Edexcel C1 — Question 9 14 marks

Exam BoardEdexcel
ModuleC1 (Core Mathematics 1)
Marks14
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTangents, normals and gradients
TypeTangent parallel to given line
DifficultyStandard +0.3 This is a straightforward C1 integration and differentiation question requiring standard techniques: integrate a polynomial, use a point to find the constant, verify a point lies on the curve, find a tangent equation, and solve for where the derivative equals a given value. All steps are routine with no novel problem-solving required, making it slightly easier than average.
Spec1.07m Tangents and normals: gradient and equations1.08b Integrate x^n: where n != -1 and sums

9. The curve \(C\) has equation \(y = \mathrm { f } ( x )\). Given that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 3 x ^ { 2 } - 20 x + 29\) and that \(C\) passes through the point \(P ( 2,6 )\),
  1. find \(y\) in terms of \(x\).
  2. Verify that \(C\) passes through the point \(( 4,0 )\).
  3. Find an equation of the tangent to \(C\) at \(P\). The tangent to \(C\) at the point \(Q\) is parallel to the tangent at \(P\).
  4. Calculate the exact \(x\)-coordinate of \(Q\).

Question 9:
Part (a)
AnswerMarks Guidance
Integrate: \(y = x^3 - 10x^2 + 29x\ (+C)\)M1 M1
\(6 = 8 - 40 + 58 + C \Rightarrow C = -20\ (y = x^3 - 10x^2 + 29x - 20)\)M1 A1 (4 marks)
Part (b)
AnswerMarks Guidance
Subs. \(x = 4\): \(64 - 160 + 116 - 20 = 0\)M1 A1 (2 marks)
Part (c)
AnswerMarks Guidance
At \(x = 2\), \(\frac{\mathrm{d}y}{\mathrm{d}x} = 12 - 40 + 29 = 1\)B1
Tangent: \(y - 6 = x - 2\) \((y = x + 4)\)M1 A1 (3 marks)
Part (d)
AnswerMarks Guidance
\(\frac{\mathrm{d}y}{\mathrm{d}x} = 1\)M1
\(3x^2 - 20x + 28 = 0\)M1
\((3x - 14)(x - 2) = 0\)M1 A1
\(x = \frac{14}{3}\)A1 (5 marks)
## Question 9:

### Part (a)
Integrate: $y = x^3 - 10x^2 + 29x\ (+C)$ | M1 M1 |
$6 = 8 - 40 + 58 + C \Rightarrow C = -20\ (y = x^3 - 10x^2 + 29x - 20)$ | M1 A1 | (4 marks)

### Part (b)
Subs. $x = 4$: $64 - 160 + 116 - 20 = 0$ | M1 A1 | (2 marks)

### Part (c)
At $x = 2$, $\frac{\mathrm{d}y}{\mathrm{d}x} = 12 - 40 + 29 = 1$ | B1 |
Tangent: $y - 6 = x - 2$ $(y = x + 4)$ | M1 A1 | (3 marks)

### Part (d)
$\frac{\mathrm{d}y}{\mathrm{d}x} = 1$ | M1 |
$3x^2 - 20x + 28 = 0$ | M1 |
$(3x - 14)(x - 2) = 0$ | M1 A1 |
$x = \frac{14}{3}$ | A1 | (5 marks)
9. The curve $C$ has equation $y = \mathrm { f } ( x )$. Given that $\frac { \mathrm { d } y } { \mathrm {~d} x } = 3 x ^ { 2 } - 20 x + 29$ and that $C$ passes through the point $P ( 2,6 )$,
\begin{enumerate}[label=(\alph*)]
\item find $y$ in terms of $x$.
\item Verify that $C$ passes through the point $( 4,0 )$.
\item Find an equation of the tangent to $C$ at $P$.

The tangent to $C$ at the point $Q$ is parallel to the tangent at $P$.
\item Calculate the exact $x$-coordinate of $Q$.
\end{enumerate}

\hfill \mbox{\textit{Edexcel C1  Q9 [14]}}