| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Find normal line equation |
| Difficulty | Standard +0.3 This is a standard C3 multi-part question covering routine differentiation (including chain rule for exponentials and power of x), finding a normal line, locating stationary points, and applying iterative formulas. While it has multiple parts (5 sub-questions), each individual step uses well-practiced techniques with no novel insight required. The algebraic manipulation is straightforward, and the iteration/convergence analysis in part (v) is a typical textbook exercise. Slightly easier than average due to the guided nature and standard methods throughout. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(\frac{dy}{dx} = \frac{1}{2}x^{-\frac{1}{2}} - 4e^{1-4x}\) | M1 A1 | |
| grad \(= -3\), grad of normal \(= \frac{1}{3}\) | M1 | |
| \(\therefore y - \frac{3}{2} = \frac{1}{3}(x - \frac{1}{4})\) \([4x - 12y + 17 = 0]\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| SP: \(\frac{1}{2}x^{-\frac{1}{2}} - 4e^{1-4x} = 0\) | ||
| \(\frac{1}{2\sqrt{x}} = 4e^{1-4x}\) | ||
| \(\frac{1}{8\sqrt{x}} = e^{1-4x}\) | M1 | |
| \(8\sqrt{x} = e^{4x-1}\) | ||
| \(4x - 1 = \ln 8\sqrt{x}\) | M1 | |
| \(x = \frac{1}{4}(1 + \ln 8\sqrt{x})\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(x_1 = 0.7699\), \(x_2 = 0.7372\), \(x_3 = 0.7317\), \(x_4 = 0.7308 = 0.731\) (3dp) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| let \(f(x) = \frac{1}{2}x^{-\frac{1}{2}} - 4e^{1-4x}\) | ||
| \(f(0.7305) = -0.00025\), \(f(0.7315) = 0.0017\) | M1 | |
| sign change, \(f(x)\) continuous \(\therefore\) root | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(x_1 = 6.304\), \(x_2 = 1.683 \times 10^{19}\) | B2 | |
| diverges rapidly away from root | (13) | |
| Total: (72) |
# Question 8:
## Part (i):
| Answer/Working | Marks | Notes |
|---|---|---|
| $\frac{dy}{dx} = \frac{1}{2}x^{-\frac{1}{2}} - 4e^{1-4x}$ | M1 A1 | |
| grad $= -3$, grad of normal $= \frac{1}{3}$ | M1 | |
| $\therefore y - \frac{3}{2} = \frac{1}{3}(x - \frac{1}{4})$ $[4x - 12y + 17 = 0]$ | A1 | |
## Part (ii):
| Answer/Working | Marks | Notes |
|---|---|---|
| SP: $\frac{1}{2}x^{-\frac{1}{2}} - 4e^{1-4x} = 0$ | | |
| $\frac{1}{2\sqrt{x}} = 4e^{1-4x}$ | | |
| $\frac{1}{8\sqrt{x}} = e^{1-4x}$ | M1 | |
| $8\sqrt{x} = e^{4x-1}$ | | |
| $4x - 1 = \ln 8\sqrt{x}$ | M1 | |
| $x = \frac{1}{4}(1 + \ln 8\sqrt{x})$ | A1 | |
## Part (iii):
| Answer/Working | Marks | Notes |
|---|---|---|
| $x_1 = 0.7699$, $x_2 = 0.7372$, $x_3 = 0.7317$, $x_4 = 0.7308 = 0.731$ (3dp) | M1 A1 | |
## Part (iv):
| Answer/Working | Marks | Notes |
|---|---|---|
| let $f(x) = \frac{1}{2}x^{-\frac{1}{2}} - 4e^{1-4x}$ | | |
| $f(0.7305) = -0.00025$, $f(0.7315) = 0.0017$ | M1 | |
| sign change, $f(x)$ continuous $\therefore$ root | A1 | |
## Part (v):
| Answer/Working | Marks | Notes |
|---|---|---|
| $x_1 = 6.304$, $x_2 = 1.683 \times 10^{19}$ | B2 | |
| diverges rapidly away from root | | **(13)** |
| | | **Total: (72)** |8. The curve $C$ has the equation $y = \sqrt { x } + \mathrm { e } ^ { 1 - 4 x } , x \geq 0$.\\
(i) Find an equation for the normal to the curve at the point $\left( \frac { 1 } { 4 } , \frac { 3 } { 2 } \right)$.
The curve $C$ has a stationary point with $x$-coordinate $\alpha$ where $0.5 < \alpha < 1$.\\
(ii) Show that $\alpha$ is a solution of the equation
$$x = \frac { 1 } { 4 } [ 1 + \ln ( 8 \sqrt { x } ) ]$$
(iii) Use the iterative formula
$$x _ { n + 1 } = \frac { 1 } { 4 } \left[ 1 + \ln \left( 8 \sqrt { x _ { n } } \right) \right]$$
with $x _ { 0 } = 1$ to find $x _ { 1 } , x _ { 2 } , x _ { 3 }$ and $x _ { 4 }$, giving the value of $x _ { 4 }$ to 3 decimal places.\\
(iv) Show that your value for $x _ { 4 }$ is the value of $\alpha$ correct to 3 decimal places.\\
(v) Another attempt to find $\alpha$ is made using the iterative formula
$$x _ { n + 1 } = \frac { 1 } { 64 } \mathrm { e } ^ { 8 x _ { n } - 2 }$$
with $x _ { 0 } = 1$. Describe the outcome of this attempt.
\hfill \mbox{\textit{OCR C3 Q8 [13]}}