If - and - are the roots of the equation- x2-x sin-2 sin-0- -0- -2- then -12-12-12-12-24 is equal toA- 212-sin-812B- 26-sin-812212C- 212-sin-412D- 212-sin-86

The correct option is A 2^12(sinθ + 8)^12x^2+xsinθ 2sinθ=0 nbsp; α +β= sinθ, α×β= 2sinθ ⇒α^12+β^12(α^ 12+β^ 12)(α β)^24 =α^12+β^12(1α^12+1β^12)(α β)^24 =( ...

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Standard XII

Mathematics

Relation between Roots and Coefficients for Quadratic

If α and β ar...

Question

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

\frac{2^{6}}{(sin θ + 8)^{12}}

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\frac{2^{12}}{(sin θ - 8)^{6}}

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\frac{2^{12}}{(sin θ + 8)^{12}}

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\frac{2^{12}}{(sin θ - 4)^{12}}

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Solution

The correct option is C $\frac{2^{12}}{(sin θ + 8)^{12}}$
$x^{2} + x sin θ - 2 sin θ = 0$
$α + β = - sin θ, α \times β = - 2 sin θ$

$\Rightarrow \frac{α^{12} + β^{12}}{(α^{- 12} + β^{- 12}) (α - β)^{24}} = \frac{α^{12} + β^{12}}{(\frac{1}{α^{12}} + \frac{1}{β^{12}}) (α - β)^{24}}$
$= \frac{(α β)^{12}}{(α - β)^{24}}$

Since,
$(α - β)^{2} = (α + β)^{2} - 4 α β = {sin}^{2} θ - 4 \times (- 2 sin θ)$

$∴ \frac{(α β)^{12}}{(α - β)^{24}} = \frac{(- 2 sin θ)^{12}}{({sin}^{2} θ + 8 sin θ)^{12}} = \frac{2^{12}}{(8 + sin θ)^{12}}$

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If α and β are the roots of the equation, x2+xsinθ−2sinθ=0,θ∈(0,π2) , then α12+β12(α−12+β−12)(α−β)24 is equal to

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