CAIE P1 2023 June — Question 11 8 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2023
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicChain Rule
TypeChain rule with single composition
DifficultyModerate -0.3 This is a straightforward chain rule application with standard follow-up tasks. Part (a) requires basic differentiation of a square root using chain rule, part (b) is routine stationary point algebra, and part (c) involves finding where dy/dx=2 then computing a normal equation—all standard A-level techniques with no novel insight required. Slightly easier than average due to the mechanical nature of all parts.
Spec1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives
The equation of a curve is $$y = k\sqrt{4x + 1} - x + 5,$$ where \(k\) is a positive constant.
  1. Find \(\frac{dy}{dx}\). [2]
  2. Find the \(x\)-coordinate of the stationary point in terms of \(k\). [2]
  3. Given that \(k = 10.5\), find the equation of the normal to the curve at the point where the tangent to the curve makes an angle of \(\tan^{-1}(2)\) with the positive \(x\)-axis. [4]
Question 11:
AnswerMarks
11(a)dy  1 1
k 4x1 2 41
AnswerMarks Guidance
dx  2 B 2,1,0 1
OE e.g. 2k4x1 2 1
B2 Three correct unsimplified { } and no others.
B1 Two correct { } or three correct { } and an additional
term e.g. +5.
B0 More than one error.
2
AnswerMarks
11(b)1 1 2k
2k4x1 2 10 leading to 4x1 2 2k or 1
1
4x1
AnswerMarks Guidance
2M1 dy 1
ak4x1
OE Equating their of the form 2 1 where
dx
a = 2 or 0.5, to 0 and dealing with the negative power
1
correctly including k not multiplied by 4x1 2.
4k2 1
x
AnswerMarks Guidance
4A1 CAO
OE simplified expression ISW.
2
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
11(c)1
210.54x1
AnswerMarks Guidance
2 12M1 dy
Putting k= 10.5 into their and equating to 2.
dx
1
AnswerMarks Guidance
74x1 2 leading to 4x149 leading to x12A1 33 1
If M1 earned SCB1 available for x from a .
64 2
y10.5 4x1x566.5 [leading to (12, 66.5)]
AnswerMarks
 A1
1
y66.5 x12
AnswerMarks Guidance
2A1 OE
4
Question 11:
--- 11(a) ---
11(a) | dy  1 1
k 4x1 2 41
dx  2  | B 2,1,0 | 1
OE e.g. 2k4x1 2 1
B2 Three correct unsimplified { } and no others.
B1 Two correct { } or three correct { } and an additional
term e.g. +5.
B0 More than one error.
2
--- 11(b) ---
11(b) | 1 1 2k
2k4x1 2 10 leading to 4x1 2 2k or 1
1
4x1
2 | M1 | dy 1
ak4x1
OE Equating their of the form 2 1 where
dx
a = 2 or 0.5, to 0 and dealing with the negative power
1
correctly including k not multiplied by 4x1 2.
4k2 1
x
4 | A1 | CAO
OE simplified expression ISW.
2
Question | Answer | Marks | Guidance
--- 11(c) ---
11(c) | 1
210.54x1
2 12 | M1 | dy
Putting k= 10.5 into their and equating to 2.
dx
1
74x1 2 leading to 4x149 leading to x12 | A1 | 33 1
If M1 earned SCB1 available for x from a .
64 2
y10.5 4x1x566.5 [leading to (12, 66.5)]
  | A1
1
y66.5 x12
2 | A1 | OE
4
The equation of a curve is
$$y = k\sqrt{4x + 1} - x + 5,$$
where $k$ is a positive constant.

\begin{enumerate}[label=(\alph*)]
\item Find $\frac{dy}{dx}$. [2]

\item Find the $x$-coordinate of the stationary point in terms of $k$. [2]

\item Given that $k = 10.5$, find the equation of the normal to the curve at the point where the tangent to the curve makes an angle of $\tan^{-1}(2)$ with the positive $x$-axis. [4]
\end{enumerate}

\hfill \mbox{\textit{CAIE P1 2023 Q11 [8]}}
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