Moderate -0.8 Part (i) requires finding a single counter example (e.g., p=11 gives 23, but p=7 gives 15=3×5), which is straightforward trial. Part (ii) is routine proof by exhaustion checking n even/odd cases with basic algebra. Both are standard textbook proof exercises requiring minimal problem-solving insight.
10. (i) Prove by counter example that the statement
"if \(p\) is a prime number then \(2 p + 1\) is also a prime number" is not true.
(ii) Use proof by exhaustion to prove that if \(n\) is an integer then
$$5 n ^ { 2 } + n + 12$$
is always even.
E.g. \(p = 7 \Rightarrow 2p + 1 = 15\), which is not a prime number (so statement is not true)
B1
Identifies a counter example with calculation and suitable conclusion showing it is not prime. Accept \(p=13\): \(2\times13+1=27=3^3\) so not prime. Accept "not prime" or showing not prime by indicating a factor.
(1 mark total)
Part (ii):
Answer
Marks
Guidance
Answer/Working
Mark
Guidance
\(n \text{ odd} \Rightarrow n = 2k+1 \Rightarrow 5n^2+n+12 = 5(2k+1)^2+2k+1+12\) or \(n \text{ even} \Rightarrow n=2k \Rightarrow 5n^2+n+12=5(2k)^2+2k+12\)
M1
Starts proof by considering \(n\) odd or \(n\) even and substituting into expression
Considers \(n\) odd and \(n\) even (both cases as above)
M1
Same criteria for each case as first M
\(n\) odd: \(20k^2+22k+18\) which is even; \(n\) even: \(20k^2+2k+12\) which is even
A1
Both cases attempted with at least one correct expression stated to be even
\(n\) odd: \(2(10k^2+11k+9)\); \(n\) even: \(2(10k^2+k+6)\); these are both even so \(5n^2+n+12\) must be even for all integers \(n\)
A1
Fully correct proof considering both \(n\) odd and \(n\) even, showing resulting expressions are even, with suitable conclusion. Must use variable other than \(n\). Note for \(n=2k-1\): \(20k^2-18k+16=2(10k^2-9k+8)\)
(4 marks total — Total 5)
# Question 10:
**Part (i):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| E.g. $p = 7 \Rightarrow 2p + 1 = 15$, which is not a prime number (so statement is not true) | B1 | Identifies a counter example with calculation and suitable conclusion showing it is not prime. Accept $p=13$: $2\times13+1=27=3^3$ so not prime. Accept "not prime" or showing not prime by indicating a factor. |
**(1 mark total)**
**Part (ii):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $n \text{ odd} \Rightarrow n = 2k+1 \Rightarrow 5n^2+n+12 = 5(2k+1)^2+2k+1+12$ **or** $n \text{ even} \Rightarrow n=2k \Rightarrow 5n^2+n+12=5(2k)^2+2k+12$ | M1 | Starts proof by considering $n$ odd or $n$ even and substituting into expression |
| Considers $n$ odd **and** $n$ even (both cases as above) | M1 | Same criteria for each case as first M |
| $n$ odd: $20k^2+22k+18$ which is even; $n$ even: $20k^2+2k+12$ which is even | A1 | Both cases attempted with at least one correct expression stated to be even |
| $n$ odd: $2(10k^2+11k+9)$; $n$ even: $2(10k^2+k+6)$; these are both even so $5n^2+n+12$ must be even for all integers $n$ | A1 | Fully correct proof considering both $n$ odd and $n$ even, showing resulting expressions are even, with suitable conclusion. **Must use variable other than $n$.** Note for $n=2k-1$: $20k^2-18k+16=2(10k^2-9k+8)$ |
**(4 marks total — Total 5)**
10. (i) Prove by counter example that the statement\\
"if $p$ is a prime number then $2 p + 1$ is also a prime number" is not true.\\
(ii) Use proof by exhaustion to prove that if $n$ is an integer then
$$5 n ^ { 2 } + n + 12$$
is always even.
\hfill \mbox{\textit{Edexcel P2 2022 Q10 [5]}}