Suppose $$f(x)$$ is a function satisfying the following conditions
(a) $$f(0)=2,f(1)=1$$,
(b) $$f$$has a minimum value at $$x=5/2$$, and
(c) for all $$x$$, $$$f'\left( x \right) = \matrix{ {2ax} & {2ax - 1} & {2ax + b + 1} \cr b & {b + 1} & { - 1} \cr {2\left( {ax + b} \right)} & {2ax + 2b + 1} & {2ax + b} \cr } $$$
where $$a,b$$ are some constants. Determine the constants $$a, b$$ and the function $$f(x)$$.

If $$\int_0^x {f\left( t \right)dt = x + \int_x^1 {t\,\,f\left( t \right)\,\,dt,} } $$ then the value of $$f(1)$$ is

Let $$f\left( x \right) = x - \left[ x \right],$$ for every real number $$x$$, where $$\left[ x \right]$$ is the integral part of $$x$$. Then $$\int_{ - 1}^1 {f\left( x \right)\,dx} $$ is

Prove that $$\int_0^1 {{{\tan }^{ - 1}}} \,\left( {{1 \over {1 - x + {x^2}}}} \right)dx = 2\int_0^1 {{{\tan }^{ - 1}}} \,x\,dx.$$
Hence or otherwise, evaluate the integral
$$\int_0^1 {{{\tan }^{ - 1}}\left( {1 - x + {x^2}} \right)dx.} $$