Class 10 Mathematics - Ch 6: Functions and Graphs
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📘 Comprehensive Syllabus & Examination Guide
Class 10 Mathematics - Ch 6: Functions and Graphs
Official curriculum roadmap, subject/topic distribution, negative marking rules, pacing guidelines, and solved sample questions.
🎯 Question Types & Curriculum Breakdown
Total Question Pool
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460 Questions
Combined Active Syllabus
Short Questions
53%
244 Questions
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Multiple Choice (MCQs)
22%
100 MCQs
Available
Fill in the Blanks
9%
40 Questions
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True / False
7%
32 Questions
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Long / Theory Questions
6%
28 Questions
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Match Columns
3%
16 Questions
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📊 Question Pool Structure
460 Solved Questions (MCQs, Short & Long Questions, Blanks, True/False).
⚡ Recommended Pacing
1 to 3 minutes per question depending on question type (MCQ, Short, Long).
⚖️ Scoring & Negative Marking
1 to 5 marks per question aligned with official board examination rubrics.
💡 Strategic Preparation & Exam Hall Guidelines
To maximize your score on Class 10 Mathematics - Ch 6: Functions and Graphs, candidates are advised to follow a structured three-pass approach. In the First Pass, solve all direct recall and formula-based questions within 30 seconds each to secure foundational marks. In the Second Pass, tackle multi-step analytical and quantitative reasoning problems. In the Third Pass, review marked questions and verify calculations.
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📝 Pre-Rendered Solved Sample Questions & Detailed Solutions
Review the solved problems below to understand question phrasing, answer choices, and step-by-step solution logic prior to starting the full interactive practice drill:
Sample Question 1
Exercise 6.1 - Identifying Functions from Ordered Pairs
MEDIUM • Short Question
If $A = \{2, 4, 6, 8\}$ and $B = \{1, 3, 5\}$, determine whether the relation $R_1 = \{(2, 3), (6, 5), (8, 3), (4, 1)\}$ is a function from $A$ to $B$.
✓ Correct Answer: Yes, $R_1$ is a function from $A$ to $B$.
📖 Step-by-Step Solution & Conceptual Rationale:
<p><strong>Given Sets and Relation:</strong></p>
<p>$$A = \{2, 4, 6, 8\}, \quad B = \{1, 3, 5\}$$</p>
<p>$$R_1 = \{(2, 3), (6, 5), (8, 3), (4, 1)\}$$</p>
<p><strong>Step 1: Determine the domain of relation $R_1$:</strong></p>
<p>The domain consists of all first elements of the ordered pairs in $R_1$:</p>
<p>$$\operatorname{Dom}(R_1) = \{2, 6, 8, 4\} = \{2, 4, 6, 8\} = A$$</p>
<p><strong>Step 2: Check for repetition of first elements:</strong></p>
<p>Each element of set $A$ appears exactly once as the first component of an ordered pair. There is no repetition of any first element.</p>
<p><strong>Step 3: Check codomain membership:</strong></p>
<p>The second elements $\{3, 5, 1\}$ are all members of set $B$, so $\operatorname{Range}(R_1) \subseteq B$.</p>
<p><strong>Conclusion:</strong></p>
<p>Since $\operatorname{Dom}(R_1) = A$ and every element in $A$ is associated with a unique element in $B$, $R_1$ satisfies both criteria of a function.</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{\text{Yes, } R_1 \text{ is a function from } A \text{ to } B.}$$</p>
<p>$$A = \{2, 4, 6, 8\}, \quad B = \{1, 3, 5\}$$</p>
<p>$$R_1 = \{(2, 3), (6, 5), (8, 3), (4, 1)\}$$</p>
<p><strong>Step 1: Determine the domain of relation $R_1$:</strong></p>
<p>The domain consists of all first elements of the ordered pairs in $R_1$:</p>
<p>$$\operatorname{Dom}(R_1) = \{2, 6, 8, 4\} = \{2, 4, 6, 8\} = A$$</p>
<p><strong>Step 2: Check for repetition of first elements:</strong></p>
<p>Each element of set $A$ appears exactly once as the first component of an ordered pair. There is no repetition of any first element.</p>
<p><strong>Step 3: Check codomain membership:</strong></p>
<p>The second elements $\{3, 5, 1\}$ are all members of set $B$, so $\operatorname{Range}(R_1) \subseteq B$.</p>
<p><strong>Conclusion:</strong></p>
<p>Since $\operatorname{Dom}(R_1) = A$ and every element in $A$ is associated with a unique element in $B$, $R_1$ satisfies both criteria of a function.</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{\text{Yes, } R_1 \text{ is a function from } A \text{ to } B.}$$</p>
Sample Question 2
Exercise 6.1 - Identifying Functions from Ordered Pairs
MEDIUM • Short Question
If $A = \{2, 4, 6, 8\}$ and $B = \{1, 3, 5\}$, determine whether the relation $R_2 = \{(2, 3), (6, 1), (8, 3), (6, 5)\}$ is a function from $A$ to $B$.
✓ Correct Answer: No, $R_2$ is not a function.
📖 Step-by-Step Solution & Conceptual Rationale:
<p><strong>Given Sets and Relation:</strong></p>
<p>$$A = \{2, 4, 6, 8\}, \quad B = \{1, 3, 5\}$$</p>
<p>$$R_2 = \{(2, 3), (6, 1), (8, 3), (6, 5)\}$$</p>
<p><strong>Step 1: Check the first elements of the ordered pairs:</strong></p>
<p>Notice that the element $6 \in A$ appears as the first element in two distinct ordered pairs:</p>
<p>$$(6, 1) \in R_2 \quad \text{and} \quad (6, 5) \in R_2$$</p>
<p><strong>Step 2: Apply the definition of a function:</strong></p>
<p>By definition, a function cannot assign multiple distinct outputs to the same input element. Here, the input $6$ is mapped to two different images, $1$ and $5$.</p>
<p><strong>Step 3: Check domain completeness:</strong></p>
<p>Furthermore, element $4 \in A$ has no corresponding ordered pair in $R_2$, so $\operatorname{Dom}(R_2) = \{2, 6, 8\} \neq A$.</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{\text{No, } R_2 \text{ is not a function from } A \text{ to } B.}$$</p>
<p>$$A = \{2, 4, 6, 8\}, \quad B = \{1, 3, 5\}$$</p>
<p>$$R_2 = \{(2, 3), (6, 1), (8, 3), (6, 5)\}$$</p>
<p><strong>Step 1: Check the first elements of the ordered pairs:</strong></p>
<p>Notice that the element $6 \in A$ appears as the first element in two distinct ordered pairs:</p>
<p>$$(6, 1) \in R_2 \quad \text{and} \quad (6, 5) \in R_2$$</p>
<p><strong>Step 2: Apply the definition of a function:</strong></p>
<p>By definition, a function cannot assign multiple distinct outputs to the same input element. Here, the input $6$ is mapped to two different images, $1$ and $5$.</p>
<p><strong>Step 3: Check domain completeness:</strong></p>
<p>Furthermore, element $4 \in A$ has no corresponding ordered pair in $R_2$, so $\operatorname{Dom}(R_2) = \{2, 6, 8\} \neq A$.</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{\text{No, } R_2 \text{ is not a function from } A \text{ to } B.}$$</p>
Sample Question 3
Exercise 6.1 - Identifying Functions from Ordered Pairs
MEDIUM • Short Question
If $A = \{2, 4, 6, 8\}$ and $B = \{1, 3, 5\}$, determine whether the relation $R_3 = \{(2, 3), (6, 5), (8, 3)\}$ is a function from $A$ to $B$.
✓ Correct Answer: No, $R_3$ is not a function.
📖 Step-by-Step Solution & Conceptual Rationale:
<p><strong>Given Sets and Relation:</strong></p>
<p>$$A = \{2, 4, 6, 8\}, \quad B = \{1, 3, 5\}$$</p>
<p>$$R_3 = \{(2, 3), (6, 5), (8, 3)\}$$</p>
<p><strong>Step 1: Determine the domain of relation $R_3$:</strong></p>
<p>Extracting the first elements of the ordered pairs:</p>
<p>$$\operatorname{Dom}(R_3) = \{2, 6, 8\}$$</p>
<p><strong>Step 2: Compare $\operatorname{Dom}(R_3)$ with set $A$:</strong></p>
<p>For a relation to be a function from set $A$ to set $B$, its domain must be strictly equal to set $A$ ($\operatorname{Dom}(f) = A$).</p>
<p>Here, $4 \in A$ but $4 \notin \operatorname{Dom}(R_3)$. The element $4$ has no image in $B$.</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{\text{No, } R_3 \text{ is not a function from } A \text{ to } B.}$$</p>
<p>$$A = \{2, 4, 6, 8\}, \quad B = \{1, 3, 5\}$$</p>
<p>$$R_3 = \{(2, 3), (6, 5), (8, 3)\}$$</p>
<p><strong>Step 1: Determine the domain of relation $R_3$:</strong></p>
<p>Extracting the first elements of the ordered pairs:</p>
<p>$$\operatorname{Dom}(R_3) = \{2, 6, 8\}$$</p>
<p><strong>Step 2: Compare $\operatorname{Dom}(R_3)$ with set $A$:</strong></p>
<p>For a relation to be a function from set $A$ to set $B$, its domain must be strictly equal to set $A$ ($\operatorname{Dom}(f) = A$).</p>
<p>Here, $4 \in A$ but $4 \notin \operatorname{Dom}(R_3)$. The element $4$ has no image in $B$.</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{\text{No, } R_3 \text{ is not a function from } A \text{ to } B.}$$</p>
Sample Question 4
Exercise 6.1 - Identifying Functions from Ordered Pairs
MEDIUM • Short Question
If $A = \{2, 4, 6, 8\}$ and $B = \{1, 3, 5\}$, determine whether the relation $R_4 = \{(2, 3), (6, 3), (8, 3), (4, 1)\}$ is a function from $A$ to $B$.
✓ Correct Answer: Yes, $R_4$ is a function from $A$ to $B$.
📖 Step-by-Step Solution & Conceptual Rationale:
<p><strong>Given Sets and Relation:</strong></p>
<p>$$A = \{2, 4, 6, 8\}, \quad B = \{1, 3, 5\}$$</p>
<p>$$R_4 = \{(2, 3), (6, 3), (8, 3), (4, 1)\}$$</p>
<p><strong>Step 1: Determine the domain:</strong></p>
<p>$$\operatorname{Dom}(R_4) = \{2, 6, 8, 4\} = \{2, 4, 6, 8\} = A$$</p>
<p><strong>Step 2: Check uniqueness of images:</strong></p>
<p>Every element in $A$ appears exactly once as the first component of an ordered pair. There is no repetition in the first coordinates.</p>
<p><em>Note:</em> The second coordinate $3$ is shared by inputs $2, 6, 8$. In a function, multiple distinct inputs may produce the same output (many-to-one mapping). This is completely valid.</p>
<p><strong>Step 3: Verify range:</strong></p>
<p>$$\operatorname{Range}(R_4) = \{1, 3\} \subseteq B$$</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{\text{Yes, } R_4 \text{ is a function from } A \text{ to } B.}$$</p>
<p>$$A = \{2, 4, 6, 8\}, \quad B = \{1, 3, 5\}$$</p>
<p>$$R_4 = \{(2, 3), (6, 3), (8, 3), (4, 1)\}$$</p>
<p><strong>Step 1: Determine the domain:</strong></p>
<p>$$\operatorname{Dom}(R_4) = \{2, 6, 8, 4\} = \{2, 4, 6, 8\} = A$$</p>
<p><strong>Step 2: Check uniqueness of images:</strong></p>
<p>Every element in $A$ appears exactly once as the first component of an ordered pair. There is no repetition in the first coordinates.</p>
<p><em>Note:</em> The second coordinate $3$ is shared by inputs $2, 6, 8$. In a function, multiple distinct inputs may produce the same output (many-to-one mapping). This is completely valid.</p>
<p><strong>Step 3: Verify range:</strong></p>
<p>$$\operatorname{Range}(R_4) = \{1, 3\} \subseteq B$$</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{\text{Yes, } R_4 \text{ is a function from } A \text{ to } B.}$$</p>
Sample Question 5
Exercise 6.1 - Function Mapping & Classification via Venn Diagrams
MEDIUM • Short Question
Identify whether the mapping $f: A \to B$ shown in the arrow diagram is a function. If so, specify its type: Set $A = \{2, 3, 5\}$ maps to Set $B = \{1, 3, 5\}$ with arrows $2 \to 3$, $3 \to 5$, and $5 \to 1$.
✓ Correct Answer: Yes, Bijective Function (One-to-One and Onto).
📖 Step-by-Step Solution & Conceptual Rationale:
<p><strong>Given Arrow Diagram:</strong></p>
<p>$$A = \{2, 3, 5\}, \quad B = \{1, 3, 5\}$$</p>
<p>$$f = \{(2, 3), (3, 5), (5, 1)\}$$</p>
<p><strong>Step 1: Check if $f$ is a function:</strong></p>
<p>$\operatorname{Dom}(f) = \{2, 3, 5\} = A$. Every element in $A$ has exactly one outgoing arrow (a unique image in $B$). Therefore, $f$ is a function.</p>
<p><strong>Step 2: Check if $f$ is One-to-One (Injective):</strong></p>
<p>Distinct elements of set $A$ have distinct images in set $B$ ($f(2) = 3$, $f(3) = 5$, $f(5) = 1$). No two arrows point to the same element in $B$. Thus, $f$ is one-to-one (injective).</p>
<p><strong>Step 3: Check if $f$ is Onto (Surjective):</strong></p>
<p>$$\operatorname{Range}(f) = \{1, 3, 5\} = B$$</p>
<p>Every element of the codomain $B$ has at least one pre-image in $A$. Thus, $f$ is onto (surjective).</p>
<p><strong>Conclusion:</strong></p>
<p>A function that is both one-to-one and onto is called a <strong>bijective function</strong>.</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{\text{Yes, Bijective Function (One-to-One and Onto)}}$$</p>
<p>$$A = \{2, 3, 5\}, \quad B = \{1, 3, 5\}$$</p>
<p>$$f = \{(2, 3), (3, 5), (5, 1)\}$$</p>
<p><strong>Step 1: Check if $f$ is a function:</strong></p>
<p>$\operatorname{Dom}(f) = \{2, 3, 5\} = A$. Every element in $A$ has exactly one outgoing arrow (a unique image in $B$). Therefore, $f$ is a function.</p>
<p><strong>Step 2: Check if $f$ is One-to-One (Injective):</strong></p>
<p>Distinct elements of set $A$ have distinct images in set $B$ ($f(2) = 3$, $f(3) = 5$, $f(5) = 1$). No two arrows point to the same element in $B$. Thus, $f$ is one-to-one (injective).</p>
<p><strong>Step 3: Check if $f$ is Onto (Surjective):</strong></p>
<p>$$\operatorname{Range}(f) = \{1, 3, 5\} = B$$</p>
<p>Every element of the codomain $B$ has at least one pre-image in $A$. Thus, $f$ is onto (surjective).</p>
<p><strong>Conclusion:</strong></p>
<p>A function that is both one-to-one and onto is called a <strong>bijective function</strong>.</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{\text{Yes, Bijective Function (One-to-One and Onto)}}$$</p>
Sample Question 6
Exercise 6.1 - Function Mapping & Classification via Venn Diagrams
MEDIUM • Short Question
Identify whether the mapping $g: P \to Q$ shown in the arrow diagram is a function: Set $P = \{0, 2, 4\}$ and Set $Q = \{5, 7\}$ with arrows $0 \to 5$ and $2 \to 7$, while element $4$ has no outgoing arrow.
✓ Correct Answer: No, $g$ is not a function.
📖 Step-by-Step Solution & Conceptual Rationale:
<p><strong>Given Arrow Diagram:</strong></p>
<p>$$P = \{0, 2, 4\}, \quad Q = \{5, 7\}$$</p>
<p>$$g = \{(0, 5), (2, 7)\}$$</p>
<p><strong>Step 1: Examine the domain of mapping $g$:</strong></p>
<p>The domain of $g$ consists of all elements in $P$ that have outgoing arrows:</p>
<p>$$\operatorname{Dom}(g) = \{0, 2\}$$</p>
<p><strong>Step 2: Compare $\operatorname{Dom}(g)$ with domain set $P$:</strong></p>
<p>$$\operatorname{Dom}(g) = \{0, 2\} \neq P = \{0, 2, 4\}$$</p>
<p>The element $4 \in P$ is not mapped to any element in set $Q$ (it has no image).</p>
<p><strong>Conclusion:</strong></p>
<p>By definition, a function from set $P$ to set $Q$ requires that every single element in set $P$ must have an image in $Q$. Since element $4$ has no image, $g$ is not a function.</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{\text{No, } g \text{ is not a function.}}$$</p>
<p>$$P = \{0, 2, 4\}, \quad Q = \{5, 7\}$$</p>
<p>$$g = \{(0, 5), (2, 7)\}$$</p>
<p><strong>Step 1: Examine the domain of mapping $g$:</strong></p>
<p>The domain of $g$ consists of all elements in $P$ that have outgoing arrows:</p>
<p>$$\operatorname{Dom}(g) = \{0, 2\}$$</p>
<p><strong>Step 2: Compare $\operatorname{Dom}(g)$ with domain set $P$:</strong></p>
<p>$$\operatorname{Dom}(g) = \{0, 2\} \neq P = \{0, 2, 4\}$$</p>
<p>The element $4 \in P$ is not mapped to any element in set $Q$ (it has no image).</p>
<p><strong>Conclusion:</strong></p>
<p>By definition, a function from set $P$ to set $Q$ requires that every single element in set $P$ must have an image in $Q$. Since element $4$ has no image, $g$ is not a function.</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{\text{No, } g \text{ is not a function.}}$$</p>
Sample Question 7
Exercise 6.1 - Function Mapping & Classification via Venn Diagrams
MEDIUM • Short Question
Identify whether the mapping $h: X \to Y$ shown in the arrow diagram is a function. If so, specify its type: Set $X = \{0, 2, 4\}$ and Set $Y = \{a, b, c\}$ with arrows $0 \to b$, $2 \to c$, and $4 \to c$.
✓ Correct Answer: Yes, Into Function.
📖 Step-by-Step Solution & Conceptual Rationale:
<p><strong>Given Arrow Diagram:</strong></p>
<p>$$X = \{0, 2, 4\}, \quad Y = \{a, b, c\}$$</p>
<p>$$h = \{(0, b), (2, c), (4, c)\}$$</p>
<p><strong>Step 1: Check if $h$ is a function:</strong></p>
<p>$$\operatorname{Dom}(h) = \{0, 2, 4\} = X$$</p>
<p>Every element in $X$ has exactly one outgoing arrow to set $Y$. Hence, $h$ is a function.</p>
<p><strong>Step 2: Find the range of $h$:</strong></p>
<p>$$\operatorname{Range}(h) = \{b, c\}$$</p>
<p><strong>Step 3: Compare range with codomain set $Y$:</strong></p>
<p>Notice that $\operatorname{Range}(h) = \{b, c\} \subset Y = \{a, b, c\}$.</p>
<p>Element $a \in Y$ has no pre-image in $X$ (no arrow points to $a$), which means $\operatorname{Range}(h) \neq Y$.</p>
<p><strong>Conclusion:</strong></p>
<p>A function $f: X \to Y$ is called an <strong>into function</strong> if there is at least one element in set $Y$ which is not the image of any element of set $X$ ($\operatorname{Range}(f) \subset Y$).</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{\text{Yes, Into Function.}}$$</p>
<p>$$X = \{0, 2, 4\}, \quad Y = \{a, b, c\}$$</p>
<p>$$h = \{(0, b), (2, c), (4, c)\}$$</p>
<p><strong>Step 1: Check if $h$ is a function:</strong></p>
<p>$$\operatorname{Dom}(h) = \{0, 2, 4\} = X$$</p>
<p>Every element in $X$ has exactly one outgoing arrow to set $Y$. Hence, $h$ is a function.</p>
<p><strong>Step 2: Find the range of $h$:</strong></p>
<p>$$\operatorname{Range}(h) = \{b, c\}$$</p>
<p><strong>Step 3: Compare range with codomain set $Y$:</strong></p>
<p>Notice that $\operatorname{Range}(h) = \{b, c\} \subset Y = \{a, b, c\}$.</p>
<p>Element $a \in Y$ has no pre-image in $X$ (no arrow points to $a$), which means $\operatorname{Range}(h) \neq Y$.</p>
<p><strong>Conclusion:</strong></p>
<p>A function $f: X \to Y$ is called an <strong>into function</strong> if there is at least one element in set $Y$ which is not the image of any element of set $X$ ($\operatorname{Range}(f) \subset Y$).</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{\text{Yes, Into Function.}}$$</p>
Sample Question 8
Exercise 6.1 - Evaluating Algebraic Functions
MEDIUM • Short Question
If $f(x) = x^2 - \frac{1}{2}x + 3$, evaluate $f(2)$.
✓ Correct Answer: $6$
📖 Step-by-Step Solution & Conceptual Rationale:
<p><strong>Given Function:</strong></p>
<p>$$f(x) = x^2 - \frac{1}{2}x + 3$$</p>
<p><strong>Step 1: Substitute $x = 2$ into the function:</strong></p>
<p>$$f(2) = (2)^2 - \frac{1}{2}(2) + 3$$</p>
<p><strong>Step 2: Simplify each term:</strong></p>
<p>$$(2)^2 = 4$$</p>
<p>$$\frac{1}{2}(2) = 1$$</p>
<p><strong>Step 3: Perform addition and subtraction:</strong></p>
<p>$$f(2) = 4 - 1 + 3 = 3 + 3 = 6$$</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{6}$$</p>
<p>$$f(x) = x^2 - \frac{1}{2}x + 3$$</p>
<p><strong>Step 1: Substitute $x = 2$ into the function:</strong></p>
<p>$$f(2) = (2)^2 - \frac{1}{2}(2) + 3$$</p>
<p><strong>Step 2: Simplify each term:</strong></p>
<p>$$(2)^2 = 4$$</p>
<p>$$\frac{1}{2}(2) = 1$$</p>
<p><strong>Step 3: Perform addition and subtraction:</strong></p>
<p>$$f(2) = 4 - 1 + 3 = 3 + 3 = 6$$</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{6}$$</p>
Sample Question 9
Exercise 6.1 - Evaluating Algebraic Functions
MEDIUM • Short Question
If $f(x) = x^2 - \frac{1}{2}x + 3$, evaluate $f(-1)$.
✓ Correct Answer: $\frac{9}{2}$
📖 Step-by-Step Solution & Conceptual Rationale:
<p><strong>Given Function:</strong></p>
<p>$$f(x) = x^2 - \frac{1}{2}x + 3$$</p>
<p><strong>Step 1: Substitute $x = -1$ into the function:</strong></p>
<p>$$f(-1) = (-1)^2 - \frac{1}{2}(-1) + 3$$</p>
<p><strong>Step 2: Simplify arithmetic operations:</strong></p>
<p>$$(-1)^2 = 1$$</p>
<p>$$-\frac{1}{2}(-1) = +\frac{1}{2}$$</p>
<p>$$f(-1) = 1 + \frac{1}{2} + 3 = 4 + \frac{1}{2}$$</p>
<p><strong>Step 3: Express as an improper fraction:</strong></p>
<p>$$4 + \frac{1}{2} = \frac{4 \times 2 + 1}{2} = \frac{9}{2}$$</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{\frac{9}{2}}$$</p>
<p>$$f(x) = x^2 - \frac{1}{2}x + 3$$</p>
<p><strong>Step 1: Substitute $x = -1$ into the function:</strong></p>
<p>$$f(-1) = (-1)^2 - \frac{1}{2}(-1) + 3$$</p>
<p><strong>Step 2: Simplify arithmetic operations:</strong></p>
<p>$$(-1)^2 = 1$$</p>
<p>$$-\frac{1}{2}(-1) = +\frac{1}{2}$$</p>
<p>$$f(-1) = 1 + \frac{1}{2} + 3 = 4 + \frac{1}{2}$$</p>
<p><strong>Step 3: Express as an improper fraction:</strong></p>
<p>$$4 + \frac{1}{2} = \frac{4 \times 2 + 1}{2} = \frac{9}{2}$$</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{\frac{9}{2}}$$</p>
Sample Question 10
Exercise 6.1 - Evaluating Algebraic Functions
MEDIUM • Short Question
If $f(x) = x^2 - \frac{1}{2}x + 3$, evaluate $f\left(\frac{2}{3}\right)$.
✓ Correct Answer: $\frac{28}{9}$
📖 Step-by-Step Solution & Conceptual Rationale:
<p><strong>Given Function:</strong></p>
<p>$$f(x) = x^2 - \frac{1}{2}x + 3$$</p>
<p><strong>Step 1: Substitute $x = \frac{2}{3}$ into the function:</strong></p>
<p>$$f\left(\frac{2}{3}\right) = \left(\frac{2}{3}\right)^2 - \frac{1}{2}\left(\frac{2}{3}\right) + 3$$</p>
<p><strong>Step 2: Simplify each term:</strong></p>
<p>$$\left(\frac{2}{3}\right)^2 = \frac{4}{9}$$</p>
<p>$$\frac{1}{2}\left(\frac{2}{3}\right) = \frac{1}{3} = \frac{3}{9}$$</p>
<p>$$3 = \frac{27}{9}$$</p>
<p><strong>Step 3: Combine numerators over the common denominator $9$:</strong></p>
<p>$$f\left(\frac{2}{3}\right) = \frac{4}{9} - \frac{3}{9} + \frac{27}{9} = \frac{4 - 3 + 27}{9} = \frac{28}{9}$$</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{\frac{28}{9}}$$</p>
<p>$$f(x) = x^2 - \frac{1}{2}x + 3$$</p>
<p><strong>Step 1: Substitute $x = \frac{2}{3}$ into the function:</strong></p>
<p>$$f\left(\frac{2}{3}\right) = \left(\frac{2}{3}\right)^2 - \frac{1}{2}\left(\frac{2}{3}\right) + 3$$</p>
<p><strong>Step 2: Simplify each term:</strong></p>
<p>$$\left(\frac{2}{3}\right)^2 = \frac{4}{9}$$</p>
<p>$$\frac{1}{2}\left(\frac{2}{3}\right) = \frac{1}{3} = \frac{3}{9}$$</p>
<p>$$3 = \frac{27}{9}$$</p>
<p><strong>Step 3: Combine numerators over the common denominator $9$:</strong></p>
<p>$$f\left(\frac{2}{3}\right) = \frac{4}{9} - \frac{3}{9} + \frac{27}{9} = \frac{4 - 3 + 27}{9} = \frac{28}{9}$$</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{\frac{28}{9}}$$</p>
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