Question

The number of real values of $m$ for which $A \cup B$ has exactly three distinct elements. Given that
$A =\left\{ x : x ^2+( m -1) x -2( m +1)=0, x \in R \right\} $
$B =\left\{ x :( m -1) x ^2+ mx +1=0, x \in R \right\}$

Answer 1

$A=\{x:(x-2)(x+m+1)=0, x \in R\} $
$B=\left\{x:(m-1) x^2+m x+1=0, x \in R\right\}$
$\text { Case-I : When } m -1=0 m =1 $
$A=\{2,-2\} \text { and } B=\{-1\} $
$\therefore A \cup B =\{2,-2,-1\} \Rightarrow \text { exactly } 3 \text { elements } $
$\therefore m =1 $
$\text { Case-II : When } m \neq 1 $
$A=\{2,-m-1\}, B=\left\{-1, \frac{-1}{m-1}\right\}$
$2=- m -1 \Rightarrow m =-3$
$A =\{2\}, B \left\{-1, \frac{1}{4}\right\} $
$\therefore m =-3 $
$2=\frac{-1}{m-1} \Rightarrow m =\frac{1}{2} $
$A =\left\{2,-\frac{3}{2}\right\}, B =\{-1,2\} $
$\therefore m =\frac{1}{2} $
$- m -1=\frac{-1}{ m -1} \Rightarrow m ^2-1=1 $
$\therefore m = \pm \sqrt{2}$
$\therefore - m -1=-1 \Rightarrow m =0 $
$-1=\frac{-1}{m-1} \Rightarrow m -1=1 $
$\therefore m =2$
Number of values of $m=7$.