If α and β are the roots of the quadratic equation, x2 +x sinθ - 2 sinθ =0, θ ∈ (0 , (π/2)), then (α12 +β12/(α-12 +β-12)(α -β)24) is equal to :

Question

If α and β are the roots of the quadratic equation, x2 +x sinθ - 2 sinθ =0, θ ∈ (0 , (π/2)), then (α12 +β12/(α-12 +β-12)(α -β)24) is equal to : (A) (26/( sin θ + 8)12) (B) (212/( sin θ - 8)6) (C) (212/( sin θ - 4)12) (D) (212/( sin θ + 8)12)

Tardigrade Admin · Accepted Answer

(α12 +β12/((1)α12 + (1)β12) (α -β)24 = ((αβ)12/(α-β)24) = ((αβ)12/[(α+β)2 -4 αβ]12) = [(αβ/(α+β)2 -4αβ)]/12) = ((-2 sinθ/ sin2θ+8 sinθ))12 = (212/( sinθ+8)12)

Q. If $α$ and $β$ are the roots of the quadratic 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 :

$\frac{2 ^{6}}{( s i n θ + 8 ) ^{12}}$

$\frac{2 ^{12}}{( s i n θ - 8 ) ^{6}}$

$\frac{2 ^{12}}{( s i n θ - 4 ) ^{12}}$

$\frac{2 ^{12}}{( s i n θ + 8 ) ^{12}}$

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

(sinθ+8)1226​

(sinθ−8)6212​

(sinθ−4)12212​

(sinθ+8)12212​

(α121​+β121​)(α−β)24α12+β12​=(α−β)24(αβ)12​ =[(α+β)2−4αβ]12(αβ)12​=[(α+β)2−4αβαβ​]12 =(sin2θ+8sinθ−2sinθ​)12=(sinθ+8)12212​

Q. If $α$ and $β$ are the roots of the quadratic 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 :

$\frac{2 ^{6}}{( s i n θ + 8 ) ^{12}}$

$\frac{2 ^{12}}{( s i n θ - 8 ) ^{6}}$

$\frac{2 ^{12}}{( s i n θ - 4 ) ^{12}}$

$\frac{2 ^{12}}{( s i n θ + 8 ) ^{12}}$