| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2016 |
| Session | October |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discriminant and conditions for roots |
| Type | Show discriminant inequality, then solve |
| Difficulty | Standard +0.3 This is a standard discriminant problem requiring rearrangement to standard form, applying b²-4ac > 0, and solving a quadratic inequality. While it involves multiple steps, each is routine for C1/C2 level 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 |
| \((13k-5)x^2 - 12kx - 6 = 0\) or \((5-13k)x^2 + 12kx + 6 = 0\) | B1 | Expresses equation as three term quadratic in \(x\); equals 0 may be implied |
| Uses \(b^2 - 4ac\) with \(a = \pm 13k \pm 5\), \(b = \pm 12k\), \(c = \pm 6\) | M1 | Or uses discriminant on two sides of equation/inequation |
| States \(b^2 - 4ac > 0\) with \(a = \pm(13k-5)\), \(b = \pm 12k\), \(c = \pm 6\) | A1ft | Uses discriminant condition \(b^2 - 4ac > 0\) or \(b^2 > 4ac\) |
| Proceeds correctly with no errors to \(6k^2 + 13k - 5 > 0\) | A1* | Condone missing \(= 0\) on equation; watch for \(a = 13k-5\), \(b = +12k\), \(c = -6\) which loses final A1* |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts to solve \(6k^2 + 13k - 5 = 0\) | M1 | Uses factorisation, formula, or completing the square to find two values of \(k\) |
| Critical values \(k = \frac{1}{3}\), \(\frac{-5}{2}\) | A1 | Accept \(-2.5\), \(0.333\) (awrt) here; also condone \(x = \frac{1}{3}\), \(\frac{-5}{2}\) for this mark |
| \(6k^2 + 13k - 5 > 0\) gives \(k > \frac{1}{3}\) or \(k < \frac{-5}{2}\) | M1 A1 | M1: chooses outside region; do not award simply for diagram or table. A1: must be exact, must be \(k\); must be two separate inequalities, not "and" |
## Question 11:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(13k-5)x^2 - 12kx - 6 = 0$ or $(5-13k)x^2 + 12kx + 6 = 0$ | B1 | Expresses equation as three term quadratic in $x$; equals 0 may be implied |
| Uses $b^2 - 4ac$ with $a = \pm 13k \pm 5$, $b = \pm 12k$, $c = \pm 6$ | M1 | Or uses discriminant on two sides of equation/inequation |
| States $b^2 - 4ac > 0$ with $a = \pm(13k-5)$, $b = \pm 12k$, $c = \pm 6$ | A1ft | Uses discriminant condition $b^2 - 4ac > 0$ or $b^2 > 4ac$ |
| Proceeds correctly with no errors to $6k^2 + 13k - 5 > 0$ | A1* | Condone missing $= 0$ on equation; watch for $a = 13k-5$, $b = +12k$, $c = -6$ which loses final A1* |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts to solve $6k^2 + 13k - 5 = 0$ | M1 | Uses factorisation, formula, or completing the square to find two values of $k$ |
| Critical values $k = \frac{1}{3}$, $\frac{-5}{2}$ | A1 | Accept $-2.5$, $0.333$ (awrt) here; also condone $x = \frac{1}{3}$, $\frac{-5}{2}$ for this mark |
| $6k^2 + 13k - 5 > 0$ gives $k > \frac{1}{3}$ or $k < \frac{-5}{2}$ | M1 A1 | M1: chooses outside region; do not award simply for diagram or table. A1: must be exact, must be $k$; must be two separate inequalities, not "and" |
---11. The equation $5 x ^ { 2 } + 6 = k \left( 13 x ^ { 2 } - 12 x \right)$, where $k$ is a constant, has two distinct real roots.
\begin{enumerate}[label=(\alph*)]
\item Show that $k$ satisfies the inequality
$$6 k ^ { 2 } + 13 k - 5 > 0$$
\item Find the set of possible values for $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2016 Q11 [8]}}