First slide

Theory of equations

Question

If $α$ and $β$ are the roots of the quadratic equation, $x^{2} + xsin θ - 2 \sin θ = 0, θ \in (0, \frac{π}{2}), then$ $\frac{α^{12} + β^{12}}{(α^{- 12} + β^{- 12}) (α - β)^{24}} isequalto$

Moderate

A

$\frac{2^{12}}{(\sin θ + 8)^{12}}$
B

$\frac{2^{6}}{(\sin θ + 8)^{12}}$
C

$\frac{2^{12}}{(\sin θ - 4)^{12}}$
D

$\frac{2^{12}}{(\sin θ - 8)^{6}}$

Solution

Given quadratic equation is $x^{2} + xsin θ - 2 \sin θ = 0, θ \in (0, \frac{π}{2})$
and its roots are $α$ and $β$
So, sum of roots $= α + β = - \sin θ$
product or roots $= αβ = - 2 \sin θ$
$\Rightarrow αβ = 2 (α + β)$ ……(i)
Now, the given express is $\frac{α^{12} + β^{12}}{(α^{- 12} + β^{- 12}) (α - β)^{24}}$
$= \frac{α^{12} + β^{12}}{(\frac{1}{α^{12}} + \frac{1}{β^{12}}) (α - β)^{24}} = \frac{α^{12} + β^{12}}{(\frac{β^{12} + α^{12}}{α^{12} β^{12}}) (α - β)^{24}}$
$\begin{array}{l} = {[\frac{α β}{(α - β)^{2}}]}^{12} = {(\frac{α β}{(α + β)^{2} - 4 α β})}^{12} \\ = {[\frac{2 (α + β)}{(α + β)^{2} - 8 (α + β)}]}^{12} [from Eq. (i)] \\ = {(\frac{2}{(α + β) - 8})}^{12} = {(\frac{2}{- \sin θ - 8})}^{12} \\ [∵ α + β = - \sin θ] \\ = \frac{2^{12}}{(\sin θ + 8)^{12}} \end{array}$

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