Question

Let $f\left(x\right)=ax+b,a<0$, then $f^{-1}\left(x\right)=f\left(x\right)$, $\forall x$ if and only if

• A. $a=-1,b\in R$
• B. $a=-1,b=4$
• C. $a=3,b\in R$
• D. None of these

Answer 1

The correct option is A $a=-1,b\in R$

Let $f\left(x\right)=ax+b,a<0$

$y=ax+b$

$x=\frac{y-b}{a}$

then$f^{-1}\left(x\right)=f\left(x\right)\forall x$ if and only if

$\frac{x-b}{a}=ax+bx-b=a^{2}x+ab$

$(1-a^2)x=b(1+a)$

$(1+a)(1-a)x=b(1+a)$

From above equation, we can see that if $a=-1,b\in R$ and the equation will be satisfied for all $x\in R$.

Hence, option $A$ is correct.