JEE Main 201910 Apr, Morning shiftMathematicsMediumComplex Numbers and Quadratic Equations264× since 2002If α\alphaα and β\betaβ are the roots of the quadratic equation, x² + x sin θ\thetaθ - 2 sin θ\thetaθ = 0, θ∈(0,π2)\theta \in \left( {0,{\pi \over 2}} \right)θ∈(0,2π), then α12+β12(α−12+β−12).(α−β)24{{{\alpha ^{12}} + {\beta ^{12}}} \over {\left( {{\alpha ^{ - 12}} + {\beta ^{ - 12}}} \right).{{\left( {\alpha - \beta } \right)}^{24}}}}(α−12+β−12).(α−β)24α12+β12 is equal to :AOption A. 212(sinθ−8)6{{{2^{12}}} \over {{{\left( {\sin \theta - 8} \right)}^6}}}(sinθ−8)6212BOption B. 26(sinθ+4)12{{{2^6}} \over {{{\left( {\sin \theta + 4} \right)}^{12}}}}(sinθ+4)1226COption C. 212(sinθ+8)12{{{2^{12}}} \over {{{\left( {\sin \theta + 8} \right)}^{12}}}}(sinθ+8)12212DOption D. 212(sinθ−4)12{{{2^{12}}} \over {{{\left( {\sin \theta - 4} \right)}^{12}}}}(sinθ−4)12212CheckSkip · show solution
