JEE MAIN - Mathematics (2021 - 20th July Evening Shift - No. 4)
If the real part of the complex number $${(1 - \cos \theta + 2i\sin \theta )^{ - 1}}$$ is $${1 \over 5}$$ for $$\theta \in (0,\pi )$$, then the value of the integral $$\int_0^\theta {\sin x} dx$$ is equal to:
1
2
$$-$$1
0
Explanation
$$z = {1 \over {1 - \cos \theta + 2i\sin \theta }}$$
$$ = {{2{{\sin }^2}{\theta \over 2} - 2i\sin \theta } \over {{{(1 - \cos \theta )}^2} + 4{{\sin }^2}\theta }}$$
$$ = {{\sin {\theta \over 2} - 2i\cos {\theta \over 2}} \over {2\sin {\theta \over 2}\left( {{{\sin }^2}{\theta \over 2} + 4{{\cos }^2}{\theta \over 2}} \right)}}$$
$${\mathop{\rm Re}\nolimits} (z) = {1 \over {2\left( {{{\sin }^2}{\theta \over 2} + 4{{\cos }^2}{\theta \over 2}} \right)}} = {1 \over 5}$$
$$\sin {{2\theta } \over 2} + 4{\cos ^2}{\theta \over 2} = {5 \over 2}$$$$1 - {\cos ^2}{\theta \over 2} + 4\cos {\theta \over 2} = {5 \over 2}$$
$$3{\cos ^2}{\theta \over 2} = {1 \over 2}$$
$${\theta \over 2} = n\pi \pm {\pi \over 4}$$
$$\theta = 2n\pi \pm {\pi \over 2}$$
$$\theta \in (0,\pi )$$
$$ \therefore $$ $$\theta = {\pi \over 2}$$
$$\int\limits_0^{{\pi \over 2}} {\sin \theta d\theta = {{[ - \cos \theta ]}_0}^{{\pi \over 2}}} $$
$$ = - (0 - 1)$$
$$ = 1$$
$$ = {{2{{\sin }^2}{\theta \over 2} - 2i\sin \theta } \over {{{(1 - \cos \theta )}^2} + 4{{\sin }^2}\theta }}$$
$$ = {{\sin {\theta \over 2} - 2i\cos {\theta \over 2}} \over {2\sin {\theta \over 2}\left( {{{\sin }^2}{\theta \over 2} + 4{{\cos }^2}{\theta \over 2}} \right)}}$$
$${\mathop{\rm Re}\nolimits} (z) = {1 \over {2\left( {{{\sin }^2}{\theta \over 2} + 4{{\cos }^2}{\theta \over 2}} \right)}} = {1 \over 5}$$
$$\sin {{2\theta } \over 2} + 4{\cos ^2}{\theta \over 2} = {5 \over 2}$$$$1 - {\cos ^2}{\theta \over 2} + 4\cos {\theta \over 2} = {5 \over 2}$$
$$3{\cos ^2}{\theta \over 2} = {1 \over 2}$$
$${\theta \over 2} = n\pi \pm {\pi \over 4}$$
$$\theta = 2n\pi \pm {\pi \over 2}$$
$$\theta \in (0,\pi )$$
$$ \therefore $$ $$\theta = {\pi \over 2}$$
$$\int\limits_0^{{\pi \over 2}} {\sin \theta d\theta = {{[ - \cos \theta ]}_0}^{{\pi \over 2}}} $$
$$ = - (0 - 1)$$
$$ = 1$$
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