)\ \Rightarrow \ \ \ y = 2x + 1$
This tangent meets $x$-axis
$\therefore\ \ y = 0$
$\Rightarrow \ \ 0 = 2x + 1\ \ \Rightarrow \ \ x = - \frac{1}{2}$
$\therefore$ Coordinates of the point on $x$-axis is $\left( - \frac{1}{2},\ 0 \right)$