Let $$f\left( x \right) = \int\limits_1^x {\sqrt {2 - {t^2}} \,dt.} $$ Then the real roots of the equation
$${x^2} - f'\left( x \right) = 0$$ are

If $$g\left( x \right) = \int_0^x {{{\cos }^4}t\,dt,} $$ then $$g\left( {x + \pi } \right)$$ equals

The area bounded by the curves $$y=f(x)$$, the $$x$$-axis and the ordinates $$x=1$$ and $$x=b$$ is $$(b-1)$$ sin $$(3b+4)$$. Then $$f(x)$$ is