Edexcel PMT Mocks — Question 3 9 marks

Exam BoardEdexcel
ModulePMT Mocks (PMT Mocks)
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicChain Rule
TypeFind stationary points and nature
DifficultyModerate -0.3 This is a straightforward stationary points question requiring standard differentiation of power functions (rewriting surds as fractional powers), setting the derivative to zero, and using the second derivative test. All steps are routine A-level techniques with no conceptual challenges, making it slightly easier than average.
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

3. The curve \(C\) has equation $$y = 8 \sqrt { x } + \frac { 18 } { \sqrt { x } } - 20 \quad x > 0$$ a. Find
i) \(\frac { d y } { d x }\) ii) \(\frac { d ^ { 2 } y } { d x ^ { 2 } }\) b. Use calculus to find the coordinates of the stationary point of \(C\).
c. Determine whether the stationary point is a maximum or minimum, giving a reason for your answer.

Part a.i:
AnswerMarks Guidance
\(\frac{dy}{dx} = 4x^{-\frac{1}{2}} - 9x^{-\frac{3}{2}}\)A1 Achieves a correct \(\frac{dy}{dx}\)
M1Differentiates to \(\frac{dy}{dx} = Ax^{-\frac{1}{2}} + Bx^{-\frac{3}{2}}\)
Part a.ii:
AnswerMarks Guidance
\(\frac{d^2y}{dx^2} = -2x^{-\frac{3}{2}} + \frac{27}{2}x^{-\frac{5}{2}}\)B1 Achieves a correct \(\frac{d^2y}{dx^2}\) for their \(\frac{dy}{dx}\)
Part b:
AnswerMarks Guidance
\(\left(\frac{9}{4}, 4\right)\)A1 Correct answer only
M1Rearrange to find \(x\), correct answer only
M1Substitutes their \(x\) value(s) into the given equation
Part c:
AnswerMarks Guidance
\(\frac{d^2y}{dx^2} > 0\), so is a minimumA1 Fully correct solution including a correct numerical second derivative (awrt) and a reference to positive or \(> 0\) and a conclusion
M1Substitutes their \(x\) value(s) into their second derivative
### Part a.i:

$\frac{dy}{dx} = 4x^{-\frac{1}{2}} - 9x^{-\frac{3}{2}}$ | A1 | Achieves a correct $\frac{dy}{dx}$

| M1 | Differentiates to $\frac{dy}{dx} = Ax^{-\frac{1}{2}} + Bx^{-\frac{3}{2}}$

### Part a.ii:

$\frac{d^2y}{dx^2} = -2x^{-\frac{3}{2}} + \frac{27}{2}x^{-\frac{5}{2}}$ | B1 | Achieves a correct $\frac{d^2y}{dx^2}$ for their $\frac{dy}{dx}$

### Part b:

$\left(\frac{9}{4}, 4\right)$ | A1 | Correct answer only

| M1 | Rearrange to find $x$, correct answer only

| M1 | Substitutes their $x$ value(s) into the given equation

### Part c:

$\frac{d^2y}{dx^2} > 0$, so is a minimum | A1 | Fully correct solution including a correct numerical second derivative (awrt) and a reference to positive or $> 0$ and a conclusion

| M1 | Substitutes their $x$ value(s) into their second derivative
3. The curve $C$ has equation

$$y = 8 \sqrt { x } + \frac { 18 } { \sqrt { x } } - 20 \quad x > 0$$

a. Find\\
i) $\frac { d y } { d x }$\\
ii) $\frac { d ^ { 2 } y } { d x ^ { 2 } }$\\
b. Use calculus to find the coordinates of the stationary point of $C$.\\
c. Determine whether the stationary point is a maximum or minimum, giving a reason for your answer.\\

\hfill \mbox{\textit{Edexcel PMT Mocks  Q3 [9]}}