Question

At what point p(x,y)of the curve $y=e^{-\left|x\right|}$ should a tangent be drawn so that area of the triangle bounded by the tangent and the co-ordinate axes be greatest ?

• A. $(\pm e,1)$
• B. $(\pm 1,\frac{1}{e})$
• C. $(1,\pm \frac{1}{e})$
• D. $(\pm 1,e)$

Answer 1

The correct option is B $(\pm 1,\frac{1}{e})$

The equation of the curve does not change if $x$ be + ive or-ive.
Hence we consider only +ive values of $x$ i.e $x>0,$
$\therefore \left|x\right|=x\therefore y=-x$
Consider any point (x,y) tangent at which is $Y-y=-e^{-x}(X-x)$
It meets the axis $Y=0at(x=y{e}^x,0)$
It meets the axis $X=0at(0,y+\frac{x}{e^x})$
Area of the triangle $=\frac{1}{2}AB$
$\Delta =\frac{1}{2}(x=y{e}^x)(y+\frac{x}{e^x})$
$=\frac{1}{2}(xy+xy+y^2{e}^x+\frac{x^2}{e^x})$ Substitute $y=\frac{1}{e^x}$
$\Delta =\frac{1}{2}[\frac{2x}{e^x}+\frac{e^x}{e^{2x}}+\frac{x^2}{e^x}]=\frac{1}{2}{e}^{-x}{(x+1)}^2$
For max. value of $\Delta$
we have$\frac{d\Delta }{dx}=0\therefore \frac{1}{2}{e}^{-x[-{(x+1)}^2+2(x+1)]}=0$
or $\frac{1}{2}(x+1)(1-x)e^{-x}=\frac{1}{2}{e}^{-x}(1-x^2)$
$\therefore \frac{d\Delta }{dx}=0\Rightarrow x=1-1.$
We shall consider only x=1as x is +ive.
$\frac{d^2\Delta }{d{x}^2}=\frac{1}{2}{e}^x[-1(1-x^2)-2x]$
$=-\frac{x}{e^x}=-\frac{1}{e}=-ive$ at $x=1$
Hence $\Delta$ is max.
When x=1 $\therefore y=e^{-1}=\frac{1}{e}.$
$\therefore$
Required point is $(1,\frac{1}{e})$. we have already stated that if $x$ be changed to $-x$ the equation of the curve does not change.
Hence the required points are $(\pm 1,\frac{1}{e})$