The rate of charge of the surface area of a sphere of radius r, when the radius is inacreasing at the rate of 2cm/s is proportional to

• r

• r²

The vector b = 3j + 4k is to be written as the sum of a vector b₁ parallel to a = i + j and a vector b₂ perpendicular to a. Then, b₁ is equal to

• $\frac{3}{2}$(i + j)

• $\frac{2}{3}$(i + J)

• $\frac{1}{2}$(i + j)

• $\frac{1}{3}$(i + j)

$\int \left(1 + \mathrm{x} - {\mathrm{x}}^{- 1}\right){\mathrm{e}}^{\mathrm{x} + {\mathrm{x}}^{- 1}}\mathrm{dx}$ is equal to

• $\left(\mathrm{x} + 1\right){\mathrm{e}}^{\mathrm{x} + {\mathrm{x}}^{- 1}}$ + C

• $\left(\mathrm{x} - 1\right){\mathrm{e}}^{\mathrm{x} + {\mathrm{x}}^{- 1}} + \mathrm{C}$

• ${\mathrm{xe}}^{\mathrm{x} + {\mathrm{x}}^{- 1}} + \mathrm{C}$

• ${\mathrm{xe}}^{\mathrm{x} + {\mathrm{x}}^{- 1}} \mathrm{x} + \mathrm{C}$

If f(x) = x - [x], for every real number x, where [x] is the integral part of x. Then ${\int }_{- 1}^1\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}$ is equal to

• 1

• 2

• 0

• $\frac{1}{2}$

35.

The value of integral

${\int }_{- 1}^1{\left[{\left(\frac{\mathrm{x} + 1}{\mathrm{x} - 1}\right)}^2 + {\left(\frac{\mathrm{x} + 1}{\mathrm{x} - 1}\right)}^2 - 2\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\mathrm{dx}$ is

• $\log \left(\frac{4}{3}\right)$

• $4\log \left(\frac{3}{4}\right)$

• $4\log \left(\frac{4}{3}\right)$

• $\log \left(\frac{3}{4}\right)$

For which of the following values of m, the area of the region bounded by the curve y = x - x² and the line y = mx equals 9/2

• - 4

• - 2

• 2

• 4

The solution of the differential equation

$\left(1 + {\mathrm{y}}^2\right) + \left(\mathrm{x} - {\mathrm{e}}^{{\tan }^{-1}\left(\mathrm{y}\right)}\right)\frac{d\mathrm{y}}{d\mathrm{x}} = 0$, is

• $\left(\mathrm{x} - 2\right) = {\mathrm{Ke}}^{{\tan }^{-1}\left(\mathrm{y}\right)}$

• $2{\mathrm{xe}}^{{\tan }^{-1}\left(\mathrm{y}\right)} = {\mathrm{e}}^{2{\tan }^{-1}\left(\mathrm{y}\right)} + \mathrm{K}$

• ${\mathrm{xe}}^{{\tan }^{-1}\left(\mathrm{y}\right)} = {\tan }^{-1}\left(\mathrm{y}\right) +\hspace{0.17em}\mathrm{K}$

• ${\mathrm{xe}}^{2{\tan }^{-1}\left(\mathrm{y}\right)} = {\mathrm{e}}^{{\tan }^{-1}\left(\mathrm{y}\right)} + \mathrm{K}$

If the gradient of the tangent at any point (x, y) of acurve which passes through the point $\left(1, \frac{\mathrm{\pi }}{4}\right)$ is $\left\{\frac{\mathrm{y}}{\mathrm{x}} - {\sin }^2\left(\frac{{\displaystyle \mathrm{y}}}{{\displaystyle \mathrm{x}}}\right)\right\}$, then the equation of the curve is

• $\mathrm{y} = {\cot }^{-1}\left({\log }_{\mathrm{e}}\left(\mathrm{x}\right)\right)$

• $\mathrm{y} = {\cot }^{-1}\left({\log }_{\mathrm{e}}\left(\frac{\mathrm{x}}{\mathrm{e}}\right)\right)$

• $\mathrm{y} = {\mathrm{xcot}}^{-1}\left({\log }_{\mathrm{e}}\left(\mathrm{ex}\right)\right)$

• $\mathrm{y} = {\cot }^{-1}\left({\log }_{\mathrm{e}}\left(\frac{\mathrm{e}}{\mathrm{x}}\right)\right)$

The solution of the differential equation $\frac{d\mathrm{y}}{d\mathrm{x}} = \frac{\mathrm{yf}\text{'}\left(\mathrm{x}\right) - {\mathrm{y}}^2}{\mathrm{f}\left(\mathrm{x}\right)}$ is

• f(x) = y + C

• f(x) = y(x + C)

• f(x) = x + C

• None of the above

If two events A and B. If odds against A are 2 : 1 and those infavour of A $\cup$ B areas 3 : 1, then

• $\frac{1}{2} \le \mathrm{P}\left(\mathrm{B}\right) \le \frac{3}{4}$

• $\frac{5}{12} \le \mathrm{P}\left(\mathrm{B}\right) \le \frac{3}{4}$

• $\frac{1}{4} \le \mathrm{P}\left(\mathrm{B}\right) \le \frac{3}{5}$

• None of these