If f(x) = $\mathrm{Asin}\left(\frac{\mathrm{\pi x}}{2}\right) + \mathrm{B}, \mathrm{f}\text{'}\left(\frac{1}{2}\right) = \sqrt{2}$ and ${\int }_0^1\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx} = \frac{2\mathrm{A}}{\mathrm{\pi }}$, then A and B are

• $\frac{\mathrm{\pi }}{2}, \frac{\mathrm{\pi }}{2}$

• $\frac{2}{\mathrm{\pi }}, \frac{3}{\mathrm{\pi }}$

• $0, - 4\mathrm{\pi }$

• $\frac{4}{\mathrm{\pi }}, 0$

Let g(x) = ${\int }_0^{\mathrm{x}}\mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}$, where f is such that $\frac{1}{2} \le \mathrm{f}\left(\mathrm{x}\right) \le 1$ for t $\in$ [0, 1] and $0 \le \mathrm{f}\left(\mathrm{t}\right) \le \frac{1}{2}$ for t $\in$ [1, 2]. Then, g(2) satisfies the inequality

• $- \frac{3}{2} \le \mathrm{g}\left(2\right) < \frac{1}{2}$

• $0 \le \mathrm{g}\left(2\right) < 2$

• $\frac{1}{2} \le \mathrm{g}\left(2\right) < \frac{3}{2}$

• 2 < g(2) < 4

For which of the followmg value of m, is the area of the region bounded by the curve y = x - x² and the line y = mx equals to $\frac{9}{2}$ ?

• - 4

• - 2

• 2

• - 4

Solution of $2\mathrm{ysin}\left(\mathrm{x}\right)\frac{\mathrm{dy}}{\mathrm{dx}} = 2\sin \left(\mathrm{x}\right)\cos \left(\mathrm{x}\right) - {\mathrm{y}}^2\cos \left(\mathrm{x}\right), \mathrm{x} = \frac{\mathrm{\pi }}{2}$, y = 1 is given by

• y² = sin(x)

• y = sin²(x)

• y² = cos(x) + 1

• None of these

Solution of ${\mathrm{x}}^2\frac{\mathrm{dy}}{\mathrm{dx}} - \mathrm{xy} = 1 +\hspace{0.17em}\cos \left(\frac{\mathrm{y}}{\mathrm{x}}\right)$ is

• $\tan \left(\frac{\mathrm{y}}{2\mathrm{x}}\right) = \mathrm{C} - \frac{1}{2{\mathrm{x}}^2}$

• $\tan \left(\frac{\mathrm{y}}{\mathrm{x}}\right) = \mathrm{C} + \frac{1}{\mathrm{x}}$

• $\cos \left(\frac{\mathrm{y}}{\mathrm{x}}\right) = 1 + \frac{\mathrm{C}}{\mathrm{x}}$

• ${\mathrm{x}}^2 = \left(\mathrm{C} + {\mathrm{x}}^2\right)\tan \left(\raisebox{1ex}{$\mathrm{y}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{x}$}\right.\right)$

16.

If a and b are unit vectors and $\mathrm{\theta }$ is the angle between them, then la + bl < 1, if

• $\mathrm{\theta } = \frac{\mathrm{\pi }}{2}$

• $\mathrm{\theta } < \frac{\mathrm{\pi }}{3}$

• $\mathrm{\pi } \ge \mathrm{\theta } > \frac{2\mathrm{\pi }}{3}$

• $\frac{\mathrm{\pi }}{3} < \mathrm{\theta } < \frac{2\mathrm{\pi }}{3}$

The points, whose position vectors are 60i + 3j, 40i - 8j and ai - 52j collinear, if

• a = 40

• a = - 40

• a = 20

• a = - 20

In a $∆$ABC, AB = ri + j, AC = si - j if the area of triangle is of unit magnitude, then

• $\left|\mathrm{r} - \mathrm{s}\right| = 2$

• $\left|\mathrm{r} + \mathrm{s}\right| = 1$

• $\left|\mathrm{r} + \mathrm{s}\right| = 2$

• $\left|\mathrm{r} - \mathrm{s}\right| = 1$

If a = i - j + k, a b = 0, a x b = c, where c = - 2i - j + k, then b 1s equal to

• (1, 0, - 1)

• (0, 1, 1)

• (- 1, - 1, 0)

• (- 1, 0, 1)

If P(A) = 65, P(B) = 80, then P(A $\cap$ B) lies in the interval

• [30, 80]

• [35, 75]

• [4, 70]

• [0.45, 0.65]