Question

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

• A. $2x-3$
• B. $2x+3$
• C. $2x^2+3x+1$
• D. $2x^2-3x-1$

Answer 1

Answer: (A)

$2x-3$

We have,

$g\left(x\right)=x^2+x-2$ $...........\left(1\right)$

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

$\left(gof\right)\left(x\right)=4x^2-10x+4$ $.............\left(2\right)$

Therefore,

$g\left(f\right)=f^2+f-2=4x^2-10x+4$

$f^2+f-(4x^2-10x-6)=0$

$f\left(x\right)=\frac{-1\pm \sqrt{1-4\times 1\times -(4x^2-10x-6)}}{2}$

$f\left(x\right)=\frac{-1\pm \sqrt{(4x-5{)}^2}}{2}$

$f\left(x\right)=\frac{-1\pm (4x-5)}{2}$

On solving, we get

$f\left(x\right)=2x-3$ or $f\left(x\right)=-2x+2$

Hence, this is the answer.