Question

Consider the curve $y=e^{2x}$.Where does the tangent to the curve at (0, 1) meet the x-axis ?

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

Answer 1

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

Slope of tangent at any point to the curve is given by ${\left|\frac{df\left(x\right)}{dx}\right|}_{(x_0,y_0)}$

Here $f\left(x\right)=y=e^{2x}$

$\Rightarrow \frac{dy}{dx}=2e^{2x}$

At point $(0,1)$

$\frac{dy}{dx}=2e^{2\left(0\right)}$

$\frac{dy}{dx}=2$

Equation of tangent is at point $(x_0,y_0)$

$y-y_0=m(x-x_0)$

So, tangent at (0,1) is

$y-1=2(x-0)$

$\therefore y=2x+1$

When this tangent meets x-axis, $y$ co-ordinate becomes zero

Putting $y$ in above equation as zero, we get

$0=2x+1$

$\Rightarrow x=\frac{-1}{2}$

Thus, the tangent to the curve meets at $(-\frac{1}{2},0)$