Types of Functions in Maths \u2013 Domain and Range<\/a><\/p>\nExample<\/strong><\/span> : Let f : A \\(\\rightarrow\\) B be a function represented by the following diagram :<\/p>\n<\/p>\n
Solution<\/strong><\/span> : Clearly, f is a bijection since it is both one-one (injective) and onto (surjective).<\/p>\nExample<\/strong><\/span> : Prove that the function f : Q \\(\\rightarrow\\) Q given by f(x) = 2x – 3 for all x \\(\\in\\) Q is a bijection.<\/p>\nSolution<\/strong><\/span> : We observe the following properties of f.<\/p>\nOne-One (Injective)<\/strong> : Let x, y be two arbitrary elements in Q. Then,<\/p>\nf(x) = f(y) \\(\\implies\\) 2x – 3 = 2y – 3 \\(\\implies\\) 2x = 2y \\(\\implies\\) x = y<\/p>\n
Thus, f(x) = f(y) \\(\\implies\\) x = y for all x, y \\(\\in\\) Q.<\/p>\n
So, f is one-one.<\/p>\n
Onto (Surjective) :\u00a0<\/strong>Let y be an arbitrary element of Q. Then,<\/p>\nf(x) = y \\(\\implies\\) 2x – 3 = y \\(\\implies\\) x = \\(y + 3\\over 2\\)<\/p>\n
Clearly, for all y \\(\\in\\) Q, x = \\(y + 3\\over 2\\) \\(\\in\\) Q. Thus, for all y \\(\\in\\) Q (co-domain) there exist x \\(\\in\\) Q (domain) given by x = \\(y + 3\\over 2\\) such that f(x) = f(\\(y + 3\\over 2\\)) = 2(\\(y + 3\\over 2\\)) – 3 = y. That is every element in the co-domain has its pre-image in x.<\/p>\n
So, f is onto function.<\/p>\n
Hence, f : Q \\(\\rightarrow\\) Q is a bijection.<\/p>\n
Example<\/strong><\/span> : Let f : R \\(\\rightarrow\\) R be a function defined as f(x) = \\(2x^3 + 6x^2\\) + 12x +3cosx – 4sinx; then f is-<\/p>\nSolution<\/strong><\/span> : We have f(x) = \\(2x^3 – 6x^2\\) + 12x + 3cosx – 4sinx<\/p>\n\\(\\implies\\) f'(x) = \\(6x^2 – 12x\\) + 12 – 3sinx – 4cosx<\/p>\n
\\(\\implies\\) \\(6x^2 – 12x\\) + 12 = 6(\\(x-1)^2\\) + 6 = g(x) and 3sinx + 4cosx = h(x)<\/p>\n
range of g(x) = [6, \\(\\infty\\))<\/p>\n
range of h(x) = [-5, 5]<\/p>\n
hence f'(x) always lies in the interval [1, \\(\\infty\\))<\/p>\n
\\(\\implies\\) f'(x) > 0<\/p>\n
Hence f(x) is increasing i.e. one-one<\/p>\n
Now x \\(\\rightarrow\\) \\(\\infty\\) \\(\\implies\\) f \\(\\rightarrow\\) \\(\\infty\\) & x \\(\\rightarrow\\) -\\(\\infty\\) \\(\\implies\\) f \\(\\rightarrow\\) -\\(\\infty\\) & f(x) is continous.<\/p>\n
hence its range is R \\(\\implies\\) f is onto so f is bijective.<\/p>\n
Note<\/strong> :<\/p>\n(i)\u00a0 If a line parallel to x-axis cuts the graph of the functions atmost at one point, then the f is one-one.<\/p>\n
(ii)\u00a0 If any line parallel to x-axis cuts the graph of the functions atleast at two points, then f is many-one.<\/p>\n
(iii)\u00a0 If continous functions f(x) is always increasing or decreasing in whole domain, then f(x) is one-one.<\/p>\n
(iv)\u00a0 All linear functions are one-one.<\/p>\n
(v)\u00a0 All trigonometric functions in their domain are many one.<\/p>\n
(vi)\u00a0 All even degree polynomials are many one.<\/p>\n
(vii)\u00a0 Linear by linear is one-one.<\/p>\n
(viii)\u00a0 Quadratic by quadratic with no common factor is many one.<\/p>\n
(ix)\u00a0 \u00a0A polynomial function of degree even define from R \\(\\rightarrow\\) R will always be into.<\/p>\n
(x)\u00a0 A polynomial function of degree odd defined from R \\(\\rightarrow\\) R will always be onto<\/p>\n
(xi)\u00a0 Quadratic by quadratic without any common factor define from R \\(\\rightarrow\\) R is always an into function.<\/p>\n
Thus a function can be of these four types :<\/p>\n
(i)\u00a0 one-one onto (Injective and Surjective)(Also known as Bijective mapping)<\/p>\n
(ii)\u00a0 one-one into (Injective but not surjective)<\/p>\n
(iii)\u00a0 many-one onto (surjective but not injective)<\/p>\n
(iv)\u00a0 many-one into (neither surjective nor injective)<\/p>\n
\u00a0<\/p>\n\n\n