# Let A= { x | x^2 + (m-1)x-2(m+1)=0, x in R} B= { x | ((m-1)x^2)+ m x +1=0, x in R} Number of values of m such that A uu B has exactly 3 distinct elements, is? A) 4 B) 5 C) 6 D) 7

Jul 16, 2017

Consider the set $A$:

$A = \left\{x \in \mathbb{R} | {x}^{2} + \left(m - 1\right) x - 2 \left(m + 1\right) = 0\right\}$

We know that $x \in \mathbb{R} \implies {\Delta}_{A} \ge 0$,and so:

${\Delta}_{A} = {\left(m - 1\right)}^{2} - 4 \left(1\right) \left(- 2 \left(m + 1\right)\right)$
$\setminus \setminus \setminus \setminus \setminus = {m}^{2} - 2 m + 1 + 8 m + 8$
$\setminus \setminus \setminus \setminus \setminus = {\left(m - 3\right)}^{2}$

${\Delta}_{A} = 0 \implies m = 3 \implies 1$ solution
${\Delta}_{A} > 0 \implies m \ne 3 \implies 2$ solutions

And for set $B$, we have:

$B = \left\{x \in \mathbb{R} | \left(\left(m - 1\right) {x}^{2}\right) + m x + 1 = 0\right\}$

Similarly, We know that $x \in \mathbb{R} \implies {\Delta}_{B} \ge 0$,and so:

${\Delta}_{B} = {m}^{2} - 4 \left(m - 1\right) \left(1\right)$
$\setminus \setminus \setminus \setminus \setminus = {m}^{2} - 4 m + 4$
$\setminus \setminus \setminus \setminus \setminus = {\left(m - 2\right)}^{2}$

${\Delta}_{B} = 0 \implies m = 2 \implies 1$ solution
${\Delta}_{B} > 0 \implies m \ne 2 \implies 2$ solutions

Now we want $A \cup B$ to have $3$ distinct elements, this requires

• One element from A, two elements from B:
$\implies {\Delta}_{A} = 0 , {\Delta}_{B} > 0$
$\implies \left(m = 3\right) \cap \left(m \ne 2\right) \implies m = 3$

• One element from B, two elements from A  => Delta_B=0, Delta_A gt 0
$\implies \left(m = 2\right) \cap \left(m \ne 3\right) \implies m = 2$

Therefore there are $2$ values of $m$ that satisfy the specified criteria