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JEE Mains · Maths · STD 11 - 4.1 complex nubers

જો \(\alpha\) અને \(\beta\) સમીકરણ \(2 z^2-3 z-2 \mathrm{i}=0\) ના બીજ હોય, જ્યાં \(\mathrm{i}=\sqrt{-1}\), તો \(16 \cdot \operatorname{Re}\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right) \cdot \operatorname{lm}\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right)\) = __________

  1. A \(441\)
  2. B \(398\)
  3. C \(312\)
  4. D \(409\)
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Answer & Solution

Correct Answer

(A) \(441\)

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Detailed explanation

\(2 z^2-3 z-2 i=0\) ...(i) \(2\left(z-\frac{i}{z}\right)=3\) જેમ કે \(\alpha, \beta\) એ (i) ના બીજ છે,…
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