| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2013 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Inequalities |
| Type | Quadratic equation real roots |
| Difficulty | Moderate -0.3 This is a standard C1 discriminant question requiring students to apply b²-4ac > 0 for two distinct real roots, then rearrange and solve a quadratic inequality. While it involves multiple steps (rearranging to standard form, applying discriminant condition, factorising), these are all routine techniques with no novel insight required. Slightly easier than average due to the straightforward algebraic manipulation and the 'show that' scaffold in part (a). |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02g Inequalities: linear and quadratic in single variable |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Method 1: Attempts \(b^2 - 4ac\) for \(a=(k+3)\), \(b=6\), \(c \neq k\) | M1 | Or uses quadratic formula with expression under square root |
| \(b^2 - 4ac = 6^2 - 4(k+3)(k-5)\) | A1 | Correct expression, need not be simplified |
| \((b^2-4ac=)\; -4k^2+8k+96\) or \(4k^2-8k-96\) (no prior algebraic errors) | B1 | Uses algebra to manipulate without error into one of these three-term quadratics |
| As \(b^2-4ac>0\), then \(-4k^2+8k+96>0\), so \(k^2-2k-24<0\) | A1* | Applies \(b^2-4ac>0\) correctly to achieve result given; no errors |
| Method 2: \(6^2 > 4(k+3)(k-5)\) | M1 | |
| \(4k^2-8k-96<0\) or \(-4k^2+8k+96>0\) or \(9>(k+3)(k-5)\) | B1 | Without error |
| \(k^2-2k-24<0\) following correct work | A1* |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts to solve \(k^2-2k-24=0 \Rightarrow k=6,-4\) | M1 | Uses factorisation, formula, or completing the square to find two values; or finds two correct answers |
| Their Lower Limit \(< k <\) Their Upper Limit | M1 | Allow \(\leq\) |
| \(k^2-2k-24<0\) gives \(-4 < k < 6\) | A1 | Lose mark for \(\leq\); must be and not or |
## Question 9:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| **Method 1:** Attempts $b^2 - 4ac$ for $a=(k+3)$, $b=6$, $c \neq k$ | M1 | Or uses quadratic formula with expression under square root |
| $b^2 - 4ac = 6^2 - 4(k+3)(k-5)$ | A1 | Correct expression, need not be simplified |
| $(b^2-4ac=)\; -4k^2+8k+96$ or $4k^2-8k-96$ (no prior algebraic errors) | B1 | Uses algebra to manipulate without error into one of these three-term quadratics |
| As $b^2-4ac>0$, then $-4k^2+8k+96>0$, so $k^2-2k-24<0$ | A1* | Applies $b^2-4ac>0$ correctly to achieve result given; no errors |
| **Method 2:** $6^2 > 4(k+3)(k-5)$ | M1 | |
| $4k^2-8k-96<0$ or $-4k^2+8k+96>0$ or $9>(k+3)(k-5)$ | B1 | Without error |
| $k^2-2k-24<0$ following correct work | A1* | |
**[4 marks]**
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts to solve $k^2-2k-24=0 \Rightarrow k=6,-4$ | M1 | Uses factorisation, formula, or completing the square to find two values; or finds two correct answers |
| Their Lower Limit $< k <$ Their Upper Limit | M1 | Allow $\leq$ |
| $k^2-2k-24<0$ gives $-4 < k < 6$ | A1 | Lose mark for $\leq$; must be **and** not **or** |
**[3 marks] — Total: 7 marks**
---9. The equation
$$( k + 3 ) x ^ { 2 } + 6 x + k = 5 , \text { where } k \text { is a constant, }$$
has two distinct real solutions for $x$.
\begin{enumerate}[label=(\alph*)]
\item Show that $k$ satisfies
$$k ^ { 2 } - 2 k - 24 < 0$$
\item Hence find the set of possible values of $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2013 Q9 [7]}}