QJEEAdv20Ma7c50: JEE Advanced 2020 Mathematics - Class 12 JEEAdv20Ma7c50 Solution: Let the functions f : R R and g : R R be defined by f(x) = e x…

Question

Verified Solution • 2025–26 Session moderate

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) =

{1 \over 2}

(e^{x $-$ 1} + e^{1 $-$ x}).

The the area of the region in the first quadrant bounded by the curves y = f(x), y = g(x) and x = 0 is

Accepted Answer

The given functions f : R

\to

R and g : R

\to

R be defined by

f(x) = {e^{x - 1}} - {e^{ - |x - 1|}}

= | \begin{matrix} e^{x - 1} - e^{1 - x}, x \geq 1 \\ 0, x < 1 \end{matrix}

and

g(x) = {1 \over 2}({e^{x - 1}} + {e^{1 - x}})

For point of intersection of curves f(x) and g(x) put f(x) = g(x)

for

x\, \ge \,1,\,{e^{x - 1}} - {e^{1 - x}} = {1 \over 2}({e^{x - 1}} + {e^{1 - x}})

\Rightarrow {e^{x - 1}} = 3{e^{1 - x}} \Rightarrow {e^{2x}} = 3{e^2}

\Rightarrow x = {1 \over 2}\log _e^3 + 1

So, required area is

\int_0^{1/2\log _e^3 + 1} {(g(x) - f(x))dx}

= \int_0^{1/2\log _e^3 + 1} {g(x)dx - \int_1^{1/2\log _e^3 + 1} {f(x)dx} }

= {1 \over 2}\int_0^{1/2\log _e^3 + 1} {({e^{x - 1}} + {e^{1 - x}})dx - } \int_1^{1/2\log _e^3 + 1} {({e^{x - 1}} - {e^{1 - x}})dx}

= {1 \over 2}[{e^{x - 1}} - {e^{1 - x}}]_0^{{1 \over 2}\log _e^3 + 1} - [{e^{x - 1}} + {e^{1 - x}}]_1^{{1 \over 2}\log _e^3 + 1}

= {1 \over 2}\left[ {\sqrt 3 - {1 \over {\sqrt 3 }} - {e^{ - 1}} + e} \right] - \left[ {\sqrt 3 + {1 \over {\sqrt 3 }} - 1 - 1} \right]

= {1 \over {\sqrt 3 }} + {1 \over 2}(e - {e^{ - 1}}) - {4 \over {\sqrt 3 }} + 2

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

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