Consider the two sets :
A = m ∈ R : both the roots of
x² – (m + 1)x + m + 4 = 0 are real}
and B = [–3, 5).
Which of the following is not true?

Solution

Maths sir · 3 steps

Find A from the discriminant, then watch the endpoints -3 and 5 very carefully.

1. Real roots condition. Both roots are real when the discriminant is non-negative. Equal roots are still real, so the condition is D ≥ 0.

   D = (m+1)² - 4(m+4) = m² - 2m - 15 = (m-5)(m+3) ≥ 0

2. Set A. The product is non-negative outside the roots, ends included. So A = (-∞, -3] ∪ [5, ∞), while B = [-3, 5).

   Option	Correct set	Statement
   (a) A ∩ B	{-3}, since -3 is in both	true
   (b) B - A	(-3, 5), since -3 is removed by A	true
   (c) A ∪ B	A covers outside, B fills [-3, 5), so ℝ	true
   (d) A - B	(-∞, -3) ∪ [5, ∞)	false: it leaves out 5

3. So. 5 belongs to A (there D = 0) but not to B, because B is open at 5. So 5 ∈ A - B, and option (d), which writes (5, ∞), is the statement that is not true.

The idea to keep. In set questions built on intervals, everything hinges on the endpoints. Check each endpoint separately: is it in A, is it in B? Remember D = 0 still gives real roots.

Answer from the NTA final answer key. Solution by PrepWiser.