If the area of the bounded region $R=\left\{(x, y): \max \left\{0, \log _{e} x\right\} \leq y \leq 2^{x}, \frac{1}{2} \leq x \leq 2\right\}$ is $\alpha\left(\log _{e} 2\right)^{-1}+\beta\left(\log _{e} 2\right)+\gamma$, then the value of $(\alpha+\beta-2 \gamma)^{2}$ is equal to :

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Area Under Curves

If the area of the bounded region

R = {(x, y) : max {0, lo g_{e} x} \leq y \leq 2^{x}, \frac{1}{2} \leq x \leq 2}

α (lo g_{e} 2)^{- 1} + β (lo g_{e} 2) + γ

, then the value of

(α + β - 2 γ)^{2}

is equal to :

Answer: (C)

[\frac{1}{2}, 1)

the lower boundary

max {0, lo g_{e} x}

0

, and on

[1, 2]

it is

lo g_{e} x

; the upper boundary is

y = 2^{x}

throughout. So

Area = \int_{1/2}^{2} 2^{x} d x - \int_{1}^{2} lo g_{e} x d x

\int_{1/2}^{2} 2^{x} d x = [\frac{2 ^{x}}{lo g _{e} 2}]_{1/2}^{2} = \frac{4 - 2}{lo g _{e} 2}, \int_{1}^{2} lo g_{e} x d x = [x lo g_{e} x - x]_{1}^{2} = 2 lo g_{e} 2 - 1

Area = (4 - 2) (lo g_{e} 2)^{- 1} - 2 lo g_{e} 2 + 1

Comparing coefficients,

α = 4 - 2

β = - 2

γ = 1

, and

(α + β - 2 γ)^{2} = (4 - 2 - 2 - 2)^{2} = 2

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Single correct option