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Relations and Functions - Types of Functions

Grade 12CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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One-to-One (Injective) Function: A function f:X→Yf: X \to Y is called injective if the images of distinct elements of XX under ff are distinct. Mathematically, f(x1)=f(x2)  ⟹  x1=x2f(x_1) = f(x_2) \implies x_1 = x_2 for all x1,x2∈Xx_1, x_2 \in X. Geometrically, any horizontal line intersects the graph of a one-to-one function at most once.

Mapping diagram showing a one-to-one function where distinct elements in domain map to distinct elements in codomain.
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Onto (Surjective) Function: A function f:X→Yf: X \to Y is surjective if every element of YY is the image of at least one element of XX under ff. In other words, the Range of ff is equal to the Codomain YY.

Mapping diagram showing a surjective function where all elements in the codomain are mapped to.
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Many-to-One Function: A function f:X→Yf: X \to Y is many-to-one if two or more distinct elements in XX have the same image in YY. For example, f(x)=x2f(x) = x^2 for x∈Rx \in \mathbb{R} is many-to-one because f(1)=f(−1)=1f(1) = f(-1) = 1.

Graph of y=x^2 showing a horizontal line intersecting the curve at two points, indicating many-to-one nature.
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Bijective Function: A function f:X→Yf: X \to Y is bijective (or a one-to-one correspondence) if it is both injective (one-to-one) and surjective (onto). A bijection ensures that every element in YY is paired with exactly one element in XX.

📐Formulae

f(x1)=f(x2)  ⟹  x1=x2 (Condition for Injectivity)f(x_1) = f(x_2) \implies x_1 = x_2 \text{ (Condition for Injectivity)}

∀y∈Y,∃x∈X s.t. f(x)=y (Condition for Surjectivity)\forall y \in Y, \exists x \in X \text{ s.t. } f(x) = y \text{ (Condition for Surjectivity)}

Total functions from A to B=[n(B)]n(A)\text{Total functions from } A \text{ to } B = [n(B)]^{n(A)}

Number of onto functions from A to B (if n(A)=n,n(B)=2)=2n−2\text{Number of onto functions from } A \text{ to } B \text{ (if } n(A)=n, n(B)=2\text{)} = 2^n - 2

💡Examples

Problem 1:

Show that the function f:R→Rf: \mathbb{R} \to \mathbb{R}, defined as f(x)=2xf(x) = 2x, is bijective.

Solution:

  1. Injectivity: Let f(x1)=f(x2)f(x_1) = f(x_2) for x1,x2∈Rx_1, x_2 \in \mathbb{R}. Then 2x1=2x2  ⟹  x1=x22x_1 = 2x_2 \implies x_1 = x_2. Thus, ff is injective.
  2. Surjectivity: Let y∈Ry \in \mathbb{R} (codomain). We need to find x∈Rx \in \mathbb{R} such that f(x)=yf(x) = y. So, 2x=y  ⟹  x=y22x = y \implies x = \frac{y}{2}. Since yy is a real number, y2\frac{y}{2} is also a real number. Therefore, for every y∈Ry \in \mathbb{R}, there exists x=y2∈Rx = \frac{y}{2} \in \mathbb{R} such that f(x)=yf(x) = y. Thus, ff is surjective. Since ff is both injective and surjective, it is bijective.

Explanation:

To prove bijectivity, we must algebraically prove the function is both one-to-one (by equating images) and onto (by expressing xx in terms of yy).

Problem 2:

Check the injectivity of the function f:N→Nf: \mathbb{N} \to \mathbb{N} defined by f(x)=x2f(x) = x^2.

Solution:

Let f(x1)=f(x2)f(x_1) = f(x_2) where x1,x2∈Nx_1, x_2 \in \mathbb{N}. x12=x22x_1^2 = x_2^2 x12−x22=0x_1^2 - x_2^2 = 0 (x1−x2)(x1+x2)=0(x_1 - x_2)(x_1 + x_2) = 0 Since x1,x2∈Nx_1, x_2 \in \mathbb{N}, x1+x2x_1 + x_2 cannot be zero. Therefore, x1−x2=0  ⟹  x1=x2x_1 - x_2 = 0 \implies x_1 = x_2. Hence, the function is injective.

Explanation:

In the domain of Natural Numbers (N\mathbb{N}), x2x^2 is injective because we cannot have negative values that would produce the same square as a positive value.

Problem 3:

Is the function f:R→Rf: \mathbb{R} \to \mathbb{R} defined by f(x)=x2f(x) = x^2 surjective?

Solution:

The codomain is R\mathbb{R} (all real numbers). The range of f(x)=x2f(x) = x^2 is [0,∞)[0, \infty). Since there are negative numbers in the codomain (e.g., −2∈R-2 \in \mathbb{R}) that have no pre-image (since x2=−2x^2 = -2 has no real solution), the range is not equal to the codomain. Range(f)≠Codomain(f)\text{Range}(f) \neq \text{Codomain}(f). Therefore, ff is not surjective.

Explanation:

Surjectivity fails if any element in the codomain lacks a corresponding value in the domain. Squares of real numbers are never negative.

Problem 4:

Check the injectivity and surjectivity of the function f:R→Rf: \mathbb{R} \to \mathbb{R} defined by f(x)=x3f(x) = x^3.

Graph of f(x) = x^3 showing it passes the horizontal line test and covers all y-values.

Solution:

  1. Injectivity: Let x1,x2∈Rx_1, x_2 \in \mathbb{R}. If f(x1)=f(x2)f(x_1) = f(x_2), then x13=x23x_1^3 = x_2^3. Taking the cube root of both sides, we get x1=x2x_1 = x_2. Thus, ff is injective.
  2. Surjectivity: Let y∈Ry \in \mathbb{R} (codomain). We need to find x∈Rx \in \mathbb{R} such that f(x)=yf(x) = y. Since x3=y  ⟹  x=y3x^3 = y \implies x = \sqrt[3]{y}, and every real number has a real cube root, x∈Rx \in \mathbb{R}. Thus, ff is surjective. Since ff is both injective and surjective, it is bijective.

Explanation:

The cubic function is strictly increasing, which ensures each y-value is mapped to exactly once. The range is the entire real line.

Problem 5:

Determine if the function f:R→Rf: \mathbb{R} \to \mathbb{R} defined by f(x)=cos⁡(x)f(x) = \cos(x) is onto.

Graph of cos(x) showing that values outside [-1, 1] are not reached by the function.

Solution:

Consider the codomain R\mathbb{R} and the range of the function. We know that for any x∈Rx \in \mathbb{R}, −1≤cos⁡(x)≤1-1 \le \cos(x) \le 1. Therefore, the Range of ff is [−1,1][-1, 1]. Since the Range [−1,1][-1, 1] is not equal to the Codomain R\mathbb{R} (elements like y=2y=2 have no pre-image), the function is not surjective (not onto).

Explanation:

A function is onto only if its range equals its codomain. For trigonometric functions, the range is often a subset of real numbers.