Question

The co−ordinates of the point $M(x,y)$ of $y=e^{-\left|x\right|}$ so that the area formed by the co−ordinate axes and the tangent at $M$ is the greatest, are:

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

Answer 1

The correct option is B

$\left(1,\frac{1}{e}\right)\text{or}\left(-1,\frac{1}{e}\right)$

Explanation for the correct option:

Step 1: Finding the equation of tangent to the curve $f\left(x\right)=y=e^{-\left|x\right|}$

The given curve,

$f\left(x\right)=y=e^{-\left|x\right|}$

We know that the graph $e^{-\left|x\right|}$ is symmetrical about $y$ axis.

Considering the point of contact of the tangent be $(\mathrm{a},{\mathrm{e}}^{-\mathrm{a}})$

Therefore,

${\frac{dy}{dx}}_{\left(a,e^{-a}\right)}=-e^{-a}$

Hence the equation of the tangent will be

$$\begin{gathered} \Rightarrow y-e^{-a}=-e^{-a}(x-a) \\ \Rightarrow y-e^{-a}=-e^{-a}x+a{e}^{-a} \\ \Rightarrow y+e^{-a}x=e^{-a}(1+a) \\ \Rightarrow \frac{y}{e^{-a}(1+a)}+\frac{x}{1+a}=1 \end{gathered}$$

Step 2: Finding the area formed by the tangent and the coordinate axes

Therefore,$x-\text{intercept}=1+a\&y-\text{intercept}=e^{-a}(1+a)$

Then the area formed by the tangent and the coordinate axes will be

$$\begin{gathered} \Rightarrow \frac{(x-\text{intercept)}(y-\text{intercept})}{2} \\ \Rightarrow A=\frac{e^{-a}{(1+a)}^2}{2} \end{gathered}$$

Step 3: Finding critical points using derivative of the area

Differentiating it with respect to $a$, we get

$$\begin{gathered} \frac{dA}{da}=\frac{1}{2}\left[2(1+a)e^{-a}-{(1+a)}^2{e}^{-a}\right] \\ =\frac{(1+a)e^{-a}}{2}[2-(1+a\left)\right] \\ =\frac{(1+a)e^{-a}}{2}(1-a) \\ =0ifa=\pm 1 \end{gathered}$$

Thus we get the points as $\left(1,\frac{1}{e}\right)or\left(-1,\frac{1}{e}\right)$

Hence, option (B) is the correct answer.