Let the solution curve of the differential equation

$$x{{dy} \over {dx}} - y = \sqrt {{y^2} + 16{x^2}} $$, $$y(1) = 3$$ be $$y = y(x)$$. Then y(2) is equal to:

Let $$f:R \to R$$ be a function defined by :

$$f(x) = \left\{ {\matrix{ {\max \,\{ {t^3} - 3t\} \,t \le x} & ; & {x \le 2} \cr {{x^2} + 2x - 6} & ; & {2 < x < 3} \cr {[x - 3] + 9} & ; & {3 \le x \le 5} \cr {2x + 1} & ; & {x > 5} \cr } } \right.$$

where [t] is the greatest integer less than or equal to t. Let m be the number of points where f is not differentiable and $$I = \int\limits_{ - 2}^2 {f(x)\,dx} $$. Then the ordered pair (m, I) is equal to :

Let $$\overrightarrow a = \alpha \widehat i + 3\widehat j - \widehat k$$, $$\overrightarrow b = 3\widehat i - \beta \widehat j + 4\widehat k$$ and $$\overrightarrow c = \widehat i + 2\widehat j - 2\widehat k$$ where $$\alpha ,\,\beta \in R$$, be three vectors. If the projection of $$\overrightarrow a $$ on $$\overrightarrow c $$ is $${{10} \over 3}$$ and $$\overrightarrow b \times \overrightarrow c = - 6\widehat i + 10\widehat j + 7\widehat k$$, then the value of $$\alpha + \beta $$ is equal to :

The area enclosed by y² = 8x and y = $$\sqrt2$$ x that lies outside the triangle formed by y = $$\sqrt2$$ x, x = 1, y = 2$$\sqrt2$$, is equal to: