Edexcel C3 — Question 4 10 marks

Exam BoardEdexcel
ModuleC3 (Core Mathematics 3)
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProduct & Quotient Rules
TypeFind stationary points and nature
DifficultyStandard +0.3 This is a slightly above-average C3 question requiring multiple standard techniques: finding x-intercepts (straightforward), differentiation using product rule, finding normal equations, and locating stationary points. Part (c) requires solving dy/dx=0 which involves some algebraic manipulation but follows standard procedures. The multi-part structure and combination of techniques elevates it slightly above average difficulty.
Spec1.06d Natural logarithm: ln(x) function and properties1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives1.07q Product and quotient rules: differentiation

4. The curve with equation \(y = x ^ { \frac { 5 } { 2 } } \ln \frac { x } { 4 } , x > 0\) crosses the \(x\)-axis at the point \(P\).
  1. Write down the coordinates of \(P\). The normal to the curve at \(P\) crosses the \(y\)-axis at the point \(Q\).
  2. Find the area of triangle \(O P Q\) where \(O\) is the origin. The curve has a stationary point at \(R\).
  3. Find the \(x\)-coordinate of \(R\) in exact form.

AnswerMarks Guidance
(a)\((4, 0)\) B1
(b)\(\frac{dy}{dx} = \frac{3}{2}x^{1/2} \times \ln\frac{4}{5} + x^{1/2} \times \frac{1}{5} = \frac{1}{5}x^{1/2}(5\ln\frac{4}{5} + 2)\) M1 A1
Grad \(= 8\), grad of normal \(= -\frac{1}{8}\)A1
\(y - 0 = -\frac{1}{8}(x-4)\)M1
At \(Q\), \(x = 0\): \(y = \frac{1}{2}\)A1
Area \(= \frac{1}{2} \times \frac{1}{2} \times 4 = 1\)A1
(c)\(\frac{1}{5}x^{1/2}(5\ln\frac{4}{5} + 2) = 0\)
\(\ln\frac{4}{5} = -\frac{2}{5}\)M1
\(x = 4e^{-\frac{4}{5}}\)M1 A1
(10)
(a) | $(4, 0)$ | B1 |

(b) | $\frac{dy}{dx} = \frac{3}{2}x^{1/2} \times \ln\frac{4}{5} + x^{1/2} \times \frac{1}{5} = \frac{1}{5}x^{1/2}(5\ln\frac{4}{5} + 2)$ | M1 A1 |
| Grad $= 8$, grad of normal $= -\frac{1}{8}$ | A1 |
| $y - 0 = -\frac{1}{8}(x-4)$ | M1 |
| At $Q$, $x = 0$: $y = \frac{1}{2}$ | A1 |
| Area $= \frac{1}{2} \times \frac{1}{2} \times 4 = 1$ | A1 |

(c) | $\frac{1}{5}x^{1/2}(5\ln\frac{4}{5} + 2) = 0$ | |
| $\ln\frac{4}{5} = -\frac{2}{5}$ | M1 |
| $x = 4e^{-\frac{4}{5}}$ | M1 A1 |
| | (10) |
4. The curve with equation $y = x ^ { \frac { 5 } { 2 } } \ln \frac { x } { 4 } , x > 0$ crosses the $x$-axis at the point $P$.
\begin{enumerate}[label=(\alph*)]
\item Write down the coordinates of $P$.

The normal to the curve at $P$ crosses the $y$-axis at the point $Q$.
\item Find the area of triangle $O P Q$ where $O$ is the origin.

The curve has a stationary point at $R$.
\item Find the $x$-coordinate of $R$ in exact form.
\end{enumerate}

\hfill \mbox{\textit{Edexcel C3  Q4 [10]}}