Consider the two sets: A = {$m ∈ R$: 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?
(A)
A ∩ B = {–$3$}
(B)
B – A = (–$3$, $5$)
(C)
A ∪ B = R
(D)
A - B = ($-∝$, –$3$) ∪ ($5$, $∝$)
MEDIUM
Correct Answer:D
Explanation
For the roots to be real, the discriminant $D ≥ 0$.
$(m+1)^2 - 4(m+4) ≥ 0$
$m^2 + 2m + 1 - 4m - 16 ≥ 0$
$m^2 - 2m - 15 ≥ 0$
$(m-5)(m+3) ≥ 0$
$m ∈ (-∞, -3] ∪ [5, ∞)$
Therefore, $A = (-∞, -3] ∪ [5, ∞)$. Given $B = [-3, 5)$.
Analyzing the options:
Option A: $A ∩ B = {-3}$. This is true since the intersection contains only $-3$.
Option B: $B – A = (-3, 5)$. This is true because removing $A$ from $B$ leaves the interval $(-3, 5)$.
Option C: $A ∪ B = R$. This is true, as the union covers all real numbers.
Option D: $A - B = (-∞, -3) ∪ (5, ∞)$. This is incorrect. $A - B$ should be $(-∞, -3) ∪ [5, ∞)$. The given interval does not include $5$.
Thus, option D is not true.