# Question 9:

## Part (a)(i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Straight line with negative gradient | B1 | Straight line with negative gradient anywhere even with no axes |
| Straight line with $y$-intercept at $(0,c)$ or just $c$ marked on positive $y$-axis, line passing through positive $y$-axis | B1 | Allow $(c,0)$ as long as marked in correct place. Allow $(0,c)$ in body of script but sketch has precedence. **Ignore any intercepts with the $x$-axis.** |

## Part (a)(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Shape of $y = \frac{1}{x}$ curve in any position (two branches, asymptotic horizontally and vertically, no obvious "overlap" with asymptotes) | B1 | Either: shape of $y=\frac{1}{x}$ curve in any position with two branches, asymptotic both ways, no obvious overlap, branches approach same asymptote. **Or** the equation $y=5$ seen independently (whether sketch has asymptote or not). Do not allow $y \neq 5$ or $x = 5$. |
| Fully correct graph with horizontal asymptote $y = 5$ on positive $y$-axis, equation $y = 5$ must be seen | B1 | Shape reasonably accurate, "ends" not bending away significantly from asymptotes, branches approach same asymptote. Ignore $x=0$ given as asymptote. |

**Total: 4 marks**

## Question (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Sets $\frac{1}{x}+5=-3x+c$, attempts to multiply by $x$ and collect terms: $\frac{1}{x}+5=-3x+c \Rightarrow 1+5x=-3x^2+cx \Rightarrow 3x^2+5x-cx+1=0$ | M1 | Sets equal and multiplies by $x$, collects terms to one side. Allow ">" or "<" or "=". At least 3 terms multiplied by $x$. Allow one slip. The "$=0$" may be implied by subsequent work. |
| $b^2-4ac=(5-c)^2-4\times1\times3$ | M1 | Attempts to use $b^2-4ac$ with their $a$, $b$ and $c$ from their equation where $a=\pm3$, $b=\pm5\pm c$ and $c=\pm1$. Could be part of quadratic formula or as $b^2<4ac$ or $b^2>4ac$ or as $\sqrt{b^2-4ac}$. There must be no $x$'s. |
| $(5-c)^2>12$ | A1* | Completes proof with **no errors or incorrect statements** and with ">" appearing correctly before the final answer. Note $3x^2+5x-cx+1>0$ or starting with $\frac{1}{x}+5>-3x+c$ would be an error. |

**Note: Minimum for (b):** $\frac{1}{x}+5=-3x+c \Rightarrow 3x^2+5x-cx+1(=0)$ (M1); $b^2>4ac \Rightarrow (5-c)^2>12$ (M1A1). If $b^2>4ac$ is not seen then $4\times3\times1$ needs to be seen explicitly.

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## Question (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $(5-c)^2=12 \Rightarrow (c=)5\pm\sqrt{12}$, or $(5-c)^2=12 \Rightarrow c^2-10c+13=0 \Rightarrow (c=)\frac{-10\pm\sqrt{(-10)^2-4\times13}}{2}$ | M1A1 | M1: Attempts to find at least one critical value using result in (b) or by expanding and solving a 3TQ. A1: Correct critical values in any form. Note $\sqrt{12}$ may be seen as $2\sqrt{3}$. |
| $c<"5-\sqrt{12}"$, $c>"5+\sqrt{12}"$ | M1 | Chooses outside region. The "$0<$" can be ignored for this mark. Look for $c$ less than $5-\sqrt{12}$, $c$ greater than $5+\sqrt{12}$. Evidence from answers not diagram. |
| $0<c<5-\sqrt{12}$, $c>5+\sqrt{12}$ | A1 | Correct ranges including "$0<$". e.g. regions written separately $(0,5-\sqrt{12})$, $(5+\sqrt{12},\infty)$. Critical values may be unsimplified but must be at least $\frac{10+\sqrt{48}}{2}$, $\frac{10-\sqrt{48}}{2}$. Note $0<c<5-\sqrt{12}$ **and** $c>5+\sqrt{12}$ would score M1A0. |

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