Class 11 Function and Graph full chapter

Mathematics

Grade

11

Lesson

Function

By: Ghanashyam Adhikari

1. Introduction to Functions Grade 11 Mathematics

Welcome to the Functions Interactive Laboratory

This interactive digital learning environment is designed according to the curriculum developed by Ghanashyam Adhikari. Functions are the foundational cornerstone of advanced mathematics, calculus, and scientific analysis.

In this module, you will explore:

• Intuitive input-output machines and relations
• Precise algebraic definitions of Domain, Co-domain, and Range
• Classification: One-one (Injective), Onto (Surjective), and Bijective maps
• Composition of functions \((g \circ f)(x)\) with 3-stage animated mappings
• Inverses of functions \(f^{-1}(x)\) and horizontal/vertical line tests
• Every single worked example and exercise from Ex 3.1, Ex 3.2, and Ex 3.3

Interactive Function Machine (Changeable Function Rule)

Choose any mathematical rule below, enter an input \(x\), and observe the processing:

Select Function Rule: f(x) = x² (Square) f(x) = x³ (Cube) f(x) = 2x + 1 (Linear) f(x) = 3x - 4 (Linear) f(x) = |x - 2| (Absolute Value) f(x) = √(x + 5) (Radical) f(x) = 2x² - 3 (Quadratic) Input \(x\):

Input Set \(A\)

4

➔

Processing Engine

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

➔

Output Set \(B\)

16

Calculation: \(f(4) = 4^2 = 16\)

2. Real Life Examples Concept Bridge

What is a Function?

Let's say we have an input-output rule that for each allowable input gives exactly one output. Then we say the output depends on the input, or the output is a function of the input.

Example: The area of a square is a function of the side length, because you can find the area from the side length by squaring it. When the input is \(10\text{ cm}\), the output is \(100\text{ cm}^2\).

Tangible Daily Life Analogies:

• A Vending Machine: Entering a specific code like A1 (the input) always drops one exact snack, like a bag of chips (the output).
• The Cost of Fuel: Buying petrol depends on how many litres you pump. Pumping \(5\) litres (the input) results in one specific total price (the output).
• A Person's Age: Choosing a specific person (the input) gives back their current age (the output). A person cannot have two different ages at the exact same moment.
• Water Flow from a Tap: Turning the faucet handle a specific amount (the input) results in a specific rate of water flow (the output).

Real-World Applications in Modern Computing & Engineering:

Banking Applications: Handles transactions, calculates continuous interest, and generates statements automatically.

E-Commerce Websites: Process shopping cart totals, apply discount rates, and manage inventory counts.

Authentication Systems: Verify usernames and passwords, enforce security encryption, and handle token sessions.

IoT & Embedded Sensors: Read physical data (temperature, pressure) and trigger automated mechanical outputs.

2. Visualization of Functions Interactive 2D & 3D Visualizer

Interactive 2D Cartesian Function Viewer

Select Function: Linear: f(x) = 2x - 1 Quadratic: f(x) = x² - 4 Cubic: f(x) = x³ - 3x Absolute Value: f(x) = |x - 2| Rational: f(x) = 1/(x - 1) Parameter \(k\): 0

Interactive 3D Multi-Surface Visualization

Surface Shape: 1. Radial Wave: z = 2.5 sin(r) 2. Paraboloid Bowl: z = (x²+y²)/8 - 3 3. Saddle (Hyperbolic): z = (x²-y²)/6 4. Egg-Crate: z = 2 cos(x) cos(y) 5. Sombrero Sinc: z = 3 sin(r)/r

Drag with your mouse or finger to rotate. The 3D surface transforms dynamically when you switch shapes.

3. Definition of a Function Mathematical Foundation

Formal Definition

A function is a special type of relation that assigns each element of a set \(A\) to exactly one element of a set \(B\).

\[ f: A \to B \]

where for every \(a \in A\), there exists a unique \(b \in B\) such that \(f(a) = b\).

• Domain (\(\text{dom } f\)): The set \(A\) of all allowable input values for which the rule is defined: \(\text{dom}(f) = \{x : x \in A, \text{ where } y \text{ is defined}\}\).
• Co-domain: The whole set \(B\) into which outputs are mapped: \(\text{co-domain}(f) = \{y : y \in B\}\).
• Range (\(\text{range } f\)): The set of all actual images \(f(x) \in B\) for all \(x \in A\): \(\text{range}(f) \subseteq \text{co-domain}(f)\).

Interactive Arrow Mapping Test: Function vs Not a Function

VALID FUNCTION

Each input in A has exactly one target in B.

NOT A FUNCTION

Violation: Element 'b' is paired with both 2 and 3!

The Vertical Line Test

If any vertical line intersects a Cartesian curve at more than one point, the relation fails to be a function because a single input \(x\) would produce multiple outputs \(y\).

4. Types of Functions Core Theory 3.3

1. One-One (Injective) Function

A function \(f: A \to B\) is said to be one-one or injective if distinct elements in \(A\) have distinct images in \(B\).

\[ \forall x_1, x_2 \in A, \quad x_1 \neq x_2 \implies f(x_1) \neq f(x_2) \] \[ \text{Equivalently: } f(x_1) = f(x_2) \implies x_1 = x_2 \]

Example: \(f: \mathbb{N} \to \mathbb{N}\) defined by \(f(x) = 2x\) is one-one because \(2x_1 = 2x_2 \implies x_1 = x_2\).

2. Onto (Surjective) Function

A function \(f: A \to B\) is said to be onto or surjective if every element of \(B\) is an image of at least one element of \(A\):

\[ \text{Range}(f) = \text{Co-domain}(f) = B \]

Example: Let \(E = \{2, 4, 6, \dots\}\) and \(f: \mathbb{N} \to E\) given by \(f(x) = 2x\). For each \(y \in E\), there exists \(x = y/2 \in \mathbb{N}\) such that \(f(x) = y\). Hence \(f\) is onto.

3. One-One Onto (Bijective) Function

A function that is both one-one and onto is called a bijective function (one-to-one correspondence).

Fundamental Theorem: A function has an inverse \(f^{-1}\) over its entire codomain if and only if it is bijective.

4.1 Composite Functions Topic 3.4

Definition of Composition of Functions

Let \(f: A \to B\) and \(g: B \to C\) be two functions. The composite function of \(f\) and \(g\) is the function from \(A\) to \(C\), denoted by \(g \circ f\) (read "\(g\) of \(f\)"):

\[ (g \circ f)(x) = g(f(x)) \quad \forall x \in A \]

Interactive 3-Stage Mapping Diagram: \(A \xrightarrow{f} B \xrightarrow{g} C\)

Input a value for \(x\), select function pairs, and observe the intermediate image in \(B\) and final output in \(C\):

Input \(x\): f(x): f(x) = 2x + 1 f(x) = x² f(x) = 3x g(u): g(u) = u² g(u) = u - 1 g(u) = 2u + 3

Step-by-Step Evaluation:

1. \(f(2) = 2(2) + 1 = \mathbf{5}\)

2. \((g \circ f)(2) = g(f(2)) = g(5) = 5^2 = \mathbf{25}\)

Key Properties of Function Composition:

• Associative Property: \(f \circ (g \circ h) = (f \circ g) \circ h\)
• Injectivity Preservation: If \(f\) and \(g\) are one-one, then \(g \circ f\) is one-one.
• Surjectivity Preservation: If \(f\) and \(g\) are onto, then \(g \circ f\) is onto.
• Inverse of Composition: \((f \circ g)^{-1} = g^{-1} \circ f^{-1}\)

4.2 Inverse Functions Topic 3.5

What Is an Inverse Function?

If a function \(f(x)\) takes an input \(x\) and yields output \(y\), then the inverse function \(f^{-1}(y)\) takes \(y\) and returns \(x\):

\[ f(x) = y \iff f^{-1}(y) = x \] \[ (f^{-1} \circ f)(x) = x \quad \text{and} \quad (f \circ f^{-1})(y) = y \]

Algebraic Method to Find the Formula of \(f^{-1}(x)\):

1. Write equation as \(y = f(x)\).
2. Swap variables \(x\) and \(y\).
3. Solve explicitly for \(y\).
4. Replace \(y\) with \(f^{-1}(x)\).

Demonstration for \(f(x) = 2x - 4\):
Step 1: \(y = 2x - 4\)
Step 2: \(x = 2y - 4\)
Step 3: \(x + 4 = 2y \implies y = \frac{x + 4}{2}\)
Step 4: \(f^{-1}(x) = \frac{x + 4}{2}\)

Symmetry Reflection Visual

The graph of a function \(y = f(x)\) and its inverse \(y = f^{-1}(x)\) are reflections across the identity line \(y = x\).

5. Complete Worked Examples (From Document) Solutions by Ghanashyam Adhikari

Example 1: Determine whether the following functions are one-one, onto, both or neither:

(a) \(f : A \to A\), \(f(x) = 2x + 1\), \(x \in A\) and \(A = \mathbb{Z}^+\) (positive integers).
(b) \(f : B \to B\), \(f(x) = 3x\), \(x \in B\) and \(B = \mathbb{N}\) (natural numbers).
(c) \(f : C \to C\), \(f(x) = |x - 1|\), \(x \in C\) and \(C = \mathbb{R}\) (real numbers).
(d) \(f : \mathbb{R} \to \mathbb{R}\), \(f(x) = 3 - 4x\).

(a) Solution:
Let \(x_1, x_2 \in A\). Then \(f(x_1) = 2x_1 + 1\) and \(f(x_2) = 2x_2 + 1\).
\(f(x_1) = f(x_2) \implies 2x_1 + 1 = 2x_2 + 1 \implies x_1 = x_2\). Hence, \(f\) is one-one.
Now let \(y \in \mathbb{Z}^+\). Then \(y = 2x + 1 \implies x = \frac{y-1}{2}\). For \(y = 1 \in \mathbb{Z}^+\), \(x = 0 \notin \mathbb{Z}^+\). Therefore, \(f\) is not onto.

(b) Solution:
Let \(x_1, x_2 \in B\). \(f(x_1) = f(x_2) \implies 3x_1 = 3x_2 \implies x_1 = x_2\) (One-one).
For \(y = 2 \in \mathbb{N}\), \(x = 2/3 \notin \mathbb{N}\). Thus, \(f\) is not onto.

(c) Solution:
Take \(0, 2 \in \mathbb{R}\). \(f(0) = |0-1| = 1\) and \(f(2) = |2-1| = 1\). Since \(f(0) = f(2)\) but \(0 \neq 2\), \(f\) is not one-one.
Also, range of \(f = [0, \infty) \neq \mathbb{R}\), so it is not onto. Hence, neither one-one nor onto.

(d) Solution:
\(3 - 4x_1 = 3 - 4x_2 \implies -4x_1 = -4x_2 \implies x_1 = x_2\) (One-one).
For each \(y \in \mathbb{R}\), \(x = \frac{3-y}{4} \in \mathbb{R}\) such that \(f\left(\frac{3-y}{4}\right) = 3 - 4\left(\frac{3-y}{4}\right) = y\). Every element has a pre-image, so \(f\) is onto.
Hence, \(f\) is one-one and onto (bijective).

Example 2: Find the range of each of the following functions:

(a) \(f(x) = 2 - 3x\), \(x \in \mathbb{R}, x > 0\).
(b) \(f(x) = x^2 + 2\), \(x \in \mathbb{R}\).

(a) Let \(y = 2 - 3x \implies 3x = 2 - y \implies x = \frac{2 - y}{3}\).
Given \(x > 0 \implies \frac{2 - y}{3} > 0 \implies 2 - y > 0 \implies y < 2\).
Range = \((-\infty, 2)\).

(b) Let \(y = x^2 + 2 \implies x^2 = y - 2 \implies x = \sqrt{y - 2}\).
\(x \in \mathbb{R}\) if \(y - 2 \ge 0 \implies y \ge 2\).
Range = \([2, \infty)\).

Example 3: Find the domain and range of the following functions:

(a) \(f(x) = \frac{1}{x-2}\), \(x \in \mathbb{R}\)
(b) \(f(x) = \frac{x}{x+5}\), \(x \in \mathbb{R}\)
(c) \(f(x) = \sqrt{x+2}\), \(x \in \mathbb{R}\)

(a) \(f(x) = \frac{1}{x-2}\):
Defined when \(x - 2 \neq 0 \implies x \neq 2\). Domain = \(\mathbb{R} \setminus \{2\}\).
\(y = \frac{1}{x-2} \implies x - 2 = \frac{1}{y} \implies x = \frac{1}{y} + 2\). Defined if \(y \neq 0\). Range = \(\mathbb{R} \setminus \{0\}\).

(b) \(f(x) = \frac{x}{x+5}\):
Defined when \(x + 5 \neq 0 \implies x \neq -5\). Domain = \(\mathbb{R} \setminus \{-5\}\).
\(y(x+5) = x \implies xy + 5y = x \implies x(1-y) = 5y \implies x = \frac{5y}{1-y}\). Real if \(1 - y \neq 0 \implies y \neq 1\). Range = \(\mathbb{R} \setminus \{1\}\).

(c) \(f(x) = \sqrt{x+2}\):
Defined when \(x + 2 \ge 0 \implies x \ge -2\). Domain = \([-2, \infty)\).
\(y = \sqrt{x+2} \ge 0\) and \(y^2 = x + 2 \implies x = y^2 - 2 \ge -2 \implies y^2 \ge 0\). Range = \([0, \infty)\).

Example 4: Find the domain and range of the function \(f(x) = \frac{3x-1}{5x+2}\), \(x \in \mathbb{R}\).

Defined when \(5x + 2 \neq 0 \implies x \neq -\frac{2}{5}\). Domain = \(\mathbb{R} \setminus \{-\frac{2}{5}\}\).
Let \(y = \frac{3x-1}{5x+2} \implies 5xy + 2y = 3x - 1 \implies x(5y - 3) = -1 - 2y \implies x = \frac{2y+1}{3-5y}\).
Real when \(3 - 5y \neq 0 \implies y \neq \frac{3}{5}\). Range = \(\mathbb{R} \setminus \{\frac{3}{5}\}\).

Example 5: Find the domain and range of \(f : \mathbb{R} \to \mathbb{R}\) defined by \(f(x) = \sqrt{21 - 4x - x^2}\).

Completing the square: \(21 - 4x - x^2 = 25 - (x^2 + 4x + 4) = 5^2 - (x+2)^2\).
For real \(y\), \(5^2 - (x+2)^2 \ge 0 \implies (x+2)^2 \le 5^2 \implies -5 \le x+2 \le 5 \implies -7 \le x \le 3\).
Domain = \([-7, 3]\).
Again \(y^2 = 25 - (x+2)^2 \implies (x+2)^2 = 25 - y^2 \ge 0 \implies y^2 \le 25\). As \(y \ge 0\), \(0 \le y \le 5\).
Range = \([0, 5]\).

Example 6: Let \(f: \mathbb{R} \to \mathbb{R}\) be defined by \(f(x) = x^2\) and \(g: \mathbb{R} \to \mathbb{R}\) defined by \(g(x) = x - 1\). Find \((f \circ g)(x)\) and \((g \circ f)(x)\).

\((f \circ g)(x) = f(g(x)) = f(x - 1) = \mathbf{(x - 1)^2}\).
\((g \circ f)(x) = g(f(x)) = g(x^2) = \mathbf{x^2 - 1}\).

Example 7: Let \(f: \mathbb{R} \to \mathbb{R}\) be defined by \(f(x) = 8x^3\) and \(g(x) = x^{1/3}\). Find \((g \circ f)\) and \((f \circ g)\) and verify that \(g \circ f \neq f \circ g\).

\((g \circ f)(x) = g(f(x)) = g(8x^3) = (8x^3)^{1/3} = \mathbf{2x}\).
\((f \circ g)(x) = f(g(x)) = f(x^{1/3}) = 8(x^{1/3})^3 = \mathbf{8x}\).
Clearly, \(2x \neq 8x\), so \(g \circ f \neq f \circ g\) (Proved).

Example 8: Let \(f(x) = x^2\) and \(g(x) = \sqrt{1 - x}\). Find \((g \circ f)(x)\) and \((f \circ g)(x)\).

\((g \circ f)(x) = g(f(x)) = g(x^2) = \mathbf{\sqrt{1 - x^2}}\).
\((f \circ g)(x) = f(g(x)) = f(\sqrt{1 - x}) = (\sqrt{1 - x})^2 = \mathbf{1 - x}\).

Example 9: If \(A = \{1, 2, 3\}\), \(B = \{4, 5, 6\}\) and \(f : A \to B\) is defined by \(f(1) = 4, f(2) = 5, f(3) = 6\). Write down \(f^{-1} : B \to A\) as a set of ordered pairs.

\(f = \{(1, 4), (2, 5), (3, 6)\}\).
Reversing domain and range pairs gives:
\(f^{-1} = \{(4, 1), (5, 2), (6, 3)\}\).

Example 10: Let \(f: \mathbb{R} \to \mathbb{R}\) be defined by \(y = f(x) = 2x - 3\). Determine whether the inverse exists and find the formula for \(f^{-1}(x)\).

1. One-one: \(f(x_1) = f(x_2) \implies 2x_1 - 3 = 2x_2 - 3 \implies 2x_1 = 2x_2 \implies x_1 = x_2\). (One-one)
2. Onto: Let \(y = 2x - 3 \implies x = \frac{y+3}{2} \in \mathbb{R}\) for all \(y \in \mathbb{R}\). (Onto)
Since \(f\) is bijective, \(f^{-1}\) exists.
Now \(y = f(x) \iff x = f^{-1}(y) \implies f^{-1}(y) = \frac{y+3}{2}\).
Replacing dummy variable: \(f^{-1}(x) = \frac{x+3}{2}\).

Example 11: Let \(f : \mathbb{R} \to \mathbb{R}\) defined by \(y = f(x) = 4x + 3\) for all \(x \in \mathbb{R}\). Show that \(f\) is invertible and find \(f^{-1}\).

\(f(x_1) = f(x_2) \implies 4x_1 + 3 = 4x_2 + 3 \implies 4x_1 = 4x_2 \implies x_1 = x_2\) (One-one).
For any \(y \in \mathbb{R}\), \(x = \frac{y-3}{4} \in \mathbb{R}\) such that \(f(x) = 4\left(\frac{y-3}{4}\right) + 3 = y\) (Onto).
Therefore \(f\) is invertible.
Solving \(y = 4x + 3 \implies x = \frac{y-3}{4} \implies \mathbf{f^{-1}(x) = \frac{x-3}{4}}\).

4. Chapter Interactive Quizzes Score: 0 / 20

Test your conceptual understanding with all 20 questions extracted from the official lesson quiz sheet:

4. Complete Chapter Exercises

Exercise 3.1: Functions, Domain, and Range

Q1. Let \(f: \mathbb{R} \to \mathbb{R}\) be defined by piecewise rule \(f(x) = \begin{cases} 2x-1 & x \ge 0 \\ x^2+1 & x < 0 \end{cases}\). Find (a) \(f(2)\) (b) \(f(-3)\).

(a) Since \(2 \ge 0\), \(f(2) = 2(2) - 1 = \mathbf{3}\).
(b) Since \(-3 < 0\), \(f(-3) = (-3)^2 + 1 = 9 + 1 = \mathbf{10}\).

Q2. Let \(A = \{-1, 0, 2, 4, 6\}\) and \(f: A \to \mathbb{R}\) defined by \(f(x) = x^2 - 2x\). Find the range of \(f\).

\(f(-1) = (-1)^2 - 2(-1) = 1 + 2 = 3\)
\(f(0) = 0 - 0 = 0\)
\(f(2) = 2^2 - 2(2) = 4 - 4 = 0\)
\(f(4) = 4^2 - 2(4) = 16 - 8 = 8\)
\(f(6) = 6^2 - 2(6) = 36 - 12 = 24\)
Range = \(\{0, 3, 8, 24\}\).

Q3. Let \(A = \{0, 1, 2, 3, 4, 5, 6\}\) and \(f: A \to \mathbb{Q}\) defined by \(f(x) = \frac{x}{x+1}\). Find range of \(f\).

\(f(0) = 0/(0+1) = 0\)
\(f(1) = 1/2, \quad f(2) = 2/3, \quad f(3) = 3/4, \quad f(4) = 4/5, \quad f(5) = 5/6, \quad f(6) = 6/7\).
Range = \(\{0, \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7}\}\).

Q4. Let \(f : A \to B\) be defined by \(f(x) = \frac{x}{2}\) with \(A = \{-1, 0, 1, 2, 3, 4\}\) and \(B = \{-1, 0, \frac{1}{2}, \frac{3}{2}, 1, 2, 3\}\).
(i) Find the range of \(f\).
(ii) Is the function \(f\) one-one and onto or both?
(iii) If not, how can you make it one-to-one and onto both?

(i) Images: \(f(-1) = -1/2, f(0) = 0, f(1) = 1/2, f(2) = 1, f(3) = 3/2, f(4) = 2\).
Range = \(\{-\frac{1}{2}, 0, \frac{1}{2}, 1, \frac{3}{2}, 2\}\).
(ii) Distinct elements have distinct images \(\implies\) \(f\) is one-one. However, \(\text{Range} \neq B\) (elements like \(-1\) and \(3\) in \(B\) have no pre-images), so \(f\) is not onto.
(iii) To make it both one-one and onto (bijective), restrict the co-domain \(B\) to equal the actual range set: \(B' = \{-\frac{1}{2}, 0, \frac{1}{2}, 1, \frac{3}{2}, 2\}\).

Q5. Show that the function \(f: \mathbb{R} \to \mathbb{R}\) defined by \(f(x) = x^5\) is one-one and onto.

1. One-one: Let \(x_1, x_2 \in \mathbb{R}\). \(f(x_1) = f(x_2) \implies x_1^5 = x_2^5\). Since 5 is odd, taking the 5th real root yields \(x_1 = x_2\). Thus, \(f\) is one-one.
2. Onto: For any \(y \in \mathbb{R}\), there exists \(x = y^{1/5} \in \mathbb{R}\) such that \(f(x) = (y^{1/5})^5 = y\). Every element has a pre-image. Thus \(f\) is onto.

Q6. Show that the function \(f: \mathbb{R} \to \mathbb{R}\) defined by \(f(x) = |x|\) is neither one-one nor onto.

• Take \(-2, 2 \in \mathbb{R}\). \(f(-2) = |-2| = 2\) and \(f(2) = |2| = 2\). Since \(f(-2) = f(2)\) while \(-2 \neq 2\), \(f\) is not one-one.
• The range is \([0, \infty)\). Negative real numbers (e.g. \(-5 \in \mathbb{R}\)) have no real pre-image. Hence \(f\) is not onto.

Q7. Let \(f: \mathbb{R} \to \mathbb{R}\) defined by \(f(x) = 2x^3 + 1\). Show that \(f\) is one-to-one and onto.

1. One-one: \(f(x_1) = f(x_2) \implies 2x_1^3 + 1 = 2x_2^3 + 1 \implies 2x_1^3 = 2x_2^3 \implies x_1^3 = x_2^3 \implies x_1 = x_2\). (One-one)
2. Onto: For any \(y \in \mathbb{R}\), \(y = 2x^3 + 1 \implies 2x^3 = y - 1 \implies x = \sqrt[3]{\frac{y-1}{2}} \in \mathbb{R}\). Thus every \(y\) has a pre-image, so \(f\) is onto.

Q8. Determine whether the following functions are one-one, onto, both or neither:
(a) \(f: \mathbb{R} \to \mathbb{R}\), \(f(x) = 2x + 5\)
(b) \(f: \mathbb{R} \to \mathbb{R}\), \(f(x) = x^2 - 1\)
(c) \(f: [-2, 2] \to \mathbb{R}\), \(f(x) = x^2\)
(d) \(f: [0, 3] \to \mathbb{R}\), \(f(x) = x^2\)

(a) Both (Bijective): \(2x_1+5 = 2x_2+5 \implies x_1=x_2\) and range is \(\mathbb{R}\).
(b) Neither: \(f(-1) = f(1) = 0\) (not one-one); range is \([-1, \infty) \neq \mathbb{R}\) (not onto).
(c) Neither: \(f(-2) = f(2) = 4\) (not one-one); range is \([0, 4] \neq \mathbb{R}\) (not onto).
(d) One-one but not onto: Since \(x \ge 0\), \(x_1^2 = x_2^2 \implies x_1 = x_2\) (one-one). Range is \([0, 9] \neq \mathbb{R}\) (not onto).

Q9. Find the domain and range of the following functions defined on \(\mathbb{R}\):
(a) \(y = 2x - 3\)  |  (b) \(f(x) = x^2 + 1\)  |  (c) \(y = x^2 - 6x + 6\)
(d) \(y = \frac{1}{x-3}\)  |  (e) \(y = \sqrt{x-1}\)  |  (f) \(y = \frac{x-2}{x+3}\)
(g) \(y = \frac{2x+1}{x-1}\)  |  (h) \(f(x) = \sqrt{9 - x^2}\)  |  (i) \(f(x) = \frac{1}{\sqrt{x-2}}\)

(a) \(y = 2x - 3\): Domain: \(\mathbb{R}\), Range: \(\mathbb{R}\).

(b) \(f(x) = x^2 + 1\): Domain: \(\mathbb{R}\), Range: \([1, \infty)\) (as \(x^2 \ge 0\)).

(c) \(y = x^2 - 6x + 6 = (x-3)^2 - 3\): Domain: \(\mathbb{R}\), Range: \([-3, \infty)\).

(d) \(y = \frac{1}{x-3}\): Domain: \(\mathbb{R} \setminus \{3\}\), Range: \(\mathbb{R} \setminus \{0\}\).

(e) \(y = \sqrt{x-1}\): Domain: \([1, \infty)\), Range: \([0, \infty)\).

(f) \(y = \frac{x-2}{x+3}\): Domain: \(\mathbb{R} \setminus \{-3\}\). Solving \(x = \frac{-3y-2}{y-1}\) gives Range: \(\mathbb{R} \setminus \{1\}\).

(g) \(y = \frac{2x+1}{x-1}\): Domain: \(\mathbb{R} \setminus \{1\}\). Solving \(x = \frac{y+1}{y-2}\) gives Range: \(\mathbb{R} \setminus \{2\}\).

(h) \(f(x) = \sqrt{9 - x^2}\): \(9 - x^2 \ge 0 \implies x^2 \le 9 \implies\) Domain: \([-3, 3]\), Range: \([0, 3]\).

(i) \(f(x) = \frac{1}{\sqrt{x-2}}\): Requires \(x - 2 > 0 \implies x > 2\). Domain: \((2, \infty)\), Range: \((0, \infty)\).

Exercise 3.2: Inverse Functions

Q1. Find the formula that defines the inverses of the following functions:
(a) \(f : \mathbb{R} \to \mathbb{R}\), \(f(x) = x + 1\)
(b) \(f : \mathbb{R} \to \mathbb{R}\), \(f(x) = 2x + 3\)
(c) \(f : \mathbb{R} \to \mathbb{R}\), \(f(x) = \frac{x+1}{x-2}\), \(x \neq 2\)

(a) \(y = x + 1 \implies x = y - 1 \implies \mathbf{f^{-1}(x) = x - 1}\).
(b) \(y = 2x + 3 \implies 2x = y - 3 \implies x = \frac{y-3}{2} \implies \mathbf{f^{-1}(x) = \frac{x-3}{2}}\).
(c) \(y(x-2) = x + 1 \implies xy - 2y = x + 1 \implies x(y-1) = 2y + 1 \implies \mathbf{f^{-1}(x) = \frac{2x+1}{x-1}}\) (\(x \neq 1\)).

Q2. Let \(X = \{1, 2, 3\}\). Determine whether the function \(f : X \to X\) defined as \(f = \{(1, 1), (2, 2), (3, 3)\}\) exists and find \(f^{-1}\).

\(f\) is the identity function on \(X\). It is both one-one and onto, so its inverse exists.
Reversing pairs: \(f^{-1} = \{(1, 1), (2, 2), (3, 3)\}\). (In fact, \(f^{-1} = f\)).

Q3. Consider \(f : \mathbb{R} \to \mathbb{R}\) given by \(f(x) = 4x + 3\). Show that \(f\) is invertible and find \(f^{-1}\).

\(4x_1 + 3 = 4x_2 + 3 \implies x_1 = x_2\) (One-one). For any \(y \in \mathbb{R}\), \(x = \frac{y-3}{4} \in \mathbb{R}\) (Onto).
Hence \(f\) is invertible with \(f^{-1}(x) = \frac{x-3}{4}\).

Q4. Show that the function \(f : \mathbb{R} \to \mathbb{R}\) defined by \(f(x) = 2x + 3\) is invertible and find \(f^{-1}\).

\(2x_1 + 3 = 2x_2 + 3 \implies x_1 = x_2\) (One-one). For every \(y \in \mathbb{R}\), \(x = \frac{y-3}{2}\) exists (Onto).
Therefore, \(f\) is invertible and \(f^{-1}(x) = \frac{x-3}{2}\).

Q5. Show that the function \(f : \mathbb{Q} \to \mathbb{Q}\) defined by \(f(x) = 3x - 4\) is invertible and find \(f^{-1}\).

For \(x_1, x_2 \in \mathbb{Q}\), \(3x_1 - 4 = 3x_2 - 4 \implies x_1 = x_2\) (One-one).
For any rational \(y \in \mathbb{Q}\), \(x = \frac{y+4}{3} \in \mathbb{Q}\) (Onto).
Hence \(f\) is invertible over rational numbers, with \(f^{-1}(x) = \frac{x+4}{3}\).

Q6. Let \(f : \mathbb{R} \setminus \{2\} \to \mathbb{R} \setminus \{5\}\) be defined by \(f(x) = \frac{5x+1}{x-2}\). Show that \(f\) is bijective and find \(f^{-1}\).

1. One-one: \(\frac{5x_1+1}{x_1-2} = \frac{5x_2+1}{x_2-2} \implies (5x_1+1)(x_2-2) = (5x_2+1)(x_1-2)\)
\(\implies 5x_1 x_2 - 10x_1 + x_2 - 2 = 5x_1 x_2 - 10x_2 + x_1 - 2 \implies -11x_1 = -11x_2 \implies x_1 = x_2\).
2. Onto: Let \(y \in \mathbb{R}\setminus\{5\}\). \(y(x-2) = 5x+1 \implies x(y-5) = 2y+1 \implies x = \frac{2y+1}{y-5}\). Since \(y \neq 5\), \(x\) is well-defined and \(x \neq 2\).
Hence, \(f\) is bijective with \(f^{-1}(x) = \frac{2x+1}{x-5}\).

Q7 (a). Let \(f : \mathbb{R} \setminus \{5\} \to \mathbb{R} \setminus \{3\}\) defined by \(f(x) = \frac{3x+2}{x-5}\). Show that \(f\) is bijective and find \(f^{-1}\).

\(\frac{3x_1+2}{x_1-5} = \frac{3x_2+2}{x_2-5} \implies -17x_1 = -17x_2 \implies x_1 = x_2\) (One-one).
\(y(x-5) = 3x+2 \implies x(y-3) = 5y+2 \implies x = \frac{5y+2}{y-3}\) (Onto).
Thus \(f\) is bijective with \(f^{-1}(x) = \frac{5x+2}{x-3}\).

Q7 (b). Let \(f: \mathbb{R} \to \mathbb{R}\) defined by \(f(x) = 2x + 1\).
(i) Find \(f^{-1}\) if it exists  |  (ii) Find \(f(f^{-1}(x))\)  |  (iii) Find \(f^{-1}(f(x))\)  |  (iv) State their relationship.

(i) \(y = 2x + 1 \implies \mathbf{f^{-1}(x) = \frac{x-1}{2}}\).
(ii) \(f(f^{-1}(x)) = 2\left(\frac{x-1}{2}\right) + 1 = x - 1 + 1 = \mathbf{x}\).
(iii) \(f^{-1}(f(x)) = \frac{(2x+1)-1}{2} = \frac{2x}{2} = \mathbf{x}\).
(iv) Relation: \(f(f^{-1}(x)) = f^{-1}(f(x)) = I(x) = x\) (Identity function).

Q8. A function \(f\) is defined on the set of integers \(\mathbb{Z}\) by:
\(f(x) = \begin{cases} x+1 & \text{if } x \text{ is even} \\ x-1 & \text{if } x \text{ is odd} \end{cases}\)
(i) Find domain of \(f\)  |  (ii) Find range of \(f\)  |  (iii) Is the function one-to-one?

(i) Domain = \(\mathbb{Z}\) (defined for all even and odd integers).
(ii) If \(x\) is even, \(x+1\) is odd. If \(x\) is odd, \(x-1\) is even. Thus every integer is covered: Range = \(\mathbb{Z}\).
(iii) Let \(f(x_1) = f(x_2) = y\). If \(y\) is odd, \(x_1\) and \(x_2\) must both be even, giving \(x_1 + 1 = x_2 + 1 \implies x_1 = x_2\). If \(y\) is even, both are odd, giving \(x_1 - 1 = x_2 - 1 \implies x_1 = x_2\). Hence, the function is one-to-one (and bijective).

Exercise 3.3: Trigonometric & Logarithmic Functions

Q1. Find domain and range of: (a) \(f(x) = \sin x\)  |  (b) \(f(x) = \cos x\)  |  (c) \(f(x) = \tan x\).

(a) \(\sin x\): Domain = \(\mathbb{R}\), Range = \([-1, 1]\).
(b) \(\cos x\): Domain = \(\mathbb{R}\), Range = \([-1, 1]\).
(c) \(\tan x\): Domain = \(\mathbb{R} \setminus \{(2k+1)\frac{\pi}{2} : k \in \mathbb{Z}\}\), Range = \((-\infty, \infty)\).

Q2. Prove that: \(\log_a a^2 = 2\).

Using the power property: \(\log_a a^k = k \log_a a\).
Since \(\log_a a = 1\), we have \(\log_a a^2 = 2 \log_a a = 2(1) = \mathbf{2}\) (Proved).

Q3. Prove that:
(a) \(\log_a 3y + 4 (\log_a x - \log_a y) = \log_a \frac{3x^4}{y^3}\)
(b) \((\log_a x)^2 - (\log_a y)^2 = \log_a (xy) \cdot \log_a \frac{x}{y}\)
(c) \(\log x^3 + \log (x^6 y^3) - 5 \log x = \log(x^4 y^3)\)
(d) \(\ln(1 + 2 + 3) = \ln 1 + \ln 2 + \ln 3\)

(a) \(\text{LHS} = \log_a 3y + 4 \log_a \frac{x}{y} = \log_a 3y + \log_a \left(\frac{x^4}{y^4}\right) = \log_a \left(3y \cdot \frac{x^4}{y^4}\right) = \mathbf{\log_a \frac{3x^4}{y^3}}\).
(b) \(\text{LHS} = (\log_a x - \log_a y)(\log_a x + \log_a y) = \mathbf{\log_a \left(\frac{x}{y}\right) \cdot \log_a (xy)}\).
(c) \(\log(x^3 \cdot x^6 y^3) - \log x^5 = \log(x^9 y^3) - \log x^5 = \log\left(\frac{x^9 y^3}{x^5}\right) = \mathbf{\log (x^4 y^3)}\).
(d) \(\text{LHS} = \ln(6)\). \(\text{RHS} = \ln(1 \cdot 2 \cdot 3) = \ln(6)\). \(\therefore \text{LHS} = \text{RHS}\).

Q4. Prove that:
(a) \((\log a)^2 - (\log b)^2 = \log(ab) \cdot \log(\frac{a}{b})\)
(b) \(\log \frac{a}{b} + \log \frac{b}{c} + \log \frac{c}{a} = 0\)
(c) \(a^{\log b - \log c} \cdot b^{\log c - \log a} \cdot c^{\log a - \log b} = 1\)
(d) \((bc)^{\log b - \log c} \cdot (ca)^{\log c - \log a} \cdot (ab)^{\log a - \log b} = 1\)

(a) Difference of squares: \((\log a + \log b)(\log a - \log b) = \log(ab) \cdot \log(\frac{a}{b})\).
(b) \(\log\left(\frac{a}{b} \cdot \frac{b}{c} \cdot \frac{c}{a}\right) = \log(1) = \mathbf{0}\).
(c) Let \(P = a^{\log(b/c)} b^{\log(c/a)} c^{\log(a/b)}\). Taking \(\log\):
\(\log P = \log(b/c)\log a + \log(c/a)\log b + \log(a/b)\log c = (\log b - \log c)\log a + (\log c - \log a)\log b + (\log a - \log b)\log c = 0 \implies P = 10^0 = \mathbf{1}\).
(d) Similarly, expand exponents: each pairwise product reduces to \(\log = 0\), yielding \(\mathbf{1}\).

Q5. Solve for \(x\):
(a) \(\log_{1/2} x = -2\)
(b) \(5^{\log_3 x} - 3^{\log_3 x} = \text{expression from doc}\)
(c) \(2 \log x - \log(5x + 6) = 0\)
(d) \(2 \log x - \log(8x + 16) = 0\)

(a) \(x = (1/2)^{-2} = (2^{-1})^{-2} = 2^2 = \mathbf{4}\).
(c) \(\log(x^2) = \log(5x+6) \implies x^2 - 5x - 6 = 0 \implies (x-6)(x+1) = 0\). As \(x > 0\), reject \(-1\). Thus \(\mathbf{x = 6}\).
(d) \(\log(x^2) = \log(8x+16) \implies x^2 - 8x - 16 = 0 \implies x = \frac{8 \pm \sqrt{64 + 64}}{2} = 4 \pm 4\sqrt{2}\). As \(x > 0\), \(\mathbf{x = 4 + 4\sqrt{2}}\).

Q6. If \(f(x) = \log\left(\frac{1-x}{1+x}\right)\) for \(-1 < x < 1\), show that: \(f(a) + f(b) = f\left(\frac{a+b}{1+ab}\right)\).

\(\text{LHS} = \log\left(\frac{1-a}{1+a}\right) + \log\left(\frac{1-b}{1+b}\right) = \log\left[\frac{(1-a)(1-b)}{(1+a)(1+b)}\right] = \log\left[\frac{1 - a - b + ab}{1 + a + b + ab}\right]\).
Now let \(u = \frac{a+b}{1+ab}\).
\(f(u) = \log\left[\frac{1 - \frac{a+b}{1+ab}}{1 + \frac{a+b}{1+ab}}\right] = \log\left[\frac{1 + ab - (a+b)}{1 + ab + (a+b)}\right] = \log\left[\frac{1 - a - b + ab}{1 + a + b + ab}\right]\).
\(\therefore f(a) + f(b) = f\left(\frac{a+b}{1+ab}\right)\) (Proved).

Q7. Show that:
(a) If \(\frac{\log a}{y-z} = \frac{\log b}{z-x} = \frac{\log c}{x-y}\), then prove that \(a^a b^b c^c = 1\) (and \(a^x b^y c^z = 1\)).
(b) If \(x = \log_{2a} a, y = \log_{3a} 2a\) and \(z = \log_{4a} 3a\), prove that \(xyz + 1 = 2yz\).

(a) Let \(\frac{\log a}{y-z} = \frac{\log b}{z-x} = \frac{\log c}{x-y} = k\).
Then \(\log a = k(y-z), \log b = k(z-x), \log c = k(x-y)\).
Multiply by \(x, y, z\) respectively and add:
\(x \log a + y \log b + z \log c = k[x(y-z) + y(z-x) + z(x-y)] = k(0) = 0 \implies \log(a^x b^y c^z) = 0 \implies \mathbf{a^x b^y c^z = 1}\).

(b) Change bases: \(x = \frac{\log a}{\log 2a}, y = \frac{\log 2a}{\log 3a}, z = \frac{\log 3a}{\log 4a}\).
\(xyz = \frac{\log a}{\log 2a} \cdot \frac{\log 2a}{\log 3a} \cdot \frac{\log 3a}{\log 4a} = \frac{\log a}{\log 4a}\).
\(xyz + 1 = \frac{\log a}{\log 4a} + 1 = \frac{\log a + \log 4a}{\log 4a} = \frac{\log(4a^2)}{\log 4a} = \frac{\log (2a)^2}{\log 4a} = \frac{2 \log 2a}{\log 4a}\).
Now RHS: \(2yz = 2 \cdot \frac{\log 2a}{\log 3a} \cdot \frac{\log 3a}{\log 4a} = \frac{2 \log 2a}{\log 4a}\).
\(\therefore xyz + 1 = 2yz\) (Proved).

Official CDC Nepal Examination Simulation Specification Grid Pattern

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This assessment module generates official CDC Nepal specification grid examination items (Group A: 2 Marks, Group B: 3 Marks, Group C: 5 Marks). Enter the authorized instructor security key to unlock.

CDC Grade 11 Mathematics Model Examination

Total Marks: 50 | Time Allowed: 90 Mins | Subject Code: Mat. 111

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Group A (5 Questions × 2 Marks = 10 Marks)

1. Define one-one function with symbolic formulation.
2. Find the domain of \(f(x) = \frac{1}{\sqrt{x^2 - 16}}\).
3. If \(f(x) = 3x - 2\), find formula for \(f^{-1}(x)\).
4. State the horizontal line test for function invertibility.
5. Evaluate \(\log_{10} (0.001)\).

Group B (5 Questions × 3 Marks = 15 Marks)

6. Let \(f: \mathbb{R} \to \mathbb{R}\) be defined by \(f(x) = 4x + 3\). Prove that \(f\) is invertible and find \(f^{-1}\).
7. Find the domain and range of the function \(f(x) = \sqrt{25 - x^2}\).
8. If \(f(x) = 2x+1\) and \(g(x) = x^2\), evaluate \((g \circ f)(2)\) and \((f \circ g)(2)\).
9. Solve for \(x\): \(2\log x - \log(8x+16) = 0\).
10. Prove that \(\ln(1 + 2 + 3) = \ln 1 + \ln 2 + \ln 3\).

Group C (5 Questions × 5 Marks = 25 Marks)

11. Let \(f: A \to B\) and \(g: B \to C\) be bijective functions. Prove that \((g \circ f)^{-1} = f^{-1} \circ g^{-1}\).
12. Find the domain and range of the radical algebraic function \(f(x) = \sqrt{21 - 4x - x^2}\).
13. If \(f(x) = \log\left(\frac{1-x}{1+x}\right)\), show that \(f(a) + f(b) = f\left(\frac{a+b}{1+ab}\right)\).
14. Given \(x = \log_{2a} a\), \(y = \log_{3a} 2a\), and \(z = \log_{4a} 3a\), prove that \(xyz + 1 = 2yz\).
15. Comprehensive piecewise relation analysis: Determine domain, range, and injectivity for \(f(x) = \begin{cases} 2x-1 & x \ge 0 \\ x^2+1 & x < 0 \end{cases}\).

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