| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2021 |
| Session | January |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof |
| Type | Algebraic inequality proof |
| Difficulty | Moderate -0.8 Part (i) is a routine algebraic inequality proof using standard techniques (rearranging to perfect square form or squaring both sides), requiring only basic manipulation skills. Part (ii) is a simple disproof by counterexample requiring minimal work. Both parts are straightforward applications of basic techniques with no novel insight needed, making this easier than average for A-level. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps1.01c Disproof by counter example |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (As \(x\geqslant0\) so \(\sqrt{3x}\) exists and) \((\sqrt{3x}-1)^2\geqslant0\) | M1 | Uses \(x\geqslant0\) and squares are non-negative to set up suitable equation |
| Hence \(3x-2\sqrt{3x}+1\geqslant0\) | M1 | Squares to achieve 3 terms |
| \(\Rightarrow3x+1\geqslant2\sqrt{3x}\) * | A1* | Rearranges correctly to given result |
| Alt 1: \(3x+1\geqslant2\sqrt{3x}\Leftrightarrow(3x+1)^2\geqslant12x\Leftrightarrow9x^2-6x+1\geqslant0\) | M1 | Starting with given statement, squares both sides, expands \((3x+1)^2\) (three terms), collects terms |
| \(9x^2-6x+1\geqslant0\Leftrightarrow(3x-1)^2\geqslant0\) | M1 | Completes square/factorises to perfect square |
| Square numbers \(\geqslant0\) so \((3x-1)^2\geqslant0\) is true hence \(3x+1\geqslant2\sqrt{3x}\) * | A1* | Achieves \((3x-1)^2\geqslant0\) with suitable conclusion |
| Alt 2: If \(3x+1<2\sqrt{3x}\) then \(3x-2\sqrt{3x}+1<0\) | M1 | Starts with negation |
| So \((...\sqrt{3x}\pm...)^2<0\) or \((\sqrt{x}\pm...)^2<0\) | M1 | Attempts to factorise to perfect square |
| But \((\sqrt{3x}-1)^2\geqslant0\) for all \(x\geqslant0\) so \(3x+1\geqslant2\sqrt{3x}\) | A1 | Contradicts assumption; states original result holds |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Shows not true for three consecutive primes e.g. \(5+7+11=23\), not divisible by \(5\) | B1 | Must evaluate the sum correctly; conclusion required; \(1+2+3\) not valid as \(1\) is not prime |
## Question 5:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| (As $x\geqslant0$ so $\sqrt{3x}$ exists and) $(\sqrt{3x}-1)^2\geqslant0$ | M1 | Uses $x\geqslant0$ and squares are non-negative to set up suitable equation |
| Hence $3x-2\sqrt{3x}+1\geqslant0$ | M1 | Squares to achieve 3 terms |
| $\Rightarrow3x+1\geqslant2\sqrt{3x}$ * | A1* | Rearranges correctly to given result |
| **Alt 1:** $3x+1\geqslant2\sqrt{3x}\Leftrightarrow(3x+1)^2\geqslant12x\Leftrightarrow9x^2-6x+1\geqslant0$ | M1 | Starting with given statement, squares both sides, expands $(3x+1)^2$ (three terms), collects terms |
| $9x^2-6x+1\geqslant0\Leftrightarrow(3x-1)^2\geqslant0$ | M1 | Completes square/factorises to perfect square |
| Square numbers $\geqslant0$ so $(3x-1)^2\geqslant0$ is true hence $3x+1\geqslant2\sqrt{3x}$ * | A1* | Achieves $(3x-1)^2\geqslant0$ with suitable conclusion |
| **Alt 2:** If $3x+1<2\sqrt{3x}$ then $3x-2\sqrt{3x}+1<0$ | M1 | Starts with negation |
| So $(...\sqrt{3x}\pm...)^2<0$ or $(\sqrt{x}\pm...)^2<0$ | M1 | Attempts to factorise to perfect square |
| But $(\sqrt{3x}-1)^2\geqslant0$ for all $x\geqslant0$ so $3x+1\geqslant2\sqrt{3x}$ | A1 | Contradicts assumption; states original result holds |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Shows not true for three consecutive primes e.g. $5+7+11=23$, not divisible by $5$ | B1 | Must evaluate the sum correctly; conclusion required; $1+2+3$ not valid as $1$ is not prime |
---5. (i) Use algebra to prove that for all $x \geqslant 0$
$$3 x + 1 \geqslant 2 \sqrt { 3 x }$$
(ii) Show that the following statement is not true.\\
"The sum of three consecutive prime numbers is always a multiple of 5 "\\
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\hfill \mbox{\textit{Edexcel P2 2021 Q5 [4]}}