Question

If α and β are the roots of the quadratic equation, \({{\rm{x}}^2} + {\rm{x\;sin\;\theta }} - 2{\rm{\;sin\;\theta }} = 0,{\rm{\theta }} \in \left( {0,\frac{{\rm{\pi }}}{2}} \right)\) then \(\frac{{{{\rm{\alpha }}^{12}} + {{\rm{\beta }}^{12}}}}{{\left( {{{\rm{\alpha }}^{ - 12}} + {{\rm{\beta }}^{ - 12}}} \right)\cdot{{({\rm{\alpha }} - {\rm{\beta }})}^{24}}}}\) is equal to:
1. \(\frac{{{2^{12}}}}{{{{({\rm{sin\theta }} - 4)}^{12}}}}\)
2. \(\frac{{{2^{12}}}}{{{{({\rm{sin\theta }} + 8)}^{12}}}}\)
3. \(\frac{{{2^{12}}}}{{{{({\rm{sin\theta }} - 8)}^6}}}\)
4. \(\frac{{{2^6}}}{{{{({\rm{sin\theta }} + 8)}^{12}}}}\)

Answer 1

Correct Answer - Option 2 : \(\frac{{{2^{12}}}}{{{{({\rm{sin\theta }} + 8)}^{12}}}}\)

The given quadratic equation is,

x² + x sin θ – 2 sin θ = 0

Now, the sum of roots is:

α + β = -sin θ

Now, the product of roots is:

αβ = -2 sin θ

From question,

\(\Rightarrow \frac{{{{\rm{\alpha }}^{12}} + {{\rm{\beta }}^{12}}}}{{\left( {{{\rm{\alpha }}^{ - 12}} + {{\rm{\beta }}^{ - 12}}} \right){{({\rm{\alpha }} - {\rm{\beta }})}^{24}}}} = \frac{{{{\rm{\alpha }}^{12}} + {{\rm{\beta }}^{12}}}}{{\left( {\frac{1}{{{{\rm{\alpha }}^{12}}}} + \frac{1}{{{{\rm{\beta }}^{12}}}}} \right){{({\rm{\alpha }} - {\rm{\beta }})}^{24}}}}\)

\(= \frac{{{{\rm{\alpha }}^{12}} + {{\rm{\beta }}^{12}}}}{{\frac{{\left( {{{\rm{\alpha }}^{12}} + {{\rm{\beta }}^{12}}} \right)}}{{\left( {{{\rm{\alpha }}^{12}}{{\rm{\beta }}^{12}}} \right)}}{{({\rm{\alpha }} - {\rm{\beta }})}^{24}}}}\)

\(= \frac{{\left( {{{\rm{\alpha }}^{12}} + {{\rm{\beta }}^{12}}} \right){{\rm{\alpha }}^{12}}{{\rm{\beta }}^{12}}}}{{\left( {{{\rm{\alpha }}^{12}} + {{\rm{\beta }}^{12}}} \right){{({\rm{\alpha }} - {\rm{\beta }})}^{24}}}}\)

\(= \frac{{{{({\rm{\alpha \beta }})}^{12}}}}{{{{({\rm{\alpha }} - {\rm{\beta }})}^{24}}}}\)

\(\therefore \frac{{{{\rm{\alpha }}^{12}} + {{\rm{\beta }}^{12}}}}{{\left( {{{\rm{\alpha }}^{ - 12}} + {{\rm{\beta }}^{ - 12}}} \right){{({\rm{\alpha }} - {\rm{\beta }})}^{24}}}} = \frac{{{{({\rm{\alpha \beta }})}^{12}}}}{{({{\left( {{\rm{\alpha }} - {\rm{\beta }}{)^2}} \right)}^{12}}}}\)

We know that,

We know that, (a + b)² = a² + b² + 2ab

⇒ a² + b² = (a + b)² – 2ab

(a – b)² = a² + b² – 2ab

(a – b)² = ((a + b)² – 2ab) – 2ab

(a – b)² = (a + b)² – 4ab

Now,

\(\Rightarrow \frac{{{{({\rm{\alpha \beta }})}^{12}}}}{{{{\left( {{{({\rm{\alpha }} - {\rm{\beta }})}^2}} \right)}^{12}}}} = \frac{{{{({\rm{\alpha \beta }})}^{12}}}}{{{{\left( {{{\left( {{\rm{\alpha }} + {\rm{\beta }}} \right)}^2} - 4{\rm{\alpha \beta }}} \right)}^{12}}}}\)

\(= \frac{{{{\left( { - 2{\rm{sin\theta }}} \right)}^{12}}}}{{{{\left( {{{\left( { - {\rm{sin\theta }}} \right)}^2} - 4\left( { - 2{\rm{sin\theta }}} \right)} \right)}^{12}}}}\)

\(= \frac{{{2^{12}}{{\sin }^{12}}{\rm{\theta }}}}{{{{\left( {{\rm{si}}{{\rm{n}}^2}{\rm{\theta }} + 8{\rm{sin\theta }}} \right)}^{12}}}} = \frac{{{2^{12}}{{\sin }^{12}}{\rm{\theta }}}}{{\left( {(\sin {\rm{\theta }} + 8} \right)\sin {\rm{\theta }}{)^{12}}}} = \frac{{{2^{12}}{{\sin }^{12}}{\rm{\theta }}}}{{{{\left( {\sin {\rm{\theta }} + 8} \right)}^{12}}{{\sin }^{12}}{\rm{\theta }}}}\)

\(\therefore \frac{{{{({\rm{\alpha \beta }})}^{12}}}}{{{{({\rm{\alpha }} - {\rm{\beta }})}^{24}}}} = \frac{{{2^{12}}}}{{{{({\rm{sin\theta }} + 8)}^{12}}}}\)