The notion of sets provides the stimulus for learning higher concepts in mathematics. A set is a collection of well-defined objects. This means that a set is merely a collection of something which we may recognize. In this chapter, we try to extend the concept of sets in two forms called Relations and Functions. For doing this, we need to first know about cartesian products that can be defined between two non-empty sets.
Question-5: Let A,B,C ⊂ N and a function f: A → B be defined by f(x) = 2x + 1 and g: B → C be defined by g(x) = x2 . Find the range of fog and gof.
Solution: f(x) = 2x + 1 ; g(x) = x2 fog = f[g(x)] = f(x2) = 2x2 + 1 2x2 + 1 ∈ N g o f = g [f(x)] = g (2x + 1) g o f = (2x + 1)2 (2x + 1)2 ∈ N Range = {y/y = 2x2 + 1, x ∈ N}; {y/y = (2x + 1)2, x ∈ N)
(ii) fofof = f o f(f(x)) = f of (x4 – 2x2) = f(f(x4 – 2x2)) = (x4 – 2x2)2 – 1 = x8 – 4x6 + 4x4 – 1
Question-7: If f : R → R and g : R → R are defined by f(x) = x5 and g(x) = x4 then check if f, g are one – one and fog is one – one?
Solution: f(x) = x5 – It is one – one function g(x) = x4 – It is one – one function fog = f[g(x)] = f(x4) = (x4)5 fag = x20 It is also one-one function.
Question-8: Consider the functions f(x), g(x), h(x) as given below. Show that (fog)oh = fo(goh) in each case. (i) f(x) = x – 1, g(x) = 3x + 1 and h(x) = x2 (ii) f(x) = x2, g(x) = 2x and h(x) = x + 4 (iii) f(x) = x – 4, g(x) = x2 and h(x) = 3x – 5
Question-9: Let f = {(-1, 3), (0, -1), (2, -9)} be a linear function from Z into Z. Find f(x).
Solution: The linear equation is f(x) = ax + b f(-1) = 3 a(-1) + b = 3 -a + b = 3 ….(1) f(0) = -1 a(0) + b = -1 0 + b = -1 b = -1 Substitute the value of b = -1 in (1) -a – 1 = 3 -a = 3 + 1 -a = 4 a = -4 ∴ The linear equation is -4(x) -1 = -4x – 1 (or) – (4x + 1)
Question-10: In electrical circuit theory, a circuit C(t) is called a linear circuit if it satisfies the superposition principle given by C(at1 + bt2) = aC(t1) + bC(t2), where a,b are constants. Show that the circuit C(t) = 31 is linear.
Solution: Given C(t) = 3t. To prove that the function is linear C(at1) = 3a(t1) C(bt2) = 3 b(t2) C(at1 + bt2) = 3 [at1 + bt2] = 3at1 + 3bt2 = a(3t1) + b(3t2) = a[C(t1) + b(Ct2)] ∴ Superposition principle is satisfied. Hence C(t) = 3t is linear function.
