Question

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

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

Answer 1

The correct option is A $2x-3$

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

It is clear that degree of $f\left(x\right)\le 1$ because

It degree of $f\left(x\right)>1$ then degree of $g(+(n\left)\right)$ will

be greater then 2

Now let us assume $f\left(n\right)=ax+b$

$\frac{g\left(f\right(x\left)\right)}{2}=\frac{(ax+b{)}^2+(ax+b)-2}{2}$

$\frac{g\left(f\right(x\left)\right)}{2}=\frac{a^2{x}^2+b^2+20bx+ox+b-2}{2}$

$\frac{g(+(x\left)\right)}{2}=\left(\frac{a^2}{2}\right)x^2+\left(\frac{a+2ab}{2}\right)x+\left(\frac{b^2+b-2}{2}\right)$

Now using coefficient companion method

$\frac{a^2}{2}=2,\frac{a+2ab}{2}=-5,\frac{b^2+b-2}{2}=2$

$a^2=4,a(1+2b)=-10,b^2+b-6=0$

$(b+3)(b-2)=0$

$a=+2,-2,$ if $a=2$ $b=-3,+2$

$1+2b=-5$

$b=-3$

if $a=-2$

$1+2b=5$

$b=2$

Thus 2 Possibilities of f(x)

$a=2,b=-3$ $a=-2,$ $b=2$

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