AQA C1 2015 June — Question 8 8 marks

Exam BoardAQA
ModuleC1 (Core Mathematics 1)
Year2015
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicInequalities
TypeLine-curve intersection conditions
DifficultyModerate -0.3 This is a standard C1 question on quadratic intersection conditions requiring substitution, algebraic manipulation, and discriminant application. While it involves multiple steps (parts a, b(i), b(ii)), each step follows routine procedures: equating expressions, using b²-4ac<0 for no intersection, and solving a quadratic inequality. The techniques are well-practiced at this level, making it slightly easier than average despite being multi-part.
Spec1.02d Quadratic functions: graphs and discriminant conditions1.02g Inequalities: linear and quadratic in single variable1.02h Express solutions: using 'and', 'or', set and interval notation
8 A curve has equation \(y = x ^ { 2 } + ( 3 k - 4 ) x + 13\) and a line has equation \(y = 2 x + k\), where \(k\) is a constant.
  1. Show that the \(x\)-coordinate of any point of intersection of the line and curve satisfies the equation $$x ^ { 2 } + 3 ( k - 2 ) x + 13 - k = 0$$
  2. Given that the line and the curve do not intersect:
    1. show that \(9 k ^ { 2 } - 32 k - 16 < 0\);
    2. find the possible values of \(k\).
      \includegraphics[max width=\textwidth, alt={}]{c7f38f7e-75aa-4b41-96fd-f38f968c225c-18_1657_1714_1050_153}
Question 8:
Part (a):
\(y = x^2 + (3k-4)x + 13\) and \(y = 2x + k\)
AnswerMarks Guidance
Setting equal: \(x^2 + (3k-4)x + 13 = 2x + k\)B1 Equating and rearranging to given form
\(x^2 + (3k-4-2)x + 13 - k = 0\)
AnswerMarks Guidance
\(x^2 + 3(k-2)x + 13 - k = 0\) ✓ Shown
Part (b)(i):
AnswerMarks Guidance
No intersection \(\Rightarrow\) discriminant \(< 0\)M1 Using \(b^2 - 4ac < 0\)
\([3(k-2)]^2 - 4(1)(13-k) < 0\)M1 Correct substitution
\(9(k-2)^2 - 4(13-k) < 0\)
\(9(k^2 - 4k + 4) - 52 + 4k < 0\)
\(9k^2 - 36k + 36 - 52 + 4k < 0\)
AnswerMarks Guidance
\(9k^2 - 32k - 16 < 0\) ✓A1 Shown
Part (b)(ii):
AnswerMarks Guidance
\((9k + 4)(k - 4) < 0\)M1 Factorising
\(k = -\frac{4}{9}\) or \(k = 4\)A1 Both critical values
\(-\frac{4}{9} < k < 4\)A1 ft Correct inequality 
## Question 8:

### Part (a):

$y = x^2 + (3k-4)x + 13$ and $y = 2x + k$

Setting equal: $x^2 + (3k-4)x + 13 = 2x + k$ | **B1** | Equating and rearranging to given form

$x^2 + (3k-4-2)x + 13 - k = 0$

$x^2 + 3(k-2)x + 13 - k = 0$ ✓ | | Shown

### Part (b)(i):

No intersection $\Rightarrow$ discriminant $< 0$ | **M1** | Using $b^2 - 4ac < 0$

$[3(k-2)]^2 - 4(1)(13-k) < 0$ | **M1** | Correct substitution

$9(k-2)^2 - 4(13-k) < 0$

$9(k^2 - 4k + 4) - 52 + 4k < 0$

$9k^2 - 36k + 36 - 52 + 4k < 0$

$9k^2 - 32k - 16 < 0$ ✓ | **A1** | Shown

### Part (b)(ii):

$(9k + 4)(k - 4) < 0$ | **M1** | Factorising

$k = -\frac{4}{9}$ or $k = 4$ | **A1** | Both critical values

$-\frac{4}{9} < k < 4$ | **A1** ft | Correct inequality
8 A curve has equation $y = x ^ { 2 } + ( 3 k - 4 ) x + 13$ and a line has equation $y = 2 x + k$, where $k$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Show that the $x$-coordinate of any point of intersection of the line and curve satisfies the equation

$$x ^ { 2 } + 3 ( k - 2 ) x + 13 - k = 0$$
\item Given that the line and the curve do not intersect:
\begin{enumerate}[label=(\roman*)]
\item show that $9 k ^ { 2 } - 32 k - 16 < 0$;
\item find the possible values of $k$.

\begin{center}
\includegraphics[max width=\textwidth, alt={}]{c7f38f7e-75aa-4b41-96fd-f38f968c225c-18_1657_1714_1050_153}
\end{center}
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{AQA C1 2015 Q8 [8]}}
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Q1 8 Q2 5 Q3 12 Q4 11 Q5 8 Q6 11 Q7 12 Q8 8