JEE MAIN - Mathematics (2019 - 9th January Morning Slot - No. 2)
Let
A = $$\left\{ {\theta \in \left( { - {\pi \over 2},\pi } \right):{{3 + 2i\sin \theta } \over {1 - 2i\sin \theta }}is\,purely\,imaginary} \right\}$$
. Then the sum of the elements in A is :
A = $$\left\{ {\theta \in \left( { - {\pi \over 2},\pi } \right):{{3 + 2i\sin \theta } \over {1 - 2i\sin \theta }}is\,purely\,imaginary} \right\}$$
. Then the sum of the elements in A is :
$${5\pi \over 6}$$
$$\pi $$
$${3\pi \over 4}$$
$${{2\pi } \over 3}$$
Explanation
Given complex number,
$${{3 + 2i\sin \theta } \over {1 - 2i\sin \theta }}$$
$$ = {{\left( {3 + 2i\sin \theta } \right)\left( {1 + 2i\sin \theta } \right)} \over {1 + 4{{\sin }^2}\theta }}$$
$$ = {{3 + 6i\sin \theta + 2i\sin \theta - 4{{\sin }^2}\theta } \over {1 + 4{{\sin }^2}\theta }}$$
$$ = {{\left( {3 - 4{{\sin }^2}\theta } \right) + i\left( {8\sin \theta } \right)} \over {1 + 4{{\sin }^2}\theta }}$$
As complex number is purely imaginary, So real part of this complex number is zero.
$$ \therefore $$ $${{3 - 4{{\sin }^2}\theta } \over {1 + 4{{\sin }^2}\theta }}$$ = 0
$$ \Rightarrow $$ $$3 - 4{\sin ^2}\theta = 0$$
$$ \Rightarrow $$ $$\sin \theta = \pm {{\sqrt 3 } \over 2}$$
as $$\theta $$ $$ \in $$ $$\left( { - {\pi \over 2},\pi } \right)$$
$$ \therefore $$ $$\theta $$ $$=$$ $$-$$ $${\pi \over 3},{\pi \over 3},{{2\pi } \over 3}$$
$$ \therefore $$ Sum of those values of A is
$$ = - {\pi \over 3} + {\pi \over 3} + {{2\pi } \over 3}$$
$$ = {{2\pi } \over 3}$$
$${{3 + 2i\sin \theta } \over {1 - 2i\sin \theta }}$$
$$ = {{\left( {3 + 2i\sin \theta } \right)\left( {1 + 2i\sin \theta } \right)} \over {1 + 4{{\sin }^2}\theta }}$$
$$ = {{3 + 6i\sin \theta + 2i\sin \theta - 4{{\sin }^2}\theta } \over {1 + 4{{\sin }^2}\theta }}$$
$$ = {{\left( {3 - 4{{\sin }^2}\theta } \right) + i\left( {8\sin \theta } \right)} \over {1 + 4{{\sin }^2}\theta }}$$
As complex number is purely imaginary, So real part of this complex number is zero.
$$ \therefore $$ $${{3 - 4{{\sin }^2}\theta } \over {1 + 4{{\sin }^2}\theta }}$$ = 0
$$ \Rightarrow $$ $$3 - 4{\sin ^2}\theta = 0$$
$$ \Rightarrow $$ $$\sin \theta = \pm {{\sqrt 3 } \over 2}$$
as $$\theta $$ $$ \in $$ $$\left( { - {\pi \over 2},\pi } \right)$$
$$ \therefore $$ $$\theta $$ $$=$$ $$-$$ $${\pi \over 3},{\pi \over 3},{{2\pi } \over 3}$$
$$ \therefore $$ Sum of those values of A is
$$ = - {\pi \over 3} + {\pi \over 3} + {{2\pi } \over 3}$$
$$ = {{2\pi } \over 3}$$
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