Question 8 - A-Level Maths

Edexcel C1 2014 January — Question 8 7 marks

Exam Board	Edexcel
Module	C1 (Core Mathematics 1)
Year	2014
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 solve the resulting quadratic inequality. While it involves multiple steps (forming discriminant, simplifying to given form, solving inequality), these are routine techniques with no novel insight required, making it slightly easier than average.
Spec	1.02d Quadratic functions: graphs and discriminant conditions 1.02g Inequalities: linear and quadratic in single variable

The equation $2 x ^ { 2 } + 2 k x + ( k + 2 ) = 0$, where $k$ is a constant, has two distinct real roots.
1. Show that $k$ satisfies
$$k ^ { 2 } - 2 k - 4 > 0$$
Find the set of possible values of $k$.

Show mark scheme Show mark scheme source

Answer	Marks	Guidance
(a)	$b^2 - 4ac = (2k)^2 - 4 \times 2 \times (k + 2)$	M1 A1
$b^2 - 4ac > 0 \Rightarrow 4k^2 - 4 \times 2 \times (k + 2) > 0 \Rightarrow k^2 - 2k - 4 > 0$	A1*	Full proof, no errors, this is a given answer. It must be stated or implied that $b^2 - 4ac > 0$. Do not accept recovery from poor or incorrect bracketing or incorrect inequalities. Do not accept the answer written down without seeing an intermediate line such as $4k^2 - 4 \times 2 \times (k + 2) > 0 \Rightarrow k^2 - 2k - 4 > 0$. Or $4k^2 - 8k - 8 > 0 \Rightarrow k^2 - 2k - 4 > 0$. The inequality must have been seen at least once before the final line for this mark to have been awarded. Eg accept $D = 4k^2 - 8k - 8 \Rightarrow 4k^2 - 8k - 8 > 0 \Rightarrow k^2 - 2k - 4 > 0$.
(b)	$k^2 - 2k - 4 = 0 \Rightarrow (k-1)^2 = 5$	M1
$k = 1 \pm \sqrt{5}$ oe	A1	Obtains critical values of $1 \pm \sqrt{5}$. Accept $\frac{2 \pm \sqrt{20}}{2}$.
$k > 1 + \sqrt{5}, \quad k < 1 - \sqrt{5}$	dM1 A1	Outsides of their values chosen. It is dependent upon the previous M mark having been awarded. States $k >$ their largest value, $k <$ their smallest value. Do not award simply for a diagram or a table- they must have chosen their 'outside regions'. Correct answer only. Accept $k > 1 + \sqrt{5}$ or $k < 1 - \sqrt{5}$, $k > 1 + \sqrt{5}$ $k < 1 - \sqrt{5}$, $(-\infty, 1-\sqrt{5}) \cup (1+\sqrt{5}, \infty)$, but not $k > 1 + \sqrt{5}$ and $k < 1 - \sqrt{5}, 1 + \sqrt{5} < k < 1 - \sqrt{5}$.
Alt (a) $b^2 > 4ac \Rightarrow (2k)^2 > 4 \times 2 \times (k + 2)$	M1 A1	$\Rightarrow k^2 - 2k - 4 > 0$.
Also accept exact alternatives as a simplified form is not explicitly asked for in the question. Accept versions such as $k > \frac{2 + \sqrt{20}}{2}$ or $k < \frac{2 - \sqrt{20}}{2}$.

(a) | $b^2 - 4ac = (2k)^2 - 4 \times 2 \times (k + 2)$ | M1 A1 | For attempting to use $b^2 - 4ac$ with the values of $a$, $b$ and $c$ from the given equation. Condone invisible brackets. $2k^2 - 4 \times 2 \times k + 2$ could be evidence. Fully correct (unsimplified) expression for $b^2 - 4ac = (2k)^2 - 4 \times 2 \times (k + 2)$. The bracketing must be correct. You can accept with or without any inequality signs. Accept $a = 2, b = 2k, c = k + 2 \Rightarrow b^2 - 4ac = (2k)^2 - 4 \times 2 \times (k + 2)$.

| $b^2 - 4ac > 0 \Rightarrow 4k^2 - 4 \times 2 \times (k + 2) > 0 \Rightarrow k^2 - 2k - 4 > 0$ | A1* | Full proof, no errors, this is a given answer. It must be stated or implied that $b^2 - 4ac > 0$. Do not accept recovery from poor or incorrect bracketing or incorrect inequalities. Do not accept the answer written down without seeing an intermediate line such as $4k^2 - 4 \times 2 \times (k + 2) > 0 \Rightarrow k^2 - 2k - 4 > 0$. Or $4k^2 - 8k - 8 > 0 \Rightarrow k^2 - 2k - 4 > 0$. The inequality must have been seen at least once before the final line for this mark to have been awarded. Eg accept $D = 4k^2 - 8k - 8 \Rightarrow 4k^2 - 8k - 8 > 0 \Rightarrow k^2 - 2k - 4 > 0$.

(b) | $k^2 - 2k - 4 = 0 \Rightarrow (k-1)^2 = 5$ | M1 | Attempt to solve the given 3 term quadratic (= 0) by formula or completing the square. Do NOT accept an attempt to factorise in this question. If the formula is given it must be correct. It can be implied by seeing either $\frac{-(-2) \pm \sqrt{(-2)^2 - 4 \times 1 \times -4}}{2 \times 1}$ or $-- 2 \pm \sqrt{-2^2 - 4 \times 1 \times -4}$ or if completing the square is used it can be implied by $(k-1)^2 \pm 1 - 4 = 0 \Rightarrow k = ...$.

| $k = 1 \pm \sqrt{5}$ oe | A1 | Obtains critical values of $1 \pm \sqrt{5}$. Accept $\frac{2 \pm \sqrt{20}}{2}$.

| $k > 1 + \sqrt{5}, \quad k < 1 - \sqrt{5}$ | dM1 A1 | Outsides of their values chosen. It is dependent upon the previous M mark having been awarded. States $k >$ their largest value, $k <$ their smallest value. Do not award simply for a diagram or a table- they must have chosen their 'outside regions'. Correct answer only. Accept $k > 1 + \sqrt{5}$ or $k < 1 - \sqrt{5}$, $k > 1 + \sqrt{5}$ $k < 1 - \sqrt{5}$, $(-\infty, 1-\sqrt{5}) \cup (1+\sqrt{5}, \infty)$, but not $k > 1 + \sqrt{5}$ and $k < 1 - \sqrt{5}, 1 + \sqrt{5} < k < 1 - \sqrt{5}$.

| | | Alt (a) $b^2 > 4ac \Rightarrow (2k)^2 > 4 \times 2 \times (k + 2)$ | M1 A1 | $\Rightarrow k^2 - 2k - 4 > 0$.

| | | Also accept exact alternatives as a simplified form is not explicitly asked for in the question. Accept versions such as $k > \frac{2 + \sqrt{20}}{2}$ or $k < \frac{2 - \sqrt{20}}{2}$.

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Show LaTeX source

\begin{enumerate}
  \item The equation $2 x ^ { 2 } + 2 k x + ( k + 2 ) = 0$, where $k$ is a constant, has two distinct real roots.\\
(a) Show that $k$ satisfies
\end{enumerate}

$$k ^ { 2 } - 2 k - 4 > 0$$

(b) Find the set of possible values of $k$.\\

\hfill \mbox{\textit{Edexcel C1 2014 Q8 [7]}}

This paper (10 questions)

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Q1 4 Q2 5 Q3 7 Q4 4 Q5 5 Q6 11 Q7 10 Q8 7 Q9 12 Q10 10

Answer	Marks	Guidance
(a)	\(b^2 - 4ac = (2k)^2 - 4 \times 2 \times (k + 2)\)	M1 A1
\(b^2 - 4ac > 0 \Rightarrow 4k^2 - 4 \times 2 \times (k + 2) > 0 \Rightarrow k^2 - 2k - 4 > 0\)	A1*	Full proof, no errors, this is a given answer. It must be stated or implied that \(b^2 - 4ac > 0\). Do not accept recovery from poor or incorrect bracketing or incorrect inequalities. Do not accept the answer written down without seeing an intermediate line such as \(4k^2 - 4 \times 2 \times (k + 2) > 0 \Rightarrow k^2 - 2k - 4 > 0\). Or \(4k^2 - 8k - 8 > 0 \Rightarrow k^2 - 2k - 4 > 0\). The inequality must have been seen at least once before the final line for this mark to have been awarded. Eg accept \(D = 4k^2 - 8k - 8 \Rightarrow 4k^2 - 8k - 8 > 0 \Rightarrow k^2 - 2k - 4 > 0\).
(b)	\(k^2 - 2k - 4 = 0 \Rightarrow (k-1)^2 = 5\)	M1
\(k = 1 \pm \sqrt{5}\) oe	A1	Obtains critical values of \(1 \pm \sqrt{5}\). Accept \(\frac{2 \pm \sqrt{20}}{2}\).
\(k > 1 + \sqrt{5}, \quad k < 1 - \sqrt{5}\)	dM1 A1	Outsides of their values chosen. It is dependent upon the previous M mark having been awarded. States \(k >\) their largest value, \(k <\) their smallest value. Do not award simply for a diagram or a table- they must have chosen their 'outside regions'. Correct answer only. Accept \(k > 1 + \sqrt{5}\) or \(k < 1 - \sqrt{5}\), \(k > 1 + \sqrt{5}\) \(k < 1 - \sqrt{5}\), \((-\infty, 1-\sqrt{5}) \cup (1+\sqrt{5}, \infty)\), but not \(k > 1 + \sqrt{5}\) and \(k < 1 - \sqrt{5}, 1 + \sqrt{5} < k < 1 - \sqrt{5}\).
Alt (a) \(b^2 > 4ac \Rightarrow (2k)^2 > 4 \times 2 \times (k + 2)\)	M1 A1	\(\Rightarrow k^2 - 2k - 4 > 0\).
Also accept exact alternatives as a simplified form is not explicitly asked for in the question. Accept versions such as \(k > \frac{2 + \sqrt{20}}{2}\) or \(k < \frac{2 - \sqrt{20}}{2}\).