JEE MAIN - Mathematics (2019 - 10th April Morning Slot - No. 6)
If $$\alpha $$ and $$\beta $$ are the roots of the quadratic equation,
x2 + x sin $$\theta $$ - 2 sin $$\theta $$ = 0, $$\theta \in \left( {0,{\pi \over 2}} \right)$$, then
$${{{\alpha ^{12}} + {\beta ^{12}}} \over {\left( {{\alpha ^{ - 12}} + {\beta ^{ - 12}}} \right).{{\left( {\alpha - \beta } \right)}^{24}}}}$$ is equal to :
x2 + x sin $$\theta $$ - 2 sin $$\theta $$ = 0, $$\theta \in \left( {0,{\pi \over 2}} \right)$$, then
$${{{\alpha ^{12}} + {\beta ^{12}}} \over {\left( {{\alpha ^{ - 12}} + {\beta ^{ - 12}}} \right).{{\left( {\alpha - \beta } \right)}^{24}}}}$$ is equal to :
$${{{2^{12}}} \over {{{\left( {\sin \theta - 8} \right)}^6}}}$$
$${{{2^6}} \over {{{\left( {\sin \theta + 4} \right)}^{12}}}}$$
$${{{2^{12}}} \over {{{\left( {\sin \theta + 8} \right)}^{12}}}}$$
$${{{2^{12}}} \over {{{\left( {\sin \theta - 4} \right)}^{12}}}}$$
Explanation
Given $$\alpha + \beta = - \sin \theta $$ and$$\alpha \beta = - 2\sin \theta $$
$${{\left( {{\alpha ^{12}} + {\beta ^{12}}} \right){\alpha ^{12}}{\beta ^{12}}} \over {\left( {{\alpha ^{12}} + {\beta ^{12}}} \right){{\left( {\alpha - \beta } \right)}^{24}}}} = {{{{\left( {\alpha \beta } \right)}^{12}}} \over {{{\left( {\alpha - \beta } \right)}^{24}}}}$$
$$\left| {\alpha - \beta } \right| = \sqrt {{{\left( {\alpha + \beta } \right)}^2} - 4\alpha \beta } = \sqrt {{{\sin }^2}\theta + 8\sin \theta } $$
Hence required quantity
$${{{{\left( {\alpha \beta } \right)}^{12}}} \over {{{\left( {\alpha - \beta } \right)}^{24}}}} = {{{{\left( {2\sin \theta } \right)}^{12}}} \over {{{\sin }^{12}}\theta {{\left( {\sin \theta + 8} \right)}^{12}}}} = {{{2^{12}}} \over {{{\left( {\sin \theta + 8} \right)}^{12}}}}$$
$${{\left( {{\alpha ^{12}} + {\beta ^{12}}} \right){\alpha ^{12}}{\beta ^{12}}} \over {\left( {{\alpha ^{12}} + {\beta ^{12}}} \right){{\left( {\alpha - \beta } \right)}^{24}}}} = {{{{\left( {\alpha \beta } \right)}^{12}}} \over {{{\left( {\alpha - \beta } \right)}^{24}}}}$$
$$\left| {\alpha - \beta } \right| = \sqrt {{{\left( {\alpha + \beta } \right)}^2} - 4\alpha \beta } = \sqrt {{{\sin }^2}\theta + 8\sin \theta } $$
Hence required quantity
$${{{{\left( {\alpha \beta } \right)}^{12}}} \over {{{\left( {\alpha - \beta } \right)}^{24}}}} = {{{{\left( {2\sin \theta } \right)}^{12}}} \over {{{\sin }^{12}}\theta {{\left( {\sin \theta + 8} \right)}^{12}}}} = {{{2^{12}}} \over {{{\left( {\sin \theta + 8} \right)}^{12}}}}$$
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