Let A = { 1967 + 1686 ? i sin 7 - 3 i cos } . If A contains exactly one positive integer n , then the value of n is [2023]

Question

Let $A = {\frac{1967 + 1686 i \sin θ}{7 - 3 i \cos θ} : θ \in ℝ} .$ If A contains exactly one positive integer $n$ , then the value of $n$ is [2023]

Smart Achievers · Accepted Answer

(281)

$\frac{281 (49 + 18 \sin θ \cos θ + i (21 \cos θ + 42 \sin θ))}{49 + 9 \cos^{2} θ}$

is a positive integer. For positive integer $Im (z) = 0$

$\Rightarrow 21 \cos θ + 42 \sin θ = 0$

$\Rightarrow \tan θ = - \frac{1}{2}, \sin 2 θ = - \frac{4}{5}, \cos^{2} θ = \frac{4}{5}$

$Now R e (2) = \frac{281 (49 - 9 \sin 2 θ)}{49 + 9 \cos^{2} θ}$

$= \frac{281 (49 - 9 \times (- \frac{4}{5}))}{49 + 9 \times \frac{4}{5}}$

$= \frac{281 (49 + \frac{36}{5})}{49 + \frac{36}{5}}$

$= 281$

Solution

Q.
Let $A = {\frac{1967 + 1686 i \sin θ}{7 - 3 i \cos θ} : θ \in ℝ} .$ If A contains exactly one positive integer $n$ , then the value of $n$ is [2023]

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Solution

Q. Let A={1967+1686 isinθ7-3icosθ:θ∈ℝ}. If A contains exactly one positive integer n, then the value of n is [2023]

Ans. (281) 281(49+18sinθcosθ+i(21cosθ+42sinθ))49+9cos2θ is a positive integer. For positive integer Im(z)=0 ⇒21cosθ+42sinθ=0 ⇒tanθ=-12, sin2θ=-45, cos2θ=45 Now Re(2)=281(49-9sin2θ)49+9cos2θ =281(49-9×(-45))49+9×45 =281(49+365)49+365 = 281

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Q.
Let $A = {\frac{1967 + 1686 i \sin θ}{7 - 3 i \cos θ} : θ \in ℝ} .$ If A contains exactly one positive integer $n$ , then the value of $n$ is [2023]