If$$\,\,\,$$ $$y = {{a{x^2}} \over {\left( {x - a} \right)\left( {x - b} \right)\left( {x - c} \right)}} + {{bx} \over {\left( {x - b} \right)\left( {x - c} \right)}} + {c \over {x - c}} + 1$$,
prove that $${{y'} \over y} = {1 \over x}\left( {{a \over {a - x}} + {b \over {b - x}} + {c \over {c - x}}} \right)$$.

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

A curve $$C$$ has the property that if the tangent drawn at any point $$P$$ on $$C$$ meets the co-ordinate axes at $$A$$ and $$B$$, then $$P$$ is the mid-point of $$AB$$. The curve passes through the point $$(1, 1)$$. Determine the equation of the curve.

Let $$h\left( x \right) = f\left( x \right) - {\left( {f\left( x \right)} \right)^2} + {\left( {f\left( x \right)} \right)^3}$$ for every real number $$x$$. Then