Let the functions f: {R}\to{R} and {g}: {R} \to{R} be defined by f(x)=e^{x-1}-e^{-|x-1|} and g(x)= \frac{1}{2} (e^{x-1}+e^{1-x} ) Then the area of the region in the first quadrant bounded by the curves y=f(x), y=g(x) and x=0 is [JEE Adv 2020 P1]

Correct Answer is : (2−√3)+12(e−e−1)

(2-\\sqrt{3})+ \frac{1}{2} ( {e}- {e}^{-1} )

(2+\\sqrt{3})+ \frac{1}{2} ( {e}- {e}^{-1} )

(2-\\sqrt{3})+ \frac{1}{2} ( {e}+ {e}^{-1} )

(2+\sqrt{3})+ \frac{1}{2} ( {e}+ {e}^{-1} )

Application of Integration

Section: Mathematics

Question : 10 of 28

Marks: +1, -0

Let the functions f:R→R and g:R→R be defined by
f(x)=ex−1−e−|x−1| and g(x)=

(ex−1+e1−x)
Then the area of the region in the first quadrant bounded by the curves y=f(x),y=g(x) and x=0 is

[JEE Adv 2020 P1]

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0,	x<1
ex−1−e−(x−1),	x≥1