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