Question:
Consider the two sets :
\(A = \{m \in \mathbb{R} : \text{both the roots of } x^2 - (m + 1)x + m + 4 = 0 \text{ are real}\}\)
and \(B = [-3, 5)\).
Which of the following is not true?
Consider the two sets :
\(A = \{m \in \mathbb{R} : \text{both the roots of } x^2 - (m + 1)x + m + 4 = 0 \text{ are real}\}\)
and \(B = [-3, 5)\).
Which of the following is not true?
\(A = \{m \in \mathbb{R} : \text{both the roots of } x^2 - (m + 1)x + m + 4 = 0 \text{ are real}\}\)
and \(B = [-3, 5)\).
Which of the following is not true?
Show Hint
Find A from the discriminant condition \(D \ge 0\), then test each set statement against \(B = [-3, 5)\).
Updated On: Nov 5, 2026
- \(A \cap B = \{-3\}\)
- \(B - A = (-3, 5)\)
- \(A \cup B = \mathbb{R}\)
- \(A - B = (-\infty, -3) \cup (5, \infty)\)
Show Solution
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The Correct Option is D
Solution and Explanation
Real roots: \(D = (m + 1)^2 - 4(m + 4) \ge 0\) \(\Rightarrow m^2 - 2m - 15 \ge 0\) \(\Rightarrow (m - 5)(m + 3) \ge 0\).
So \(A = (-\infty, -3] \cup [5, \infty)\) and \(B = [-3, 5)\).
(1) \(A \cap B = \{-3\}\) : true.
(2) \(B - A = (-3, 5)\) : true.
(3) \(A \cup B = \mathbb{R}\) : true.
(4) \(A - B = (-\infty, -3) \cup [5, \infty)\), because 5 belongs to A and not to B. The option has \((5, \infty)\), which omits 5, so it is not true.
Answer: (4) \(A - B = (-\infty, -3) \cup (5, \infty)\)
So \(A = (-\infty, -3] \cup [5, \infty)\) and \(B = [-3, 5)\).
(1) \(A \cap B = \{-3\}\) : true.
(2) \(B - A = (-3, 5)\) : true.
(3) \(A \cup B = \mathbb{R}\) : true.
(4) \(A - B = (-\infty, -3) \cup [5, \infty)\), because 5 belongs to A and not to B. The option has \((5, \infty)\), which omits 5, so it is not true.
Answer: (4) \(A - B = (-\infty, -3) \cup (5, \infty)\)
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