• $\log \left(\frac{4}{3}\right)$

• $\log \left(\frac{8}{3}\right)$

• $\log \left(\frac{1}{4}\right)$

${\int }_{\raisebox{1ex}{$- \mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\sin }^4\left(\mathrm{x}\right){\cos }^6\left(\mathrm{x}\right)\mathrm{dx} \mathrm{is} \mathrm{equal} \mathrm{to}$

• $\frac{3\mathrm{\pi }}{128}$

• $\frac{3\mathrm{\pi }}{256}$

• $\frac{3\mathrm{\pi }}{572}$

• $\frac{3\mathrm{\pi }}{64}$

The approximate value of ${\int }_2^9 {\mathrm{x}}^2\mathrm{dx}$ by using trapezoidal rule with 4 equal intervals, is

• 248

• 242.5

• 242.8

• 243

Order of the differential equation of the family of all concentric circles centered at (h, k) is

• 1

• 2

• 3

• 5

The solution of $\frac{\mathrm{dy}}{\mathrm{dx}} + \frac{1}{3}\mathrm{y} = 1 \mathrm{is}$

• $\mathrm{y} = 3 + {\mathrm{ce}}^{\raisebox{1ex}{$\mathrm{x}$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

• $\mathrm{y} = 3 + {\mathrm{ce}}^{\raisebox{1ex}{$ - \mathrm{x}$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

• $3\mathrm{y} = \mathrm{c} + {\mathrm{e}}^{\raisebox{1ex}{$ \mathrm{x}$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

• ${\mathrm{y}}^2 + \mathrm{x} + {\mathrm{x}}^2 + 2 = {\mathrm{ce}}^{2\mathrm{x}}$

$\mathrm{y} + {\mathrm{x}}^2 =\frac{\mathrm{dy}}{\mathrm{dx}}$ has the solution

• $\mathrm{y} + {\mathrm{x}}^2 + 2\mathrm{x} + 2 = {\mathrm{ce}}^{\mathrm{x}}$

• $\mathrm{y} + \mathrm{x} + 2{\mathrm{x}}^2 + 2 = {\mathrm{ce}}^{\mathrm{x}}$

• ${\mathrm{y}}^2 + \mathrm{x} + {\mathrm{x}}^2 + 2 = {\mathrm{ce}}^{2\mathrm{x}}$

• $\mathrm{y} + \mathrm{x} + {\mathrm{x}}^2 + 2 = {\mathrm{ce}}^{2\mathrm{x}}$

$\mathrm{The} \mathrm{solution} \mathrm{of} \frac{\mathrm{dy}}{\mathrm{dx}} = {\left(\frac{\mathrm{x}}{\mathrm{y}}\right)}^{ - \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} \mathrm{is}$

• ${\mathrm{x}}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} + {\mathrm{y}}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} = \mathrm{c}$

• ${\mathrm{y}}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} - {\mathrm{x}}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} = \mathrm{c}$

• ${\mathrm{x}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} + {\mathrm{y}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} = \mathrm{c}$

• ${\mathrm{y}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} - {\mathrm{x}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} = \mathrm{c}$

If the pair of straight lines xy - x - y + 1 = 0 and the line ax + 2y - 3a = 0 are concurrent, then a is equal to

• 0

• 1

• - 1

• 3

99.

The radius of the larger circle lying in the first quadrant and touching the line 4x + 3y - 12 =0 and the co-ordinate axes,is

• 5

• 6

• 7

• 8

$$\begin{gathered} \mathrm{The} \mathrm{quadratic} \mathrm{equation} \mathrm{whose} \mathrm{roots} \mathrm{are} \mathrm{l} \mathrm{and} \mathrm{m},\mathrm{where} \\ \mathrm{l} = \underset{\mathrm{\theta } \to 0}{\lim } \left(\frac{3\sin \left(\mathrm{\theta }\right) - 4{\sin }^2\left(\mathrm{\theta }\right)}{\mathrm{\theta }}\right), \\ \mathrm{m} = \underset{\mathrm{\theta } \to 0}{\lim } \left(\frac{2\tan \left(\mathrm{\theta }\right)}{\mathrm{\theta } \left(1 - {\tan }^2\left(\mathrm{\theta }\right) \right)}\right) \mathrm{is} \end{gathered}$$

• ${\mathrm{x}}^2 + 5\mathrm{x} + 6$

• ${\mathrm{x}}^2 - 5\mathrm{x} + 6$

• ${\mathrm{x}}^2 - 5\mathrm{x} - 6$

• ${\mathrm{x}}^2 + 5\mathrm{x} - 6$