Edexcel C1 — Question 10 13 marks

Exam BoardEdexcel
ModuleC1 (Core Mathematics 1)
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSimultaneous equations
TypeLine intersecting quadratic curve
DifficultyStandard +0.3 This is a standard C1 question on line-curve intersection with differentiation for tangents. Part (a) requires solving a quadratic equation, parts (b-c) use basic differentiation to find tangent equations, and part (d) solves simultaneous linear equations. All techniques are routine for C1 with no novel insight required, making it slightly easier than average.
Spec1.02c Simultaneous equations: two variables by elimination and substitution1.02q Use intersection points: of graphs to solve equations1.07m Tangents and normals: gradient and equations

10. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{ddc2483c-fc21-4d6f-9e5b-7c48339dbc88-4_647_775_879_475} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} Figure 1 shows the curve \(y = x ^ { 2 } - 3 x + 5\) and the straight line \(y = 2 x + 1\). The curve and line intersect at the points \(P\) and \(Q\).
  1. Using algebra, show that \(P\) has coordinates \(( 1,3 )\) and find the coordinates of \(Q\).
  2. Find an equation for the tangent to the curve at \(P\).
  3. Show that the tangent to the curve at \(Q\) has the equation \(y = 5 x - 11\).
  4. Find the coordinates of the point where the tangent to the curve at \(P\) intersects the tangent to the curve at \(Q\).

AnswerMarks Guidance
(a) \(x^2 - 3x + 5 = 2x + 1\)
\(x^2 - 5x + 4 = 0\)M1
\((x - 1)(x - 4) = 0\)M1
\(x = 1, 4\)A1
when \(x = 1\), \(y = 2(1) + 1 = 3\) ⇒ \(P(1, 3), Q(4, 9)\)A1
(b) \(\frac{dy}{dx} = 2x - 3\)M1
grad \(= -1\)A1
⇒ \(y - 3 = -(x - 1)\) [\(y = 4 - x\)]M1 A1
(c) grad \(= 5\)
⇒ \(y - 9 = 5(x - 4)\)M1
\(y - 9 = 5x - 20\)
\(y = 5x - 11\)A1
(d) \(4 - x = 5x - 11\)M1
\(x = \frac{5}{2}\)A1
⇒ \(\left(\frac{5}{2}, \frac{3}{2}\right)\)A1 (13 marks)
Total: 75 marks
**(a)** $x^2 - 3x + 5 = 2x + 1$ | |
$x^2 - 5x + 4 = 0$ | M1 |
$(x - 1)(x - 4) = 0$ | M1 |
$x = 1, 4$ | A1 |
when $x = 1$, $y = 2(1) + 1 = 3$ ⇒ $P(1, 3), Q(4, 9)$ | A1 |

**(b)** $\frac{dy}{dx} = 2x - 3$ | M1 |
grad $= -1$ | A1 |
⇒ $y - 3 = -(x - 1)$ [$y = 4 - x$] | M1 A1 |

**(c)** grad $= 5$ | |
⇒ $y - 9 = 5(x - 4)$ | M1 |
$y - 9 = 5x - 20$ | |
$y = 5x - 11$ | A1 |

**(d)** $4 - x = 5x - 11$ | M1 |
$x = \frac{5}{2}$ | A1 |
⇒ $\left(\frac{5}{2}, \frac{3}{2}\right)$ | A1 | (13 marks)

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**Total: 75 marks**
10.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{ddc2483c-fc21-4d6f-9e5b-7c48339dbc88-4_647_775_879_475}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}

Figure 1 shows the curve $y = x ^ { 2 } - 3 x + 5$ and the straight line $y = 2 x + 1$. The curve and line intersect at the points $P$ and $Q$.
\begin{enumerate}[label=(\alph*)]
\item Using algebra, show that $P$ has coordinates $( 1,3 )$ and find the coordinates of $Q$.
\item Find an equation for the tangent to the curve at $P$.
\item Show that the tangent to the curve at $Q$ has the equation $y = 5 x - 11$.
\item Find the coordinates of the point where the tangent to the curve at $P$ intersects the tangent to the curve at $Q$.
\end{enumerate}

\hfill \mbox{\textit{Edexcel C1  Q10 [13]}}