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

Answer: 7

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