If $\mathrm{y} = \frac{\mathrm{A}}{\mathrm{x}} + {\mathrm{Bx}}^2,$ then ${\mathrm{x}}^2\frac{d^2\mathrm{y}}{d{\mathrm{x}}^2}$ is equal to

• 2y

• y²

• y³

• y⁴

The area enclosed between y² = x and y = x is

• $\frac{2}{3}$ sq unit

• $\frac{1}{2}$ unit

• $\frac{1}{3}$ unit

• $\frac{1}{6}$

The area bounded by y² = 4x and x = 4y is

• $\frac{20}{3}$ sq units

• $\frac{16}{3}$ sq units

• $\frac{14}{3}$

• $\frac{10}{3}$

The solution of $\frac{d\mathrm{y}}{d\mathrm{x}} = \frac{\mathrm{y}}{\mathrm{x}} + \tan \left(\frac{{\displaystyle \mathrm{y}}}{{\displaystyle \mathrm{x}}}\right)$ is

• x = c$\sin \left(\raisebox{1ex}{$\mathrm{y}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{x}$}\right.\right)$

• x = c$\sin \left(\mathrm{xy}\right)$

• y = c$\sin \left(\raisebox{1ex}{$\mathrm{y}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{x}$}\right.\right)$

• xy = c$\sin \left(\raisebox{1ex}{$\mathrm{x}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{y}$}\right.\right)$

Integrating Factor (IF) of the differential equation $\frac{d\mathrm{y}}{d\mathrm{x}} - \frac{3{\mathrm{x}}^2\mathrm{y}}{1 + {\mathrm{x}}^3} = \frac{{\displaystyle {\sin }^2\left(\mathrm{x}\right)}}{1 + \mathrm{x}}$

• ${\mathrm{e}}^{1 + {\mathrm{x}}^3}$

• $\log \left(1 + {\mathrm{x}}^3\right)$

• 1 + x³

• $\frac{1}{1 + {\mathrm{x}}^3}$

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

• yy₁ = y₂²

• yy₂ = y₁²

• yy₁² = y₂

• yy₂² = y₁

The value of ${\int }_0^{\mathrm{\pi }}{\sin }^{50}\left(\mathrm{x}\right){\cos }^{49}\left(\mathrm{x}\right)d\mathrm{x}$ is

• 0

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

• $\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

• 1

$\int 2^{\mathrm{x}}\left(\mathrm{f}\text{'}\left(\mathrm{x}\right) + \mathrm{f}\left(\mathrm{x}\right)\log \left(2\right)\right)\mathrm{dx}$ is

• 2^xf'(x) + C

• 2^xf(x) + C

• 2^x(log(2))f(x) + C

• log(2)f(x) + C

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

80.

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