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

$2x-3$

The correct answer is Option (1) → $2x-3$

We have,

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

$⇒g(f(x)) = 4x^2-10x+4$

$⇒(f(x))^2+f (x) − 2 = 4 x^2 - 10x+4$

$⇒(f(x))^2+f (x) −(4x^2-10x+6) = 0$

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

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

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

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