If $$\tan \left( {{\pi \over 9}} \right),x,\tan \left( {{{7\pi } \over {18}}} \right)$$ are in arithmetic progression and $$\tan \left( {{\pi \over 9}} \right),y,\tan \left( {{{5\pi } \over {18}}} \right)$$ are also in arithmetic progression, then $$|x - 2y|$$ is equal to :

Let the mean and variance of the frequency distribution

$$\matrix{ {x:} & {{x_1} = 2} & {{x_2} = 6} & {{x_3} = 8} & {{x_4} = 9} \cr {f:} & 4 & 4 & \alpha & \beta \cr } $$

be 6 and 6.8 respectively. If x₃ is changed from 8 to 7, then the mean for the new data will be :

The area of the region bounded by y $$-$$ x = 2 and x² = y is equal to :

Let y = y(x) be the solution of the differential

equation (x $$-$$ x³)dy = (y + yx² $$-$$ 3x⁴)dx, x > 2. If y(3) = 3, then y(4) is equal to :