CIE AS/A Level Maths-1.2 Function- Study Notes- New Syllabus - 2026-2027
CIE AS/A Level Maths-1.2 Function- Study Notes- New Syllabus
Ace AS/A Level Maths Exam with CIE AS/A Level Maths-1.2 Function- Study Notes
Key Concepts:
- Functions
- Domain of a Function
- Range of a Function
- One-One (Injective) Functions
- Inverse Function
- Composition of Functions
- Transformations of Graphs
Domain of a Function
Function
A function is a rule that maps each input (from a set called the domain) to exactly one output (from a set called the codomain).
Notation: \( f: x \mapsto f(x) \).
Example: \( f(x) = x^2 + 1 \).
The set of all possible outputs is called the range.
Domain of a Function
The domain of a function is the set of all possible input values (values of \(x\)) for which the function is defined.
When finding the domain, we look for restrictions that make the function undefined. Common restrictions include:
- Division by zero: A denominator cannot be zero.
- Square roots: The expression under a square root must be non-negative (for real functions).
- Logarithms: The argument of a logarithm must be positive.
- Trig functions: Exclude values where the function is undefined (like \(\tan x\) where \(\cos x = 0\)).
Example:
Find the domain of \( f(x) = \dfrac{1}{x-2} \).
▶️ Answer/Explanation
The denominator \(x – 2 = 0 \Rightarrow x = 2\) is not allowed.
Domain: \( \mathbb{R} – \{2\} \).
Example:
Find the domain of \( f(x) = \sqrt{x – 3} \).
▶️ Answer/Explanation
Restriction: \(x – 3 \geq 0 \Rightarrow x \geq 3\).
Domain: \([3, \infty)\).
Example:
Find the domain of \( f(x) = \ln(x+1) \).
▶️ Answer/Explanation
Logarithm requires argument \(x + 1 > 0 \Rightarrow x > -1\).
Domain: \((-1, \infty)\).
Range of a Function
Range of a Function
The range of a function is the set of all possible output values (values of \(y\)) that the function can take for inputs in its domain.
To find the range:
- Start with the domain and consider how the function transforms those inputs.
- For simple functions, apply algebraic manipulations (e.g., inequalities).
- For quadratic or trigonometric functions, use completing the square, symmetry, or known function behaviour.
- Graphical reasoning is often helpful.
Example:
Find the range of \( f(x) = x^2 \), where \( x \in \mathbb{R} \).
▶️ Answer/Explanation
Since \(x^2 \geq 0\) for all real \(x\), the function never gives negative values.
Range: \([0, \infty)\).
Example:
Find the range of \( f(x) = \dfrac{1}{x} \), where \( x \in \mathbb{R} \setminus \{0\} \).
▶️ Answer/Explanation
The function can take any real value except \(0\), since \( \dfrac{1}{x} \neq 0 \).
Range: \( \mathbb{R} \setminus \{0\} \).
Example :
Find the range of \( f(x) = \sin x \), where \( x \in \mathbb{R} \).
▶️ Answer/Explanation
The sine function oscillates between \(-1\) and \(+1\).
Range: \([-1, 1]\).
One-One (Injective) Functions
One-One (Injective) Functions
A function is called one-one (or injective) if different inputs give different outputs. In other words:
\( f(x_1) = f(x_2) \Rightarrow x_1 = x_2 \).
This means no two distinct values of \(x\) map to the same value of \(y\). Graphically, a function is one-one if any horizontal line cuts the graph at most once (called the Horizontal Line Test).
Key Notes:
- Linear functions like \(f(x) = 2x + 3\) are one-one.
- Quadratic functions like \(f(x) = x^2\) are not one-one on \(\mathbb{R}\) (since \(f(2) = f(-2)\)), but can be made one-one if the domain is restricted (e.g., \(x \geq 0\)).
- Exponential functions are always one-one on their domains.
Example :
Show that \( f(x) = 3x + 2 \) is a one-one function.
▶️ Answer/Explanation
Suppose \( f(x_1) = f(x_2) \).
\( 3x_1 + 2 = 3x_2 + 2 \)
\( 3x_1 = 3x_2 \Rightarrow x_1 = x_2 \).
Thus, \(f(x)\) is one-one.
Example :
Is \( f(x) = x^2 \) a one-one function on \( \mathbb{R} \)?
▶️ Answer/Explanation
Check: \( f(2) = 4 \), \( f(-2) = 4 \). Different inputs gave the same output.
Therefore, \( f(x) = x^2 \) is not one-one on \( \mathbb{R} \).
But if we restrict the domain to \( x \geq 0 \) or \( x \leq 0 \), then it becomes one-one.
Inverse Function
Inverse Function
An inverse function essentially “reverses” the effect of the original function. If \( f : A \to B \) is one-one and onto, then its inverse \( f^{-1} : B \to A \) is defined by:
\( f^{-1}(y) = x \iff f(x) = y \).
Key Properties:
- A function must be one-one and onto to have an inverse.
- \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \).
- To find an inverse algebraically:
- Replace \( f(x) \) with \( y \).
- Solve for \( x \) in terms of \( y \).
- Interchange \( x \) and \( y \).
- Graphically, the graph of \( y = f^{-1}(x) \) is the reflection of the graph of \( y = f(x) \) in the line \( y = x \).
Example :
Find the inverse of \( f(x) = 2x + 3 \).
▶️ Answer/Explanation
Step 1: Let \( y = 2x + 3 \).
Step 2: Solve for \(x\): \( x = \dfrac{y – 3}{2} \).
Step 3: Swap variables: \( f^{-1}(x) = \dfrac{x – 3}{2} \).
Final Answer: \( f^{-1}(x) = \dfrac{x – 3}{2} \).
Graphically: The line \( y = 2x + 3 \) and its inverse \( y = \dfrac{x – 3}{2} \) are reflections in the line \( y = x \).
Example:
Find the inverse of \( f(x) = x^2 \), restricting the domain to \( x \geq 0 \).
▶️ Answer/Explanation
Step 1: Let \( y = x^2 \), with \( x \geq 0 \).
Step 2: Solve for \( x \): \( x = \sqrt{y} \).
Step 3: Swap variables: \( f^{-1}(x) = \sqrt{x} \).
Final Answer: \( f^{-1}(x) = \sqrt{x} \), with domain \( x \geq 0 \).
Graphically: The parabola \( y = x^2 \) (for \( x \geq 0 \)) and the curve \( y = \sqrt{x} \) are reflections in the line \( y = x \).
Example:
Find the inverse of \( f(x) = \sin x \), restricting the domain to \( -\dfrac{\pi}{2} \leq x \leq \dfrac{\pi}{2} \).
▶️ Answer/Explanation
Step 1: Let \( y = \sin x \), with \( -\dfrac{\pi}{2} \leq x \leq \dfrac{\pi}{2} \).
Step 2: Solve for \(x\): \( x = \sin^{-1}(y) \).
Step 3: Swap variables: \( f^{-1}(x) = \sin^{-1}(x) \).
Final Answer: \( f^{-1}(x) = \sin^{-1}(x) \), with domain \( -1 \leq x \leq 1 \).
Graphically: The graph of \( y = \sin x \) (restricted) and \( y = \sin^{-1} x \) are reflections in the line \( y = x \).
Composition of Functions
Composition of Functions
The composition of functions means applying one function after another. If \( f \) and \( g \) are two functions, then the composition is defined as:
\( (f \circ g)(x) = f(g(x)) \).
This means we first apply \( g \) to \( x \), then apply \( f \) to the result.
Key Notes:
- In general, \( f \circ g \neq g \circ f \). Order matters.
- The domain of \( f \circ g \) is all \( x \) in the domain of \( g \) such that \( g(x) \) lies in the domain of \( f \).
- If \( f \) and \( g \) are inverses, then:
- \( f(g(x)) = x \)
- \( g(f(x)) = x \)
Example:
Let \( f(x) = 2x + 1 \), \( g(x) = x^2 \). Find \( (f \circ g)(x) \) and \( (g \circ f)(x) \).
▶️ Answer/Explanation
\( (f \circ g)(x) = f(g(x)) = f(x^2) = 2(x^2) + 1 = 2x^2 + 1 \).
\( (g \circ f)(x) = g(f(x)) = g(2x+1) = (2x+1)^2 = 4x^2 + 4x + 1 \).
Conclusion: \( f \circ g \neq g \circ f \).
Example:
If \( f(x) = \sqrt{x} \) (domain \( x \geq 0 \)), and \( g(x) = x – 4 \), find \( (f \circ g)(x) \).
▶️ Answer/Explanation
\( (f \circ g)(x) = f(g(x)) = f(x – 4) = \sqrt{x – 4} \).
Domain Restriction: Since square roots require non-negative input, \( x – 4 \geq 0 \Rightarrow x \geq 4 \).
Final Answer: \( (f \circ g)(x) = \sqrt{x – 4}, \; x \geq 4 \).
Example:
Let \( f(x) = 3x – 2 \). Show that \( g(x) = \dfrac{x+2}{3} \) is its inverse using composition.
▶️ Answer/Explanation
\( (f \circ g)(x) = f\!\left(\dfrac{x+2}{3}\right) = 3\!\left(\dfrac{x+2}{3}\right) – 2 = x+2-2 = x \).
\( (g \circ f)(x) = g(3x – 2) = \dfrac{(3x-2)+2}{3} = \dfrac{3x}{3} = x \).
Conclusion: Both compositions give \( x \). So, \( g(x) \) is indeed the inverse of \( f(x) \).
Transformations of Graphs
Transformations of Graphs
Given a graph of \( y = f(x) \), we can transform it by shifting, stretching, or reflecting. These transformations affect the shape and position of the graph but do not change its basic nature.
Key Terms:
- Translation: Shifts the graph horizontally or vertically without changing its shape.
- Reflection: Flips the graph across a line (e.g. the \(x\)-axis, \(y\)-axis, or line \(y=x\)).
- Stretch: Enlarges or compresses the graph either vertically or horizontally.
Table of Transformations
| Transformation | Equation | Effect on Graph | Description |
|---|---|---|---|
| Vertical translation | \( y = f(x) + a \) | Graph moves up if \(a>0\), down if \(a<0\) | Shift by \(a\) units vertically |
| Horizontal translation | \( y = f(x + a) \) | Graph moves left if \(a>0\), right if \(a<0\) | Shift by \(a\) units horizontally |
| Vertical stretch | \( y = af(x) \) | Stretches graph away from \(x\)-axis by factor \(|a|\) | If \(0<a<1\), it compresses vertically |
| Horizontal stretch | \( y = f(ax) \) | Stretches graph away from \(y\)-axis by factor \(\tfrac{1}{|a|}\) | If \(|a|>1\), compressed horizontally; if \(0<|a|<1\), stretched |
| Reflection in \(x\)-axis | \( y = -f(x) \) | Flips graph upside down | Mirror image across \(x\)-axis |
| Reflection in \(y\)-axis | \( y = f(-x) \) | Flips graph left-to-right | Mirror image across \(y\)-axis |
Example:
The graph of \( y = x^2 \) is transformed to \( y = (x – 3)^2 + 2 \). Describe the transformation and sketch the new graph.
▶️ Answer/Explanation
Step 1: Compare with \( y = f(x) \) form.
\( y = (x – 3)^2 + 2 \) comes from \( y = f(x) = x^2 \).
Step 2: Identify transformations.
\( (x – 3)^2 \) means a horizontal translation 3 units to the right.
\(+2\) means a vertical translation 2 units up.
Step 3: Describe graph.
Original parabola \( y = x^2 \) with vertex at \((0,0)\).
New vertex is at \((3,2)\).
Shape is unchanged (still U-shaped).
Answer: The graph is the parabola shifted right by 3 units and up by 2 units, with vertex at \((3,2)\).
Example:
The graph of \( y = \sin x \) is transformed to \( y = -2\sin(x + \tfrac{\pi}{4}) \). Describe the transformation and sketch the new graph.
▶️ Answer/Explanation
Step 1: Compare with \( y = f(x) = \sin x \).
New function: \( y = -2\sin(x + \tfrac{\pi}{4}) \).
Step 2: Identify transformations.
\( (x + \tfrac{\pi}{4}) \) → horizontal translation \(\tfrac{\pi}{4}\) units left.
\(-\sin x\) → reflection in the \(x\)-axis.
\(2\sin x\) → vertical stretch by factor 2 (amplitude = 2).
Step 3: Describe graph.
Original sine wave has amplitude 1, period \(2\pi\), passing through \((0,0)\).
New graph has amplitude 2, reflected, and starts shifted left by \(\tfrac{\pi}{4}\).
Answer: The graph is a sine wave with amplitude 2, reflected in the \(x\)-axis, and shifted \(\tfrac{\pi}{4}\) units left.