JEE MAIN - Mathematics (2021 - 27th July Morning Shift - No. 5)
If the area of the bounded region
$$R = \left\{ {(x,y):\max \{ 0,{{\log }_e}x\} \le y \le {2^x},{1 \over 2} \le x \le 2} \right\}$$ is ,
$$\alpha {({\log _e}2)^{ - 1}} + \beta ({\log _e}2) + \gamma $$, then the value of $${(\alpha + \beta - 2\lambda )^2}$$ is equal to :
$$R = \left\{ {(x,y):\max \{ 0,{{\log }_e}x\} \le y \le {2^x},{1 \over 2} \le x \le 2} \right\}$$ is ,
$$\alpha {({\log _e}2)^{ - 1}} + \beta ({\log _e}2) + \gamma $$, then the value of $${(\alpha + \beta - 2\lambda )^2}$$ is equal to :
8
2
4
1
Explanation
$$R = \left\{ {(x,y):\max \{ 0,{{\log }_e}x\} \le y \le {2^x},{1 \over 2} \le x \le 2} \right\}$$
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$$\int\limits_{{1 \over 2}}^2 {{2^x}dx} - \int\limits_1^2 {\ln xdx} $$
$$ \Rightarrow \left[ {{{{2^x}} \over {\ln 2}}} \right]_{1/2}^2 - [x\ln x - x]_1^2$$
$$ \Rightarrow {{({2^2}) - {2^{1/2}}} \over {{{\log }_e}2}} - (2\ln 2 - 1)$$
$$ \Rightarrow {{\left( {{2^2} - \sqrt 2 } \right)} \over {{{\log }_e}2}} - 2\ln 2 + 1$$
$$\therefore$$ $$\alpha = {2^2} - \sqrt 2 $$, $$\beta = - 2$$, $$\gamma = 1$$
$$ \therefore {(\alpha + \beta + 2\gamma )^2}$$
$$ = {({2^2} - \sqrt 2 - 2 - 2)^2}$$
$$ = {(\sqrt 2 )^2} = 2$$
$$\int\limits_{{1 \over 2}}^2 {{2^x}dx} - \int\limits_1^2 {\ln xdx} $$
$$ \Rightarrow \left[ {{{{2^x}} \over {\ln 2}}} \right]_{1/2}^2 - [x\ln x - x]_1^2$$
$$ \Rightarrow {{({2^2}) - {2^{1/2}}} \over {{{\log }_e}2}} - (2\ln 2 - 1)$$
$$ \Rightarrow {{\left( {{2^2} - \sqrt 2 } \right)} \over {{{\log }_e}2}} - 2\ln 2 + 1$$
$$\therefore$$ $$\alpha = {2^2} - \sqrt 2 $$, $$\beta = - 2$$, $$\gamma = 1$$
$$ \therefore {(\alpha + \beta + 2\gamma )^2}$$
$$ = {({2^2} - \sqrt 2 - 2 - 2)^2}$$
$$ = {(\sqrt 2 )^2} = 2$$
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