Question

The tangent drawn at the point $\left(0,1\right)$ on the curve $y={\mathrm{e}}^{2x}$ meets $x$-axis at the point

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

Answer 1

The correct option is B

$\left(-\frac{1}{2},0\right)$

Explanation of the correct option:

Step 1: Determine the slope of the curve

The slope of a curve at a point is obtained by differentiating the curve and substituting the value of the $x$-coordinate of the point into the derivative.

Given curve is, $y={\mathrm{e}}^{2\mathrm{x}}$.

Differentiating,

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{\mathrm{d}}{dx}{\mathrm{e}}^{2\mathrm{x}}\\ & \Rightarrow & \frac{d\mathrm{y}}{d\mathrm{x}}=2{\mathrm{e}}^{2\mathrm{x}}\end{array}$

Thus, the slope of the curve at the given point $\left(0,1\right)$ is,

${\left(\frac{\mathrm{dy}}{\mathrm{d}x}\right)}_{\left(0,1\right)}=2{\mathrm{e}}^{2\times 0}=2$

Step 2: Determine the equation of the tangent

The equation of the tangent can be determined by the point-slope formula which is given as,

$\left(y-y_1\right)=m\left(x-x_1\right)$

where $\left(x_1,y_1\right)$ are the points the tangent passes through and $m$ is the slope of the tangent.

Here, $m=2$ as determined in the previous step.

Thus, the equation of the tangent is,

$\begin{array}{cc}& \left(y-1\right)=2\left(\mathrm{x}-0\right)\\ \Rightarrow & y=2\mathrm{x}+1\end{array}$

Step 3: Find the $x$-intercept of the tangent

The $x$-intercept of a line can be found by substituting $y=0$ in the equation and solving for $x$.

Thus, the $x$-intercept of the tangent is,

$\begin{array}{cc}& 0=2x+1\\ \Rightarrow & x=-\frac{1}{2}\end{array}$

Therefore, the point at which the tangent meets the $x$-axis is $\left(-\frac{1}{2},0\right)$

Hence, option B is correct.