Short Answer TypeLet * be a binary operation on the set Q of rational numbers given as a * b = (2a – b)2, a, b ∈ Q. Find 3 * 5 and 5 * 3. Is 3 * 5 = 5 * 3 ?
 where R+ is the set of all non-negative real numbers.
f : R+ → [4, ⊂) is given by f(x) = x2 + 4.
Let x1, x2 ∈ Df = R+ = [0, ⊂) such that f(x1) = f(x2)
∴ x12 + 4 = x 22 + 4 ⇒ x12 = x22 ⇒ | x1 | = | x2 |
⇒    x1 = x2    [∵ x1x2 ≥ 0]
∴ f(x1) = f(x2) ⇒ x1 = x2 ∴ f is one-one.
Let y ∈ Rf then y = f(x), x ∈ Df = R+
⇒    y = x2 + 4 ⇒ x2 = y – 4
Now y – 4 ≥ 0 as x ≥ 0
∴ y ≥ 4    ⇒ Rf = [4, ⊂).
∴ Rf = co-domain    ⇒ f is onto.
∴ f is both one-one and onto and so invertible.
Let    y = f(x)
∴ y = x2 + 4
Let f : R → R be defined as f (x) = 3 x. Choose the correct answer. (A) f is one-one onto  (B) f is many-one onto  (C) f is one-one but not onto (D) f is neither onc-one nor onto.
how that the function f : R* → R* defined by
 is one-one and onto, where R* is the set of all non-zero real numbers. Is the result true, if the domain R* is replaced by N with co-domain being same as R*?
Show that the modulus function f : R → R, given by f (x) = | x |, is neither one-one nor onto, where |x| is x, if x is positive or 0 and | x | is —x, if. x negative.
Â
Long Answer TypeLet A= R × R and * be a binary operation on A defined by
(a, b) * (c, d) = (a+c, b+d)
Show that * is commutative and associative. Find the identity element for *
on A. Also find the inverse of every element (a, b) ∈ A.
Let A = Q × Q, where Q is the set of all rational numbers, and * be a binary operation on A defined by (a, b) * (c, d) = (ac, b+ad) for (a, b), (c, d) 
 A. Then find
(i) The identify element of * in A.
(ii) Invertible elements of A, and write the inverse of elements (5, 3) and 
Switch
