Consider the two sets:A = {m in R : text{both | vedclass

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(A) For the roots of the quadratic equation $x^{2} - (m+1)x + m+4 = 0$ to be real,the discriminant $D$ must be greater than or equal to $0$.$D = (m+1)

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

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(A) For the roots of the quadratic equation $x^{2} - (m+1)x + m+4 = 0$ to be real,the discriminant $D$ must be greater than or equal to $0$.$D = (m+1)^{2} - 4(1)(m+4) \geq 0$$m^{2} + 2m + 1 - 4m - 16 \geq 0$$m^{2} - 2m - 15 \geq 0$$(m-5)(m+3) \geq 0$Thus,$m \in (-\infty, -3] \cup [5, \infty)$.So,$A = (-\infty, -3] \cup [5, \infty)$.Given $B = [-3, 5)$.$1$. $A - B$: Elements in $A$ that are not in $B$. Since $B$ includes $-3$ and goes up to $5$,$A - B = (-\infty, -3) \cup [5, \infty)$. (True)$2$. $A \cap B$: Intersection of $A$ and $B$. The only common element is $\{-3\}$. (True)$3$. $B - A$: Elements in $B$ that are not in $A$. Since $A$ contains $-3$ and values $\geq 5$,$B - A = (-3, 5)$. (True)$4$. $A \cup B$: Union of $A$ and $B$. $A \cup B = (-\infty, -3] \cup [5, \infty) \cup [-3, 5) = (-\infty, \infty) = R$. (True)Wait,checking the options again: Option $A$ states $A-B = (-\infty, -3) \cup [5, \infty)$. This is correct. All options provided are actually true. However,based on standard competitive exam patterns for this specific question,if one must be chosen as 'not true',there might be a typo in the provided options. Given the logic,all statements are mathematically correct.

Consider the two sets:
$A = \{m \in R : \text{both the roots of } x^{2} - (m+1)x + m+4 = 0 \text{ are real}\}$ and $B = [-3, 5)$.
Which of the following is not true?

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Consider the two sets:$A = \{m \in R : \text{both the roots of } x^{2} - (m+1)x + m+4 = 0 \text{ are real}\}$ and $B = [-3, 5)$.Which of the following is not true?