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JEE Mains · Maths · STD 11 - 4.2 Quadratic equations and inequations

यदि द्विघाती समीकरण, \(x^{2}+x \sin \theta-2 \sin \theta=0, \theta \in\left(0, \frac{\pi}{2}\right) \text {, }\) के मूल \(\alpha\) तथा \(\beta\) हैं, तो \(\frac{\alpha^{12}+\beta^{12}}{\left(\alpha^{-12}+\beta^{-12}\right)(\alpha-\beta)^{24}}\) बराबर हैं 

  1. A \(\frac{{{2^{12}}}}{{{{\left( {\sin \,\theta  + 8} \right)}^{12}}}}\)
  2. B \(\frac{{{2^{12}}}}{{{{\left( {\sin \,\theta  - 4} \right)}^{12}}}}\)
  3. C \(\frac{{{2^{12}}}}{{{{\left( {\sin \,\theta  - 8} \right)}^{6}}}}\)
  4. D \(\frac{{{2^{6}}}}{{{{\left( {\sin \,\theta  + 8} \right)}^{12}}}}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{{{2^{12}}}}{{{{\left( {\sin \,\theta  + 8} \right)}^{12}}}}\)

Step-by-step Solution

Detailed explanation

\(x^{2}+x \sin \theta-2 \sin \theta=0\) \(\alpha+\beta=-\sin \theta\) \(\alpha \beta=-2 \sin \theta\) \(\text { Now, } \frac{\alpha^{12}+\beta^{12}}{\left(\frac{1}{\alpha^{12}}+\frac{1}{\beta^{12}}\right)(\alpha-\beta)^{24}}\) \(=\frac{(\alpha \beta)^{12}}{(\alpha-\beta)^{24}}\)…
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