## Question 10:

### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Correct shape | B1 | Shape anywhere on axes; condone pen slips toward asymptotes as long as not clear turning points |
| $x$-intercepts at $\left(-\frac{1}{3}, 0\right)$ and $\left(\frac{1}{3}, 0\right)$ | B1 | May be marked as $-\frac{1}{3}$ and $\frac{1}{3}$. Do not allow $\left(0, -\frac{1}{3}\right)$ or $\left(0, \frac{1}{3}\right)$. Accept $0.\dot{3}$ and $-0.\dot{3}$. Cannot be scored without a corresponding sketch |
| Asymptote $x = 0$ or $y = -9$ | B1 | As long as it is an asymptote for their sketch |
| Both asymptotes $x = 0$ and $y = -9$ | B1 | As long as they are asymptotes for their sketch |
| | | **Total: 4** |

### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $k < 0$ | B1 | Which may be part of their final answer |
| $\frac{1}{x^2} - 9 = kx^2 \Rightarrow 1 - 9x^2 = kx^4$ | M1 | Sets $C = D$ and multiplies through by $x^2$ to achieve a quartic; does not need to be a 3TQ in $x^2$ for this mark; terms do not need to be collected on one side |
| $kx^4 + 9x^2 - 1 = 0$; require $b^2 - 4ac \Rightarrow 81 - 4 \times k \times -1$ | M1 | Considers discriminant of their 3TQ in $x^2$; condone sign slips in rearrangement before attempting discriminant |
| Critical value $k = -\frac{81}{4}$ | A1 | Ignore use of equals or inequality sign for this mark; or $x$ used instead of $k$. Note $1 - 9x^2 = kx^4 \Rightarrow kx^4 - 9x^2 - 1 = 0 \Rightarrow -\frac{81}{4}$ is A0A0 (sign error) |
| $-\frac{81}{4} < k < 0$ | A1cso | Equivalent expressions such as $k > -\frac{81}{4} \cap k < 0$ or $0 > k > -\frac{81}{4}$ accepted. **Must be in terms of $k$**. Do not accept "$k > -\frac{81}{4}$ or $k < 0$", "$k > -\frac{81}{4}$, $k < 0$", or "$k > -\frac{81}{4} \cup k < 0$" |
| | | **Total: 5** |