A particle is projected from a point O with velocity u at an angle of 60^o with the horizontal. When it is moving in a direction at right angles to its direction at O, its velocity then is given by

• u/3

• u/2

• 2u/3

• 2u/3

If f : R S, defined by  , is onto, then the interval of S is

• [0, 3]

• [-1, 1]

• [0, 1]

• [0, 1]

Let $f\left(x\right) = x^2 + \frac{1}{x^2} and g \left(x\right) = x - \frac{1}{x}, x {\rm E} R - \{-1,0,1\}.$
if $h\left(x\right) = \frac{f\left(x\right)}{g\left(x\right)}$, then the local minimum value of h(x) is:

• $2\sqrt{2}$

• 3

• -3

• $-2\sqrt{2}$

The possible values of x, which satisfy the trigonometric equation

${\tan }^{-1}\left(\frac{\mathrm{x} - 1}{\mathrm{x} - 2}\right) + {\tan }^{-1}\left(\frac{\mathrm{x} + 1}{\mathrm{x} + 2}\right) = \frac{\mathrm{\pi }}{4}$ are

• $\pm \frac{1}{\sqrt{2}}$

• $\pm \sqrt{2}$

• $\pm \frac{1}{2}$

• $\pm 2$

If ${\sin }^{- 1}\left(\mathrm{x} - \frac{{\mathrm{x}}^2}{2} + \frac{{\mathrm{x}}^3}{4} - \frac{{\mathrm{x}}^4}{8} + \right) = \frac{\mathrm{\pi }}{6}, \mathrm{where} \left|\mathrm{x}\right| < 2$ then the value of x is

• $\frac{2}{3}$

• $\frac{3}{2}$

• - $\frac{2}{3}$

• - $\frac{3}{2}$

The trigonometric equation ${\sin }^{-1}\left(\mathrm{x}\right) = 2{\sin }^{-1}\left(2\mathrm{a}\right)$ has a real solution, if

• $\left|\mathrm{a}\right| > \frac{1}{\sqrt{2}}$

• $\frac{1}{2\sqrt{2}} < \left|\mathrm{a}\right| < \frac{1}{\sqrt{2}}$

• $\left|\mathrm{a}\right| > \frac{1}{2\sqrt{2}}$

• $\left|\mathrm{a}\right| \le \frac{1}{2\sqrt{2}}$

The value of $2{\cot }^{-1}\left(\frac{1}{2}\right) - {\cot }^{-1}\left(\frac{4}{3}\right)$ is

• $- \frac{\mathrm{\pi }}{8}$

• $\frac{3\mathrm{\pi }}{2}$

• $\frac{\mathrm{\pi }}{4}$

• $\frac{\mathrm{\pi }}{2}$

If ${\sin }^{-1}\left(\frac{\mathrm{x}}{13}\right) + {\csc }^{-1}\left(\frac{13}{12}\right) = \frac{\mathrm{\pi }}{2}$, then the value of x is

• 5

• 4

• 12

• 11

${\tan }^{-1}\left(\frac{\mathrm{x}}{\mathrm{y}}\right) - {\tan }^{-1}\left(\frac{\mathrm{x} - \mathrm{y}}{\mathrm{x} + \mathrm{y}}\right)$ is equal to (where x > y > 0)

• $- \frac{\mathrm{\pi }}{4}$

• $\frac{\mathrm{\pi }}{4}$

• $\frac{3\mathrm{\pi }}{4}$

• None of these

If ${\sin }^{-1}\left(\mathrm{x}\right) - {\cos }^{-1}\left(\mathrm{x}\right) = \frac{\mathrm{\pi }}{6}$, then x is

• $\frac{1}{2}$

• $\frac{\sqrt{3}}{2}$

• $\frac{- 1}{2}$

• None of these