JEE Mains · Maths · STD 11 - 4.2 Quadratic equations and inequations
If \(\alpha \) and \(\beta \) are the roots of the quadratic equation, \(x^2 + x\, sin\,\theta -2sin\,\theta = 0\), \(\theta \in \left( {0,\frac{\pi }{2}} \right)\) then \(\frac{{{\alpha ^{12}} + {\beta ^{12}}}}{{\left( {{\alpha ^{ - 12}} + {\beta ^{ - 12}}} \right){{\left( {\alpha - \beta } \right)}^{24}}}}\) is equal to
- A \(\frac{{{2^{12}}}}{{{{\left( {\sin \,\theta + 8} \right)}^{12}}}}\)
- B \(\frac{{{2^{12}}}}{{{{\left( {\sin \,\theta - 4} \right)}^{12}}}}\)
- C \(\frac{{{2^{12}}}}{{{{\left( {\sin \,\theta - 8} \right)}^{6}}}}\)
- D \(\frac{{{2^{6}}}}{{{{\left( {\sin \,\theta + 8} \right)}^{12}}}}\)
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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