Consider the two sets :


Consider the two sets :

$A=\{m \in R:$ both the roots of

$x^{2}-(m+1) x+m+4=0$ are real $\}$ and


Which of the following is not true?

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

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

  3. $\mathrm{B}-\mathrm{A}=(-3,5)$

  4. $A \cup B=R$

Correct Option: 1


$\mathrm{A}: \mathrm{D} \geq 0$

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

$\Rightarrow \mathrm{m}^{2}+2 \mathrm{~m}+1-4 \mathrm{~m}-16 \geq 0$

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

$\Rightarrow(\mathrm{m}-5)(\mathrm{m}+3) \geq 0$

$\Rightarrow m \in(-\infty,-3] \cup[5, \infty)$

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


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

$A \cap B=\{-3\}$


$A \cup B=R$

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