If $g(x)=x^2+x-2$ and $\frac{1}{2}(g \circ f)(x)=2 x^2-5 x+2$, then one such function $f(x)=$

Given, $$ g(x)=x^2+x-2 $$ and $$ \begin{aligned} & \frac{1}{2}(g \circ f)(x)=2 x^2-5 x+2 \\ & \Rightarrow \quad g(f(x))=4 x^2-10 x+4 \\ & \Rightarrow(f(x))^2+(f(x))-2=4 x^2-10 x+4 \\ & \end{aligned} $$ Now, it is necessary that $f(x)$ should be linear polynomial expression, so let $f(x)=a x+b$, then $$ \begin{array}{r} (a x+b)^2+(a x+b)-2=4 x^2-10 x+4 \\ \Rightarrow \quad a^2 x^2+(2 a b+a) x+\left(b^2+b-2\right) \\ =4 x^2-10 x+4 \end{array} $$ On comparing the coefficient of different kind of terms, we are getting $$ \begin{aligned} & a^2=4 \text {, } \\ & \Rightarrow \quad 2 a b+a=-10 \\ & \text { and } \quad b^2+b-2=4 \\ & \end{aligned} $$ So, then $$ \begin{aligned} & a= \pm 2, \\ & b=\left\{\begin{aligned} -3 ; & \text { if } a=2 \\ 2 ; & \text { if } a-2^{\prime} \end{aligned}\right. \end{aligned} $$ and these value satisfy the all above relations, so $$ f(x)=2 x-3 \text { or }-2 x+2 $$