CAIE P1 2024 June — Question 10 10 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2024
SessionJune
Marks10
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Mark schemeDownload PDF ↗
TopicConnected Rates of Change
TypeCurve motion: find x-coordinate
DifficultyStandard +0.3 Part (a) is a standard connected rates of change problem requiring chain rule application with given dy/dt to find dx/dt. Part (b) involves finding coordinates from gradient condition, computing midpoint, and forming perpendicular bisector equation. Both parts use routine A-level techniques with straightforward algebra, making this slightly easier than average but requiring multiple steps across both parts.
Spec1.07m Tangents and normals: gradient and equations1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates
The equation of a curve is \(y = (5-2x)^{\frac{1}{2}} + 5\) for \(x < \frac{5}{2}\).
  1. A point \(P\) is moving along the curve in such a way that the \(y\)-coordinate of point \(P\) is decreasing at 5 units per second. Find the rate at which the \(x\)-coordinate of point \(P\) is increasing when \(y = 32\). [4]
  2. Point \(A\) on the curve has \(y\)-coordinate 32. Point \(B\) on the curve is such that the gradient of the curve at \(B\) is \(-3\). Find the equation of the perpendicular bisector of \(AB\). Give your answer in the form \(ax + by + c = 0\), where \(a\), \(b\) and \(c\) are integers. [6]
Question 10:
AnswerMarks Guidance
10(a)x2 B1
dy 1  3 1
k52x 2  2 52x 2
AnswerMarks Guidance
dx  2 M1* OE
1
Differentiating to get k52x 2 only.
dy dy dt  dt
  leading to 95
 
AnswerMarks Guidance
dx dt dx  dxDM1 Correct statement linking their numerical expression
dy dt
for with and 5.
dx dx
5
or 0.556 =
AnswerMarks Guidance
9A1 AWRT
4
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks Guidance
10(b)1
k52x 2 3M1 dy 1
Equating their of the form k52x 2 to 3.
dx
AnswerMarks
[B is] 2, 6A1
326 1 4
Gradient ABm  , gradient of perpendicular  
1 22 m 26
AnswerMarks Guidance
1M1* For A, y must be 32.
difference in y co-ordinates
Clear use of for points
difference in x co-ordinates
A and B, condone inconsistent order, and using
m m =1.
1 2
If incorrect values or another complete method used,
then working must be clear.
22 632
Mid point is , 0,19
 
AnswerMarks Guidance
 2 2 M1* Finding the midpoint of AB using A and B. If
incorrect values used then all working must be clear.
For A, y must be 32.
2
y19 x0
AnswerMarks Guidance
13DM1 Finding the equation of the perpendicular bisector
using their midpoint and their perpendicular
gradient.
AnswerMarks
2x13y2470 or  integer multiples of this.A1
6
Question 10:
--- 10(a) ---
10(a) | x2 | B1
dy 1  3 1
k52x 2  2 52x 2
dx  2  | M1* | OE
1
Differentiating to get k52x 2 only.
dy dy dt  dt
  leading to 95
 
dx dt dx  dx | DM1 | Correct statement linking their numerical expression
dy dt
for with and 5.
dx dx
5
or 0.556 =
9 | A1 | AWRT
4
Question | Answer | Marks | Guidance
--- 10(b) ---
10(b) | 1
k52x 2 3 | M1 | dy 1
Equating their of the form k52x 2 to 3.
dx
[B is] 2, 6 | A1
326 1 4
Gradient ABm  , gradient of perpendicular  
1 22 m 26
1 | M1* | For A, y must be 32.
difference in y co-ordinates
Clear use of for points
difference in x co-ordinates
A and B, condone inconsistent order, and using
m m =1.
1 2
If incorrect values or another complete method used,
then working must be clear.
22 632
Mid point is , 0,19
 
 2 2  | M1* | Finding the midpoint of AB using A and B. If
incorrect values used then all working must be clear.
For A, y must be 32.
2
y19 x0
13 | DM1 | Finding the equation of the perpendicular bisector
using their midpoint and their perpendicular
gradient.
2x13y2470 or  integer multiples of this. | A1
6
The equation of a curve is $y = (5-2x)^{\frac{1}{2}} + 5$ for $x < \frac{5}{2}$.

\begin{enumerate}[label=(\alph*)]
\item A point $P$ is moving along the curve in such a way that the $y$-coordinate of point $P$ is decreasing at 5 units per second.

Find the rate at which the $x$-coordinate of point $P$ is increasing when $y = 32$. [4]

\item Point $A$ on the curve has $y$-coordinate 32. Point $B$ on the curve is such that the gradient of the curve at $B$ is $-3$.

Find the equation of the perpendicular bisector of $AB$. Give your answer in the form $ax + by + c = 0$, where $a$, $b$ and $c$ are integers. [6]
\end{enumerate}

\hfill \mbox{\textit{CAIE P1 2024 Q10 [10]}}
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