If a < b < c < d, then the roots of the equation (x - a) (x - c) + 2 ( x - b) (x - d) = 0 are real and distinct.

For real $$x$$, the function $$\,{{\left( {x - a} \right)\left( {x - b} \right)} \over {x - c}}$$ will assume all real values provided

If $$\,{a^2} + {b^2} + {c^2} = 1$$, then ab + bc + ca lies in the interval

The equation $$x - {2 \over {x - 1}} = 1 - {2 \over {x - 1}}$$ has