Let the functions $f: \mathbb{R} \rightarrow \mathbb{R}$ and $g: \mathbb{R} \rightarrow \mathbb{R}$ be defined by $$f(x)=e^{x-1}-e^{-|x-1|} \text{ and } g(x)=\frac{1}{2}\left(e^{x-1}+e^{1-x}\right).$$ Then the area of the region in the first quadrant bounded by the curves $y=f(x)$, $y=g(x)$ and $x=0$ is

(A) $(2-\sqrt{3})+\frac{1}{2}\left(e-e^{-1}\right)$

(B) $(2+\sqrt{3})+\frac{1}{2}\left(e-e^{-1}\right)$

(C) $(2-\sqrt{3})+\frac{1}{2}\left(e+e^{-1}\right)$

(D) $(2+\sqrt{3})+\frac{1}{2}\left(e+e^{-1}\right)$

JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · Area Under Curves

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

Answer: (A)

We have

f (x) = {0 e^{x - 1} - e^{1 - x} x \leq 1 x \geq 1

and

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

.

Solving

f (x) = g (x)

for

x \geq 1

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

gives

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

, so

e^{2 (x - 1)} = 3

, thus

x = 1 + ln 3

.

The bounded area is:

\int_{0}^{1} \frac{1}{2} (e^{x - 1} + e^{1 - x}) d x + \int_{1}^{1 + l n 3} [\frac{1}{2} (e^{x - 1} + e^{1 - x}) - (e^{x - 1} - e^{1 - x})] d x

= \frac{1}{2} [e^{x - 1} - e^{1 - x}]_{0}^{1} + [- \frac{1}{2} e^{x - 1} - \frac{3}{2} e^{1 - x}]_{1}^{1 + l n 3}

= \frac{1}{2} (e - \frac{1}{e}) + [(- \frac{3}{2} - \frac{3}{2}) + 2] = 2 - 3 + \frac{1}{2} (e - \frac{1}{e})

JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · Single correct option