## Types of Functions:

(1) ** One-to-one Functions-** A function, f: A

**â†’**B, is said to be one-to-one or injective if for distinct elements x

_{1}, x

_{2}âˆˆ A, there exist f(x

_{1}), f(x

_{2}) respectively in B such that for x

_{1}â‰ x

_{2}, f( x

_{1}) â‰ f(x

_{2}). Linear functions like f(x) = 2x – 1, f(x) = x + 2, f(x) = (2x + 1)/3 etc., are all one-to-one functions. The graphs of one-to-one functions denote straight lines. The mapping has been shown in the figure below.

(2) ** Many-to-one Functions-** Functions that are not one-to-one are said to be many-to-one functions. Functions like y – x2, y = x3, y = | x | are many-to-one functions. The mapping has been shown in the figure below. The codomain of the function, in this case, is set B but the range is {y

_{1}, y

_{2}, y

_{3}}.

(3) ** Onto-(Surjective) Functions-** A function, f: A

**â†’**B, is said to be an onto or surjective function if each element y of B has a pre-image x in A. The codomain in this case is called the Range of f. Set B in this case must also be exhausted as shown in the figure below.

Thus, Range of f = f(A) = {f(x): x âˆˆ A} Obviously, f(A) âŠ† B |

(4) ** Into Functions-** A function, f: A

**â†’**B, is said to be an into-function if there exists at least one element in B having no pre-image in A. In the figure below, there are two elements y

_{4}and y

_{5}in set B for which there is no pre-image in A. The function in the case is an into-function.

(5) ** Bijection-** A function, f: A

**â†’**B, is a bijection if-

- It is one-one i.e. f(x
_{1}) = f(x_{2}) â‡’ x_{1}= x_{2}for all x_{1}, x_{2}âˆˆ A. - It is onto i.e. for all y âˆˆ B, there exists x âˆˆ A such f(x) = y.

(6) ** Identity Function-** Let A be a non-void set. A function, f: A

**â†’**A, is said to be the identity function on set A if f associates every element of set A to the element itself. Identity function is given by f(x) = x.

(7) ** Constant Function-** A function, f: A

**â†’**B, is a constant function if every member of A has the same image in B under the function f. Thus, in this case f(x) = c, for all x âˆˆ A, c âˆˆ B.

(8) ** Exponential Function-** The function is given by f(x) = e

^{x}. It lies completely above the X-axis and from left to right it rises exponentially.

(9) ** Logarithmic Function-** It is given by f(x) = log

_{e}x. The graph of the function lies completely on the right side of the Y-axis and it rises above from left to right as shown in figure (a).

When f(x) = e^{x }and f(x) = log_{e} x are drawn on the same scale, it would be clear that one is reflection of the others in the line y = x (the identity function) as shown in figure (b).

(10) ** The Greatest Integer Function-** It is defined as f(x) = [x], where [x] refers to the greatest integer contained in x. It is sometimes referred to as the

**or the**

*Integral value function***(as the graph of the function looks like a staircase). The graph of the function consists of many broken pieces, each being coincident with the graph of a constant function as shown in the figure below.**

*Staircase function*(11) ** Trigonometric Functions-** The graphs of f(x) = sin x, f(x) = cos x and f(x) = tan x are shown below.

(12) ** Signum Function-** The function is defined as f(x) = | x |/x, when x â‰ 0 and f(x) = 0, when x = 0. Clearly, (0, 0) is a point on the graph and | x |/x = 1, when x > 0, | x |/x = -1, when x < 0 are the two branches of the graph as shown in the figure below.

(13) ** Reciprocal Function-** The function is given by f(x) = 1/x, x â‰ 0. It is symmetrical about the origin as shown in the figure below.

(14) *Absolute Value Function-*

It is also known as the ** Modulus function**. The graph is composed of two rays originating from the same point and is symmetrical with respect to the Y-axis. It lies completely above the X-axis.

Find whether the function, f: I Example-â†’ I defined by f(x) = x^{2} + 5, for all x âˆˆ I is one-one or not. Let xSolution-_{1}, x_{2} âˆˆ I (domain) such that f(x_{1}) = f(x_{2})â‡’ x _{1}^{2} + 5 = x_{2}^{2} + 5â‡’ x _{1}^{2} = x_{2}^{2}â‡’ x _{1} = Â± x_{2}â‡’ x _{1} â‰ x_{2}âˆ´ f is not one-one. |

State whether the following functions are onto-functions:Example- (i) f: R â†’ R given by f(x) = x^{2} + 3 for all x âˆˆ R Let y âˆˆ R (co-domain) such that f(x) = ySolution-â‡’ x ^{2} + 3 = yâ‡’ x ^{2} = y – 3â‡’ x = âˆš(y – 3) For all values of y < 3, x âˆ‰ R (domain). âˆ´ the function is into. (ii) f: I â†’ I defined by f(x) = 5x + 2 for all x âˆˆ I Let y âˆˆ I (co-domain) such that f(x) = ySolution-â‡’ 5x + 2 = y â‡’ 5x = y -2 â‡’ x = (y – 2)/5 For some values of y, x âˆ‰ I. âˆ´ f is an into function. (iii) f: Q â†’ Q given by f(x) = 2x – 5 for all x âˆˆ Q Let y âˆˆ Q (co-domain) such that f(x) = ySolution-â‡’ 2x – 5 = y â‡’ 2x = y + 5 â‡’ x = (y + 5)/2 For each y âˆˆ Q (co-domain), x âˆˆ Q (domain). âˆ´ f is an onto function. |

Show that g: R Example-â†’ R defined by g(x) = 2x^{3} + 8 for all x âˆˆ R is a bijection.Solution-g is one-one:Let x _{1}, x_{2} âˆˆ R (Domain) such that g(x_{1}) = g(x_{2})â‡’ 2x _{1}^{3} + 8 = 2x_{2}^{3} + 8â‡’ 2x _{1}^{3} = 2x_{2}^{3}â‡’ x _{1}^{3} = x_{2}^{3}â‡’ x _{1} = x_{2}âˆ´ g is one-one. g is onto:Let y âˆˆ R (co-domain) such that g(x) = y â‡’ 2x ^{3} + 8 = yâ‡’ 2x ^{3} = y -8â‡’ x ^{3} = (y -8)/2â‡’ x = [(y -8)/2] ^{1/3}For each y âˆˆ R (co-domain), x âˆˆ R (domain). âˆ´ g is onto. Thus, g is a bijection. |

Show that the function, g: A Example-â†’ B defined by g(x) = (x – 3)/(x – 4) where A = {x: x âˆˆ R, x â‰ 3} and B = {y: y âˆˆ R, y â‰ 1}, is a bijection.Solution-g is one-one:Let x _{1}, x_{2} âˆˆ A (Domain) such that g(x_{1}) = g(x_{2})â‡’ (x _{1} – 3)/(x_{1} – 4) = (x_{2} – 3)/(x_{2} – 4)â‡’ x _{1}x_{2} – 4x_{1} – 3x_{2} + 12 = x_{1}x_{2} – 3x_{1} – 4x_{2} + 12â‡’ – 4x _{1} – 3x_{2} = – 3x_{1} – 4x_{2}â‡’ -x _{1} = -x_{2}â‡’ x _{1} = x_{2}âˆ´ g is one-one. g is onto:Let y âˆˆ B (co-domain) such that g(x) = y â‡’ (x – 3)/(x – 4) = y Apply Componendo and Dividendo: â‡’ 2x – 7 = (y + 1)/(y – 1) â‡’ 2x = (y + 1)/(y – 1) + 7 â‡’ 2x = (y + 1 + 7y -7)/(y – 1) â‡’ 2x = (8y – 6)/(y – 1) â‡’ x = (4y – 3)/(y – 1) For each y âˆˆ B = R – {1}, x âˆˆ A = R – {4} âˆ´ g is onto. Thus, g is a bijection. |

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