CAIE P1 2023 November — Question 10 7 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2023
SessionNovember
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStationary points and optimisation
TypeClassify nature of stationary points
DifficultyStandard +0.3 This is a straightforward stationary points question requiring differentiation of a fractional power and a linear term, solving a simple equation, and using the second derivative test. The algebra is routine and the question follows a standard template, making it slightly easier than average for A-level.
Spec1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx
The equation of a curve is \(y = f(x)\), where \(f(x) = (4x - 3)^{\frac{3}{4}} - \frac{20}{3}x\).
  1. Find the \(x\)-coordinates of the stationary points of the curve and determine their nature. [6]
  2. State the set of values for which the function f is increasing. [1]
Question 10:
AnswerMarks
10(a)dy 5 2  20
= (4x−3) 3 4 − 
AnswerMarks Guidance
dx 3   3 B2,1,0 B2 Three correct unsimplified { } and no others.
B1 Two correct { }or three correct { } and an additional term
e.g. + c.
B0 More than one error.
20 2 20 
 (4x−3) 3 − =0  leading to (4x−3)2 =k, k0 leading to
 3 3 
AnswerMarks Guidance
4x−3=mM1 dy
Equating their to 0 and using a valid method to arrive at
dx
2 answers.
1
 4x−3=1  x= ,1
AnswerMarks
2A1
d2y 40 (4x−3)− 1
= 34
AnswerMarks Guidance
dx2 9B1 OE
 1 d2y  160 (4x−3)− 1 160
 x=  =   3 0 or − or−17.8 so max
 2 dx2  9  9
x=1  d2y =   160  (4x−3)− 1 3 0 or 160 or 1 7.8 so min
AnswerMarks Guidance
dx2  9  9B1 d2y
If evaluated the answers for both must be correct OR
dx2
dy
Clear use of change in sign of correctly for both B1’s.
dx
If B1M1A0B0B0 scored then SCB1 can be awarded for:
dy 5 2 20
= (4x−3) 3−  leading to (4x−3)2 =64 leading
dx 3   3 
5 11
to x=− , .
4 4
d2y 10 (4x−3)− 1 5 d2y
= 3 , x=− , 0 so max,
dx2 9 4 dx2
11 d2y
x= , 0 so min.
4 dx2
6
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks Guidance
10(b)x 1, x1
2B1 Allow ⩽ and/or ⩾. FT only from special case x−5, x11
4 4
Condone: 1x 1 .
2
1
AnswerMarks Guidance
QuestionAnswer Marks 
Question 10:
--- 10(a) ---
10(a) | dy 5 2  20
= (4x−3) 3 4 − 
dx 3   3  | B2,1,0 | B2 Three correct unsimplified { } and no others.
B1 Two correct { }or three correct { } and an additional term
e.g. + c.
B0 More than one error.
20 2 20 
 (4x−3) 3 − =0  leading to (4x−3)2 =k, k0 leading to
 3 3 
4x−3=m | M1 | dy
Equating their to 0 and using a valid method to arrive at
dx
2 answers.
1
 4x−3=1  x= ,1
2 | A1
d2y 40 (4x−3)− 1
= 34
dx2 9 | B1 | OE
 1 d2y  160 (4x−3)− 1 160
 x=  =   3 0 or − or−17.8 so max
 2 dx2  9  9
x=1  d2y =   160  (4x−3)− 1 3 0 or 160 or 1 7.8 so min
dx2  9  9 | B1 | d2y
If evaluated the answers for both must be correct OR
dx2
dy
Clear use of change in sign of correctly for both B1’s.
dx
If B1M1A0B0B0 scored then SCB1 can be awarded for:
dy 5 2 20
= (4x−3) 3−  leading to (4x−3)2 =64 leading
dx 3   3 
5 11
to x=− , .
4 4
d2y 10 (4x−3)− 1 5 d2y
= 3 , x=− , 0 so max,
dx2 9 4 dx2
11 d2y
x= , 0 so min.
4 dx2
6
Question | Answer | Marks | Guidance
--- 10(b) ---
10(b) | x 1, x1
2 | B1 | Allow ⩽ and/or ⩾. FT only from special case x−5, x11
4 4
Condone: 1x 1 .
2
1
Question | Answer | Marks | Guidance
The equation of a curve is $y = f(x)$, where $f(x) = (4x - 3)^{\frac{3}{4}} - \frac{20}{3}x$.

\begin{enumerate}[label=(\alph*)]
\item Find the $x$-coordinates of the stationary points of the curve and determine their nature. [6]

\item State the set of values for which the function f is increasing. [1]
\end{enumerate}

\hfill \mbox{\textit{CAIE P1 2023 Q10 [7]}}
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