IGCSE · Practice questions · Functions, domain and range

Every input, one answer.

Six original Cambridge-style questions on functions: notation with \(e^{x}\) and \(\lg x\), the largest domain, ranges written correctly in terms of \(f(x)\), one-one and many-one functions, and a quadratic on a restricted domain.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Check the inputs, then trace the outputs.

Function? Every input in the domain gives exactly one output.
One-one or many-one? Look for two inputs with the same output, e.g. either side of a vertex.
Largest domain: square roots need \(\ge 0\), logarithms need \(> 0\), denominators \(\ne 0\).
Range of a quadratic: complete the square, check if the vertex is in the domain, then check both ends.
Write the range in terms of \(f(x)\), choosing \(<\) or \(\le\) by whether the value is reached.

Recall

[2 marks]

The function f is defined by \(f(x) = 4 - 3x\) for \(-2 \le x \le 3\). Find the range of f.

\(f\) is decreasing: \(f(-2) = 4 + 6 = 10\) and \(f(3) = 4 - 9 = -5\). ✓
Range: \(-5 \le f(x) \le 10\). ✓

Calculation

[4 marks]

The function f is defined by \(f(x) = 2e^{x} - 1\) for \(x \in \mathbb{R}\).

(a) Find \(f(0)\).

(b) Solve \(f(x) = 7\), giving your answer in the form \(\ln a\).

(a) \(f(0) = 2e^{0} - 1 = 2 - 1 = 1\). ✓
(b) \(2e^{x} - 1 = 7 \Rightarrow 2e^{x} = 8 \Rightarrow e^{x} = 4\), so \(x = \ln 4\). ✓
(c) \(e^{x} > 0\), so \(2e^{x} - 1 > -1\), and \(-1\) is never reached. ✓
Range: \(f(x) > -1\). ✓

Calculation

[3 marks]

The function g is defined by \(g : x \mapsto \lg(x + 5)\) for \(x > -5\).

(a) Find \(g(95)\).

(b) Solve \(g(x) = -1\).

(a) \(g(95) = \lg 100 = 2\). ✓
(b) \(\lg(x + 5) = -1 \Rightarrow x + 5 = 10^{-1} = 0.1 \Rightarrow x = -4.9\). ✓
(c) \(g(-5) = \lg 0\), and the logarithm of 0 does not exist (no power of 10 equals 0), so \(x = -5\) has no output. ✓

Analysis

[5 marks]

The function f is defined by \(f(x) = 2x^{2} - 12x + 13\) for \(1 \le x \le 7\).

(a) Express \(f(x)\) in the form \(a(x + b)^{2} + c\).

(b) Find the range of f.

(a) \(2(x^{2} - 6x) + 13 = 2\big[(x - 3)^{2} - 9\big] + 13 = 2(x - 3)^{2} - 5\). ✓
(b) The vertex is at \(x = 3\), inside the domain, so the least value is \(-5\). ✓
Ends: \(f(1) = 2(4) - 5 = 3\) and \(f(7) = 2(16) - 5 = 27\). Range: \(-5 \le f(x) \le 27\). ✓
(c) \(f(1) = 3\) and \(f(5) = 2(4) - 5 = 3\). ✓
Two different inputs in the domain give the same output, so f is many-one, not one-one. ✓

Proof

[3 marks]

(a) Explain why the rule 'x maps to each number whose square is x', for \(x \ge 0\), does not define a function.

(b) The function h is defined by \(h : x \mapsto \sqrt{x}\) for \(x \ge 0\). Explain why h is a function, and state whether it is one-one or many-one.

(a) The input 9 maps to both 3 and \(-3\), since \(3^{2} = (-3)^{2} = 9\). One input with two outputs is not allowed, so the rule is not a function. ✓
(b) \(\sqrt{x}\) means the non-negative square root only, so every \(x \ge 0\) gives exactly one output: h is a function. ✓
\(\sqrt{x}\) increases as \(x\) increases, so different inputs give different outputs: h is one-one. ✓

Analysis

[5 marks]

The function f is defined by \(f(x) = x^{2} - 4x + k\) for \(0 \le x \le 5\), where \(k\) is a constant. The least value of \(f(x)\) is 1.

(a) By completing the square, find the value of \(k\).

(b) Find the range of f.

(a) \(f(x) = (x - 2)^{2} - 4 + k\). The vertex \(x = 2\) is in the domain, so the least value is \(k - 4\). ✓
\(k - 4 = 1\), so \(k = 5\) and \(f(x) = (x - 2)^{2} + 1\). ✓
(b) Ends: \(f(0) = 5\) and \(f(5) = 10\). Range: \(1 \le f(x) \le 10\). ✓
(c) \((x - 2)^{2} + 1 = 5 \Rightarrow (x - 2)^{2} = 4 \Rightarrow x - 2 = \pm 2\), so \(x = 0\) or \(x = 4\). ✓
Both lie in the domain and give the same output 5, so f is many-one. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.