Edexcel PURE 2024 October — Question 4

Exam BoardEdexcel
ModulePURE
Year2024
SessionOctober
PaperDownload PDF ↗
TopicImplicit equations and differentiation
TypeTangent/normal with axis intercepts
DifficultyStandard +0.8 This question combines implicit differentiation with exponential terms (requiring chain rule and knowledge that d/dx(8^x) = 8^x ln 8), finding a normal line equation, and expressing the final answer in exact logarithmic form. While the individual techniques are A-level standard, the combination of implicit differentiation with exponential functions, plus the algebraic manipulation to reach the specific form a + b ln 2, makes this moderately challenging and above average difficulty.
Spec1.03g Parametric equations: of curves and conversion to cartesian1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07s Parametric and implicit differentiation
4. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{fa121449-492f-4737-a9eb-a14a62ced47b-10_634_638_255_717} \captionsetup{labelformat=empty} \caption{Figure 2}
\end{figure} Figure 2 shows a sketch of the curve \(C\) with equation $$3 x ^ { 2 } + 2 y ^ { 2 } - 4 x y + 8 ^ { x } - 11 = 0$$ The point \(P\) has coordinates ( 1,2 ).
  1. Verify that \(P\) lies on \(C\).
  2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\). The normal to \(C\) at \(P\) crosses the \(x\)-axis at a point \(Q\).
  3. Find the \(x\) coordinate of \(Q\), giving your answer in the form \(a + b \ln 2\) where \(a\) and \(b\) are integers.
Question 4:
AnswerMarks Guidance
4(a)x = 1 , y = 2  3 x 2 + 2 y 2 − 4 x y + 8 x − 1 1 = 3 + 8 − 8 + 8 − 1 1 = 0 * B1*
(1)
AnswerMarks Guidance
(b)8x →8x ln8 B1
d y d y
4 x y → 4 x + 4 y or y 2 → 2 y
AnswerMarks
d x d xM1
d y d y
6 x + 4 y − 4 y − 4 x + 8 x ln 8 = 0 o.e.
AnswerMarks
d x d xA1
d y d y
( 4 y − 4 x ) = 4 y − 6 x − 8 x l n 8  = ...
AnswerMarks
d x d xM1
d y 4 y − 6 x − 8 x ln 8 d y 6 x + 8 x ln 8 − 4 y dy 4y−6x−exln8ln8
= o.e. e.g = or e.g =
AnswerMarks
d x 4 y − 4 x d x 4 x − 4 y dx 4y−4xA1
(5)
AnswerMarks
(c)d y 8 − 6 − 8 l n 8 4
( 1 , 2 ) → =  y−2= (x−1)
AnswerMarks
d x 8 − 4 8ln8−2M1
4
y = 0  0 − 2 = ( x − 1 )  x = . . .
AnswerMarks
8 l n 8 − 2M1
x = 2 − 1 2 l n 2A1
(3)

Total 9

M1: Attempts to make
d
d
y
x
dy
the subject with 2 terms in coming from the correct places. i.e. they
dx
must have two terms (one of each) of the form x
d
d
y
x
dy
 and y where
dx
,  are non-zero
constants.
A1: Any correct expression for
d
d
y
x
. Allow y=... o.e. isw after a correct answer is seen
Condone poor dividing lines which do not go quite far enough provided an incorrect method
4y−6x −8x ln8
was not seen. E.g.
4y−4x
(c) Note that an incorrect derivative in part (b) can score maximum 110 in part (c)
dy
M1: Attempts to use x = 1 and y = 2 in their to find the gradient at P (condone if this is a
dx
decimal) and attempts to form the equation of the normal at P using the negative reciprocal
gradient. Condone one sign slip when substituting in the coordinates into the straight line
equation.
They may set y = 0 at this stage such that − 2 =
8 l n
4
8 − 2
( x − 1 ) which is acceptable.
If they use y = m x + c they must proceed as far as c = ...
PMT
M1: States y = 0 or substitutes y = 0 into their straight-line equation using a changed gradient from
dy
their and rearranges to find x. Do not be concerned by the mechanics of their
dx
rearrangement.
A1: Correct expression in the required form
Question 4:
--- 4(a) ---
4(a) | x = 1 , y = 2  3 x 2 + 2 y 2 − 4 x y + 8 x − 1 1 = 3 + 8 − 8 + 8 − 1 1 = 0 * | B1*
(1)
(b) | 8x →8x ln8 | B1
d y d y
4 x y → 4 x + 4 y or y 2 → 2 y
d x d x | M1
d y d y
6 x + 4 y − 4 y − 4 x + 8 x ln 8 = 0 o.e.
d x d x | A1
d y d y
( 4 y − 4 x ) = 4 y − 6 x − 8 x l n 8  = ...
d x d x | M1
d y 4 y − 6 x − 8 x ln 8 d y 6 x + 8 x ln 8 − 4 y dy 4y−6x−exln8ln8
= o.e. e.g = or e.g =
d x 4 y − 4 x d x 4 x − 4 y dx 4y−4x | A1
(5)
(c) | d y 8 − 6 − 8 l n 8 4
( 1 , 2 ) → =  y−2= (x−1)
d x 8 − 4 8ln8−2 | M1
4
y = 0  0 − 2 = ( x − 1 )  x = . . .
8 l n 8 − 2 | M1
x = 2 − 1 2 l n 2 | A1
(3)
Total 9
M1: Attempts to make
d
d
y
x
dy
the subject with 2 terms in coming from the correct places. i.e. they
dx
must have two terms (one of each) of the form x
d
d
y
x
dy
 and y where
dx
,  are non-zero
constants.
A1: Any correct expression for
d
d
y
x
. Allow y=... o.e. isw after a correct answer is seen
Condone poor dividing lines which do not go quite far enough provided an incorrect method
4y−6x −8x ln8
was not seen. E.g.
4y−4x
(c) Note that an incorrect derivative in part (b) can score maximum 110 in part (c)
dy
M1: Attempts to use x = 1 and y = 2 in their to find the gradient at P (condone if this is a
dx
decimal) and attempts to form the equation of the normal at P using the negative reciprocal
gradient. Condone one sign slip when substituting in the coordinates into the straight line
equation.
They may set y = 0 at this stage such that − 2 =
8 l n
4
8 − 2
( x − 1 ) which is acceptable.
If they use y = m x + c they must proceed as far as c = ...
PMT
M1: States y = 0 or substitutes y = 0 into their straight-line equation using a changed gradient from
dy
their and rearranges to find x. Do not be concerned by the mechanics of their
dx
rearrangement.
A1: Correct expression in the required form
4.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{fa121449-492f-4737-a9eb-a14a62ced47b-10_634_638_255_717}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}

Figure 2 shows a sketch of the curve $C$ with equation

$$3 x ^ { 2 } + 2 y ^ { 2 } - 4 x y + 8 ^ { x } - 11 = 0$$

The point $P$ has coordinates ( 1,2 ).
\begin{enumerate}[label=(\alph*)]
\item Verify that $P$ lies on $C$.
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$ and $y$.

The normal to $C$ at $P$ crosses the $x$-axis at a point $Q$.
\item Find the $x$ coordinate of $Q$, giving your answer in the form $a + b \ln 2$ where $a$ and $b$ are integers.
\end{enumerate}

\hfill \mbox{\textit{Edexcel PURE 2024 Q4}}
This paper (10 questions)
View full paper
Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10