Can anyone solve 48 number?
Answers
EXPLANATION.
⇒ A = {x ∈ R : x² + 6x - 7 < 0}. - - - - - (1).
⇒ B = {x ∈ R : x² + 9x + 14 > 0}. - - - - - (2).
As we know that,
From equation (1), we get.
⇒ A = {x ∈ R : x² + 6x - 7 < 0}. - - - - - (1).
⇒ x² + 6x - 7 < 0.
Factorizes the equation into middle term splits, we get.
⇒ x² + 7x - x - 7 < 0.
⇒ x(x + 7) - 1(x + 7) < 0.
⇒ (x - 1)(x + 7) < 0.
Put this point on wavy curve method, we get.
⇒ x ∈ (-7,1). - - - - - (a).
From equation (2), we get.
⇒ B = {x ∈ R : x² + 9x + 14 > 0}. - - - - - (2).
⇒ x² + 9x + 14 > 0.
Factorizes the equation into middle term splits, we get.
⇒ x² + 7x + 2x + 14 > 0.
⇒ x(x + 7) + 2(x + 7) > 0.
⇒ (x + 2)(x + 7) > 0.
Put this point on wavy curve method, we get.
⇒ x ∈ (-∞,-7) ∪ (-2,∞). - - - - - (b).
(1) = A ∩ B.
⇒ x ∈ (-7,1) ∩ (-∞,-7) ∪ (-2,∞).
Put this point on number line, we get.
⇒ x ∈ (-2,1).
(2) = (A \ B).
⇒ (A \ B) = (A - B) = {x ∈ R : -7 < x < -2}.
⇒ (A \ B) = (-7,-2).
Option [C] is correct answer.
MORE INFORMATION.
Number of elements in different sets.
If A,B & C are finite sets and U be the finite universal set, then.
(1) = n(A ∪ B) = n(A) + n(B) - n(A ∩ B).
(2) = n(A ∪ B) = n(A) + n(B) [if A & B are disjoint sets].
(3) = n(A - B) = n(A) - n(A ∩ B).
(4) = n(A Δ B) = n[(A - B) ∪ (B - A)] = n(A) + n(B) - 2n(A ∩ B).
(5) = n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(A ∩ C) + n(A ∩ B ∩ C).
(6) = n(A' ∪ B') = n(A ∩ B)' = n(U) - n(A ∩ B).
(7) = n(A' ∩ B') = n(A ∪ B)' = n(U) - n(A ∪ B).