68. Given a non-empty set X, let *: P(X) * P(X) → P(X) be defined as A * B = (A − B) ∪ (B − A), &mn For E; A, B ∈ P(X). Show that the empty set Φ is the identity for the operation * and all the elements A of P(X) are invertible with A−1 = A. (Hint: (A − Φ) ∪ (Φ − A) = A and (A − A) ∪ (A − A) = A * A = Φ).
68. Given a non-empty set X, let *: P(X) * P(X) → P(X) be defined as A * B = (A − B) ∪ (B − A), &mn For E; A, B ∈ P(X). Show that the empty set Φ is the identity for the operation * and all the elements A of P(X) are invertible with A−1 = A. (Hint: (A − Φ) ∪ (Φ − A) = A and (A − A) ∪ (A − A) = A * A = Φ).
It is given that ∗: P (X) * P (X) → P (X) be defined as
A * B = (A – B) ∪ (B – A), A, B ∈ P (X).
Now, let A? P (X). Then, we get,
A *? = (A –? ) ∪ (? –A) = A∪? = A
? * A = (? - A) ∪ (A -? ) =? ∪A = A
A *? = A =? * A, A? P (X)
Therefore? is the identity element for the given operation *.
Now, an eleme
Similar Questions for you
R1 = { (1, 1) (1, 2), (1, 3)., (1, 20), (2, 2), (2, 4). (2, 20), (3, 3), (3, 6), . (3, 18),
(4, 4), (4, 8), . (4, 20), (5, 5), (5, 10), (5, 15), (5, 20), (6, 6), (6, 12), (6, 18), (7. 7),
(7, 14), (8, 8), (8, 16), (9, 9), (9, 18), (10, 10), (10, 20), (11, 11), (12, 12)


⇒ (y, x) ∈ R V (x, y) ∈ R
(x, y) ∈ R ⇒ 2x = 3y and (y, x) ∈ R ⇒ 3x = 2y
Which holds only for (0, 0)
Which does not belongs to R.
∴ Value of n = 0
f is increasing function
x < 5x < 7x

f (x) < f (5x) < f (7x)
->
Given f (k) =
Case I : If x is even then g (x) = x . (i)
Case II : If x is odd then g (x + 1) = x + 1 . (ii)
From (i) & (ii), g (x) = x, when x is even
So total no. of functions = 105 × 1 = 105
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Maths Relations and Functions 2025
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