If $\alpha $ and $\beta $ are the roots of the quadratic equation, $x^2 + x\, sin\,\theta -2sin\,\theta = 0$, $\theta \in \left( {0,\frac{\pi }{2}} \right)$ then $\frac{{{\alpha ^{12}} + {\beta ^{12}}}}{{\left( {{\alpha ^{ - 12}} + {\beta ^{ - 12}}} \right){{\left( {\alpha - \beta } \right)}^{24}}}}$ is equal to
$\frac{{{2^{12}}}}{{{{\left( {\sin \,\theta + 8} \right)}^{12}}}}$
$\frac{{{2^{12}}}}{{{{\left( {\sin \,\theta - 4} \right)}^{12}}}}$
$\frac{{{2^{12}}}}{{{{\left( {\sin \,\theta - 8} \right)}^{6}}}}$
$\frac{{{2^{6}}}}{{{{\left( {\sin \,\theta + 8} \right)}^{12}}}}$
Consider the equation $(1+a+b)^2=3\left(1+a^2+b^{2})\right.$ where $a, b$ are real numbers. Then,
Let $[t]$ denote the greatest integer $\leq t .$ Then the equation in $x ,[ x ]^{2}+2[ x +2]-7=0$ has
If $\alpha , \beta , \gamma $ are roots of equation ${x^3} + a{x^2} + bx + c = 0$, then ${\alpha ^{ - 1}} + {\beta ^{ - 1}} + {\gamma ^{ - 1}} = $
If $a < 0$ then the inequality $a{x^2} - 2x + 4 > 0$ has the solution represented by
The value of $x$ in the given equation ${4^x} - {3^{x\,\; - \;\frac{1}{2}}} = {3^{x + \frac{1}{2}}} - {2^{2x - 1}}$is