The tangent to the curve y=e2x at the point (0, l) meets x-axis at

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

A
(0, 2)
B
(2, 0)
C
(-1/2, 0)
D
none of these

Solution:

We have,

$y = e^{2 x} \Rightarrow \frac{d y}{d x} = 2 e^{2 x} \Rightarrow {(\frac{d y}{d x})}_{(0, 1)} = 2$

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)

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