Solution of the equation $\sqrt {x + 3 - 4\sqrt {x - 1} } + \sqrt {x + 8 - 6\sqrt {x - 1} } = 1$ is
$x \in \left[ {4,9} \right]$
$x \in \left[ {3,8} \right]$
$x \in \left[ {5,10} \right]$
$x \in \left[ {4,7} \right]$
Number of natural solutions of the equation $xyz = 2^5 \times 3^2 \times 5^2$ is equal to
Let $\alpha, \beta$ be roots of $x^2+\sqrt{2} x-8=0$. If $\mathrm{U}_{\mathrm{n}}=\alpha^{\mathrm{n}}+\beta^{\mathrm{n}}$, then $\frac{\mathrm{U}_{10}+\sqrt{2} \mathrm{U}_9}{2 \mathrm{U}_8}$ is equal to ............
If $x$ is real, the function $\frac{{(x - a)(x - b)}}{{(x - c)}}$ will assume all real values, provided
Number of positive integral values of $'K'$ for which the equation $k = \left| {x + \left| {2x - 1} \right|} \right| - \left| {x - \left| {2x - 1} \right|} \right|$ has exactly three real solutions, is
If the equation $\frac{1}{x} + \frac{1}{{x - 1}} + \frac{1}{{x - 2}} = 3{x^3}$ has $k$ real roots, then $k$ is equal to -