Question

If $f:B\to A$ is defined by $f\left(x\right)=\frac{3x+4}{5x-7}$ and $g:A\to B$ is defined by $g\left(x\right)=\frac{7x+4}{5x-3}$, where $A=R-\left\{\frac{3}{5}\right\}$ and $B=R-\left\{\frac{7}{5}\right\}$ and $I_A$ is an identity function on A and $I_B$ is identity function on B, then

• A. $fog=I_A$ and $gof=I_A$
• B. $fog=I_A$ and $gof=I_B$
• C. $fog=I_B$ and $gof=I_B$
• D. $fog=I_B$ and $gof=I_A$

Answer 1

Answer: (B)

We have, $gof\left(x\right)=g\left(\frac{3x+4}{5x-7}\right)=\frac{7\left(\frac{\left(3x+4\right)}{\left(5x-7\right)}\right)+4}{5\left(\frac{\left(3x+4\right)}{\left(5x-7\right)}\right)-3}$
$=\frac{21x+28+20x-28}{15x+20-15x+21}=\frac{41x}{41}=x$
Similarly, $fog\left(x\right)=f\left(\frac{7x+4}{5x-3}\right)$
$=\frac{3\left(\frac{\left(7x+4\right)}{\left(5x-3\right)}\right)+4}{5\left(\frac{\left(7x+4\right)}{\left(5x-3\right)}\right)-7}$
$=\frac{21x+12+20x-12}{35x+20-35x+21}=\frac{41x}{41}=x$
Thus, $gof\left(x\right)=x,\forall x\in B$ and $fog\left(x\right)=x,\forall x\in A$, which implies that $gof=I_B$ and $fog=I_A.$