| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2014 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discriminant and conditions for roots |
| Type | Show discriminant inequality, then solve |
| Difficulty | Moderate -0.3 This is a straightforward discriminant question requiring students to rearrange to standard form, apply b²-4ac > 0, simplify the resulting inequality, and solve a quadratic inequality. All steps are routine C1/C2 techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02g Inequalities: linear and quadratic in single variable |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(kx^2 + 4x + k = 2 \Rightarrow kx^2 + 4x + k - 2 = 0\) | — | Rearrangement |
| Attempts \(b^2 - 4ac\) with \(a = k\), \(b = 4\), \(c = k \pm 2\) | M1 | Condone poor/incomplete bracketing |
| \(b^2 - 4ac = 4^2 - 4 \times k \times (k-2)\) | A1 | Correct unsimplified. Bracketing must be correct |
| Sets \(b^2 - 4ac > 0 \Rightarrow 16 - 4k(k-2) > 0\) | dM1 | Dependent on previous M |
| \(4k^2 - 8k - 16 < 0\) | — | |
| \(k^2 - 2k - 4 < 0\) | A1* | Proceeds correctly to given answer; check for sign errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Solves \((k-1)^2 - 5 = 0 \Rightarrow k = 1 \pm \sqrt{5}\) | M1 | Formula or completing the square. If formula quoted must be correct. Accept awrt 3.24, -1.24 |
| \(1 - \sqrt{5} < k < 1 + \sqrt{5}\) | M1A1 | Chooses inside region. Accept \(k > 1-\sqrt{5}\) and \(k < 1+\sqrt{5}\). Do not accept \(k > 1-\sqrt{5}\) or \(k < 1+\sqrt{5}\), or decimals |
# Question 10:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $kx^2 + 4x + k = 2 \Rightarrow kx^2 + 4x + k - 2 = 0$ | — | Rearrangement |
| Attempts $b^2 - 4ac$ with $a = k$, $b = 4$, $c = k \pm 2$ | M1 | Condone poor/incomplete bracketing |
| $b^2 - 4ac = 4^2 - 4 \times k \times (k-2)$ | A1 | Correct unsimplified. Bracketing must be correct |
| Sets $b^2 - 4ac > 0 \Rightarrow 16 - 4k(k-2) > 0$ | dM1 | Dependent on previous M |
| $4k^2 - 8k - 16 < 0$ | — | |
| $k^2 - 2k - 4 < 0$ | A1* | Proceeds correctly to given answer; check for sign errors |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Solves $(k-1)^2 - 5 = 0 \Rightarrow k = 1 \pm \sqrt{5}$ | M1 | Formula or completing the square. If formula quoted must be correct. Accept awrt 3.24, -1.24 |
| $1 - \sqrt{5} < k < 1 + \sqrt{5}$ | M1A1 | Chooses inside region. Accept $k > 1-\sqrt{5}$ and $k < 1+\sqrt{5}$. Do not accept $k > 1-\sqrt{5}$ or $k < 1+\sqrt{5}$, or decimals |
---10. The equation
$$k x ^ { 2 } + 4 x + k = 2 , \text { where } k \text { is a constant, }$$
has two distinct real solutions for $x$.
\begin{enumerate}[label=(\alph*)]
\item Show that $k$ satisfies
$$k ^ { 2 } - 2 k - 4 < 0$$
\item Hence find the set of all possible values of $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2014 Q10 [7]}}