The tangent to the curve $y=e^{2 x}$ at the point (0, 1) meets x-axis at

(0, 2) (2, 0) (-1/2, 0) none of these

(-1/2, 0)

We have, $y=e^{2 x} \Rightarrow \frac{d y}{d x}=2 e^{2 x} \Rightarrow\left(\frac{d y}{d x}\right)_{(0,1)}=2$ The equation of the tangent at (0, 1) is $y-1=2(x-0) \Rightarrow 2 x-y+1=0$ This meets x-axis at $(-1 / 2,0)$