Chapter 2 — Relations and Functions

Concept Used

In mathematics, a relation between two sets associates elements of one set with elements of another set. A relation becomes a function if each element of the domain is mapped to exactly one element of the codomain.

• Domain → set of all first elements of ordered pairs
• Range → set of all second elements of ordered pairs
• A relation is not a function if one input corresponds to multiple outputs.

Thus, the key rule is:

"One input → Only one output"

---

Solution Roadmap

To determine whether a relation is a function, we follow these steps:

1. List all ordered pairs.
2. Identify the domain from first elements.
3. Check whether any element of the domain is repeated with different outputs.
4. If each input has only one output → the relation is a function.
5. Determine the range from second elements.

---

Illustration (Mapping Concept)

In a valid function, each domain element has exactly one arrow leaving it.

---

Solution

Among the given relations, (i) and (ii) are functions, whereas (iii) is not a function.

(i)

\[ \begin{aligned} R &= \{(2,1),(5,1),(8,1),(11,1),(14,1),(17,1)\} \end{aligned} \] Domain = {2,5,8,11,14,17} Range = {1}

Each element of the domain is associated with exactly one element of the range. Therefore, relation (i) is a function.

(ii)

\[ \begin{aligned} R &= \{(2,1),(4,2),(6,3),(8,4),(10,5),(12,6),(14,7)\} \end{aligned} \] Domain = {2,4,6,8,10,12,14} Range = {1,2,3,4,5,6,7}

Every element of the domain corresponds to exactly one element of the range. Hence, relation (ii) is also a function.

(iii)

The relation \[ \{(1,3),(1,5),(2,5)\} \] is not a function because the element 1 in the domain corresponds to two different outputs (3 and 5).

Since a function cannot assign multiple outputs to the same input, relation (iii) is not a function.

---

Significance for Board & Competitive Exams

• This question tests the core definition of functions from relations.
• Frequently appears in CBSE board exams as short answer questions.
• Forms the conceptual base for topics such as:
  • Functions
  • Inverse functions
  • Graphing of functions
  • Mappings and transformations
• Understanding this concept is essential for competitive exams like JEE Main, NDA, CUET, and other entrance examinations.
• It also helps in solving domain-range problems and verifying whether a relation represents a valid mathematical function.

Concept Used

For a real function \(f(x)\):

• Domain is the set of all real values of \(x\) for which the function is defined.
• Range is the set of all possible values of \(f(x)\).

While determining domain and range, the following rules are useful:

• The absolute value function is defined for all real numbers.
• The expression inside a square root must be non-negative.
• Graph interpretation helps visualize domain and range.

---

Solution Roadmap

1. Identify the type of function (absolute value or square root).
2. Apply restrictions if any.
3. Determine the possible values of \(x\) → Domain.
4. Determine the possible outputs \(f(x)\) → Range.
5. Use graphical interpretation to verify results.

---

Illustration for \(f(x)=-|x|\)

The graph opens downward because of the negative sign in front of the absolute value.

---

Solution

(i) \(f(x)=-|x|\)

\[ f(x)=-|x| \] The absolute value function is defined for all real numbers. \[ \text{Domain}=\mathbb{R} \] Since \[ |x|\ge0 \] multiplying by −1 gives \[ -|x|\le0 \] Therefore the function can take all values less than or equal to zero. \[ \text{Range}=(-\infty,0] \]

---

Illustration for \(f(x)=\sqrt{9-x^2}\)

This function represents the upper semicircle of radius 3.

---

(ii) \(f(x)=\sqrt{9-x^2}\)

For the square root to be defined, the expression inside the radical must be non-negative. \[ 9-x^2\ge0 \] \[ x^2\le9 \] \[ -3\le x\le3 \] Therefore, \[ \text{Domain}=[-3,3] \]

Since square root values are always non-negative and the maximum value occurs at \(x=0\), \[ f(0)=\sqrt{9}=3 \] Thus, \[ 0\le f(x)\le3 \] \[ \text{Range}=[0,3] \]

---

Significance for Board & Competitive Exams

• This problem tests understanding of domain and range of functions.
• It introduces restrictions involving:
  • absolute value functions
  • square root expressions
• These concepts are frequently used in:
  • CBSE Board Examinations
  • JEE Main mathematics
  • NDA and CUET entrance tests
• The second function represents a semicircle graph, which connects algebra with geometry.
• Mastering domain and range is essential for later topics such as:
  • inverse functions
  • graph transformations
  • trigonometric functions

Concept Used

A function assigns a unique output value to every input value. If a function is expressed in the form

\(f(x)=ax+b\)

it is called a linear function. The graph of a linear function is always a straight line.

• The coefficient of \(x\) represents the slope.
• The constant term represents the y-intercept.
• The value of the function at any point is obtained by substituting the value of \(x\).

---

Solution Roadmap

1. Write the given function.
2. Substitute the required value of \(x\).
3. Simplify the expression.
4. Obtain the corresponding value of \(f(x)\).

---

Graphical Illustration of the Function

The function \(f(x)=2x-5\) represents a straight line with slope 2 and y-intercept −5.

---

Solution

The given function is

\[ f(x)=2x-5 \]

(i) Value of \(f(0)\)

\[ \begin{aligned} f(0)&=2\times0-5 \ &=-5 \end{aligned} \]

(ii) Value of \(f(7)\)

\[ \begin{aligned} f(7)&=2\times7-5 \ &=14-5 \ &=9 \end{aligned} \]

(iii) Value of \(f(-3)\)

\[ \begin{aligned} f(-3)&=2\times(-3)-5 \ &=-6-5 \ &=-11 \end{aligned} \]

---

Significance for Board & Competitive Exams

• This question reinforces the concept of evaluating a function.
• It introduces students to linear functions, which are fundamental in algebra.
• Similar problems appear frequently in CBSE board examinations.
• The concept is also important for competitive exams such as:
  • JEE Main
  • NDA
  • CUET
• Understanding linear functions helps in later topics like:
  • graph plotting
  • coordinate geometry
  • linear equations

Concept Used

Temperature can be measured using different scales such as Celsius (°C) and Fahrenheit (°F). The relation between these scales is linear and can be represented by a function.

\[ t(C)=\frac{9C}{5}+32 \]

• This function converts a temperature from Celsius to Fahrenheit.
• The function is a linear function because it is of the form \(ax+b\).
• Evaluating the function involves substituting the given value of \(C\).

---

Solution Roadmap

1. Write the given conversion function.
2. Substitute the required value of \(C\).
3. Simplify the expression to obtain the Fahrenheit temperature.
4. For the inverse case, solve the equation to find \(C\).

---

Illustration: Celsius–Fahrenheit Relation

The Celsius–Fahrenheit conversion forms a straight line because the relationship between the two scales is linear.

---

Solution

The temperature conversion function is

\[ t(C)=\frac{9C}{5}+32 \]

(i) Value of \(t(0)\)

\[ \begin{aligned} t(0)&=\frac{9\times0}{5}+32 \ &=32 \end{aligned} \]

(ii) Value of \(t(28)\)

\[ \begin{aligned} t(28)&=\frac{9\times28}{5}+32 \ &=\frac{252}{5}+32 \ &=50.4+32 \ &=82.4 \end{aligned} \]

(iii) Value of \(t(-10)\)

\[ \begin{aligned} t(-10)&=\frac{9\times(-10)}{5}+32 \ &=-18+32 \ &=14 \end{aligned} \]

(iv) When \(t(C)=212\)

\[ \begin{aligned} 212&=\frac{9C}{5}+32 \ 212-32&=\frac{9C}{5} \ 180&=\frac{9C}{5} \ 9C&=900 \ C&=100 \end{aligned} \]

---

Significance for Board & Competitive Exams

• This problem demonstrates how real-life situations can be represented using functions.
• It reinforces the concept of evaluating functions and solving simple equations.
• Similar questions often appear in CBSE board examinations.
• The idea of conversion functions is useful in many competitive exams such as:
  • JEE Main
  • NDA
  • CUET
• Understanding linear conversion functions helps in later topics such as:
  • linear equations
  • graph interpretation
  • mathematical modelling

Concept Used

The range of a function is the set of all possible values that the function can take. To determine the range, we analyze how the output of the function behaves for the given domain.

• If a function increases or decreases continuously, the range can often be determined using limits.
• For quadratic expressions like \(x^2\), the minimum value occurs at the vertex.
• If a function is \(f(x)=x\), its output equals the input.

---

Solution Roadmap

1. Identify the type of function.
2. Analyze how the function behaves as \(x\) increases or decreases.
3. Determine the minimum or maximum values if they exist.
4. Express the set of possible outputs as the range.

---

Illustration for \(f(x)=2-3x,\; x>0\)

The function decreases continuously for \(x>0\). As \(x\) increases, the function values decrease without bound.

---

Solution

(i) \(f(x)=2-3x,\; x>0\)

\[ f(x)=2-3x \] Since \(x>0\), the quantity \(3x\) is positive and increases as \(x\) increases. Therefore \(2-3x\) decreases continuously. As \[ x \to 0^{+}, \quad f(x)\to 2 \] but the value \(2\) is not attained because \(x\) cannot be zero. As \[ x \to \infty,\quad f(x)\to -\infty \] Hence, \[ \text{Range}=(-\infty,2) \]

---

Illustration for \(f(x)=x^2+2\)

The graph is a parabola opening upward with minimum value at \(x=0\).

---

(ii) \(f(x)=x^2+2\)

Since \[ x^2 \ge 0 \quad \text{for all real } x \] The smallest value of \(x^2\) occurs at \(x=0\). \[ f(0)=0+2=2 \] Therefore the function cannot take values smaller than 2. \[ \text{Range}=[2,\infty) \]

---

Illustration for \(f(x)=x\)

The graph of \(f(x)=x\) is a straight line through the origin.

---

(iii) \(f(x)=x,\; x\in\mathbb{R}\)

As \(x\) takes every real value, the function also takes every real value. Therefore, \[ \text{Range}=\mathbb{R} \]

---

Significance for Board & Competitive Exams

• This problem strengthens understanding of range of functions.
• It demonstrates three important types of functions:
  • linear decreasing functions
  • quadratic functions
  • identity functions
• Questions related to range determination frequently appear in CBSE board examinations.
• The concept is essential for competitive exams such as:
  • JEE Main
  • NDA
  • CUET
• Understanding the range of functions is crucial for later topics like:
  • inverse functions
  • graph transformations
  • trigonometric functions