Short Answer TypeConsider the identity function 1N : N → N defined as lN(x) = x ∀ x ∈ N. Show that although IN is onto but IN + IN : N → N defined as (IN + IN) (x) = IN(x) + IN(x) = x + x = 2 x is not onto.
In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.
(i) f : R → R defined by f (x) = 3 – 4 x
(ii) f : R → R defined by f (x) = 1 + x2.
(i) Here f (x) =3 – 4 x ∀ x ∈ R
Let x1, x2 ∈ R be such that f (x1) = f (x2) ∴ 3 – 4 x1 = 3 – 4 x2
⇒ –4 x1 = – 4 x2 ⇒ x1= x2
∴ f is one-one.
Let y ∈ R be any real number.
Put f(x) = y , ∴ 3 – 4 x = y, i.e.,
Thus, corresponding to every y ∈ R, there exists
such that 
∴ f is onto.
∴ f is a bijection.
(ii) Let x1, x2 ∈ R be such that f (x1) = f (x2)
∴ 1 + x12 = 1 + x22 ∴ x = ∴ x1 = ± x2 ∴ f(x1) = f(–x1)
⇒ f is not one-one.
Also, range of f contains only those reals which are greater than or equal to 1.
[∵ x2 ≥ 0 ∀ x ∈ R, ∴ 1 + x2 ∁ 1 ∀ x ∈ R]
∴ Rf ≠ R,
⇒ f is not onto.
Thus, f is neither one-one nor onto.
Let f : N – {1} → N defined by f (n) = the highest prime factor of n. Show that f is neither one-to-one nor onto. Find the range of f.
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