Theory Of Equations Ques 5
- If $\alpha$ and $\beta$ are the roots of the quadratic equation, $x^{2}+x \sin \theta-2 \sin \theta=0, \theta \in 0, \frac{\pi}{2}$, then $\frac{\alpha^{12}+\beta^{12}}{\left(\alpha^{-12}+\beta^{-12}\right)(\alpha-\beta)^{24}}$ is equal to
(a) $\frac{2^{12}}{(\sin \theta+8)^{12}}$
(b) $\frac{2^{6}}{(\sin \theta+8)^{12}}$
(c) $\frac{2^{12}}{(\sin \theta-4)^{12}}$
(d) $\frac{2^{12}}{(\sin \theta-8)^{6}}$
(2019 Main, 10 April I)
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Answer:
Correct Answer: 5.(a)
Solution:
- Given quadratic equation is
$ x^{2}+x \sin \theta-2 \sin \theta=0, \theta \in (0, \frac{\pi}{2}) $
and its roots are $\alpha$ and $\beta$.
So, sum of roots $=\alpha+\beta=-\sin \theta$
and product of roots $=\alpha \beta=-2 \sin \theta$
$\Rightarrow \quad \alpha \beta=2(\alpha+\beta) ……(i)$
Now, the given expression is $\frac{\alpha^{12}+\beta^{12}}{\left(\alpha^{-12}+\beta^{-12}\right)(\alpha-\beta)^{24}}$
$ =\frac{\alpha^{12}+\beta^{12}}{(\frac{1}{\alpha^{12}}+\frac{1}{\beta^{12})}(\alpha-\beta)^{24}}=\frac{\alpha^{12}+\beta^{12}}{(\frac{\beta^{12}+\alpha^{12}}{\alpha^{12} \beta^{12})}(\alpha-\beta)^{24}} $
$ =[\frac{\alpha \beta}{(\alpha-\beta)^{2}}]^ {12}=(\frac{\alpha \beta}{(\alpha+\beta)^{2}-4 \alpha \beta})^ {12} $
$ =[\frac{2(\alpha+\beta)}{(\alpha+\beta)^{2}-8(\alpha+\beta)}]^{12} \quad \text { [from Eq. (i)] } $
$ =(\frac{2}{(\alpha+\beta)-8})^{12}=(\frac{2}{-\sin \theta-8})^{12} \quad[\because \alpha+\beta=-\sin \theta] $
$ =\frac{2^{12}}{(\sin \theta+8)^{12}}$
BETA
