Consider the two sets :
$A=\{m \in R:$ both the roots of
$x^{2}-(m+1) x+m+4=0$ are real $\}$ and
$\mathrm{B}=[-3,5)$
Which of the following is not true?
Correct Option: 1
$\mathrm{A}: \mathrm{D} \geq 0$
$\Rightarrow(\mathrm{m}+1)^{2}-4(\mathrm{~m}+4) \geq 0$
$\Rightarrow \mathrm{m}^{2}+2 \mathrm{~m}+1-4 \mathrm{~m}-16 \geq 0$
$\Rightarrow \mathrm{m}^{2}-2 \mathrm{~m}-15 \geq 0$
$\Rightarrow(\mathrm{m}-5)(\mathrm{m}+3) \geq 0$
$\Rightarrow m \in(-\infty,-3] \cup[5, \infty)$
$\therefore \quad \mathrm{A}=(-\infty,-3] \cup[5, \infty)$
$\mathrm{B}=[-3,5)$
$\mathrm{A}-\mathrm{B}=(-\infty,-3) \cup[5, \infty)$
$A \cap B=\{-3\}$
$B-A=(-3,5)$
$A \cup B=R$
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