Question

# The coordinates of the point $$M(x, y)$$ on $$\displaystyle y= e^{-\left | x \right |}$$ so that the area formed by the coordinates axes and the tangent at $$M$$ is greatest, are

A
(e,1)
B
(1,e1)
C
(1,e1)
D
(0,1)

Solution

## The correct option is C $$\displaystyle \left (-1, e^{-1} \right )$$The curve of f(x) is symmetrical about y axis.Let the point of contact of the tangent be $$(a,e^{-a})$$Therefore$$\dfrac{dy}{dx}_{(a,e^{-a})}=-e^{-a}$$Hence the equation of the tangent will be $$y-e^{-a}=-e^{-a}(x-a)$$$$y-e^{-a}=-e^{-a}x+ae^{-a}$$$$y+e^{-a}x=e^{-a}(1+a)$$$$\dfrac{y}{e^{-a}(1+a)}+\dfrac{x}{1+a}=1$$Hence the area formed by the tangent and the coordinate axes will be$$=\dfrac{(x-intercept)(y-intercept)}{2}$$$$A=\dfrac{e^{-a}(1+a)^{2}}{2}$$Differentiating with respect to a, we get$$\dfrac{dA}{da}=\dfrac{1}{2}[2(1+a)e^{-a}-(1+a)^{2}e^{-a}]$$$$=\dfrac{(1+a)e^{-a}}{2}[2-(1+a)]$$$$=\dfrac{(1+a)e^{-a}}{2}(1-a)$$$$=0$$Hence$$a=\pm1$$Thus we get the points as $$(1,e^{-1})$$ and $$(-1,e^{-1})$$.Mathematics

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