| Exam Board | OCR |
|---|---|
| Module | PURE |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Equation of normal line |
| Difficulty | Standard +0.8 This is a multi-part question requiring chain rule differentiation (with x^(-1/2)), finding a normal equation, and computing an area involving integration of x^(-1/2). Part (c) requires careful setup of the area calculation with multiple regions and exact arithmetic. More demanding than a standard tangent/normal question due to the integration component and need for detailed reasoning. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(y = 1 - x + \frac{6}{\sqrt{x}}\) leading to \(y' = \ldots\) | M1 | Derivative of the form \(-1 + kx^{-\frac{3}{2}}\) |
| \(y' = -1 - 3x^{-\frac{3}{2}}\) | A1 | |
| At \(x=1\), \(m_T = -4 \Rightarrow m_N = \frac{1}{4}\) | M1* | Substitutes \(x=1\) into their derivative and correct use of \(mm' = -1\) |
| \(y - 6 = \frac{1}{4}(x-1)\) | M1dep* | Use of \(y - 6 = m_N(x-1)\) |
| \(-x + 4y = 23\) | A1 | oe |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x=4\), \(y = 1 - 4 + \frac{6}{\sqrt{4}} = 0\) | B1 | AG – must show sufficient working and must see \(= 0\) |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\int\left(1 - x + \frac{6}{\sqrt{x}}\right)dx =\) | M1* | Attempt to integrate with at least two terms correct |
| \(= x - \frac{1}{2}x^2 + 12\sqrt{x}\) | A1 | |
| \(\left(4 - \frac{1}{2}(4^2) + 12\sqrt{4}\right) - \left(1 - \frac{1}{2} + 12\right) = \ldots\) | M1dep* | Use of correct limits (1 and 4). If correct, then expect to see 7.5 |
| \(\frac{1}{2}\left(\frac{23}{4} + 6\right)(1)\) | B1ft | Any correct numerical expression for the area of the trapezium between \(x=0\) and \(x=1\) using their result from (a) |
| \(\frac{107}{8}\) | A1 | Or exact equivalent (e.g. 13.375) |
| [5] |
## Question 8:
### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $y = 1 - x + \frac{6}{\sqrt{x}}$ leading to $y' = \ldots$ | M1 | Derivative of the form $-1 + kx^{-\frac{3}{2}}$ |
| $y' = -1 - 3x^{-\frac{3}{2}}$ | A1 | |
| At $x=1$, $m_T = -4 \Rightarrow m_N = \frac{1}{4}$ | M1* | Substitutes $x=1$ into their derivative and correct use of $mm' = -1$ |
| $y - 6 = \frac{1}{4}(x-1)$ | M1dep* | Use of $y - 6 = m_N(x-1)$ |
| $-x + 4y = 23$ | A1 | oe |
| **[5]** | | |
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $x=4$, $y = 1 - 4 + \frac{6}{\sqrt{4}} = 0$ | B1 | AG – must show sufficient working and must see $= 0$ |
| **[1]** | | |
### Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\int\left(1 - x + \frac{6}{\sqrt{x}}\right)dx =$ | M1* | Attempt to integrate with at least two terms correct |
| $= x - \frac{1}{2}x^2 + 12\sqrt{x}$ | A1 | |
| $\left(4 - \frac{1}{2}(4^2) + 12\sqrt{4}\right) - \left(1 - \frac{1}{2} + 12\right) = \ldots$ | M1dep* | Use of correct limits (1 and 4). If correct, then expect to see 7.5 |
| $\frac{1}{2}\left(\frac{23}{4} + 6\right)(1)$ | B1ft | Any correct numerical expression for the area of the trapezium between $x=0$ and $x=1$ using their result from (a) |
| $\frac{107}{8}$ | A1 | Or exact equivalent (e.g. 13.375) |
| **[5]** | | |
---8\\
\includegraphics[max width=\textwidth, alt={}, center]{31b0d5b6-1593-489b-bbcd-486e7c96ff18-06_823_588_260_242}
The diagram shows the curve $y = 1 - x + \frac { 6 } { \sqrt { x } }$ and the line $l$, which is the normal to the curve at the point (1, 6).
\begin{enumerate}[label=(\alph*)]
\item Determine the equation of $l$ in the form
$$a x + b y = c$$
where $a$, $b$ and $c$ are integers whose values are to be stated.
\item Verify that the curve intersects the $x$-axis at the point where $x = 4$.
\item In this question you must show detailed reasoning.
Determine the exact area of the shaded region enclosed between $l$, the curve, the $x$-axis and the $y$-axis.
\end{enumerate}
\hfill \mbox{\textit{OCR PURE Q8 [11]}}