Short Answer TypeLet g1 and g2 be two inverses of f for all y ∈ Y, we have, (f o g1)(y) = y = IY (y) and (f o g 2) (y) = y = IY (y)
⇒ (f o g1) (y) = (f o g2) (y) for all y ∈ Y
⇒ f(g1(y)) = f(g2(y)) for all y ∈ Y
⇒ g1 (y) = g2(y) [∵ f is one-one as f is invertible]
∴ g1 = g2 ∴ inverse of f is unique.
Let f : x → Y be an invertible function. Show that the inverse of f–1 is f,
i.e.(f–1)–1 = f.
Let f : N → Y be function defined as f (x) = 4 x + 3, where, Y = {y ∈N : y = 4 x + 3 for some x ∈ N}. Show that f is invertible. Find the inverse.
Let R be the relation in the set N given by R = {(a, b) : a = b – 2, b > 6}. Choose the correct answer. (A) (2, 4) ∈ R (B) (3, 8) ∈ R (C) (6, 8) ∈ R (D) (8, 7) ∈ R
Show 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.?
Check the injectivity and surjectivity of the following functions:
f : N → N given by f(x) = x2
Check the injectivity and surjectivity of the following functions:
f : Z → Z given by f(x) = x2
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