Let f be a real valued continuous function on [0, 1] and $$f(x) = x + \int\limits_0^1 {(x - t)f(t)dt} $$.

Then, which of the following points (x, y) lies on the curve y = f(x) ?

If $$\int\limits_0^2 {\left( {\sqrt {2x} - \sqrt {2x - {x^2}} } \right)dx = \int\limits_0^1 {\left( {1 - \sqrt {1 - {y^2}} - {{{y^2}} \over 2}} \right)dy + \int\limits_1^2 {\left( {2 - {{{y^2}} \over 2}} \right)dy + I} } } $$, then I equals

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 :

$$\int_0^5 {\cos \left( {\pi \left( {x - \left[ {{x \over 2}} \right]} \right)} \right)dx} $$,

where [t] denotes greatest integer less than or equal to t, is equal to: