## Question 2:

### Part (i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| 'if $n$ even then $n^3$ even, so $n^3 + 1$ odd' | B1 | Must mention $n^3$ is even or even$^3$ is even or even $\times$ even = even. 0 for just 'if $n$ is even, $n^3+1$ is odd'. 0 if just examples of numbers used |
| $\Leftarrow$ with if $n^3+1$ odd then $n^3$ even but if $n^3$ is even, $n$ is not necessarily an integer; **or** $\Leftrightarrow$ with '$n^3+1$ odd then $n^3$ even so $n$ even' [assuming $n$ is an integer] | B1 | or '$\Leftrightarrow$' with if $n$ is odd, $n^3$ is odd, so $n^3+1$ is even'. Condone $\leftrightarrow$ instead of $\Leftrightarrow$ etc in both parts |
| if 0 in question, allow SC1 for $\Leftrightarrow$ or $\Leftarrow$ and attempt at using general odd/even in explanation | | Must go further than restating the info in the question; please annotate as SC |
| **[2]** | | |

### Part (ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| showing $\Leftarrow$ is true | B1 | e.g. when $x > 3$, $+\text{ve} \times +\text{ve} > 0$. 0 for just example(s) or for simply stating it is true |
| $\Leftarrow$ chosen and showing that $\Rightarrow$ [and therefore $\Leftrightarrow$] is/are not true | B1 | Stating that true when $x < 2$ or giving a counterexample such as 1, 0 or a negative number [to show quadratic inequality also true for this number]. 0 for saying another solution $x > 2$ |
| allow B2 for $\Leftarrow$ and $x > 3$ and $x < 2$ shown/stated as solution or sketch showing two solutions of $x^2 - 5x + 6 > 0$ | | or B1 for this argument with another symbol |
| **[2]** | | |

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