31.

f(x) and g(x) are differentiable in the interval [0, 1] such that f(0) = 2, g(0) = 0, f(1) = 6, g(1) = 2, then Rolle's theorem is applicable for which of the following in [0, 1] ?

• f(x) - g(x)

• f(x) - 2g(x)

• f(x) + 3g(x)

• None of the above

The positive root of equation x² - 2x - 5 = 0 lies in the interval

• {0, 1}

• (1, 2)

• (2, 3)

• (3, 4)

The equation of the plane through intersection of planes x + 2y + 3z = 4 and 2x + y - z = - 5 and perpendicular to the plane 5x + 3y + 6z = - 8 is

• 23x + 14y - 9z = - 8

• 51x + 15y - 50z = - 173

• 7x - 2y + 3z = - 81

• None of the above

If l, m, n are the direction cosines of a line, then the maximum value of lmn is

• $\frac{1}{5\sqrt{3}}$

• $\frac{1}{\sqrt{3}}$

• None of the above

If the shortest distance between the lines $\frac{\mathrm{x} - 1}{2} = \frac{\mathrm{y} - 2}{3} = \frac{\mathrm{z} - 3}{4}$ and $\frac{\mathrm{x} - 2}{3} = \frac{\mathrm{y} - 4}{4} = \frac{\mathrm{z} - 5}{5}$ is d, then [d], where [.] is the greatest integer function, is equal to

• 0

• 1

• 2

• 3

The area bounded by the curves y = x², y = - x² and y² = 4x - 3 is k, then the value of 6k is

• 2

• 3

• 0

• 4

The degree of the differential equation satisfying $\sqrt{1 - {\mathrm{x}}^2} + \sqrt{1 - {\mathrm{y}}^2} = \mathrm{a}\left(\mathrm{x} - \mathrm{y}\right)$ is

• 1

• 2

• 3

• None of the above

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

• y = C(x + a)(1 - ay)

• y = C(x + a)(a + ay)

• y = C(x - a)(1 - ay)

• None of the above

The solution of differential equation (2y - 1)dx - (2x + 3)dy = 0 will be

• $\frac{2\mathrm{x} - 1}{2\mathrm{y} + 3} = \mathrm{C}$

• $\frac{2\mathrm{y} + 1}{2\mathrm{x} - 3} = \mathrm{C}$

• $\frac{2\mathrm{x} + 3}{2\mathrm{y} - 1} = \mathrm{C}$

• $\frac{2\mathrm{x} - 1}{2\mathrm{y} - 1} = \mathrm{C}$

The solution of the differential equation $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{xlog}\left({\mathrm{x}}^2\right) + \mathrm{x}}{\sin \left(\mathrm{y}\right){\displaystyle }{\displaystyle +}{\displaystyle }{\displaystyle \mathrm{ycos}}{\displaystyle \left(\mathrm{y}\right)}}$ will be

• ysin(y) = x²log(x) + C

• ysin(y) = x² + C

• ysin(y) = x² + lo(x) + C

• ysin(y) = xlog(x) + C