Edexcel PURE 2024 October — Question 4

Exam BoardEdexcel
ModulePURE
Year2024
SessionOctober
PaperDownload PDF ↗
TopicTangents, normals and gradients
TypeFind stationary points
DifficultyModerate -0.8 This is a straightforward differentiation question testing basic calculus techniques: differentiating powers (including fractional indices), finding stationary points by setting dy/dx = 0, using the second derivative test, and interpreting the sign of dy/dx. All steps are routine applications of standard methods with no problem-solving insight required. The algebra is simple (solving 2x = 9/2 gives x = 9/4). This is easier than average A-level questions which typically require combining multiple techniques or some element of problem-solving.
Spec1.07e Second derivative: as rate of change of gradient1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives
  1. In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.
The curve \(C\) has equation $$y = 4 x ^ { \frac { 1 } { 2 } } + 9 x ^ { - \frac { 1 } { 2 } } + 3 \quad x > 0$$
  1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) giving each term in simplest form.
  2. Hence find the \(x\) coordinate of the stationary point of \(C\).
    1. Find \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) giving each term in simplest form.
    2. Hence determine the nature of the stationary point of \(C\), giving a reason for your answer.
  4. State the range of values of \(x\) for which \(y\) is decreasing.
Question 4:
AnswerMarks
4(a)y = 4 x 12 + 9 x − 12 + 3
 d y  9
 = 2 x − 12 − x − 32
AnswerMarks
d x 2M1A1
(2)
AnswerMarks
(b)d y 9
= 0  2 x − 12 − x − 32 = 0  4 x − 9 = 0  x = . ..
AnswerMarks
d x 2M1
9
x = oe e.g. 2.25
AnswerMarks
4A1
(2)
4(c)(i)
AnswerMarks
(ii) d 2 y  2 7
= − x − 32 + x − 52 oe e.g. − x − 32 + 6 . 7 5 x − 52
AnswerMarks
d x 2 4B1ft
 
 d 2 y   9  − 32 2 7  9  − 52  1 6 
= − + = ( 0 . 5 9 2 5 . .. )
d x 2 4 4 4 2 7
94
x =
 d 2 y 
 0 so (local) minimum
AnswerMarks
d x 2B1
(2)
AnswerMarks
(d)9
0  x 
AnswerMarks
4B1ft
(1)

Total 7

Question 4:
--- 4(a) ---
4(a) | y = 4 x 12 + 9 x − 12 + 3
 d y  9
 = 2 x − 12 − x − 32
d x 2 | M1A1
(2)
(b) | d y 9
= 0  2 x − 12 − x − 32 = 0  4 x − 9 = 0  x = . ..
d x 2 | M1
9
x = oe e.g. 2.25
4 | A1
(2)
4(c)(i)
(ii) |  d 2 y  2 7
= − x − 32 + x − 52 oe e.g. − x − 32 + 6 . 7 5 x − 52
d x 2 4 | B1ft
 
 d 2 y   9  − 32 2 7  9  − 52  1 6 
= − + = ( 0 . 5 9 2 5 . .. )
d x 2 4 4 4 2 7
94
x =
 d 2 y 
 0 so (local) minimum
d x 2 | B1
(2)
(d) | 9
0  x 
4 | B1ft
(1)
Total 7
\begin{enumerate}
  \item In this question you must show all stages of your working.
\end{enumerate}

Solutions relying entirely on calculator technology are not acceptable.\\
The curve $C$ has equation

$$y = 4 x ^ { \frac { 1 } { 2 } } + 9 x ^ { - \frac { 1 } { 2 } } + 3 \quad x > 0$$

(a) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ giving each term in simplest form.\\
(b) Hence find the $x$ coordinate of the stationary point of $C$.\\
(c) (i) Find $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$ giving each term in simplest form.\\
(ii) Hence determine the nature of the stationary point of $C$, giving a reason for your answer.\\
(d) State the range of values of $x$ for which $y$ is decreasing.

\hfill \mbox{\textit{Edexcel PURE 2024 Q4}}
This paper (11 questions)
View full paper
Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11