JEE Advance - Mathematics (2023 - Paper 1 Online - No. 11)
Let $A=\left\{\frac{1967+1686 i \sin \theta}{7-3 i \cos \theta}: \theta \in \mathbb{R}\right\}$. If $A$ contains exactly one positive integer $n$, then the value of $n$ is
Answer
281
Explanation
$$
\begin{aligned}
& A=\frac{1967+1686 i \sin \theta}{7-3 i \cos \theta} \\\\
& =\frac{281(7+6 i \sin \theta)}{7-3 i \cos \theta} \times \frac{7+3 i \cos \theta}{7+3 i \cos \theta} \\\\
& =\frac{281(49-18 \sin \theta \cos \theta+i(21 \cos \theta+42 \sin \theta))}{49+9 \cos ^2 \theta}
\end{aligned}
$$
For positive integer
$$ \begin{aligned} & \operatorname{Im}(A)=0 \\\\ & 21 \cos \theta+42 \sin \theta=0 \\\\ & \tan \theta=\frac{-1}{2} ; \sin 2 \theta=\frac{-4}{5}, \cos ^2 \theta=\frac{4}{5} \end{aligned} $$
$$ \begin{aligned} & \operatorname{Re}(\mathrm{A})=\frac{281(49-9 \sin 2 \theta)}{49+9 \cos ^2 \theta} \\\\ & =\frac{281\left(49-9 \times \frac{-4}{5}\right)}{49+9 \times \frac{4}{5}}=281(+ \text { ve integer }) \end{aligned} $$
For positive integer
$$ \begin{aligned} & \operatorname{Im}(A)=0 \\\\ & 21 \cos \theta+42 \sin \theta=0 \\\\ & \tan \theta=\frac{-1}{2} ; \sin 2 \theta=\frac{-4}{5}, \cos ^2 \theta=\frac{4}{5} \end{aligned} $$
$$ \begin{aligned} & \operatorname{Re}(\mathrm{A})=\frac{281(49-9 \sin 2 \theta)}{49+9 \cos ^2 \theta} \\\\ & =\frac{281\left(49-9 \times \frac{-4}{5}\right)}{49+9 \times \frac{4}{5}}=281(+ \text { ve integer }) \end{aligned} $$
Comments (0)
