If $\mathrm{\alpha } \le 2{\sin }^{-1}\left(\mathrm{x}\right) + {\cos }^{-1}\left(\mathrm{x}\right) \le \mathrm{\beta }$, then

• $\mathrm{\alpha } = 0, \mathrm{\beta } = \mathrm{\pi }$

• $\mathrm{\alpha } = \frac{- \mathrm{\pi }}{2}, \mathrm{\beta } = \frac{\mathrm{\pi }}{2}$

• $\mathrm{\alpha } = 0, \mathrm{\beta } = 2\mathrm{\pi }$

• $\mathrm{\alpha } = \frac{- \mathrm{\pi }}{2}, \mathrm{\beta } = \frac{3\mathrm{\pi }}{2}$

If A = $\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$, then A² is equal to

• $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

• $\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$

• $\left[\begin{array}{cc}0& 1\\ 0& 1\end{array}\right]$

• $\left[\begin{array}{cc}1& 0\\ 1& 0\end{array}\right]$

The function f(x) = [x], where [x] denotes greatest integer function, is continuous at

• 1

• 4

• 1.5

• - 2

If y = $\log \left(\frac{1 - {\mathrm{x}}^2}{1 + {\mathrm{x}}^2}\right)$, then $\frac{\mathrm{dy}}{\mathrm{dx}}$ is equal to

• $\frac{1}{1 - {\mathrm{x}}^4}$

• $\frac{- 4\mathrm{x}}{1 - {\mathrm{x}}^4}$

• $\frac{- 4{\mathrm{x}}^3}{1 - {\mathrm{x}}^4}$

• $\frac{4{\mathrm{x}}^3}{1 - {\mathrm{x}}^4}$

The two curves x³ - 3xy² + 2 = 0 and 3x²y - y³ = 2

• cut at angle $\mathrm{\pi }$/3

• touch each other

• cut at angle $\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$

• cut at right angle

If x is real, then the minimum value of x² - 8x + 17 is

• 3

• 1

• 4

• 2

The order of differential equation of all circles of given radius 'a' is

• 1

• 4

• 3

• 2

18.

The solution of differential equation $\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}} + 2\mathrm{y} = {\mathrm{x}}^2$ is

• $\mathrm{y} = \frac{{\mathrm{x}}^4 + \mathrm{C}}{{\mathrm{x}}^2}$

• $\mathrm{y} = \frac{{\mathrm{x}}^2 + \mathrm{C}}{4{\mathrm{x}}^2}$

• $\mathrm{y} = \frac{{\mathrm{x}}^4 + \mathrm{C}}{4{\mathrm{x}}^2}$

• $\mathrm{y} = \frac{{\mathrm{x}}^2}{4} + \mathrm{C}$

If a = $\hat{\mathrm{i}} + 2\hat{\mathrm{j}} + 2\hat{\mathrm{k}}, \left|\mathrm{b}\right| = 5$ and the angle between a and b is $\frac{\mathrm{\pi }}{6}$, then the area of the triangle formed by these two vectors as two sides is

• $\frac{15}{4}$

• $\frac{15}{2}$

• $\frac{15\sqrt{3}}{2}$

• 15

Let a = $\hat{\mathrm{i}} - 2\hat{\mathrm{j}} + 3\hat{\mathrm{k}}$, if b is a vector such that $\mathrm{a} . \mathrm{b} = {\left|\mathrm{b}\right|}^2$ and $\left|\mathrm{a} - \mathrm{b}\right| = \sqrt{7}, \mathrm{then} \left|\mathrm{b}\right|$, then lbl is equal to

• $\sqrt{7}$

• 7

• 21

• 14