Question

If $f\left(x\right)=\frac{ax+b}{cx+d}$ and $\left(fof\right)x=x,$ then d=?

• A. $a$
• B. $-a$
• C. $b$
• D. $-b$

Answer 1

Answer: (B)

$-a$

$\left(fof\right)x=x\Rightarrow f\left[f\left(x\right)\right]=x$ or $f\left[\frac{ax+b}{cx+d}\right]=x$ or $\frac{a\left[\frac{ax+b}{cx+d}\right]+b}{c\left[\frac{ax+b}{cx+d}\right]+d}=x$ $\therefore \frac{x(a^2+bc)+b(a+d)}{cx(a+d)+(bc+d^2)}=x$ Clearly if $a+d=0$ or $d=-a$ then in that case $L.H.S.=\frac{x(a^2+bc)+0}{0+(bc+a^2)}=x$ $\because d=-a$