| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find equation of normal |
| Difficulty | Standard +0.8 This question requires finding where ln(x/4)=0, then differentiating a product involving x^(5/2) and ln(x/4) using both chain and product rules, finding the normal equation, and locating a stationary point. The algebraic manipulation and multi-step nature (especially part iii requiring exact form) elevate this above a standard C3 question, though the individual techniques are syllabus-standard. |
| Spec | 1.07l 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 |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \((4, 0)\) | B1 | |
| (ii) \(\frac{dy}{dx} = \frac{5}{2}x^{\frac{1}{4}} \ln \frac{4}{x} + x^{\frac{1}{4}} \times \frac{1}{x} = \frac{1}{2}x^{\frac{1}{4}}(5 \ln \frac{4}{x} + 2)\) | M1, A1 | |
| grad = 8, grad of normal = \(-\frac{1}{8}\) | A1 | |
| \(\therefore y - 0 = -\frac{1}{8}(x - 4)\) | M1 | |
| at Q, \(x = 0\), \(y = \frac{1}{2}\) | M1 | |
| area = \(\frac{1}{2} \times \frac{1}{2} \times 4 = 1\) | A1 | |
| (iii) \(\frac{1}{2}x^{\frac{1}{4}}(5 \ln \frac{4}{x} + 2) = 0\) | M1 | |
| \(x > 0 \therefore \ln \frac{4}{x} = -\frac{2}{5}\) | M1 | |
| \(x = 4e^{-\frac{2}{5}}\) | A1 | (9 marks) |
**(i)** $(4, 0)$ | B1 |
**(ii)** $\frac{dy}{dx} = \frac{5}{2}x^{\frac{1}{4}} \ln \frac{4}{x} + x^{\frac{1}{4}} \times \frac{1}{x} = \frac{1}{2}x^{\frac{1}{4}}(5 \ln \frac{4}{x} + 2)$ | M1, A1 |
grad = 8, grad of normal = $-\frac{1}{8}$ | A1 |
$\therefore y - 0 = -\frac{1}{8}(x - 4)$ | M1 |
at Q, $x = 0$, $y = \frac{1}{2}$ | M1 |
area = $\frac{1}{2} \times \frac{1}{2} \times 4 = 1$ | A1 |
**(iii)** $\frac{1}{2}x^{\frac{1}{4}}(5 \ln \frac{4}{x} + 2) = 0$ | M1 |
$x > 0 \therefore \ln \frac{4}{x} = -\frac{2}{5}$ | M1 |
$x = 4e^{-\frac{2}{5}}$ | A1 | **(9 marks)** |
---7. The curve with equation $y = x ^ { \frac { 5 } { 2 } } \ln \frac { x } { 4 } , x > 0$ crosses the $x$-axis at the point $P$.\\
(i) Write down the coordinates of $P$.
The normal to the curve at $P$ crosses the $y$-axis at the point $Q$.\\
(ii) Find the area of triangle $O P Q$ where $O$ is the origin.
The curve has a stationary point at $R$.\\
(iii) Find the $x$-coordinate of $R$ in exact form.\\
\hfill \mbox{\textit{OCR C3 Q7 [9]}}