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JEE Mains · Maths · STD 11 - 4.1 complex nubers

माना \(\mathrm{S}=\left\{\mathrm{z} \in \mathrm{C}-\{\mathrm{i}, 2 \mathrm{i}\}: \frac{\mathrm{z}^2+8 \mathrm{iz}-15}{\mathrm{z}^2-3 \mathrm{iz}-2} \in \mathrm{R}\right\}\) है। यदि \(\alpha-\frac{13}{11} \mathrm{i} \in \mathrm{S}, \alpha \in \mathbb{R}-\{0\}\) है, तो \(242 \alpha^2\) बराबर है।

  1. A \(1680\)
  2. B \(1681\)
  3. C \(1682\)
  4. D \(1683\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(1680\)

Step-by-step Solution

Detailed explanation

\(\left(\frac{z^2+81 z-15}{z^2-3 i z-2}\right) \in R\) \(\Rightarrow 1+\frac{(11 i z-13)}{\left(z^2-3 i z-2\right)} \in R\) Put \(z=\alpha-\frac{13}{11} i\) \(\Rightarrow\left(z^2-3 i z-2\right)\) is imaginary Put \(z=x+i y\)…
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