TOPICS
Miscellaneous

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.

Solution :-
    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 
    graph
   

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

Solution :-
   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).

Solution :-
   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).

Solution :-
    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 function Draw the graph of f (x).

Solution :-
   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
    graph
   
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