Edexcel C1 2007 June — Question 7 6 marks

Exam BoardEdexcel
ModuleC1 (Core Mathematics 1)
Year2007
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicInequalities
TypeQuadratic equation real roots
DifficultyModerate -0.8 This is a straightforward application of the discriminant condition for real roots (b² - 4ac > 0), followed by solving a simple quadratic inequality by factorization. Both parts are routine C1 procedures requiring no problem-solving insight, making it easier than average but not trivial since it involves two connected steps.
Spec1.02d Quadratic functions: graphs and discriminant conditions1.02g Inequalities: linear and quadratic in single variable
7. The equation \(x ^ { 2 } + k x + ( k + 3 ) = 0\), where \(k\) is a constant, has different real roots.
  1. Show that \(k ^ { 2 } - 4 k - 12 > 0\).
  2. Find the set of possible values of \(k\).
(a) M1 for use of \(b^2 - 4ac\), one of \(b\) or \(c\) must be correct. Or full attempt using completing the square that leads to a 3TQ in \(k\) e.g. \(\left(x + \frac{k}{2}\right)^2 = \frac{k^2}{4} - (k+3)\)
Attempt to use discriminant \(b^2 - 4ac\)
\(k^2 - 4(k+3) > 0 \Rightarrow k^2 - 4k - 12 > 0\) (*)
A1cso Correct argument to printed result. Need to state (or imply) that \(b^2 - 4ac > 0\) and no incorrect working seen. Must have \(> 0\). If \(> 0\) just appears with \(k^2 - 4(k+3) > 0\) that is OK. If \(> 0\) appears on last line only with no explanation give A0.
\(b^2 - 4ac\) followed by \(k^2 - 4k - 12 > 0\) only is insufficient so M0A0
e.g. \(k^2 - 4 \times 1 \times k + 3\) (missing brackets) can get M1A0 but \(k^2 + 4(k+3)\) is M0A0 (wrong formula)
Using \(b^2 - 40ac > 0\) is M0.
(b) M1 for attempting to find critical regions. Factors, formula or completing the square.
\(k^2 - 4k - 12 = 0 \Rightarrow (k \pm a)(k \pm b)\), with \(ab = 12\) or \((k =) \frac{4 \pm \sqrt{4^2 - (-4) \times 12}}{2}\) or \((k-2)^2 \pm 2^2 - 12\)
A1 for \(k = -2\) and \(k = 6\) (both)
M1 for choosing the "outside" regions
\(k < -2, k > 6\) or \((-\infty, -2); (6, \infty)\)
A1ft as printed or f.t. their (non-identical) critical values
\(6 < k < -2\) is M1A0 but ignore if it follows a correct version
\(-2 < k < 6\) is M0A0 whatever their diagram looks like
Condone use of \(x\) instead of \(k\) for critical values and final answers in (b).
Treat this question as 3 two mark parts. If part (a) is seen in (b) or vice versa marks can be awarded.
(a) M1 for use of $b^2 - 4ac$, one of $b$ or $c$ must be correct. Or full attempt using completing the square that leads to a 3TQ in $k$ e.g. $\left(x + \frac{k}{2}\right)^2 = \frac{k^2}{4} - (k+3)$

Attempt to use discriminant $b^2 - 4ac$

$k^2 - 4(k+3) > 0 \Rightarrow k^2 - 4k - 12 > 0$ (*)

A1cso Correct argument to printed result. Need to state (or imply) that $b^2 - 4ac > 0$ and no incorrect working seen. Must have $> 0$. If $> 0$ just appears with $k^2 - 4(k+3) > 0$ that is OK. If $> 0$ appears on last line only with no explanation give A0.

$b^2 - 4ac$ followed by $k^2 - 4k - 12 > 0$ only is insufficient so M0A0

e.g. $k^2 - 4 \times 1 \times k + 3$ (missing brackets) can get M1A0 but $k^2 + 4(k+3)$ is M0A0 (wrong formula)

Using $b^2 - 40ac > 0$ is M0.

(b) M1 for attempting to find critical regions. Factors, formula or completing the square.

$k^2 - 4k - 12 = 0 \Rightarrow (k \pm a)(k \pm b)$, with $ab = 12$ or $(k =) \frac{4 \pm \sqrt{4^2 - (-4) \times 12}}{2}$ or $(k-2)^2 \pm 2^2 - 12$

A1 for $k = -2$ and $k = 6$ (both)

M1 for choosing the "outside" regions

$k < -2, k > 6$ or $(-\infty, -2); (6, \infty)$

A1ft as printed or f.t. their (non-identical) critical values

$6 < k < -2$ is M1A0 but ignore if it follows a correct version

$-2 < k < 6$ is M0A0 whatever their diagram looks like

Condone use of $x$ instead of $k$ for critical values and final answers in (b).

Treat this question as 3 two mark parts. If part (a) is seen in (b) or vice versa marks can be awarded.

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7. The equation $x ^ { 2 } + k x + ( k + 3 ) = 0$, where $k$ is a constant, has different real roots.
\begin{enumerate}[label=(\alph*)]
\item Show that $k ^ { 2 } - 4 k - 12 > 0$.
\item Find the set of possible values of $k$.
\end{enumerate}

\hfill \mbox{\textit{Edexcel C1 2007 Q7 [6]}}
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