JEE Advance - Mathematics (2015 - Paper 2 Offline - No. 7)
If $$\alpha = \int\limits_0^1 {\left( {{e^{9x + 3{{\tan }^{ - 1}}x}}} \right)\left( {{{12 + 9{x^2}} \over {1 + {x^2}}}} \right)} dx$$ where $${\tan ^{ - 1}}x$$ takes only principal values, then the value of $$\left( {{{\log }_e}\left| {1 + \alpha } \right| - {{3\pi } \over 4}} \right)$$ is
Answer
9
Explanation
$$\alpha = \int\limits_0^1 {{e^{(9x + 3{{\tan }^{ - 1}}x)}}\left( {{{12 + 9{x^2}} \over {1 + {x^2}}}} \right)dx} $$
Set $$9x + 3{\tan ^{ - 1}}x = t$$
so that $${{dt} \over {dx}} = 9 + {3 \over {1 + {x^2}}} = {{12 + 9{x^2}} \over {1 + {x^2}}}$$
We have, $$\alpha = \int\limits_0^{9 + {{3\pi } \over 4}} {{e^t}dt = {e^{9 + {{3\pi } \over 4}}} - 1} $$
$$\therefore$$ $$\ln \left| {\alpha + 1} \right| = 9 + {{3\pi } \over 4}$$
Thus $$\ln \left| {\alpha + 1} \right| - {{3\pi } \over 4} = 9$$
Comments (0)
