Chapter 2 — Relations and Functions

📘 Concept Theory

A function is a special type of relation in which every element of the domain is associated with exactly one unique element of the codomain.

For piecewise defined relations, we must carefully check the boundary points where two intervals overlap.

If the overlapping input produces two different outputs, the relation violates the definition of a function.

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🧭 Solution Roadmap

To determine whether a relation is a function:

• Identify the common boundary points where intervals overlap.
• Evaluate the relation using both definitions.
• If both expressions give the same value, the relation remains a function.
• If two different outputs appear for the same input, the relation is not a function.

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✏️ Solution

The given relations are

\[ f(x)= \begin{cases} x^{2}, & 0\le x\le 3\\ 3x, & 3\le x\le 10 \end{cases} \] \[ g(x)= \begin{cases} x^{2}, & 0\le x\le 2\\ 3x, & 2\le x\le 10 \end{cases} \]

Checking relation \(f\)

The intervals overlap at \(x=3\).

Using the first rule: \[ f(3)=3^{2}=9 \] Using the second rule: \[ f(3)=3\times3=9 \]

Both definitions give the same output. Therefore every input corresponds to a unique output.

Hence,

\(f\) is a function.

Checking relation \(g\)

The intervals overlap at \(x=2\).

Using the first rule: \[ g(2)=2^{2}=4 \] Using the second rule: \[ g(2)=3\times2=6 \]

The same input \(x=2\) produces two different outputs:

• 4
• 6

This violates the definition of a function.

Therefore, \(g\) is not a function.

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📊 Illustration

The graph illustrates that \(g\) produces two different outputs at the same input, which violates the vertical line test.

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🎯 Why This Question Matters

Understanding this concept is crucial for both board exams and competitive entrance tests.

• CBSE Board Exams: Questions often test identification of functions from relations.
• JEE Main / Advanced: Piecewise functions frequently appear in calculus and limits.
• Graph Interpretation: Helps understand the vertical line test used in coordinate geometry.
• Foundation for Calculus: Piecewise functions are essential in limits, continuity, and differentiation.

📘 Concept Theory

The expression

\[ \frac{f(x+h)-f(x)}{h} \]

is known as the difference quotient. It measures the average rate of change of a function over an interval.

Geometrically, it represents the slope of the secant line joining two points on the graph of a function.

In this question, the function is

\[ f(x)=x^2 \]

and we calculate the rate of change between the points \(x=1\) and \(x=1.1\).

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🧭 Solution Roadmap

• Evaluate \(f(1)\).
• Evaluate \(f(1.1)\).
• Substitute the values into the difference quotient.
• Simplify the expression.

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✏️ Solution

Given

\[ f(x)=x^2 \] First compute the function values. \[ f(1)=1^2=1 \] \[ f(1.1)=(1.1)^2=1.21 \] Now substitute into the required expression: \[ \frac{f(1.1)-f(1)}{1.1-1} \] \[ =\frac{1.21-1}{0.1} \] \[ =\frac{0.21}{0.1} \] \[ =2.1 \]

Therefore, the required value is \(2.1\).

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📊 Geometric Illustration

The dashed line represents the secant line joining the two points on the curve. Its slope equals the value of the difference quotient.

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🎯 Why This Question Matters

• Foundation of Calculus: The difference quotient leads directly to the concept of the derivative.
• Board Exams: CBSE frequently asks numerical evaluations of functions.
• JEE Preparation: Understanding rate of change is essential for limits and derivatives.
• Graph Interpretation: Helps students connect algebraic expressions with geometric slope.

📘 Concept Theory

The domain of a function is the set of all real values of \(x\) for which the function is defined.

For a rational function

\[ f(x)=\frac{P(x)}{Q(x)} \]

the function is undefined when the denominator \(Q(x)=0\). Therefore, such values must be excluded from the domain.

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🧭 Solution Roadmap

• Identify the denominator of the function.
• Find the values of \(x\) that make the denominator zero.
• Exclude those values from the real number set.
• Express the remaining values in interval notation.

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✏️ Solution

The given function is

\[ f(x)=\frac{x^2+2x+1}{x^2-8x+12} \] To determine the domain, we find when the denominator becomes zero. \[ x^2-8x+12=0 \] Factorizing, \[ x^2-6x-2x+12=0 \] \[ x(x-6)-2(x-6)=0 \] \[ (x-6)(x-2)=0 \] \[ x=2 \quad \text{or} \quad x=6 \]

At \(x=2\) and \(x=6\), the denominator becomes zero. Hence the function is not defined at these values.

Therefore, these values must be excluded from the domain.

\[ D=(-\infty,2)\cup(2,6)\cup(6,\infty) \]

Hence, the domain of the function is all real numbers except \(2\) and \(6\).

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📊 Domain Visualization

The open circles indicate values excluded from the domain because the denominator becomes zero.

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🎯 Why This Question Matters

• CBSE Board Exams: Domain determination from rational expressions is a frequent question type.
• JEE Main: Domain restrictions appear in calculus, limits, and function analysis.
• Graph Interpretation: Undefined points correspond to vertical asymptotes or holes in graphs.
• Concept Foundation: Understanding domain restrictions is essential for higher mathematics.

📘 Concept Theory

For real-valued functions involving a square root, the expression inside the radical must be non-negative.

\[ \sqrt{A} \quad \text{is defined only when} \quad A \ge 0 \]

Thus, when determining the domain, we ensure that the quantity under the radical sign satisfies this condition.

The range is determined by analyzing the possible values of the function output once the domain restriction is applied.

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🧭 Solution Roadmap

• Apply the square root condition: radicand ≥ 0.
• Solve the inequality to obtain the domain.
• Find the minimum value of the function.
• Determine the set of possible outputs to obtain the range.

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✏️ Solution

The given function is

\[ f(x)=\sqrt{x-1} \] Finding the Domain For the square root to be defined, \[ x-1 \ge 0 \] \[ x \ge 1 \]

Therefore, the domain of the function is

\[ D=[1,\infty) \] Finding the Range

Since the square root of any non-negative number is always non-negative,

\[ f(x) \ge 0 \]

The smallest value occurs when \(x=1\).

\[ f(1)=\sqrt{0}=0 \]

As \(x\) increases, \(f(x)\) increases without bound.

\[ R=[0,\infty) \]

Domain: \(D=[1,\infty)\)
Range: \(R=[0,\infty)\)

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📊 Graphical Illustration

The graph begins at the point \((1,0)\) and increases continuously, showing that the function exists only for \(x\ge1\).

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🎯 Why This Question Matters

• CBSE Board Exams: Domain and range of radical functions is a standard question.
• JEE Preparation: Domain restrictions are essential in limits, continuity, and calculus.
• Graph Interpretation: Students learn how algebraic restrictions affect graphs.
• Concept Foundation: Helps understand transformations like \( \sqrt{x-a} \).

🧭 Solution Roadmap

• Determine whether any restrictions exist on \(x\).
• Analyze the minimum value of the absolute value expression.
• Use these observations to determine the domain and range.

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✏️ Solution

The given function is

\[ f(x)=|x-1| \] Domain

The absolute value function is defined for all real numbers. There is no restriction on \(x\).

\[ D=\mathbb{R} \] Range

The absolute value of any real number is always non-negative.

\[ |x-1|\ge0 \]

The smallest value occurs when the expression inside the modulus becomes zero.

\[ x-1=0 \] \[ x=1 \] \[ f(1)=0 \]

Thus the function cannot produce negative values and increases as \(x\) moves away from \(1\).

\[ R=[0,\infty) \]

Domain: \(D=\mathbb{R}\)
Range: \(R=[0,\infty)\)

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📊 Graphical Illustration

The graph forms a V-shape with vertex at \((1,0)\). This confirms that the minimum value is \(0\) and the function is defined for all real values of \(x\).

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🎯 Why This Question Matters

• CBSE Board Exams: Absolute value functions are frequently used to test domain and range concepts.
• JEE Preparation: Modulus functions appear in inequalities, graphs, and transformations.
• Graph Understanding: Students learn how absolute value functions produce V-shaped graphs.
• Higher Mathematics: Modulus expressions are widely used in calculus and coordinate geometry.

📘 Concept Theory

To determine the range of a function, we analyze the set of values that the function output can take.

For rational expressions involving squares such as

\[ f(x)=\frac{x^{2}}{1+x^{2}}, \]

two observations are useful:

• \(x^2\ge0\) for all real numbers.
• \(1+x^2 > x^2\), meaning the denominator is always larger than the numerator.

These properties help determine the possible values of the function.

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🧭 Solution Roadmap

• Verify that the function is defined for all real numbers.
• Determine the minimum possible value of the function.
• Determine the upper bound of the function.
• Check whether the upper bound is actually attained.

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✏️ Solution

The function is

\[ f(x)=\frac{x^2}{1+x^2}, \quad x\in\mathbb{R} \] ### Domain

For every real number \(x\),

\[ 1+x^2>0 \]

Hence the function is defined for all real numbers.

\[ D=\mathbb{R} \] ### Finding the Range Since \(x^2\ge0\), \[ \frac{x^2}{1+x^2}\ge0 \] Thus the function values are never negative. Next observe that \[ x^2 < 1+x^2 \] Dividing by \(1+x^2\), \[ \frac{x^2}{1+x^2} < 1 \] Thus the function values are always less than \(1\). Now check boundary values. At \(x=0\), \[ f(0)=0 \] Hence the value \(0\) is attained. As \(|x|\to\infty\), \[ \frac{x^2}{1+x^2}\to1 \] However the value \(1\) is never actually reached. Therefore, \[ R=[0,1) \]

Range of the function: \(R=[0,1)\)

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📊 Graphical Insight

The graph approaches the horizontal line \(y=1\) but never touches it, showing that \(1\) is not included in the range.

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🎯 Why This Question Matters

• CBSE Board Exams: Determining the range of rational functions is a common question.
• JEE Main / Advanced: Similar expressions appear in limits, inequalities, and function analysis.
• Graph Interpretation: Introduces the idea of horizontal asymptotes.
• Concept Building: Helps students understand bounded functions.

📘 Concept Theory

Functions can be combined using algebraic operations just like numbers. If \(f\) and \(g\) are functions defined on the same domain, then

\[ (f+g)(x)=f(x)+g(x) \] \[ (f-g)(x)=f(x)-g(x) \] \[ \left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}, \quad g(x)\ne0 \]

Note that for the quotient of functions, the denominator function must not become zero.

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🧭 Solution Roadmap

• Substitute the expressions of \(f(x)\) and \(g(x)\).
• Perform algebraic addition and subtraction.
• Form the quotient carefully and determine domain restrictions.

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✏️ Solution

The given functions are

\[ f(x)=x+1 \] \[ g(x)=2x-3 \]

1️⃣ Finding \(f+g\)

\[ (f+g)(x)=f(x)+g(x) \] \[ =(x+1)+(2x-3) \] \[ =3x-2 \]

Thus,

\[ f+g=3x-2 \]

2️⃣ Finding \(f-g\)

\[ (f-g)(x)=f(x)-g(x) \] \[ =(x+1)-(2x-3) \] \[ =x+1-2x+3 \] \[ =4-x \]

Thus,

\[ f-g=4-x \]

3️⃣ Finding \(\dfrac{f}{g}\)

\[ \left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)} \] \[ =\frac{x+1}{2x-3} \] For the quotient to be defined, \[ 2x-3\ne0 \] \[ x\ne\frac{3}{2} \]

\[ \frac{f}{g}(x)=\frac{x+1}{2x-3}, \quad x\ne\frac{3}{2} \]

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📊 Visual Interpretation

The sum and difference of functions correspond to algebraic combinations of their outputs, while the quotient introduces restrictions where the denominator function becomes zero.

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🎯 Why This Question Matters

• CBSE Board Exams: Operations on functions are frequently tested.
• JEE Preparation: Function operations appear in composition, limits, and calculus.
• Concept Building: Helps understand algebraic manipulation of functions.
• Domain Awareness: Students learn to identify restrictions when dividing functions.

📘 Concept Theory

A function of the form

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

is called a linear function. If the function values are known for certain inputs, we can substitute those ordered pairs into the expression and form equations in \(a\) and \(b\).

Solving these equations gives the required constants defining the function.

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🧭 Solution Roadmap

• Use the ordered pairs provided by the function.
• Substitute them into \(f(x)=ax+b\).
• Form simultaneous equations in \(a\) and \(b\).
• Solve the system to determine the values of \(a\) and \(b\).

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✏️ Solution

The function is

\[ f(x)=ax+b \] Given ordered pairs: \[ (1,1), (2,3), (0,-1), (-1,-3) \] ### Using the pair \((1,1)\) \[ f(1)=a(1)+b \] \[ a+b=1 \] ### Using the pair \((2,3)\) \[ f(2)=2a+b \] \[ 2a+b=3 \] ### Solving the equations From \[ a+b=1 \] \[ a=1-b \] Substitute into the second equation: \[ 2(1-b)+b=3 \] \[ 2-2b+b=3 \] \[ -b=1 \] \[ b=-1 \] Now substitute into \(a+b=1\): \[ a-1=1 \] \[ a=2 \] ### Verification Check remaining ordered pairs: For \(x=0\) \[ f(0)=2(0)-1=-1 \] For \(x=-1\) \[ f(-1)=2(-1)-1=-3 \] Both match the given function values.

Therefore,

\[ (a,b)=(2,-1) \]

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📊 Graphical Interpretation

All given points lie on the straight line \(f(x)=2x-1\), confirming the values of \(a\) and \(b\).

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🎯 Why This Question Matters

• CBSE Board Exams: Determining parameters of linear functions from given values is a common problem.
• JEE Preparation: Similar problems appear when identifying functions from data points.
• Coordinate Geometry: Students understand that two points uniquely determine a straight line.
• Concept Foundation: Builds algebraic reasoning for function identification.

📘 Concept Theory

Relations may possess certain important properties such as reflexive, symmetric, and transitive.

• Reflexive: \((a,a)\in R\) for every \(a\) in the set.
• Symmetric: If \((a,b)\in R\), then \((b,a)\in R\).
• Transitive: If \((a,b)\in R\) and \((b,c)\in R\), then \((a,c)\in R\).

The given relation satisfies

\[ a=b^{2} \]

which means the first element of the ordered pair must be the square of the second element.

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🧭 Solution Roadmap

• Use the defining rule \(a=b^2\).
• Substitute appropriate values to test each statement.
• Check whether the required conditions hold for all natural numbers.

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✏️ Solution

The relation is

\[ R=\{(a,b):a,b\in\mathbb{N},\ a=b^{2}\} \]

(i) Checking whether \((a,a)\in R\)

For \((a,a)\) to belong to \(R\), \[ a=a^{2} \] This equation holds only when \[ a=0 \quad \text{or} \quad a=1 \] It is not true for all natural numbers.

Therefore, statement (i) is false.

(ii) Checking symmetry

Suppose \[ (a,b)\in R \] Then \[ a=b^{2} \] For \((b,a)\in R\), \[ b=a^{2} \] In general, this does not follow from \(a=b^{2}\). Example: \[ (4,2)\in R \] because \(4=2^{2}\). But \[ (2,4)\notin R \] since \(2\ne4^{2}\).

Therefore, statement (ii) is false.

(iii) Checking transitivity

Suppose \[ (a,b)\in R \] and \[ (b,c)\in R \] Then \[ a=b^{2} \] \[ b=c^{2} \] Substitute \(b=c^{2}\): \[ a=(c^{2})^{2}=c^{4} \] For \((a,c)\in R\), \[ a=c^{2} \] But we obtained \(a=c^{4}\), which is not equal to \(c^{2}\) in general.

Therefore, statement (iii) is also false.

Final Result

• (i) False
• (ii) False
• (iii) False

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📊 Relation Illustration

Example mapping \((4,2)\) shows that \(a=b^2\), illustrating the rule of the relation.

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🎯 Why This Question Matters

• CBSE Board Exams: Properties of relations are a frequently tested concept.
• JEE Preparation: Understanding reflexive, symmetric, and transitive relations is fundamental in discrete mathematics.
• Concept Building: Helps students analyze relations using definitions rather than memorization.
• Foundation for Equivalence Relations: Important for later topics such as equivalence classes and partitions.

📘 Concept Theory

A relation from a set \(A\) to a set \(B\) is any subset of the Cartesian product \(A\times B\).

\[ R\subseteq A\times B \]

A function is a special type of relation in which each element of the domain is associated with exactly one element of the codomain.

If any element of \(A\) is related to more than one element of \(B\), the relation cannot be a function.

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🧭 Solution Roadmap

• Verify that each ordered pair belongs to \(A\times B\).
• Check whether every element of \(A\) has exactly one image.
• Identify any element of \(A\) having multiple images.

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✏️ Solution

Given sets

\[ A=\{1,2,3,4\},\qquad B=\{1,5,9,11,15,16\} \] and relation \[ f=\{(1,5),(2,9),(3,1),(4,5),(2,11)\}. \]

(i) Checking whether \(f\) is a relation from \(A\) to \(B\)

In every ordered pair \((a,b)\in f\):

• the first element belongs to \(A\)
• the second element belongs to \(B\)

Thus \[ f\subseteq A\times B \]

Therefore, \(f\) is a relation from \(A\) to \(B\).

(ii) Checking whether \(f\) is a function

For a function, every element of \(A\) must have exactly one image.

Observe the element \(2\in A\): \[ (2,9)\in f \] \[ (2,11)\in f \]

Thus the same input \(2\) corresponds to two different outputs.

Therefore, \(f\) is not a function from \(A\) to \(B\).

Final Conclusion

• (i) True — \(f\) is a relation from \(A\) to \(B\).
• (ii) False — \(f\) is not a function.

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📊 Mapping Diagram

The element \(2\) maps to both \(9\) and \(11\), showing that the relation is not a function.

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🎯 Why This Question Matters

• CBSE Board Exams: Students must distinguish between relations and functions.
• JEE Preparation: Mapping diagrams and domain-image analysis are fundamental.
• Concept Building: Reinforces the idea that each input must have a unique output.
• Foundation Topic: Essential for studying function composition and inverse functions.

📘 Concept Theory

A function from a set \(A\) to a set \(B\) assigns exactly one element of \(B\) to each element of \(A\).

When a relation is written using ordered pairs, the first component represents the input and the second component represents the output.

If the same input is associated with two different outputs, the relation cannot be a function.

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🧭 Solution Roadmap

• Identify the first component of the ordered pair.
• Check whether a single input can arise from different values of \(a\) and \(b\).
• Determine whether this produces different outputs.
• If so, the relation is not a function.

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✏️ Solution

The relation is defined by

\[ f=\{(ab,a+b):a,b\in\mathbb{Z}\} \]

Thus the ordered pair has:

• First element: \(ab\)
• Second element: \(a+b\)

### Consider the input \(ab=2\) This value can be obtained in multiple ways. \[ 2=1\times2 \] \[ 2=(-1)\times(-2) \] ### Case 1 For \(a=1,\ b=2\) \[ (ab,a+b)=(2,1+2) \] \[ (2,3) \] ### Case 2 For \(a=-1,\ b=-2\) \[ (ab,a+b)=(2,-1+(-2)) \] \[ (2,-3) \]

Thus the same input \(2\) corresponds to two different outputs.

\[ (2,3)\in f,\qquad (2,-3)\in f \]

Therefore a unique image does not exist for every input.

Hence, \(f\) is not a function from \(\mathbb{Z}\) to \(\mathbb{Z}\).

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📊 Illustration

The same input \(2\) produces two outputs \(3\) and \(-3\), which violates the definition of a function.

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🎯 Why This Question Matters

• CBSE Board Exams: Students must distinguish between relations and functions.
• JEE Preparation: Similar reasoning appears in function definition problems.
• Concept Building: Reinforces the rule that every input must have exactly one output.
• Logical Reasoning: Encourages use of counterexamples to disprove statements.

📘 Concept Theory

A prime factor of a number is a prime number that divides it exactly. Every integer greater than 1 can be expressed as a product of prime numbers according to the Fundamental Theorem of Arithmetic.

The highest prime factor of a number is the largest prime number present in its prime factorization.

The range of a function is the set of all output values obtained when the function is evaluated for every element of the domain.

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🧭 Solution Roadmap

• Write the prime factorization of each element of set \(A\).
• Identify the highest prime factor in each case.
• Collect all distinct output values to determine the range.

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✏️ Solution

Given

\[ A=\{9,10,11,12,13\} \] and the function \[ f(n)=\text{highest prime factor of } n. \] ### Evaluate the function

\(n\)	Prime factorization	Highest prime factor \(f(n)\)
9	\(3\times3\)	3
10	\(2\times5\)	5
11	Prime number	11
12	\(2\times2\times3\)	3
13	Prime number	13

### Determine the range

The outputs obtained are

\[ 3,5,11,3,13 \]

The range consists of the distinct output values.

\[ R=\{3,5,11,13\} \]

Hence, the range of the function is \(R=\{3,5,11,13\}\).

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📊 Mapping Illustration

The mapping shows each element of set \(A\) mapped to its highest prime factor.

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🎯 Why This Question Matters

• CBSE Board Exams: Students must compute ranges of functions defined using number properties.
• JEE Preparation: Combines number theory (prime factors) with function evaluation.
• Concept Building: Reinforces how range is obtained by evaluating function outputs.
• Mathematical Thinking: Encourages systematic evaluation of domain elements.