The differential equation of y = ae^bx (a and b are parameters) is

• yy₁ = y₂²

• yy₂ = y₁²

• yy₁² = y₂

• yy₂² = y₁

Let $\mathrm{f}\left(\mathrm{x}\right) = {\tan }^{-1}\left(\mathrm{x}\right)$. Then, f'(x) + f''(x) is 0, when x is equal to

• 0

• 1

• i

• - i

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

• 1/4

• 1/2

• - 1/4

• - 1/2

If $\frac{d\mathrm{y}}{d\mathrm{x}} + \sqrt{\frac{1 - {\mathrm{y}}^2}{1 - {\mathrm{x}}^2}} = 0$ prove that, $\mathrm{x}\sqrt{1 - {\mathrm{y}}^2} + \mathrm{y}\sqrt{1 - {\mathrm{x}}^2} = \mathrm{A}$ where A is constant.

If f(a) = 2, f'(a) = 1, g(a) = - 1 and g'(a) = 2, find the value of $\underset{\mathrm{x}\to \mathrm{a}}{\lim }\frac{\mathrm{g}\left(\mathrm{a}\right)\mathrm{f}\left(\mathrm{a}\right) - \mathrm{g}\left(\mathrm{a}\right)\mathrm{f}\left(\mathrm{x}\right)}{\left(\mathrm{x} - \mathrm{a}\right)}$

386.

If the displacement, velocity and acceleration of a particle at time t be x, v and f respectively, then which one is true ?

• $\mathrm{f} = {\mathrm{v}}^3\frac{d^2\mathrm{t}}{d{\mathrm{x}}^2}$

• $\mathrm{f} = - {\mathrm{v}}^3\frac{d^2\mathrm{t}}{d{\mathrm{x}}^2}$

• $\mathrm{f} = {\mathrm{v}}^2\frac{d^2\mathrm{t}}{d{\mathrm{x}}^2}$

• $\mathrm{f} = - {\mathrm{v}}^2\frac{d^2\mathrm{t}}{d{\mathrm{x}}^2}$

The displacement x of a particle at time t is given by x = At² + Bt + C where A, B, C are constants and v is velocity of a particle, then the value of 4Ax - v² is

• 4AC + B²

• 4AC - B²

• 2AC - B²

• 2AC + B²

The displacement of a particle at time t is x,  where x = t⁴ - kt³. If the velocity of the particle at time t = 2 is minimum, then 

• k = 4

• k = - 4

• k = 8

• k = - 8

The general solution of the differential equation

$100\frac{d^2\mathrm{y}}{d{\mathrm{x}}^2} - 20\frac{d\mathrm{y}}{d\mathrm{x}} + \mathrm{y} = 0$ is

• y = (c₁ + c₂x)e^x

• y = (c₁ + c₂x)e^- x

• $\mathrm{y} = \left({\mathrm{c}}_1 + {\mathrm{c}}_2\mathrm{x}\right){\mathrm{e}}^{\frac{\mathrm{x}}{10}}$

• y = c₁e^x + c₂e^{- x}

If y'' - 3y' + 2y = 0 where y(0) = 1, y'(0) = 0, then the value of y at x = log(2) is

• 1

• - 1

• 2

• 0