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QJEEAdv23Ma1d3f: JEE Advanced 2023 Mathematics - Class 11 JEEAdv23Ma1d3f Solution: Let . If contains exactly one positive integer , then the value …

Let A={1967+1686isin⁡θ7−3icos⁡θ:θ∈R}A=\left\{\frac{1967+1686 i \sin \theta}{7-3 i \cos \theta}: \theta \in \mathbb{R}\right\} . If AA contains exactly one positive integer nn , then the value of nn is

Step-by-Step Verified Solution

amp;A=1967+1686isin⁡θ7−3icos⁡θamp;=281(7+6isin⁡θ)7−3icos⁡θ×7+3icos⁡θ7+3icos⁡θamp;=281(49−18sin⁡θcos⁡θ+i(21cos⁡θ+42sin⁡θ))49+9cos⁡2θ\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

amp;Im⁡(A)=0amp;21cos⁡θ+42sin⁡θ=0amp;tan⁡θ=−12;sin⁡2θ=−45,cos⁡2θ=45\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}

amp;Re⁡(A)=281(49−9sin⁡2θ)49+9cos⁡2θamp;=281(49−9×−45)49+9×45=281(+ ve integer )\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}

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