JEE Main 2021 · 27 July · Morning Shift · Mathematics · Area Under Curves
B. 222
R={(x,y):max(0,logex)≤y≤2x,12≤x≤2}R = \left\{\left(x, y\right): max\left(0, \log_{e}x\right) \le y \le {2}^{x}, \dfrac{1}{2} \le x \le 2\right\}R={(x,y):max(0,logex)≤y≤2x,21≤x≤2}
So, the area of the region is
∫1222xdx−∫12ℓnxdx{∫}_{\dfrac{1}{2}}^{2}{2}^{x}dx - {∫}_{1}^{2}ℓnxdx∫2122xdx−∫12ℓnxdx
=[2xln2]1/22−[xlnx−x]12= {\left[\dfrac{{2}^{x}}{\ln 2}\right]}_{1/2}^{2} - {\left[xlnx - x\right]}_{1}^{2}=[ln22x]1/22−[xlnx−x]12
=(22)−21/2loge2−(2ln2−1)= \dfrac{\left({2}^{2}\right) - {2}^{1 / 2}}{\log_{e}2} - \left(2\ln 2 - 1\right)=loge2(22)−21/2−(2ln2−1)
=(22−2)loge2−2ln2+1= \dfrac{\left({2}^{2} - \sqrt{2}\right)}{\log_{e}2} - 2\ln 2 + 1=loge2(22−2)−2ln2+1
∴α=22−2,β=−2,γ=1∴ \alpha = {2}^{2} - \sqrt{2}, \beta = - 2, \gamma = 1∴α=22−2,β=−2,γ=1
So, (α+β−2γ)2{\left(\alpha + \beta - 2\gamma\right)}^{2}(α+β−2γ)2
=(22−2−2−2)2= {\left({2}^{2} - \sqrt{2} - 2 - 2\right)}^{2}=(22−2−2−2)2
=(−2)2=2= {\left(- \sqrt{2}\right)}^{2} = 2=(−2)2=2
