Question

The coordinates of the point P on the graph of the function $y=e^{-}\left|x\right|,$ where area of triangle made by tangent and the coordinate axis has the greatest area, is

• A. $(1,\frac{1}{e})$
• B. $(-1,\frac{1}{e})$
• C. $(e,e^{-e})$
• D. none

Answer 1

Answer: (A), (B)

The correct options are
A $(1,\frac{1}{e})$
B $(-1,\frac{1}{e})$

The curve of f(x) is symmetrical about y axis.
Let the point of contact of the tangent be $(a,e^{-a})$
Therefore
${\frac{dy}{dx}}_{(a,e^{-a})}=-e^{-a}$
Hence the equation of the tangent will be
$y-e^{-a}=-e^{-a}(x-a)$
$y-e^{-a}=-e^{-a}x+a{e}^{-a}$
$y+e^{-a}x=e^{-a}(1+a)$
$\frac{y}{e^{-a}(1+a)}+\frac{x}{1+a}=1$
Hence the area formed by the tangent and the coordinate axes will be
$=\frac{(x-intercept)(y-intercept)}{2}$
$A=\frac{e^{-a}(1+a{)}^2}{2}$
Differentiating with respect to a, we get
$\frac{dA}{da}=\frac{1}{2}\left[2\right(1+a)e^{-a}-(1+a{)}^2{e}^{-a}]$
$=\frac{(1+a)e^{-a}}{2}[2-(1+a\left)\right]$
$=\frac{(1+a)e^{-a}}{2}(1-a)$
$=0$
Hence
$a=\pm 1$
Thus we get the points as
$(1,e^{-1})$ and $(-1,e^{-1})$.