Complete solution set of the inequality $\left( {{{\sec }^{ - 1}}\,x - 4} \right)\left( {{{\sec }^{ 1}}\,x - 1} \right)\left( {{{\sec }^{ - 1}}\,x - 2} \right) \ge 0$ is
$\left[ {\sec 2\,,\,\sec \,1} \right]$
$\left[ {\sec 1\,,\,\sec \,2} \right]\, \cup \,\left[ {\sec \,4\,,\,\infty } \right)$
$\left( { - \infty \,,\,\sec \,2} \right]\, \cup \,\left[ {\sec \,1\,,\,\infty } \right)$
$\left( { - \infty \,,\,\sec \,4} \right]\, \cup \,\left[ {\sec \,2\,,\,\infty } \right)$
If $a$ and $b$ are the roots of equation $x^2-7 x-1=0$, then the value of $\frac{a^{21}+b^{21}+a^{17}+b^{17}}{a^{19}+b^{19}}$ is equal to $........$.
If $\alpha, \beta $ and $\gamma$ are the roots of the equation $2{x^3} - 3{x^2} + 6x + 1 = 0$, then ${\alpha ^2} + {\beta ^2} + {\gamma ^2}$ is equal to
If $x$ be real, then the minimum value of ${x^2} - 8x + 17$ is
Equation $\frac{3}{{x - {a^3}}} + \frac{5}{{x - {a^5}}} + \frac{7}{{x - {a^7}}} = 0,a > 1$ has