Edexcel P1 2021 October — Question 8 10 marks

Exam BoardEdexcel
ModuleP1 (Pure Mathematics 1)
Year2021
SessionOctober
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCurve Sketching
TypeCompleting square from standard form
DifficultyEasy -1.2 This is a routine P1 completing the square question with standard follow-up parts. Part (a)(i) is a textbook exercise in completing the square, (a)(ii) requires reading the maximum from completed square form, and (b) involves finding a tangent parallel to a given line using basic differentiation. All techniques are standard with no problem-solving insight required.
Spec1.02w Graph transformations: simple transformations of f(x)

8. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{f1e1d4f5-dd27-4839-a6f3-f6906666302c-22_657_659_214_646} \captionsetup{labelformat=empty} \caption{Figure 4}
\end{figure} Figure 4 shows a sketch of the curve \(C\) with equation $$y = 4 + 12 x - 3 x ^ { 2 }$$ The point \(M\) is the maximum turning point on \(C\).
    1. Write \(4 + 12 x - 3 x ^ { 2 }\) in the form $$a + b ( x + c ) ^ { 2 }$$ where \(a , b\) and \(c\) are constants to be found.
    2. Hence, or otherwise, state the coordinates of \(M\). The line \(l _ { 1 }\) passes through \(O\) and \(M\), as shown in Figure 4.
      A line \(l _ { 2 }\) touches \(C\) and is parallel to \(l _ { 1 }\)
  1. Find an equation for \(l _ { 2 }\)

Question 8 (Completing the Square/Tangent):
Part (a)(i):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(4+12x-3x^2 = a\pm3(x+c)^2\) or \(a+b(x\pm2)^2\)M1 Attempt to complete the square; look for \(b=\pm3\), \(c=\pm2\)
Two of: \(16-3(x-2)^2\), or \(a=16\), \(b=-3\), \(c=-2\)A1
\(16-3(x-2)^2\)A1 \((16-3(2-x)^2\) scores M1A1A0)
Part (a)(ii):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Coordinates \(M=(2,16)\)B1ft B1ft ft on \((-c,a)\) from \(a+b(x+c)^2\) where \(b\neq\pm1\); wrong order e.g. \((16,2)\) scores SC B1 B0
Part (b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
States \(l_2\) has equation \(y = "8"x + k\)M1 ft on their gradient from \(M\)
Sets \(4+12x-3x^2 = "8x"+k\) and proceeds to 3TQdM1
Correct 3TQ: \(3x^2-4x+k-4=0\)A1 "=0" may be implied
Attempts \(b^2-4ac=0\) to find \(k\)ddM1
\(k=\frac{16}{3} \Rightarrow y=8x+\frac{16}{3}\)A1 Condone just \(k=\frac{16}{3}\) if \(y=8x+k\) was stated
Alternative for part (b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Differentiates \(4+12x-3x^2\) and sets equal to 8M1
Solves for \(x\), finds coordinates of point of contactdM1 Tangent meets curve at \(\left(\frac{2}{3},\frac{32}{3}\right)\)
Substitutes \(\left(\frac{2}{3},\frac{32}{3}\right)\) into \(y="8"x+k\)ddM1
\(y=8x+\frac{16}{3}\)A1
## Question 8 (Completing the Square/Tangent):

### Part (a)(i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $4+12x-3x^2 = a\pm3(x+c)^2$ or $a+b(x\pm2)^2$ | M1 | Attempt to complete the square; look for $b=\pm3$, $c=\pm2$ |
| Two of: $16-3(x-2)^2$, or $a=16$, $b=-3$, $c=-2$ | A1 | |
| $16-3(x-2)^2$ | A1 | $(16-3(2-x)^2$ scores M1A1A0) |

### Part (a)(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Coordinates $M=(2,16)$ | B1ft B1ft | ft on $(-c,a)$ from $a+b(x+c)^2$ where $b\neq\pm1$; wrong order e.g. $(16,2)$ scores SC B1 B0 |

### Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| States $l_2$ has equation $y = "8"x + k$ | M1 | ft on their gradient from $M$ |
| Sets $4+12x-3x^2 = "8x"+k$ and proceeds to 3TQ | dM1 | |
| Correct 3TQ: $3x^2-4x+k-4=0$ | A1 | "=0" may be implied |
| Attempts $b^2-4ac=0$ to find $k$ | ddM1 | |
| $k=\frac{16}{3} \Rightarrow y=8x+\frac{16}{3}$ | A1 | Condone just $k=\frac{16}{3}$ if $y=8x+k$ was stated |

**Alternative for part (b):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Differentiates $4+12x-3x^2$ and sets equal to 8 | M1 | |
| Solves for $x$, finds coordinates of point of contact | dM1 | Tangent meets curve at $\left(\frac{2}{3},\frac{32}{3}\right)$ |
| Substitutes $\left(\frac{2}{3},\frac{32}{3}\right)$ into $y="8"x+k$ | ddM1 | |
| $y=8x+\frac{16}{3}$ | A1 | |
8.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{f1e1d4f5-dd27-4839-a6f3-f6906666302c-22_657_659_214_646}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}

Figure 4 shows a sketch of the curve $C$ with equation

$$y = 4 + 12 x - 3 x ^ { 2 }$$

The point $M$ is the maximum turning point on $C$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Write $4 + 12 x - 3 x ^ { 2 }$ in the form

$$a + b ( x + c ) ^ { 2 }$$

where $a , b$ and $c$ are constants to be found.
\item Hence, or otherwise, state the coordinates of $M$.

The line $l _ { 1 }$ passes through $O$ and $M$, as shown in Figure 4.\\
A line $l _ { 2 }$ touches $C$ and is parallel to $l _ { 1 }$
\end{enumerate}\item Find an equation for $l _ { 2 }$
\end{enumerate}

\hfill \mbox{\textit{Edexcel P1 2021 Q8 [10]}}