Question

JEE ADVANCED MATHS QUESTION SEPTEMBER 2020

Answer 1

Let the functions f : R → R and g : R → 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 by the curves y = f(x) , y = g(x) and x = 0

solution : f is modulus function so let's break it into simpler form.

for x ≤ 1

f(x) = e^(x - 1) - e^(x -1) = 0

for x > 1

f(x) = e^(x - 1) - e^-(x -1) = e^(x - 1) - e^(1 - x)

and g(x) = 1/2 (e^(x - 1) + e^(1 - x))

solving f(x) and g(x) ,

1/2 [e^(x - 1) + e^(1 - x) ] = e^(x - 1) - e^(1 - x)

⇒3e^(1 - x) = e^(x - 1)

⇒3 = e^{2(x - 1)}

⇒ln3 = 2(x - 1)

⇒x = 1 + ln√3

now see the graph of enclosed area,

so bounded area = $+\int\limits^1_0{(g(x)-f(x))}\,dx+\int\limits^{1+ln\sqrt{3}}_1{(g(x)-f(x))}\,dx$

= $\int\limits^1_0{\frac{1}{2}(e^{x-1}+e^{1-x})-0}\,dx+\int\limits^{1+ln\sqrt{3}}_1{\frac{1}{2}(e^{x-1}+e^{1-x})-(e^{x-1}-e^{1-x})}\,dx$

= $\frac{1}{2}\left[e^{x-1}-e^{1-x}\right]^1_0+\left[-\frac{1}{2}e^{x-1}-\frac{3}{2}e^{1-x}\right]^{1+ln\sqrt{3}}_1$

= $\frac{1}{2}\left[e-\frac{1}{e}\right]+\left[\left(-\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}\right)+2\right]$

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

Therefore the area bounded by the curves y = f(x), y = g(x) and x = 0 is option (A) (2 - √3) + 1/2 (e - e¯¹)