Question 10 - A-Level Maths

Edexcel P1 2019 June — Question 10 10 marks

Exam Board	Edexcel
Module	P1 (Pure Mathematics 1)
Year	2019
Session	June
Marks	10
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Curve Sketching
Type	Horizontal translation of cubic with root finding
Difficulty	Moderate -0.8 This is a straightforward P1 question testing basic differentiation and curve properties. Part (a) requires identifying where a squared factor touches the x-axis, (b) is routine product rule differentiation, (c) shows the derivative equals zero at a stationary point (standard verification), and (d) requires simple substitution. All parts are textbook exercises with no problem-solving insight required, making it easier than average.
Spec	1.02w Graph transformations: simple transformations of f(x)1.07m Tangents and normals: gradient and equations 1.07q Product and quotient rules: differentiation

A curve has equation $y = \mathrm { f } ( x )$, where

$$f ( x ) = ( x - 4 ) ( 2 x + 1 ) ^ { 2 }$$ The curve touches the $x$-axis at the point $P$ and crosses the $x$-axis at the point $Q$.

State the coordinates of the point $P$.
Find $f ^ { \prime } ( x )$.
Hence show that the equation of the tangent to the curve at the point where $x = \frac { 5 } { 2 }$ can be expressed in the form $y = k$, where $k$ is a constant to be found. The curve with equation $y = \mathrm { f } ( x + a )$, where $a$ is a constant, passes through the origin $O$.
State the possible values of $a$.
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Show mark scheme Show mark scheme source

Question 10:

Part (a)

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$P = \left(-\frac{1}{2}, 0\right)$	B1 (1)	Accept exact equivalent. Accept coordinates written separately. If both $(4,0)$ and $\left(-\frac{1}{2},0\right)$ given, $P$ must be identified as $\left(-\frac{1}{2},0\right)$

Part (b)

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$f(x) = (x-4)(2x+1)^2 \Rightarrow f(x) = ax^3 + bx^2 + cx + d$	M1	Attempts to multiply out to form 4-term cubic. Terms need not be collected.
$= 4x^3 - 12x^2 - 15x - 4$	A1	Terms do not have to be collected for this mark.
$f'(x) = 12x^2 - 24x - 15$	dM1 A1 (4)	dM1: reduces power by one in all terms, indices must be processed. A1: must be fully simplified. Cannot be awarded from incorrect $f(x)$. Alternative by product rule: $f'(x) = 1\times(2x+1)^2 + 4(x-4)(2x+1)$

Part (c)

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Attempts $f'(2.5) = 12\times2.5^2 - 24\times2.5 - 15 = 0$	M1 A1	Shows $f'(2.5) = 75 - 60 - 15 = 0$ with either embedded values shown. $f'(x)$ must be correct. Alternative: solve $f'(x)=0 \Rightarrow x=(-0.5), 2.5$
Finds $y$ coordinate for $x=2.5$: $y=-54$	A1 (3)	CSO. Equation of tangent $y=-54$, but allow $k=-54$

Part (d)

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$a = -\frac{1}{2}$	B1	For one of $-\frac{1}{2}$, $(+)4$. Alternatively score for both $a=+\frac{1}{2},-4$. Implied by $y=f\!\left(x-\frac{1}{2}\right)$ or $y=f(x+4)$ for this mark only.
$a = -\frac{1}{2}$, $(+)4$	B1 (2) (10 marks)	For both $a=-\frac{1}{2},(+)4$ and no others. Cannot be $x=\ldots$ but allow just the values $-\frac{1}{2},(+)4$

## Question 10:

### Part (a)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $P = \left(-\frac{1}{2}, 0\right)$ | B1 **(1)** | Accept exact equivalent. Accept coordinates written separately. If both $(4,0)$ and $\left(-\frac{1}{2},0\right)$ given, $P$ must be identified as $\left(-\frac{1}{2},0\right)$ |

### Part (b)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x) = (x-4)(2x+1)^2 \Rightarrow f(x) = ax^3 + bx^2 + cx + d$ | M1 | Attempts to multiply out to form 4-term cubic. Terms need not be collected. |
| $= 4x^3 - 12x^2 - 15x - 4$ | A1 | Terms do not have to be collected for this mark. |
| $f'(x) = 12x^2 - 24x - 15$ | dM1 A1 **(4)** | dM1: reduces power by one in all terms, indices must be processed. A1: must be fully simplified. Cannot be awarded from incorrect $f(x)$. Alternative by product rule: $f'(x) = 1\times(2x+1)^2 + 4(x-4)(2x+1)$ |

### Part (c)

| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts $f'(2.5) = 12\times2.5^2 - 24\times2.5 - 15 = 0$ | M1 A1 | Shows $f'(2.5) = 75 - 60 - 15 = 0$ with either embedded values shown. $f'(x)$ must be correct. Alternative: solve $f'(x)=0 \Rightarrow x=(-0.5), 2.5$ |
| Finds $y$ coordinate for $x=2.5$: $y=-54$ | A1 **(3)** | CSO. Equation of tangent $y=-54$, but allow $k=-54$ |

### Part (d)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $a = -\frac{1}{2}$ | B1 | For one of $-\frac{1}{2}$, $(+)4$. Alternatively score for both $a=+\frac{1}{2},-4$. Implied by $y=f\!\left(x-\frac{1}{2}\right)$ or $y=f(x+4)$ for this mark only. |
| $a = -\frac{1}{2}$, $(+)4$ | B1 **(2) (10 marks)** | For both $a=-\frac{1}{2},(+)4$ and no others. Cannot be $x=\ldots$ but allow just the values $-\frac{1}{2},(+)4$ |

Show LaTeX source

\begin{enumerate}
  \item A curve has equation $y = \mathrm { f } ( x )$, where
\end{enumerate}

$$f ( x ) = ( x - 4 ) ( 2 x + 1 ) ^ { 2 }$$

The curve touches the $x$-axis at the point $P$ and crosses the $x$-axis at the point $Q$.\\
(a) State the coordinates of the point $P$.\\
(b) Find $f ^ { \prime } ( x )$.\\
(c) Hence show that the equation of the tangent to the curve at the point where $x = \frac { 5 } { 2 }$ can be expressed in the form $y = k$, where $k$ is a constant to be found.

The curve with equation $y = \mathrm { f } ( x + a )$, where $a$ is a constant, passes through the origin $O$.\\
(d) State the possible values of $a$.\\
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\hfill \mbox{\textit{Edexcel P1 2019 Q10 [10]}}

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