If the area of the bounded region R- x-y-max 0-log ex - y- 2x-12- x- 2 is- - log e2-1- log e2- - then the value of - - -2- 2 is equal to

A=∫ _1/2^2 2^xdx ∫ _1^2ln x dx ⇒ A= [ 2^xln 2 ]_1/2^2 [xln x x]_1^2 ⇒ A=4 √(2)ln 2 2ln 2+1 ⇒ A=(4 √(2))(ln 2)^ 1 2ln 2+1 Given: α (log _e2)^ 1+β (log _e2)+γ ∴α ...

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Standard XII

Mathematics

Definition of Relations

If the area o...

Question

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

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Solution

The correct option is A $2$

$A = 2 \int 1 / 2 2^{x} d x - 2 \int 1 ln x d x$
$\Rightarrow A = {[\frac{2^{x}}{ln 2}]}_{1 / 2}^{2} - [x ln x - x]_{1}^{2}$
$\Rightarrow A = \frac{4 - \sqrt{2}}{ln 2} - 2 ln 2 + 1$
$\Rightarrow A = (4 - \sqrt{2}) (ln 2)^{- 1} - 2 ln 2 + 1$
Given: $α ({log}_{e} 2)^{- 1} + β ({log}_{e} 2) + γ$
$∴ α = 4 - \sqrt{2}, β = - 2$ and $γ = 1$
Now,
$(α + β - 2 γ)^{2} = ((4 - \sqrt{2}) + (- 2) - 2 (1))^{2}$
$∴ (α + β - 2 γ)^{2} = 2$

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If the area of the bounded region R={(x,y):max{0,logex}≤y≤2x,12≤x≤2} is, α(loge2)−1+β(loge2)+γ, then the value of (α+β−2γ)2 is equal to

The correct option is A 2 A=2∫1/22xdx−2∫1lnx dx ⇒A=[2xln2]21/2−[xlnx−x]21 ⇒A=4−√2ln2−2ln2+1 ⇒A=(4−√2)(ln2)−1−2ln2+1 Given: α(loge2)−1+β(loge2)+γ ∴α=4−√2,β=−2 and γ=1 Now, (α+β−2γ)2=((4−√2)+(−2)−2(1))2 ∴(α+β−2γ)2=2

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