Prepared by Muhammad Tayyab, Subject Specialist Mathematics, Govt Christian High School Daska
📌 Based on National Curriculum 2023 / PECTAA 2026 Syllabus
📖 What's Inside: This exercise covers operations on functions (addition, subtraction, multiplication, division), composite functions (\(f \circ g\), \(g \circ f\)), inverse functions \(f^{-1}(x)\), and verifying properties. Step-by-step solutions with domain considerations. Perfect for Punjab Boards exam preparation.
📚 Related Resources – Unit 4: Functions and Graphs
Class 10 Math Unit 4 Exercise 4.1 – Functions and Graphs: Complete Guide
Exercise 4.1 of Unit 4 introduces students to the fundamental concepts of functions – operations on functions, composition of functions, and inverse functions. This exercise builds the foundation for understanding how functions can be combined, transformed, and inverted, which is essential for advanced mathematics.
What You Will Learn
By working through Exercise 4.1, students master: performing operations on functions (addition \(f+g\), subtraction \(f-g\), multiplication \(fg\), division \(f/g\)), composing functions \((f \circ g)(x) = f(g(x))\) and \((g \circ f)(x) = g(f(x))\), finding inverse functions \(f^{-1}(x)\) using the swap-and-solve method, verifying that \(f(f^{-1}(x)) = f^{-1}(f(x)) = x\), and determining domain and range of inverse functions without explicitly finding them.
Topics Covered in This Exercise
- Function Operations – addition, subtraction, multiplication, division
- Composite Functions – \((f \circ g)(x)\) and \((g \circ f)(x)\)
- Inverse Functions – finding \(f^{-1}(x)\) using the swap-and-solve method
- Verification – proving \(f(f^{-1}(x)) = f^{-1}(f(x)) = x\)
- Domain and Range – finding domain and range of inverse functions
Why Exercise 4.1 Is Important
Exercise 4.1 is the foundation of Unit 4. Board exam papers frequently feature questions on function operations, composition, and inverses. This exercise builds essential skills that are used throughout the rest of the unit and in later mathematics courses.
Punjab Board Preparation
Students preparing for board exams under any of the 10 BISE Punjab boards should prioritise this exercise. Function operations, composition, and inverses are frequently tested topics. Practising every part of Q1-Q10 by hand is highly recommended.
Exam Tips for Functions
- Remember that \((f \circ g)(x) = f(g(x))\) – apply \(g\) first, then \(f\).
- For inverse functions, swap \(x\) and \(y\), then solve for \(y\).
- Always check domain restrictions when dividing functions.
- Verify inverse by checking \(f(f^{-1}(x)) = x\).
Common Mistakes Students Make
- Confusing \((f \circ g)(x)\) with \((g \circ f)(x)\) – order matters!
- Forgetting domain restrictions when dividing functions.
- Not swapping variables correctly when finding inverses.
- Forgetting to check inverse verification.
📖 Exercise 4.1 – Solved Problems
1
If \(f(x) = 2x + 5\), \(g(x) = 3x - 2\), then find
(i) \(f(x) + g(x)\)
\[ \begin{aligned}
f(x) + g(x) &= (2x + 5) + (3x - 2) \\
&= 2x + 5 + 3x - 2 \\
&= 5x + 3
\end{aligned} \]
(ii) \(f(x) - g(x)\)
\[ \begin{aligned}
f(x) - g(x) &= (2x + 5) - (3x - 2) \\
&= 2x + 5 - 3x + 2 \\
&= -x + 7
\end{aligned} \]
2
If \(f(x) = x + 2\), \(g(x) = 2x + 4\), then find
(i) \(f(x) \cdot g(x)\)
\[ \begin{aligned}
f(x) \cdot g(x) &= (x + 2)(2x + 4) \\
&= x(2x + 4) + 2(2x + 4) \\
&= 2x^2 + 4x + 4x + 8 \\
&= 2x^2 + 8x + 8 \\
&= 2(x^2 + 4x + 4)
\end{aligned} \]
(ii) \(g(x) \cdot f(x)\)
\[ \begin{aligned}
g(x) \cdot f(x) &= (2x + 4)(x + 2) \\
&= 2x(x + 2) + 4(x + 2) \\
&= 2x^2 + 4x + 4x + 8 \\
&= 2x^2 + 8x + 8 \\
&= 2(x^2 + 4x + 4)
\end{aligned} \]
(iii) \(\frac{f(x)}{g(x)}\)
\[ \begin{aligned}
\frac{f(x)}{g(x)} &= \frac{x + 2}{2x + 4} \\
&= \frac{x + 2}{2(x + 2)} = \frac{1}{2}
\end{aligned} \]
(iv) \(\frac{g(x)}{f(x)}\)
\[ \begin{aligned}
\frac{g(x)}{f(x)} &= \frac{2x + 4}{x + 2} \\
&= \frac{2(x + 2)}{x + 2} = 2
\end{aligned} \]
3
For the functions \(f\) and \(g\), find \((f \circ g)(x)\), \((g \circ f)(x)\), \((f \circ f)(x)\), \((g \circ g)(x)\)
(i) \(f(x) = 2x + 3,\; g(x) = x^3\)
\((f \circ g)(x) = f(g(x)) = f(x^3) = 2x^3 + 3\)
\((g \circ f)(x) = g(f(x)) = g(2x + 3) = (2x + 3)^3\)
\((f \circ f)(x) = f(f(x)) = f(2x + 3) = 2(2x + 3) + 3 = 4x + 9\)
\((g \circ g)(x) = g(g(x)) = g(x^3) = (x^3)^3 = x^9\)
\((f \circ g)(x) = f(g(x)) = f(x^3) = 2x^3 + 3\)
\((g \circ f)(x) = g(f(x)) = g(2x + 3) = (2x + 3)^3\)
\((f \circ f)(x) = f(f(x)) = f(2x + 3) = 2(2x + 3) + 3 = 4x + 9\)
\((g \circ g)(x) = g(g(x)) = g(x^3) = (x^3)^3 = x^9\)
(ii) \(f(x) = \frac{2}{x},\; x \neq 0;\quad g(x) = 2x^2 - 1\)
\((f \circ g)(x) = f(g(x)) = f(2x^2 - 1) = \frac{2}{2x^2 - 1}\)
\((g \circ f)(x) = g(f(x)) = g\left(\frac{2}{x}\right) = 2\left(\frac{2}{x}\right)^2 - 1 = \frac{8}{x^2} - 1 = \frac{8 - x^2}{x^2}\)
\((f \circ f)(x) = f(f(x)) = f\left(\frac{2}{x}\right) = \frac{2}{2/x} = 2 \times \frac{x}{2} = x\)
\((g \circ g)(x) = g(g(x)) = g(2x^2 - 1) = 2(2x^2 - 1)^2 - 1 = 8x^4 - 8x^2 + 1\)
\((f \circ g)(x) = f(g(x)) = f(2x^2 - 1) = \frac{2}{2x^2 - 1}\)
\((g \circ f)(x) = g(f(x)) = g\left(\frac{2}{x}\right) = 2\left(\frac{2}{x}\right)^2 - 1 = \frac{8}{x^2} - 1 = \frac{8 - x^2}{x^2}\)
\((f \circ f)(x) = f(f(x)) = f\left(\frac{2}{x}\right) = \frac{2}{2/x} = 2 \times \frac{x}{2} = x\)
\((g \circ g)(x) = g(g(x)) = g(2x^2 - 1) = 2(2x^2 - 1)^2 - 1 = 8x^4 - 8x^2 + 1\)
(iii) \(f(x) = 2x - 1,\; x \neq 0;\quad g(x) = \frac{x+1}{2}\)
\((f \circ g)(x) = f(g(x)) = f\left(\frac{x+1}{2}\right) = 2\left(\frac{x+1}{2}\right) - 1 = x + 1 - 1 = x\)
\((g \circ f)(x) = g(f(x)) = g(2x - 1) = \frac{(2x - 1) + 1}{2} = \frac{2x}{2} = x\)
\((f \circ f)(x) = f(f(x)) = f(2x - 1) = 2(2x - 1) - 1 = 4x - 3\)
\((g \circ g)(x) = g(g(x)) = g\left(\frac{x+1}{2}\right) = \frac{(x+1)/2 + 1}{2} = \frac{x+3}{4}\)
\((f \circ g)(x) = f(g(x)) = f\left(\frac{x+1}{2}\right) = 2\left(\frac{x+1}{2}\right) - 1 = x + 1 - 1 = x\)
\((g \circ f)(x) = g(f(x)) = g(2x - 1) = \frac{(2x - 1) + 1}{2} = \frac{2x}{2} = x\)
\((f \circ f)(x) = f(f(x)) = f(2x - 1) = 2(2x - 1) - 1 = 4x - 3\)
\((g \circ g)(x) = g(g(x)) = g\left(\frac{x+1}{2}\right) = \frac{(x+1)/2 + 1}{2} = \frac{x+3}{4}\)
4
Find \(k\) such that \((f \circ g)(x) = (g \circ f)(x)\) where \(f(x) = 3x + 2\), \(g(x) = 6x - k\)
\[ \begin{aligned}
(f \circ g)(x) &= f(g(x)) = f(6x - k) = 3(6x - k) + 2 = 18x - 3k + 2 \\
(g \circ f)(x) &= g(f(x)) = g(3x + 2) = 6(3x + 2) - k = 18x + 12 - k \\
\text{Given: } & 18x - 3k + 2 = 18x + 12 - k \\
& -3k + 2 = 12 - k \\
& -3k + k = 12 - 2 \\
& -2k = 10 \\
& k = -5
\end{aligned} \]
5
Given \(f(x) = 3x + 2\) and \(g(x) = 2x + 3\). Find
(i) \(g(f(4))\)
\[ \begin{aligned}
f(4) &= 3(4) + 2 = 12 + 2 = 14 \\
g(f(4)) &= g(14) = 2(14) + 3 = 28 + 3 = 31
\end{aligned} \]
(ii) \(f(f(3))\)
\[ \begin{aligned}
f(3) &= 3(3) + 2 = 9 + 2 = 11 \\
f(f(3)) &= f(11) = 3(11) + 2 = 33 + 2 = 35
\end{aligned} \]
(iii) \(f(g(-2))\)
\[ \begin{aligned}
g(-2) &= 2(-2) + 3 = -4 + 3 = -1 \\
f(g(-2)) &= f(-1) = 3(-1) + 2 = -3 + 2 = -1
\end{aligned} \]
6
Find \(f^{-1}(x)\) in each of the following:
(i) \(f(x) = 2x - 3\)
\[ \begin{aligned}
y &= 2x - 3 \implies y + 3 = 2x \implies x = \frac{y+3}{2} \\
f^{-1}(x) &= \frac{x+3}{2}
\end{aligned} \]
(ii) \(f(x) = 4x^3 - 1\)
\[ \begin{aligned}
y &= 4x^3 - 1 \implies y+1 = 4x^3 \implies x^3 = \frac{y+1}{4} \\
f^{-1}(x) &= \sqrt[3]{\frac{x+1}{4}}
\end{aligned} \]
(iii) \(f(x) = \sqrt{x-1},\; x \ge 1\)
\[ \begin{aligned}
y &= \sqrt{x-1} \implies y^2 = x-1 \implies x = y^2 + 1 \\
f^{-1}(x) &= x^2 + 1
\end{aligned} \]
(iv) \(f(x) = \frac{x+1}{3x-2},\; x \neq \frac{2}{3}\)
\[ \begin{aligned}
y &= \frac{x+1}{3x-2} \implies y(3x-2) = x+1 \\
3xy - 2y &= x+1 \implies 3xy - x = 2y + 1 \\
x(3y - 1) &= 2y + 1 \implies x = \frac{2y+1}{3y-1} \\
f^{-1}(x) &= \frac{2x+1}{3x-1}
\end{aligned} \]
7
\(f(x) = 4x + 2\), \(g(x) = 6x - 18\). Find \(f^{-1}(x)\), \(g^{-1}(x)\) and \(x\) if \(f^{-1}(x) = g^{-1}(x)\)
\(f^{-1}(x)\):
\[ \begin{aligned}
y &= 4x + 2 \implies y-2 = 4x \implies x = \frac{y-2}{4} \\
f^{-1}(x) &= \frac{x-2}{4}
\end{aligned} \]
\(g^{-1}(x)\):
\[ \begin{aligned}
y &= 6x - 18 \implies y+18 = 6x \implies x = \frac{y+18}{6} \\
g^{-1}(x) &= \frac{x+18}{6}
\end{aligned} \]
Solve \(f^{-1}(x) = g^{-1}(x)\):
\[ \begin{aligned}
\frac{x-2}{4} &= \frac{x+18}{6} \implies 6(x-2) = 4(x+18) \\
6x - 12 &= 4x + 72 \implies 2x = 84 \implies x = 42
\end{aligned} \]
8
Verify \(f(f^{-1}(x)) = f^{-1}(f(x)) = x\)
(i) \(f(x) = x - 6\)
\[ \begin{aligned}
y = x-6 &\implies x = y+6 \implies f^{-1}(x) = x+6 \\
f(f^{-1}(x)) &= f(x+6) = (x+6)-6 = x \\
f^{-1}(f(x)) &= f^{-1}(x-6) = (x-6)+6 = x
\end{aligned} \]
(ii) \(f(x) = 7x - 4\)
\[ \begin{aligned}
y = 7x-4 &\implies y+4 = 7x \implies x = \frac{y+4}{7} \implies f^{-1}(x) = \frac{x+4}{7} \\
f(f^{-1}(x)) &= f\left(\frac{x+4}{7}\right) = 7 \cdot \frac{x+4}{7} - 4 = x+4-4 = x \\
f^{-1}(f(x)) &= f^{-1}(7x-4) = \frac{(7x-4)+4}{7} = \frac{7x}{7} = x
\end{aligned} \]
(iii) \(f(x) = \frac{x-4}{x+2},\; x \neq -2\)
\[ \begin{aligned}
y &= \frac{x-4}{x+2} \implies y(x+2) = x-4 \implies xy + 2y = x-4 \\
xy - x &= -4 - 2y \implies x(y-1) = -4-2y \implies x = \frac{-4-2y}{y-1} \\
f^{-1}(x) &= \frac{-4-2x}{x-1} \\
f(f^{-1}(x)) &= f\left(\frac{-4-2x}{x-1}\right) = \frac{\frac{-4-2x}{x-1} - 4}{\frac{-4-2x}{x-1} + 2}
= \frac{-4-2x -4(x-1)}{-4-2x +2(x-1)} = \frac{-6x}{-6} = x \\
f^{-1}(f(x)) &= f^{-1}\left(\frac{x-4}{x+2}\right) = \frac{-4 - 2\left(\frac{x-4}{x+2}\right)}{\left(\frac{x-4}{x+2}\right) - 1}
= \frac{-4(x+2)-2(x-4)}{x-4 - (x+2)} = \frac{-6x}{-6} = x
\end{aligned} \]
9
Without finding \(f^{-1}(x)\), find domain and range of \(f^{-1}(x)\):
(i) \(f(x) = 12x - 3\)
Domain of \(f\): \((-\infty, \infty)\) Range of \(f\): \((-\infty, \infty)\)
Domain of \(f^{-1}\): \((-\infty, \infty)\) Range of \(f^{-1}\): \((-\infty, \infty)\)
Domain of \(f\): \((-\infty, \infty)\) Range of \(f\): \((-\infty, \infty)\)
Domain of \(f^{-1}\): \((-\infty, \infty)\) Range of \(f^{-1}\): \((-\infty, \infty)\)
(ii) \(f(x) = \frac{1}{2}x + 8\)
Domain of \(f\): \((-\infty, \infty)\) Range of \(f\): \((-\infty, \infty)\)
Domain of \(f^{-1}\): \((-\infty, \infty)\) Range of \(f^{-1}\): \((-\infty, \infty)\)
Domain of \(f\): \((-\infty, \infty)\) Range of \(f\): \((-\infty, \infty)\)
Domain of \(f^{-1}\): \((-\infty, \infty)\) Range of \(f^{-1}\): \((-\infty, \infty)\)
(iii) \(f(x) = \frac{x}{1+x},\; x \neq -1\)
Domain of \(f\): \((-\infty, -1) \cup (-1, \infty)\)
Range of \(f\): \((-\infty, 1) \cup (1, \infty)\)
Domain of \(f^{-1}\): \((-\infty, 1) \cup (1, \infty)\) Range of \(f^{-1}\): \((-\infty, -1) \cup (-1, \infty)\)
Domain of \(f\): \((-\infty, -1) \cup (-1, \infty)\)
Range of \(f\): \((-\infty, 1) \cup (1, \infty)\)
Domain of \(f^{-1}\): \((-\infty, 1) \cup (1, \infty)\) Range of \(f^{-1}\): \((-\infty, -1) \cup (-1, \infty)\)
(iv) \(f(x) = \sqrt{x-2},\; x \ge 2\)
Domain of \(f\): \([2, \infty)\)
Range of \(f\): \([0, \infty)\)
Domain of \(f^{-1}\): \([0, \infty)\) Range of \(f^{-1}\): \([2, \infty)\)
Domain of \(f\): \([2, \infty)\)
Range of \(f\): \([0, \infty)\)
Domain of \(f^{-1}\): \([0, \infty)\) Range of \(f^{-1}\): \([2, \infty)\)
10
Given \(f(x) = x^2 + 9\) and \(g(x) = x + 21\). Find values of \(a\) such that \(f(a) = g(a)\)
\[ \begin{aligned}
f(a) &= g(a) \\
a^2 + 9 &= a + 21 \\
a^2 - a - 12 &= 0 \\
a^2 - 4a + 3a - 12 &= 0 \\
a(a-4) + 3(a-4) &= 0 \\
(a-4)(a+3) &= 0 \\
a = 4 \quad \text{or} \quad a = -3
\end{aligned} \]
📝 Multiple Choice Questions (Unit 4 Review)
1. If \(f(x) = 2x + 3\) and \(g(x) = x^2\), then \((f \circ g)(x)\) is:
✅ Correct Answer: (A) \(2x^2 + 3\)
\((f \circ g)(x) = f(g(x)) = f(x^2) = 2x^2 + 3\)
\((f \circ g)(x) = f(g(x)) = f(x^2) = 2x^2 + 3\)
2. The inverse of \(f(x) = 2x - 5\) is:
✅ Correct Answer: (B) \(f^{-1}(x) = \frac{x+5}{2}\)
\(y = 2x - 5 \implies y+5 = 2x \implies x = \frac{y+5}{2} \implies f^{-1}(x) = \frac{x+5}{2}\)
\(y = 2x - 5 \implies y+5 = 2x \implies x = \frac{y+5}{2} \implies f^{-1}(x) = \frac{x+5}{2}\)
3. If \(f(x) = x^2\) and \(g(x) = x + 1\), then \((g \circ f)(2)\) is:
✅ Correct Answer: (A) 5
\((g \circ f)(2) = g(f(2)) = g(4) = 4 + 1 = 5\)
\((g \circ f)(2) = g(f(2)) = g(4) = 4 + 1 = 5\)
4. The domain of \(f(x) = \sqrt{x-3}\) is:
✅ Correct Answer: (A) \([3, \infty)\)
For \(\sqrt{x-3}\) to be defined, \(x-3 \ge 0 \implies x \ge 3\)
For \(\sqrt{x-3}\) to be defined, \(x-3 \ge 0 \implies x \ge 3\)
5. If \(f(x) = 3x + 1\) and \(g(x) = x - 2\), then \((f \circ g)(x) = (g \circ f)(x)\) for:
✅ Correct Answer: (B) No \(x\)
\((f \circ g)(x) = 3(x-2) + 1 = 3x - 5\)
\((g \circ f)(x) = (3x+1) - 2 = 3x - 1\)
\(3x - 5 = 3x - 1 \implies -5 = -1\), impossible.
\((f \circ g)(x) = 3(x-2) + 1 = 3x - 5\)
\((g \circ f)(x) = (3x+1) - 2 = 3x - 1\)
\(3x - 5 = 3x - 1 \implies -5 = -1\), impossible.
📈 Key Concepts – Functions
- Function Operations: \( (f+g)(x)=f(x)+g(x),\ (f-g)(x)=f(x)-g(x),\ (fg)(x)=f(x)g(x),\ \left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)} \)
- Composite Functions: \( (f \circ g)(x)=f(g(x)),\ (g \circ f)(x)=g(f(x)) \)
- Inverse Function: \( f(f^{-1}(x))=f^{-1}(f(x))=x \)
- Finding Inverse: Set \( y = f(x) \), swap \( x \) and \( y \), solve for \( y \).
- Domain of \(f^{-1}\): Range of \(f\), Range of \(f^{-1}\): Domain of \(f\)
❓ Frequently Asked Questions
What is taught in Exercise 4.1 of Class 10 Math Unit 4?
Exercise 4.1 covers operations on functions (addition, subtraction, multiplication, division), composite functions \((f \circ g)\) and \((g \circ f)\), and inverse functions \(f^{-1}(x)\). It includes step-by-step solutions with domain considerations.
How many questions are there in Unit 4 Exercise 4.1?
Exercise 4.1 has 10 questions covering function operations, composition of functions, finding inverse functions, verifying properties, and finding domain and range of inverse functions.
What is the composition of functions \((f \circ g)(x)\)?
The composition of functions \((f \circ g)(x)\) is defined as \(f(g(x))\), meaning we first apply function \(g\) to \(x\), then apply function \(f\) to the result.
How do you find the inverse of a function?
To find the inverse of a function \(f(x)\), set \(y = f(x)\), solve for \(x\) in terms of \(y\), then interchange \(x\) and \(y\) to get \(f^{-1}(x)\). The inverse satisfies \(f(f^{-1}(x)) = f^{-1}(f(x)) = x\).
Is this solution according to the PECTAA 2026 syllabus?
Yes, these solutions are prepared according to the PECTAA 2026 / National Curriculum 2023 syllabus for Class 10 Mathematics.
Is this Exercise 4.1 solution valid for all Punjab Boards?
Yes, the content follows the unified Punjab textbook and is applicable to students of all 10 BISE Punjab boards.
Are solved PDF notes available for Exercise 4.1?
Yes, a complete solved PDF for Exercise 4.1 is embedded on this page and available to download for free.
Can I download the Unit 4 Exercise 4.1 solution as a PDF?
Yes, use the Download PDF button on this page to save the complete solved Exercise 4.1 notes to your device.
Is Exercise 4.1 important for Class 10 board exams?
Yes, functions, composition, and inverse functions are frequently tested in board exams. Exercise 4.1 builds essential skills in understanding and working with functions.
Who prepared these Class 10 Math Unit 4 notes?
These notes were prepared by Muhammad Tayyab, Subject Specialist Mathematics at Govt Christian High School Daska, for Hira Science Academy.