Question

The equation of the tangent to the curve $y=e^{-\left|x\right|}$ at the point where the curve cuts the line $x=1$ is :

• A. $x+y=e$
• B. $y+ex=1$
• C. $x+y=\frac{1}{e}$
• D. $x+ey=2$

Answer 1

The correct option is D

$x+ey=2$

Explanation for the correct answer:

Step 1: Find the first order derivative of the equation of the given curve

Given, equation of curve $y=e^{-\left|x\right|}$

Around $x=1$, $\left|x\right|=x$

So, $y=e^{-x}...\left(\mathrm{i}\right)$

On differentiating eq. $\left(\mathrm{i}\right)$ w.r.t. $x$, we get

$\frac{dy}{dx}=-e^{-x}$

Step 2:Find the slope of the tangent at the required point

By substituting $x=1$ in above equation we get,

$y=\frac{1}{e}$

The first order derivative at a given point gives the slope of the tangent at that point

$\therefore$Slope of tangent at $\left(1,\frac{1}{e}\right)$ is ${\left(\frac{dy}{dx}\right)}_{\left(1,\frac{1}{e}\right)}=-e^{-1}$

$=-\frac{1}{e}$

Step 3: Find the equation of the tangent using slope-point form

Therefore, equation of tangent to the curve passing through $\left(1,\frac{1}{e}\right)$ is

$$\begin{gathered} y-\frac{1}{e}=-\frac{1}{e}\left(x-1\right) \\ \Rightarrow y-\frac{1}{e}=-\frac{x}{e}+\frac{1}{e} \\ \Rightarrow y+\frac{x}{e}-\frac{1}{e}-\frac{1}{e}=0 \\ \Rightarrow \frac{ey+x-2}{e}=0 \\ \Rightarrow ey+x-2=0 \\ \Rightarrow x+ey=2 \end{gathered}$$

Hence, the correct answer is option (D).