TOPICS

Miscellaneous

Relations and Functions

**Example-1 :-** Let R be the set of real numbers. Define the real function f: R → R by f(x) = x + 10 and sketch the graph of this function.

Given that f: R → R by f(x) = x + 10. Here, f(0) = 10, f(1) = 11, f(2) = 12, ..., f(10) = 20, etc., and f(–1) = 9, f(–2) = 8, ..., f(–10) = 0 and so on. Therefore, shape of the graph of the given

**Example-2 :-** Let R be a relation from Q to Q defined by R = {(a,b): a,b ∈ Q and a – b ∈ Z}. Show that

(i) (a,a) ∈ R for all a ∈ Q

(ii) (a,b) ∈ R implies that (b, a) ∈ R

(iii) (a,b) ∈ R and (b,c) ∈ R implies that (a,c) ∈ R

Given that R = {(a,b): a,b ∈ Q and a – b ∈ Z}. (i) (a,a) ∈ R for all a ∈ Q Since, a – a = 0 ∈ Z, if follows that (a, a) ∈ R. (ii) (a,b) ∈ R implies that (b, a) ∈ R Since, (a,b) ∈ R implies that a – b ∈ Z. So, b – a ∈ Z. Therefore, (b, a) ∈ R (iii) (a,b) ∈ R and (b,c) ∈ R implies that (a,c) ∈ R Since, (a, b) and (b, c) ∈ R implies that a – b ∈ Z. b – c ∈ Z. So, a – c = (a – b) + (b – c) ∈ Z. Therefore, (a,c) ∈ R

**Example-3 :-** Let f = {(1,1), (2,3), (0, –1), (–1, –3)} be a linear function from Z into Z. Find f(x).

Given that f (x) = mx + c. Since (1, 1), (0, – 1) ∈ R, f (1) = m + c = 1 and f (0) = c = –1. This gives m = 2 and f(x) = 2x – 1.

**Example-4 :-** Find the domain of the function f(x) = (x² + 3x + 5)/(x² – 5x + 4).

Since x² – 5x + 4 = (x – 4)(x –1), the function f is defined for all real numbers except at x = 4 and x = 1. Hence the domain of f is R – {1, 4}.

**Example-5 :-** The function f is defined by
Draw the graph of f (x).

Here, f(x) = 1 – x, x < 0, this gives f(– 4) = 1 – (– 4)= 5; f(– 3) =1 – (– 3) = 4, f(– 2) = 1 – (– 2)= 3 f(–1) = 1 – (–1) = 2; f(1) = 2, f (2) = 3, f (3) = 4 f(4) = 5 and so on for f(x) = x + 1, x > 0. Thus, the graph of f is

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