Consider the two sets :

Question:

Consider the two sets :

$A=\left\{m \in \mathbf{R}:\right.$ both the roots of $x^{2}-(m+1) x+m+4=0$

are real $\}$ and $B=[-3,5)$.

Which of the following is not true?

  1. (1) $A-B=(-\infty,-3) \cup(5, \infty)$

  2. (2) $A \cap B=\{-3\}$

  3. (3) $B-A=(-3,5)$

  4. (4) $A \cup B=\mathbf{R}$


Correct Option: 1

Solution:

$A=\left\{m \in \mathbf{R}: x^{2}-(m+1) x+m+4=0\right.$ has real roots $\}$

$D \geq 0$

$\Rightarrow(m+1)^{2}-4(m+4) \geq 0$

$\Rightarrow m^{2}-2 m-15 \geq 0$

$A=\{(-\infty,-3] \cup[5, \infty)\}$

$B=[-3,5) \Rightarrow A-B=(-\infty,-3) \cup[5, \infty)$

Leave a comment