If $$g\left( x \right) = {x^2} + x - 2$$    and $$\frac{1}{2}\left( {gof} \right)\left( x \right) = 2{x^2} - 5x + 2,$$      then $$f\left( x \right)$$  is equal to :

Solution :
$$\eqalign{ & g\left( x \right) = {x^2} + x - 2{\text{ and }}\frac{1}{2}\left( {gof} \right)\left( x \right) = 2{x^2} - 5x + 2 \cr & \Rightarrow g\left( {f\left( x \right)} \right) = 4{x^2} - 10x + 4 \cr & \Rightarrow {\left( {f\left( x \right)} \right)^2} + f\left( x \right) - 2 = 4{x^2} - 10x + 4 \cr & \Rightarrow {\left( {f\left( x \right)} \right)^2} + f\left( x \right) - \left( {4{x^2} - 10x + 6} \right) = 0 \cr & \Rightarrow f\left( x \right) = \frac{{ - 1 \pm \sqrt {16{x^2} - 40x + 25} }}{2} \cr & \Rightarrow f\left( x \right) = \frac{{ - 1 \pm \left( {4x - 5} \right)}}{2} \cr & \Rightarrow f\left( x \right) = \frac{{4x - 6}}{2}{\text{ }}\,{\text{or}}\,{\text{ }}\frac{{ - 4x + 4}}{2} \cr & \Rightarrow f\left( x \right) = 2x - 3{\text{ or }} - 2x + 2 \cr & {\text{Hence, }}f\left( x \right) = 2x - 3 \cr} $$