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. $e(x+y)=1$
• B. $y=e^x+1$
• C. $x+y=e$
• D. None of these

Answer 1

The correct option is D

None of these

Explanation for the correct answer:

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

Given, equation of curve $y=e^{-\left|x\right|}...\left(\mathrm{i}\right)$

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

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

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

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

Hence, the correct answer is option (D).