Verified Solution • 2025–26 Session moderate

QJEEM19Ma12f4: JEE Main 2019 Mathematics - Class 11 JEEM19Ma12f4 Solution

If α\alpha and β\beta are the roots of the quadratic equation,
x2 + x sin θ\theta - 2 sin θ\theta = 0, θ∈(0,π2)\theta \in \left( {0,{\pi \over 2}} \right) , 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}}}} is equal to :

Multiple Choice Options

  • A 212(sin⁡θ−8)6{{{2^{12}}} \over {{{\left( {\sin \theta - 8} \right)}^6}}}
  • B 26(sin⁡θ+4)12{{{2^6}} \over {{{\left( {\sin \theta + 4} \right)}^{12}}}}
  • C 212(sin⁡θ+8)12{{{2^{12}}} \over {{{\left( {\sin \theta + 8} \right)}^{12}}}}
  • D 212(sin⁡θ−4)12{{{2^{12}}} \over {{{\left( {\sin \theta - 4} \right)}^{12}}}}

Step-by-Step Verified Solution

Given α+β=−sin⁡θ\alpha + \beta = - \sin \theta and αβ=−2sin⁡θ\alpha \beta = - 2\sin \theta

(α12+β12)α12β12(α12+β12)(α−β)24=(αβ)12(α−β)24{{\left( {{\alpha ^{12}} + {\beta ^{12}}} \right){\alpha ^{12}}{\beta ^{12}}} \over {\left( {{\alpha ^{12}} + {\beta ^{12}}} \right){{\left( {\alpha - \beta } \right)}^{24}}}} = {{{{\left( {\alpha \beta } \right)}^{12}}} \over {{{\left( {\alpha - \beta } \right)}^{24}}}}

∣α−β∣=(α+β)2−4αβ=sin⁡2θ+8sin⁡θ\left| {\alpha - \beta } \right| = \sqrt {{{\left( {\alpha + \beta } \right)}^2} - 4\alpha \beta } = \sqrt {{{\sin }^2}\theta + 8\sin \theta }

Hence required quantity

(αβ)12(α−β)24=(2sin⁡θ)12sin⁡12θ(sin⁡θ+8)12=212(sin⁡θ+8)12{{{{\left( {\alpha \beta } \right)}^{12}}} \over {{{\left( {\alpha - \beta } \right)}^{24}}}} = {{{{\left( {2\sin \theta } \right)}^{12}}} \over {{{\sin }^{12}}\theta {{\left( {\sin \theta + 8} \right)}^{12}}}} = {{{2^{12}}} \over {{{\left( {\sin \theta + 8} \right)}^{12}}}}

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