${\int }_0^{10\mathrm{\pi }}\left|\sin \left(\mathrm{x}\right)\right|\mathrm{dx}$ is equal to

• 20

• 8

• 10

• 18

The general solution of the differential equation (x + y)dx + xdy = 0 is

• x² + y² = c

• 2x² - y²

• x² + 2xy = c

• y² + 2xy = c

The order and degree of the differential ${\left(1 + 3\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} = 4\frac{{\mathrm{d}}^3\mathrm{y}}{{\mathrm{dx}}^3}$ are

• 1, 2/3

• 3, 1

• 3, 3

• 1, 2

The differential equation of all straight lines passing through the point (1, - 1)is

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

• $\mathrm{y} = \left(\mathrm{x} + 1\right)\frac{\mathrm{dy}}{\mathrm{dx}} - 1$

• $\mathrm{y} = \left(\mathrm{x} - 1\right)\frac{\mathrm{dy}}{\mathrm{dx}} + 1$

• $\mathrm{y} = \left(\mathrm{x} - 1\right)\frac{\mathrm{dy}}{\mathrm{dx}} - 1$

The solution of the differential equation $\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2} = {\mathrm{e}}^{- 2\mathrm{x}}$ is

• $\mathrm{y} = \frac{{\mathrm{e}}^{- 2\mathrm{x}}}{4}$

• $\mathrm{y} = \frac{{\mathrm{e}}^{- 2\mathrm{x}}}{4} + \mathrm{cx} + \mathrm{d}$

• $\mathrm{y} = \frac{{\mathrm{e}}^{- 2\mathrm{x}}}{4} + {\mathrm{cx}}^2 + \mathrm{d}$

• $\mathrm{y} = \frac{{\mathrm{e}}^{- 2\mathrm{x}}}{4} + \mathrm{c} + \mathrm{d}$

The solution of the differential equation $\frac{\mathrm{dy}}{\mathrm{dx}} + {\sin }^2\left(\mathrm{y}\right) = 0$ is

• $\mathrm{x} = \cot \left(\mathrm{y}\right) + \mathrm{c}$

• $\mathrm{y} = \cot \left(\mathrm{x}\right) + \mathrm{c}$

• $\mathrm{x} = 2\csc \left(\mathrm{y}\right)\cot \left(\mathrm{y}\right) + \mathrm{c}$

• $\mathrm{y} = 2\sin \left(\mathrm{y}\right)\cos \left(\mathrm{y}\right) + \mathrm{c}$

47.

There are four letters and four addressed envelopes. The chance that all letters are not despatched in the right envelope is

• 19/24

• 21/23

• 23/24

• None of these

If I = ${\int }_{{\mathrm{x}}_0}^{{\mathrm{x}}_0 + \mathrm{nh}}\mathrm{ydx}$, then by Trapezoidal rule I is equal to

• $\mathrm{h}\left[\left({\mathrm{y}}_0 + {\mathrm{y}}_{\mathrm{n}}\right) + 2\left({\mathrm{y}}_1 + {\mathrm{y}}_2 + ... + {\mathrm{y}}_{\mathrm{n} - 1}\right)\right]$

• $\mathrm{h}\left[\frac{1}{2}\left({\mathrm{y}}_0 + {\mathrm{y}}_{\mathrm{n}}\right) + 2\left({\mathrm{y}}_1 + {\mathrm{y}}_2 + ... + {\mathrm{y}}_{\mathrm{n} - 1}\right)\right]$

• $\frac{\mathrm{h}}{2}\left[\left({\mathrm{y}}_0 + {\mathrm{y}}_{\mathrm{n}}\right) + 2\left({\mathrm{y}}_1 + {\mathrm{y}}_2 + ... + {\mathrm{y}}_{\mathrm{n} - 1}\right)\right]$

• $\mathrm{h}\left[\left({\mathrm{y}}_0 + {\mathrm{y}}_{\mathrm{n}}\right) + 2\left({\mathrm{y}}_1 + {\mathrm{y}}_2 + ... + {\mathrm{y}}_{\mathrm{n} - 1}\right)\right]$

By graphical method, the solution of linear programming problem maxirmze z = 3x₁ + 5x₂ subject to 3x₁ + 2x₂ $\le 18, {\mathrm{x}}_1 \le 4, {\mathrm{x}}_2 \le 6, {\mathrm{x}}_1 \ge 0, {\mathrm{x}}_2 \ge 0$

• x₁ = 2, x₂ = 0, z = 6

• x₁ = 2, x₂ = 6, z = 36

• x₁ = 4, x₂ = 3, z = 36

• x₁ = 4, x₂ = 6, z = 42