Functions
Algebra and functions
- A2
- Non-calculator
- 8 questions
- 37 marks
These are original practice questions for the Edexcel A-level Pure content, not a past paper. This site is independent and not affiliated with Pearson.
-
Solve $|2x - 6| = 4$.
3 marks
-
The function $f$ is given by $f(x) = \sqrt{x - 2}$. State its domain and range.
3 marks
-
The functions $f$ and $g$ are given by $f(x) = 3x - 1$ and $g(x) = x^2$. Find $fg(x)$ and $gf(2)$.
4 marks
-
The function $f$ is given by $f(x) = 2x + 3$. Find $f^{-1}(x)$, and find $f^{-1}(7)$.
4 marks
-
The function $f$ is given by $f(x) = |x - 2| + 1$. State the range of $f$, and solve $f(x) = 5$.
5 marks
-
The function $f$ is given by $f(x) = \dfrac{1}{x - 3}$, $x \ne 3$. Find $f^{-1}(x)$ and state its domain.
5 marks
-
The functions $f$ and $g$ are given by $f(x) = 2x - 1$ and $g(x) = x^2 + 4$. Solve $fg(x) = gf(x)$.
6 marks
-
The function $f$ is given by $f(x) = x^2 - 6x + 5$, with domain $x \ge 3$. Find the range of $f$, and find an expression for $f^{-1}(x)$, including its domain.
7 marks
Mark scheme
-
Answer. $x = 5$ or $x = 1$
- $2x - 6 = 4$ or $2x - 6 = -4$.
- $x = 5$ or $x = 1$.
-
Answer. The domain is $x \ge 2$ and the range is $f(x) \ge 0$.
- The expression under the square root must be at least $0$.
- A square root gives values that are at least $0$.
-
Answer. $fg(x) = 3x^2 - 1$ and $gf(2) = 25$.
- $fg(x) = f(x^2) = 3x^2 - 1$.
- $f(2) = 5$ and $g(5) = 25$.
-
Answer. $f^{-1}(x) = \dfrac{x - 3}{2}$ and $f^{-1}(7) = 2$.
- Let $y = 2x + 3$ and rearrange to $x = \dfrac{y - 3}{2}$.
- Substitute $7$ to get $2$.
-
Answer. The range is $f(x) \ge 1$. The solutions are $x = 6$ and $x = -2$.
- $|x - 2| \ge 0$, so $f(x) \ge 1$.
- $|x - 2| = 4$.
- $x - 2 = 4$ or $x - 2 = -4$.
-
Answer. $f^{-1}(x) = 3 + \dfrac{1}{x}$, with domain $x \ne 0$.
- Let $y = \dfrac{1}{x - 3}$ and rearrange to $x = 3 + \dfrac{1}{y}$.
- The inverse is not defined when the denominator is $0$.
-
Answer. $x = 1 \pm \sqrt{2}$
- $fg(x) = 2x^2 + 7$.
- $gf(x) = (2x - 1)^2 + 4 = 4x^2 - 4x + 5$.
- $2x^2 + 7 = 4x^2 - 4x + 5$ gives $x^2 - 2x - 1 = 0$.
- $x = 1 \pm \sqrt{2}$.
-
Answer. The range is $f(x) \ge -4$. The inverse is $f^{-1}(x) = 3 + \sqrt{x + 4}$, with domain $x \ge -4$.
- $f(x) = (x - 3)^2 - 4$. For $x \ge 3$ the least value is $-4$.
- The range is $y \ge -4$.
- $y + 4 = (x - 3)^2$, and $x - 3 \ge 0$, so $x = 3 + \sqrt{y + 4}$.
- The domain of the inverse is the range of $f$.