| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2011 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discriminant and conditions for roots |
| Type | Prove/show always positive |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question on discriminants requiring algebraic manipulation and completing the square. Part (a) is routine substitution, part (b) is standard completing the square, and part (c) follows immediately from recognizing that a squared term plus a positive constant is always positive. While it requires multiple steps, each technique is basic C1 material with no problem-solving insight needed. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02e Complete the square: quadratic polynomials and turning points |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Discriminant: \(b^2 - 4ac = (k + 3)^2 - 4k\) or equivalent | M1 A1 | If formula \(b^2 - 4ac\) is seen at least 2 of a, b and c must be correct. If formula \(b^2 - 4ac\) is not seen all 3 of a, b and c must be correct. Use of \(b^2 + 4ac\) is M0. A1: correct unsimplified. |
| (b) \((k + 3)^2 - 4k = k^2 + 2k + 9 = (k + 1)^2 + 8\) | M1 A1 | M1: Attempt at completion of square (see earlier notes). A1: both correct (no ft for this mark). |
| (c) For real roots, \(b^2 - 4ac \geq 0\) or \(b^2 - 4ac > 0\) or \((k + 1)^2 + 8 > 0\) \((k + 1)^2 \geq 0\) for all k, so \(b^2 - 4ac > 0\), so roots are real for all k (or equiv.) | M1 A1 cso | M1: States condition as on scheme or attempts to explain that their \((k + 1)^2 + 8\) is greater than 0. A1: The final mark (A1cso) requires \((k + 1)^2 \geq 0\) and conclusion. We will allow \((k + 1)^2 > 0\) (or word positive) also allow \(b^2 - 4ac \geq 0\) and conclusion. |
(a) Discriminant: $b^2 - 4ac = (k + 3)^2 - 4k$ or equivalent | M1 A1 | If formula $b^2 - 4ac$ is seen at least 2 of a, b and c must be correct. If formula $b^2 - 4ac$ is not seen all 3 of a, b and c must be correct. Use of $b^2 + 4ac$ is M0. A1: correct unsimplified.
(b) $(k + 3)^2 - 4k = k^2 + 2k + 9 = (k + 1)^2 + 8$ | M1 A1 | M1: Attempt at completion of square (see earlier notes). A1: both correct (no ft for this mark).
(c) For real roots, $b^2 - 4ac \geq 0$ or $b^2 - 4ac > 0$ or $(k + 1)^2 + 8 > 0$ $(k + 1)^2 \geq 0$ for all k, so $b^2 - 4ac > 0$, so roots are real for all k (or equiv.) | M1 A1 cso | M1: States condition as on scheme or attempts to explain that their $(k + 1)^2 + 8$ is greater than 0. A1: The final mark (A1cso) requires $(k + 1)^2 \geq 0$ and conclusion. We will allow $(k + 1)^2 > 0$ (or word positive) also allow $b^2 - 4ac \geq 0$ and conclusion.
---7.
$$\mathrm { f } ( x ) = x ^ { 2 } + ( k + 3 ) x + k$$
where $k$ is a real constant.
\begin{enumerate}[label=(\alph*)]
\item Find the discriminant of $\mathrm { f } ( x )$ in terms of $k$.
\item Show that the discriminant of $\mathrm { f } ( x )$ can be expressed in the form $( k + a ) ^ { 2 } + b$, where $a$ and $b$ are integers to be found.
\item Show that, for all values of $k$, the equation $\mathrm { f } ( x ) = 0$ has real roots.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2011 Q7 [6]}}