Moderate -0.8 Part (i) is a routine algebraic proof requiring expansion and rearrangement to show a perfect square is non-negative. Part (ii) requires finding a simple counter-example (e.g., n=3 gives 9, which is composite). Both parts are straightforward applications of basic techniques with no novel insight required, making this easier than average for A-level.
3. (i) Use algebra to prove that for all real values of \(x\)
$$( x - 4 ) ^ { 2 } \geqslant 2 x - 9$$
(ii) Show that the following statement is untrue.
$$2 ^ { n } + 1 \text { is a prime number for all values of } n , n \in \mathbb { N }$$
Attempts to expand \((x-4)^2\) and work from form \((x-4)^2 ... 2x - 9\) to form a 3TQ on one side of equation or an inequality
\(\Rightarrow (x-5)^2 ... 0\)
A1
Achieves both \(x^2 - 10x + 25\) and \((x-5)^2\). Allow \((x-5)^2\) written as \((x-5)(x-5)\)
Explains that "square numbers are greater than or equal to zero" hence (as \(x \in \mathbb{R}\)), \(\Rightarrow (x-4)^2 \geq 2x - 9\) *
A1*
For a correct proof. Eg "square numbers are greater than or equal to zero, hence (as \(x \in \mathbb{R}\)), \((x-5)^2 \geq 0\) \(\Rightarrow (x-4)^2 \geq 2x - 9\). This requires (1) Correct algebra throughout, (2) a correct explanation concerning square numbers and (3) a reference back to the original statement
(3 marks)
(ii) Shows that it is not true for a value of \(n\)
B1
Eg. When \(n = 3\), \(2^n + 1 = 8 + 1 = 9\) ✗ Not prime. Condone sloppily expressed proofs. Eg. '\(2^7 + 1 = \frac{129}{3} = 43\) which is not prime'. Condone implied proofs where candidates write \(2^5 + 1 = 33\) which has a factor of 11. If there are lots of calculations mark positively. Only one value is required to be found (with the relevant statement) to score the B1. The calculation cannot be incorrect. Eg. \(2^3 + 1 = 10\) which is not prime.
(1 mark, 4 marks total)
Additional notes for (i):
M1: Attempts to expand \((x-4)^2\) and work from form \((x-4)^2 ... 2x - 9\) to form a 3TQ on one side of an equation or an inequality
A1: Achieves both \(x^2 - 10x + 25\) and \((x-5)^2\). Allow \((x-5)^2\) written as \((x-5)(x-5)\)
A1*: For a correct proof. Requires (1) Correct algebra throughout, (2) a correct explanation concerning square numbers and (3) a reference back to the original statement
Answers via \(b^2 - 4ac\) are unlikely to be correct. Whilst it is true that there is only one root and therefore it touches the x-axis, it does not show that it is always positive. The explanation could involve a sketch of \(y = (x-5)^2\) but it must be accurate with a minimum on the +ve x axis with some statement alluding to why this shows \((x-5)^2 \geq 0\)
Approaches via odd and even numbers will usually not score anything. They would need to proceed using the main scheme via \((2m-4)^2 \geq 4m - 9\) and \((2m-1-4)^2 \geq 2(2m-1) - 9\)
Alt to (i) via contradiction:
Proof by contradiction is acceptable and marks in a similar way
M1: For setting up the contradiction 'Assume that there is an \(x\) such that \((x-4)^2 < 2x - 9 \Rightarrow x^2 - 10x + 25 ... 0\)
A1: \(\Rightarrow (x-5)^2 ... 0\) or \((x-5)(x-5) ... 0\)
A1*: This is not true as square numbers are always greater than or equal to 0, hence \((x-4)^2 \geq 2x - 9\)
Alt to part (i): States \((x-5)^2 \geq 0\) \(\Rightarrow x^2 - 10x + 25 \geq 0\) \(\Rightarrow x^2 - 8x - 16 \geq 2x - 9\) \(\Rightarrow (x-4)^2 \geq 2x - 9\)
**(i)** $(x-4)^2 \geq 2x - 9 \Rightarrow x^2 - 10x + 25 ... 0$ | M1 | Attempts to expand $(x-4)^2$ and work from form $(x-4)^2 ... 2x - 9$ to form a 3TQ on one side of equation or an inequality
$\Rightarrow (x-5)^2 ... 0$ | A1 | Achieves both $x^2 - 10x + 25$ and $(x-5)^2$. Allow $(x-5)^2$ written as $(x-5)(x-5)$
Explains that "square numbers are greater than or equal to zero" hence (as $x \in \mathbb{R}$), $\Rightarrow (x-4)^2 \geq 2x - 9$ * | A1* | For a correct proof. Eg "square numbers are **greater than or equal to zero**, hence (as $x \in \mathbb{R}$), $(x-5)^2 \geq 0$ $\Rightarrow (x-4)^2 \geq 2x - 9$. This requires (1) Correct algebra throughout, (2) a correct explanation concerning square numbers and (3) a reference back to the original statement
| (3 marks)
**(ii)** Shows that it is not true for a value of $n$ | B1 | Eg. When $n = 3$, $2^n + 1 = 8 + 1 = 9$ ✗ Not prime. Condone sloppily expressed proofs. Eg. '$2^7 + 1 = \frac{129}{3} = 43$ which is not prime'. Condone implied proofs where candidates write $2^5 + 1 = 33$ which has a factor of 11. If there are lots of calculations mark positively. Only one value is required to be found (with the relevant statement) to score the B1. The calculation cannot be incorrect. Eg. $2^3 + 1 = 10$ which is not prime.
| (1 mark, 4 marks total)
**Additional notes for (i):**
M1: Attempts to expand $(x-4)^2$ and work from form $(x-4)^2 ... 2x - 9$ to form a 3TQ on one side of an equation or an inequality
A1: Achieves both $x^2 - 10x + 25$ and $(x-5)^2$. Allow $(x-5)^2$ written as $(x-5)(x-5)$
A1*: For a correct proof. Requires (1) Correct algebra throughout, (2) a correct explanation concerning square numbers and (3) a reference back to the original statement
Answers via $b^2 - 4ac$ are unlikely to be correct. Whilst it is true that there is only one root and therefore it touches the x-axis, it does not show that it is always positive. The explanation could involve a sketch of $y = (x-5)^2$ but it must be accurate with a minimum on the +ve x axis with some statement alluding to why this shows $(x-5)^2 \geq 0$
Approaches via odd and even numbers will usually not score anything. They would need to proceed using the main scheme via $(2m-4)^2 \geq 4m - 9$ and $(2m-1-4)^2 \geq 2(2m-1) - 9$
Alt to (i) via contradiction:
Proof by contradiction is acceptable and marks in a similar way
M1: For setting up the contradiction 'Assume that there is an $x$ such that $(x-4)^2 < 2x - 9 \Rightarrow x^2 - 10x + 25 ... 0$
A1: $\Rightarrow (x-5)^2 ... 0$ or $(x-5)(x-5) ... 0$
A1*: This is not true as square numbers are always greater than or equal to 0, hence $(x-4)^2 \geq 2x - 9$
Alt to part (i): States $(x-5)^2 \geq 0$ $\Rightarrow x^2 - 10x + 25 \geq 0$ $\Rightarrow x^2 - 8x - 16 \geq 2x - 9$ $\Rightarrow (x-4)^2 \geq 2x - 9$
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3. (i) Use algebra to prove that for all real values of $x$
$$( x - 4 ) ^ { 2 } \geqslant 2 x - 9$$
(ii) Show that the following statement is untrue.
$$2 ^ { n } + 1 \text { is a prime number for all values of } n , n \in \mathbb { N }$$
\hfill \mbox{\textit{Edexcel P2 2019 Q3 [4]}}