Question

$f:C\to C,x\in C$ is defined as $f\left(x\right)=\frac{ax+b}{cx+d},bd\ne 0$ then $f$ is a constant function when

• A. $a=c$
• B. $b=d$
• C. $ad=bc$
• D. $ab=cd$

Answer 1

The correct option is C $ad=bc$

$f\left(x\right)=\frac{ax+b}{cx+d}$
For $f$ to be a constant function, $f$ should be independent of $x$.
Hence $f^{\prime }\left(x\right)$ will be $0$.
Therefore differentiating $f\left(x\right)$ with respect of x we get.
$f^{\prime }\left(x\right)=\frac{(cx+d)a-\left(c\right)(ax+b)}{(cx+d{)}^2}$
$=\frac{acx+ad-acx-bc}{(cx+d{)}^2}$
$=\frac{ad-bc}{(cx+d{)}^2}$
Now $(cx+d{)}^2$ cannot be equal to zero.
So, $ad-bc=0$
Hence, $ad=bc$