Question

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

• A. $(2,0)$
• B. $(0,1)$
• C. $(-\frac{1}{2},0)$
• D. $(0,2)$

Answer 1

The correct option is C $(-\frac{1}{2},0)$

The given equation of curve is :
$y=e^{2x}$

Differentiating the equation w.r.t. $x$,
$\Rightarrow \frac{dy}{dx}=2e^{2x}$

$\Rightarrow {\left(\frac{dy}{dx}\right)}_{(0,1)}=2\cdot e^0=2$
$\therefore$ Slope of the tangent to curve $=2$

Equation of tangent at point $(0,1)$ is
$y-1=2(x-0)$

$[\because \text{Eq. of tangent}:y-y_1=m(x-x_1)]$

$\Rightarrow y=2x+1$

For getting point of intersection with $x$ -axis, substituting $y=0$ in above equation, we get
$x=-\frac{1}{2},$

So, the required point on $x$−axis is $(-\frac{1}{2},0)$

Hence, option (b) is correct.