| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sign Change & Interval Methods |
| Type | Sign Change with Intersection Points |
| Difficulty | Standard +0.3 This is a standard C3 question combining differentiation, normal lines, sign change verification, and iteration. All techniques are routine: finding dy/dx, equation of normal, substituting endpoints to show sign change, algebraic rearrangement, and iterative formula application. The multi-part structure and 4 parts suggest moderate length, but each step follows textbook procedures without requiring novel insight or complex problem-solving. |
| Spec | 1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations1.09d Newton-Raphson method |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = 2 - \frac{3}{2x+5} \times 2 = 2 - \frac{6}{2x+5}\) | M1 | |
| \(\text{grad} = -4\), \(\text{grad of normal} = \frac{1}{4}\) | A1 | |
| \(y + 4 = \frac{1}{4}(x + 2)\) \(\left[y = \frac{1}{4}x - \frac{7}{2}\right]\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{4}x - \frac{7}{2} = 2x - 3\ln(2x+5)\) | M1 | |
| \(\frac{7}{4}x + \frac{7}{2} - 3\ln(2x+5) = 0\) | ||
| let \(f(x) = \frac{7}{4}x + \frac{7}{2} - 3\ln(2x+5)\) | ||
| \(f(1) = -0.59\), \(f(2) = 0.41\) | M1 | |
| sign change, \(f(x)\) continuous \(\therefore\) root | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{7}{4}x + \frac{7}{2} - 3\ln(2x+5) = 0\) | ||
| \(7x + 14 - 12\ln(2x+5) = 0\) | ||
| \(7x = 12\ln(2x+5) - 14\) | M1 | |
| \(x = \frac{12}{7}\ln(2x+5) - 2\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x_{n+1} = \frac{12}{7}\ln(2x_n + 5) - 2\), \(x_0 = 1.5\) | ||
| \(x_1 = 1.5648\), \(x_2 = 1.5923\), \(x_3 = 1.6039\), \(x_4 = 1.6087\), \(x_5 = 1.6107\) | M1 | |
| \(q = 1.61\) (3sf) | A1 | |
| \(f(1.605) = -0.0073\), \(f(1.615) = 0.0029\) | M1 | |
| sign change, \(f(x)\) continuous \(\therefore\) root \(\therefore q = 1.61\) (3sf) | A1 | (13) |
# Question 9:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = 2 - \frac{3}{2x+5} \times 2 = 2 - \frac{6}{2x+5}$ | M1 | |
| $\text{grad} = -4$, $\text{grad of normal} = \frac{1}{4}$ | A1 | |
| $y + 4 = \frac{1}{4}(x + 2)$ $\left[y = \frac{1}{4}x - \frac{7}{2}\right]$ | M1 A1 | |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{4}x - \frac{7}{2} = 2x - 3\ln(2x+5)$ | M1 | |
| $\frac{7}{4}x + \frac{7}{2} - 3\ln(2x+5) = 0$ | | |
| let $f(x) = \frac{7}{4}x + \frac{7}{2} - 3\ln(2x+5)$ | | |
| $f(1) = -0.59$, $f(2) = 0.41$ | M1 | |
| sign change, $f(x)$ continuous $\therefore$ root | A1 | |
## Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{7}{4}x + \frac{7}{2} - 3\ln(2x+5) = 0$ | | |
| $7x + 14 - 12\ln(2x+5) = 0$ | | |
| $7x = 12\ln(2x+5) - 14$ | M1 | |
| $x = \frac{12}{7}\ln(2x+5) - 2$ | A1 | |
## Part (iv)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x_{n+1} = \frac{12}{7}\ln(2x_n + 5) - 2$, $x_0 = 1.5$ | | |
| $x_1 = 1.5648$, $x_2 = 1.5923$, $x_3 = 1.6039$, $x_4 = 1.6087$, $x_5 = 1.6107$ | M1 | |
| $q = 1.61$ (3sf) | A1 | |
| $f(1.605) = -0.0073$, $f(1.615) = 0.0029$ | M1 | |
| sign change, $f(x)$ continuous $\therefore$ root $\therefore q = 1.61$ (3sf) | A1 | **(13)** |
**Total: (72)**9.
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{49d985bf-7c94-4a54-88c1-c0084cd94000-3_485_945_1119_447}
\end{center}
The diagram shows the curve with equation $y = 2 x - 3 \ln ( 2 x + 5 )$ and the normal to the curve at the point $P ( - 2 , - 4 )$.\\
(i) Find an equation for the normal to the curve at $P$.
The normal to the curve at $P$ intersects the curve again at the point $Q$ with $x$-coordinate $q$.\\
(ii) Show that $1 < q < 2$.\\
(iii) Show that $q$ is a solution of the equation
$$x = \frac { 12 } { 7 } \ln ( 2 x + 5 ) - 2 .$$
(iv) Use an iterative process based on the equation above with a starting value of 1.5 to find the value of $q$ to 3 significant figures and justify the accuracy of your answer.
\hfill \mbox{\textit{OCR C3 Q9 [13]}}