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

$2x-3$

We have,

$\frac{1}{2}gof (x) = 2x^2 -5x+2$

$⇒g(f(x))=4x^2-10x+4$

$⇒(f(x))^2+f(x)-2=4x^2-10x+4$

$⇒(f(x))^2+f(x)-(4x^2-10x+6)=0$

$⇒f(x)=\frac{-1±\sqrt{1+4(4x^2-10x+6)}}{2}$

$⇒f(x)=\frac{-1±\sqrt{16x^2-40x+25}}{2}$

$⇒f(x)=\frac{-1±(4x-5)}{2}=2x-3,-2x+2$

Hence, $f(x)= 2x-3$