Edexcel C1 — Question 9 11 marks

Exam BoardEdexcel
ModuleC1 (Core Mathematics 1)
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFactor & Remainder Theorem
TypeVerify factor then sketch or analyse curve
DifficultyModerate -0.3 This is a straightforward C1 question combining factorization, differentiation, and solving equations. Part (a) requires factoring a cubic (which factors easily as x(x²+3x-4)), part (b) needs basic differentiation to find the tangent at the origin, and part (c) involves solving a cubic equation by substitution. All techniques are standard C1 procedures with no novel insight required, making it slightly easier than average but still requiring multiple steps across 11 marks.
Spec1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02q Use intersection points: of graphs to solve equations1.07m Tangents and normals: gradient and equations

\includegraphics{figure_1} Figure 1 shows the curve \(C\) with the equation \(y = x^3 + 3x^2 - 4x\) and the straight line \(l\). The curve \(C\) crosses the \(x\)-axis at the origin, \(O\), and at the points \(A\) and \(B\).
  1. Find the coordinates of \(A\) and \(B\). [3]
The line \(l\) is the tangent to \(C\) at \(O\).
  1. Find an equation for \(l\). [4]
  2. Find the coordinates of the point where \(l\) intersects \(C\) again. [4]

AnswerMarks
(a) \(x(x^2 + 3x - 4) = 0\)M1
\(x(x + 4)(x - 1) = 0\)M1
\(x = 0\) (at O), \(-4, 1\)
\(\therefore (-4, 0), (1, 0)\)A1
(b) \(\frac{dy}{dx} = 3x^2 + 6x - 4\)M1 A1
grad \(= -4\)M1
\(\therefore y = -4x\)A1
(c) \(x^3 + 3x^2 - 4x = -4x\)M1
\(x^3 + 3x^2 = 0\)
\(x^2(x + 3) = 0\)M1
\(x = 0\) (at O), \(-3\)A1
\(\therefore (-3, 12)\)A1
Total: 11 marks
**(a)** $x(x^2 + 3x - 4) = 0$ | M1 | 
$x(x + 4)(x - 1) = 0$ | M1 | 
$x = 0$ (at O), $-4, 1$ | 
$\therefore (-4, 0), (1, 0)$ | A1 | 

**(b)** $\frac{dy}{dx} = 3x^2 + 6x - 4$ | M1 A1 | 
grad $= -4$ | M1 | 
$\therefore y = -4x$ | A1 | 

**(c)** $x^3 + 3x^2 - 4x = -4x$ | M1 | 
$x^3 + 3x^2 = 0$ | 
$x^2(x + 3) = 0$ | M1 | 
$x = 0$ (at O), $-3$ | A1 | 
$\therefore (-3, 12)$ | A1 | 
**Total: 11 marks**

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\includegraphics{figure_1}

Figure 1 shows the curve $C$ with the equation $y = x^3 + 3x^2 - 4x$ and the straight line $l$.

The curve $C$ crosses the $x$-axis at the origin, $O$, and at the points $A$ and $B$.

\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of $A$ and $B$. [3]
\end{enumerate}

The line $l$ is the tangent to $C$ at $O$.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find an equation for $l$. [4]
\item Find the coordinates of the point where $l$ intersects $C$ again. [4]
\end{enumerate}

\hfill \mbox{\textit{Edexcel C1  Q9 [11]}}