| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof |
| Type | Contradiction proof of irrationality |
| Difficulty | Standard +0.3 Part (i) is a straightforward inequality requiring basic reasoning about exponential functions (always true for x≥0, sometimes true overall). Part (ii) is a standard proof by contradiction that √3 is irrational—a textbook exercise requiring the template of assuming √3=p/q, squaring, and showing 3 divides both p and q. While proof questions can intimidate students, this is one of the most commonly taught proofs in A-level Further Maths with a well-established method, making it slightly easier than average overall. |
| Spec | 1.01d Proof by contradiction1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| When \(x = -1\), \(\frac{1}{3} < \frac{1}{2}\) or \(\frac{1}{3}\) is not greater than or equal to \(\frac{1}{2}\); or when \(x < 0\), \(3^x < 2^x\) or \(3^x\) is not greater than or equal to \(2^x\) | M1 | For an explanation or statement to show when the claim \(3^x \geq 2^x\) fails |
| \(x = 2\), \(9 \geq 4\) or 9 is greater than or equal to 4; or when \(x \geq 0\), \(3^x \geq 2^x\); and a correct conclusion e.g. so the claim \(3^x \geq 2^x\) is sometimes true | A1 | Followed by explanation/statement to show when claim is true, plus correct conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Assume that \(\sqrt{3}\) is a rational number. So \(\sqrt{3} = \frac{p}{q}\), where \(p\) and \(q\) are integers, \(q \neq 0\), and the HCF of \(p\) and \(q\) is 1 | M1 | Uses proof by contradiction, initially assuming \(\sqrt{3}\) is rational and expresses \(\sqrt{3}\) in the form \(\frac{p}{q}\) where \(p\) and \(q\) are correctly defined |
| \(\Rightarrow p = \sqrt{3}\,q \Rightarrow p^2 = 3q^2\) | M1 | Writes \(\sqrt{3} = \frac{p}{q}\) and rearranges to make \(p^2\) the subject |
| \(\Rightarrow p^2\) is divisible by 3 and so \(p\) is divisible by 3 | A1 | Uses a logical argument to prove that \(p\) is divisible by 3 |
| So \(p = 3c\), where \(c\) is an integer. From earlier, \(p^2 = 3q^2 \Rightarrow (3c)^2 = 3q^2\) | M1 | Uses the result that \(p\) is divisible by 3 to construct initial stage of proving \(q\) is also divisible by 3, by substituting \(p = 3c\) into expression for \(p^2\) |
| \(\Rightarrow q^2 = 3c^2 \Rightarrow q^2\) is divisible by 3 and so \(q\) is divisible by 3 | A1 | Uses correct argument, in same way as before, to deduce \(q\) is also divisible by 3 |
| As both \(p\) and \(q\) are divisible by 3 the HCF of \(p\) and \(q\) is not 1. This contradiction implies that \(\sqrt{3}\) is an irrational number | A1 | Completes the argument that \(\sqrt{3}\) is irrational. Note: All previous 5 marks must be scored to obtain this final A mark |
## Question 14:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| When $x = -1$, $\frac{1}{3} < \frac{1}{2}$ or $\frac{1}{3}$ is not greater than or equal to $\frac{1}{2}$; or when $x < 0$, $3^x < 2^x$ or $3^x$ is not greater than or equal to $2^x$ | M1 | For an explanation or statement to show when the claim $3^x \geq 2^x$ fails |
| $x = 2$, $9 \geq 4$ or 9 is greater than or equal to 4; or when $x \geq 0$, $3^x \geq 2^x$; and a correct conclusion e.g. so the claim $3^x \geq 2^x$ is sometimes true | A1 | Followed by explanation/statement to show when claim is true, plus correct conclusion |
**(2 marks)**
---
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Assume that $\sqrt{3}$ is a rational number. So $\sqrt{3} = \frac{p}{q}$, where $p$ and $q$ are integers, $q \neq 0$, and the HCF of $p$ and $q$ is 1 | M1 | Uses proof by contradiction, initially assuming $\sqrt{3}$ is rational and expresses $\sqrt{3}$ in the form $\frac{p}{q}$ where $p$ and $q$ are correctly defined |
| $\Rightarrow p = \sqrt{3}\,q \Rightarrow p^2 = 3q^2$ | M1 | Writes $\sqrt{3} = \frac{p}{q}$ and rearranges to make $p^2$ the subject |
| $\Rightarrow p^2$ is divisible by 3 and so $p$ is divisible by 3 | A1 | Uses a logical argument to prove that $p$ is divisible by 3 |
| So $p = 3c$, where $c$ is an integer. From earlier, $p^2 = 3q^2 \Rightarrow (3c)^2 = 3q^2$ | M1 | Uses the result that $p$ is divisible by 3 to construct initial stage of proving $q$ is also divisible by 3, by substituting $p = 3c$ into expression for $p^2$ |
| $\Rightarrow q^2 = 3c^2 \Rightarrow q^2$ is divisible by 3 and so $q$ is divisible by 3 | A1 | Uses correct argument, in same way as before, to deduce $q$ is also divisible by 3 |
| As both $p$ and $q$ are divisible by 3 the HCF of $p$ and $q$ is not 1. This contradiction implies that $\sqrt{3}$ is an irrational number | A1 | Completes the argument that $\sqrt{3}$ is irrational. **Note:** All previous 5 marks must be scored to obtain this final A mark |
**(6 marks)**
**Total: 8 marks**\begin{enumerate}
\item (i) Kayden claims that
\end{enumerate}
$$3 ^ { x } \geqslant 2 ^ { x }$$
Determine whether Kayden's claim is always true, sometimes true or never true, justifying your answer.\\
(ii) Prove that $\sqrt { 3 }$ is an irrational number.
\hfill \mbox{\textit{Edexcel Paper 2 Q14 [8]}}