| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2016 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discriminant and conditions for roots |
| Type | Line tangent to curve, find constant |
| Difficulty | Standard +0.3 This is a standard C1 question combining routine quadratic techniques: solving by completing the square/quadratic formula, sketching, and using the discriminant condition (b²-4ac=0) for tangency. Part (b)(i) is algebraic manipulation shown for the student, and (b)(ii) requires applying the tangent condition to find k values. All techniques are core curriculum with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02f Solve quadratic equations: including in a function of unknown1.02n Sketch curves: simple equations including polynomials |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \((x=)\ \dfrac{4\pm\sqrt{80}}{-4}\) or \((x=)\ \dfrac{-4\pm\sqrt{80}}{4}\) or \((x=)\ \dfrac{-2\pm\sqrt{20}}{2}\) | M1 | If completing the square must have at least \(x+1=\pm\sqrt{5}\) |
| \((x=)\ -1\pm\sqrt{5}\) | A1 | Do not accept \(-1\pm-\sqrt{5}\) for A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\wedge\) shape sketch | M1 | \(\wedge\) shape |
| \(y\)-intercept at 8 marked | A1 | As shown in all 4 quadrants, max to left of \(y\)-axis with \(y\)-intercept 8 stated/marked |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(2x^2 + (k+4)x + 4(k-2) = 0\) | B1 | Must expand \(k(x+4)\) and have all terms on one side with \(=0\) before final line; AG be convinced |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \((k+4)^2 - 4\times2\times4(k-2)\ (=0)\) | M1 | Correct discriminant |
| \(k^2 - 24k + 80\ (=0)\) | A1 | |
| \(k=4,\quad k=20\) | A1cso |
# Question 6:
## Part (a)(i)
| Working | Mark | Guidance |
|---------|------|----------|
| $(x=)\ \dfrac{4\pm\sqrt{80}}{-4}$ or $(x=)\ \dfrac{-4\pm\sqrt{80}}{4}$ or $(x=)\ \dfrac{-2\pm\sqrt{20}}{2}$ | M1 | If completing the square must have at least $x+1=\pm\sqrt{5}$ |
| $(x=)\ -1\pm\sqrt{5}$ | A1 | Do **not** accept $-1\pm-\sqrt{5}$ for A1 |
## Part (a)(ii)
| Working | Mark | Guidance |
|---------|------|----------|
| $\wedge$ shape sketch | M1 | $\wedge$ shape |
| $y$-intercept at 8 marked | A1 | As shown in all 4 quadrants, max to left of $y$-axis with $y$-intercept 8 stated/marked |
## Part (b)(i)
| Working | Mark | Guidance |
|---------|------|----------|
| $2x^2 + (k+4)x + 4(k-2) = 0$ | B1 | Must expand $k(x+4)$ and have all terms on one side with $=0$ before final line; AG be convinced |
## Part (b)(ii)
| Working | Mark | Guidance |
|---------|------|----------|
| $(k+4)^2 - 4\times2\times4(k-2)\ (=0)$ | M1 | Correct discriminant |
| $k^2 - 24k + 80\ (=0)$ | A1 | |
| $k=4,\quad k=20$ | A1cso | |
---6
\begin{enumerate}[label=(\alph*)]
\item A curve has equation $y = 8 - 4 x - 2 x ^ { 2 }$.
\begin{enumerate}[label=(\roman*)]
\item Find the values of $x$ where the curve crosses the $x$-axis, giving your answer in the form $m \pm \sqrt { n }$, where $m$ and $n$ are integers.
\item Sketch the curve, giving the value of the $y$-intercept.
\end{enumerate}\item A line has equation $y = k ( x + 4 )$, where $k$ is a constant.
\begin{enumerate}[label=(\roman*)]
\item Show that the $x$-coordinates of any points of intersection of the line with the curve $y = 8 - 4 x - 2 x ^ { 2 }$ satisfy the equation
$$2 x ^ { 2 } + ( k + 4 ) x + 4 ( k - 2 ) = 0$$
\item Find the values of $k$ for which the line is a tangent to the curve $y = 8 - 4 x - 2 x ^ { 2 }$.\\[0pt]
[3 marks]
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C1 2016 Q6 [8]}}