Edexcel PURE 2024 October — Question 7

Exam BoardEdexcel
ModulePURE
Year2024
SessionOctober
PaperDownload PDF ↗
TopicTangents, normals and gradients
TypeDifferentiate composite functions
DifficultyStandard +0.3 This is a straightforward multi-part differentiation question requiring product rule and chain rule (part a), solving f'(x)=0 for turning point (part b), applying a vertical stretch to find range (part c), and applying horizontal/vertical transformations to a point (part d). All techniques are standard A-level procedures with no novel problem-solving required, making it slightly easier than average.
Spec1.07n Stationary points: find maxima, minima using derivatives1.07q Product and quotient rules: differentiation
7. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{b9472037-c143-4b68-86e2-801f71029773-20_554_559_264_753} \captionsetup{labelformat=empty} \caption{Figure 3}
\end{figure} The curve \(C\) has equation \(y = \mathrm { f } ( x )\), where $$\mathrm { f } ( x ) = x ^ { 3 } \sqrt { 4 x + 7 } \quad x \geqslant - \frac { 7 } { 4 }$$
  1. Show that $$\mathrm { f } ^ { \prime } ( x ) = \frac { k x ^ { 2 } ( 2 x + 3 ) } { \sqrt { 4 x + 7 } }$$ where \(k\) is a constant to be found. The point \(P\), shown in Figure 3, is the minimum turning point on \(C\).
  2. Find the coordinates of \(P\).
  3. Hence find the range of the function g defined by $$g ( x ) = - 4 f ( x ) \quad x \geqslant - \frac { 7 } { 4 }$$ The point \(Q\) with coordinates \(\left( \frac { 1 } { 2 } , \frac { 3 } { 8 } \right)\) lies on \(C\).
  4. Find the coordinates of the point to which \(Q\) is mapped when \(C\) is transformed to the curve with equation $$y = 40 \mathrm { f } \left( x - \frac { 3 } { 2 } \right) - 8$$
Question 7:
AnswerMarks
7 (a)12
f ( x ) = x 3 4 x + 7  f ( x ) = 3 x 2 4 x + 7 + 2 x 3 ( 4 x + 7 ) −M1, A1
2x3 3x2( 4x+7 )+2x3
f(x)=3x2 4x+7+ =
AnswerMarks
4x+7 4x+7dM1
7 x 2 ( 2 x + 3 )
 f ( x ) =
AnswerMarks
4 x + 7A1
(4)
AnswerMarks
(b)3 Substitutes x = − into x 3 4 x + 7  y = ...
2M1
 3 27 
− ,−
 
AnswerMarks
 2 8 A1
(2)
AnswerMarks
(c)27
Attempts −4''− ''
AnswerMarks
8M1
2 7
y „
AnswerMarks
2A1
(2)
AnswerMarks
(d)( )
2 , 7M1, A1
(2)

Total 10

……………………………………………………………………………………………………………………………
………………….
(b)
M1: Attempts to substitute x = − 3 into x3 4x+7 to find the y value of the minimum point .
2
This may be implied by the correct y value.
It can be attempted even if part (a) is missing or incorrect
If they ignore the given form of f ( x ) and use their own version, they must
• Set their f ( x ) which must be of the form P x 2 4 x + 7 + Q x 3 ( 4 x + 7 ) −
1
2 = 0
• Solve P x 2 4 x + 7 + Q x 3 ( 4 x + 7 ) −
1
2 = 0
1
by multiplying by ( 4x+7 ) 2o.e. to achieve a non-zero
value for x
• Substitute this non zero value for x in into x 3 4 x + 7 to find y
A1:

−
3
2
, −
2
8
7 
which may be awarded separately as, for example, x = − 1 . 5 , y = − 3 . 3 7 5
Ignore (0, 0) if also given.
(c)
M1: Attempts  4  − ''
2
8
7
'' which may be seen within an inequality.
A1: y „
2 7
2
which must be correctly simplified.
27
Allow it to be written in other correct forms such as g„ ,
2
g ( x ) „
2 7
2
or ( −  , 1 3 .5 
The original function is f so f ( x ) „
2 7
2
would be M1 A0
(d)
M1: For one correct coordinate. Look for ( 2 , . . . ) or ( ...,7 ). Allow for either x =2or y = 7
If the coordinates have been built up mark the final answer.
E.g.
 1
2
,
3
8

→

2 ,
3
8

→ ( 2 , 1 5 ) → ( − 6 , 1 5 )
PMT
would score M0 A0
A1: ( 2,7 ). Allow x =2and y = 7 ISW after a correct answer
Question 7:
--- 7 (a) ---
7 (a) | 12
f ( x ) = x 3 4 x + 7  f ( x ) = 3 x 2 4 x + 7 + 2 x 3 ( 4 x + 7 ) − | M1, A1
2x3 3x2( 4x+7 )+2x3
f(x)=3x2 4x+7+ =
4x+7 4x+7 | dM1
7 x 2 ( 2 x + 3 )
 f ( x ) =
4 x + 7 | A1
(4)
(b) | 3 Substitutes x = − into x 3 4 x + 7  y = ...
2 | M1
 3 27 
− ,−
 
 2 8  | A1
(2)
(c) | 27
Attempts −4''− ''
8 | M1
2 7
y „
2 | A1
(2)
(d) | ( )
2 , 7 | M1, A1
(2)
Total 10
……………………………………………………………………………………………………………………………
………………….
(b)
M1: Attempts to substitute x = − 3 into x3 4x+7 to find the y value of the minimum point .
2
This may be implied by the correct y value.
It can be attempted even if part (a) is missing or incorrect
If they ignore the given form of f ( x ) and use their own version, they must
• Set their f ( x ) which must be of the form P x 2 4 x + 7 + Q x 3 ( 4 x + 7 ) −
1
2 = 0
• Solve P x 2 4 x + 7 + Q x 3 ( 4 x + 7 ) −
1
2 = 0
1
by multiplying by ( 4x+7 ) 2o.e. to achieve a non-zero
value for x
• Substitute this non zero value for x in into x 3 4 x + 7 to find y
A1:

−
3
2
, −
2
8
7 
which may be awarded separately as, for example, x = − 1 . 5 , y = − 3 . 3 7 5
Ignore (0, 0) if also given.
(c)
M1: Attempts  4  − ''
2
8
7
'' which may be seen within an inequality.
A1: y „
2 7
2
which must be correctly simplified.
27
Allow it to be written in other correct forms such as g„ ,
2
g ( x ) „
2 7
2
or ( −  , 1 3 .5 
The original function is f so f ( x ) „
2 7
2
would be M1 A0
(d)
M1: For one correct coordinate. Look for ( 2 , . . . ) or ( ...,7 ). Allow for either x =2or y = 7
If the coordinates have been built up mark the final answer.
E.g.
 1
2
,
3
8

→

2 ,
3
8

→ ( 2 , 1 5 ) → ( − 6 , 1 5 )
PMT
would score M0 A0
A1: ( 2,7 ). Allow x =2and y = 7 ISW after a correct answer
7.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{b9472037-c143-4b68-86e2-801f71029773-20_554_559_264_753}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}

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

$$\mathrm { f } ( x ) = x ^ { 3 } \sqrt { 4 x + 7 } \quad x \geqslant - \frac { 7 } { 4 }$$
\begin{enumerate}[label=(\alph*)]
\item Show that

$$\mathrm { f } ^ { \prime } ( x ) = \frac { k x ^ { 2 } ( 2 x + 3 ) } { \sqrt { 4 x + 7 } }$$

where $k$ is a constant to be found.

The point $P$, shown in Figure 3, is the minimum turning point on $C$.
\item Find the coordinates of $P$.
\item Hence find the range of the function g defined by

$$g ( x ) = - 4 f ( x ) \quad x \geqslant - \frac { 7 } { 4 }$$

The point $Q$ with coordinates $\left( \frac { 1 } { 2 } , \frac { 3 } { 8 } \right)$ lies on $C$.
\item Find the coordinates of the point to which $Q$ is mapped when $C$ is transformed to the curve with equation

$$y = 40 \mathrm { f } \left( x - \frac { 3 } { 2 } \right) - 8$$
\end{enumerate}

\hfill \mbox{\textit{Edexcel PURE 2024 Q7}}
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