| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Equation of tangent line |
| Difficulty | Standard +0.3 This is a standard C3 multi-part question combining differentiation (chain rule), tangent equations, and numerical methods. Part (a) is routine verification, parts (b-c) are guided algebraic manipulation, and part (d) is straightforward iteration. All steps are well-signposted with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations1.07q Product and quotient rules: differentiation1.09a Sign change methods: locate roots1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks |
|---|---|
| \(\frac{dy}{dx} = 2x - \frac{1}{2}(4 + \ln x)^{-\frac{1}{2}} \times \frac{1}{x} = 2x - \frac{1}{2x\sqrt{4 + \ln x}}\) | M1 A1 |
| \(x = 1, y = -1, \text{ grad } = \frac{7}{4}\) | A1 |
| \(\therefore y + 1 = \frac{7}{4}(x - 1)\) | M1 |
| \(4y + 4 = 7x - 7\) | |
| \(7x - 4y = 11\) | A1 |
| Answer | Marks |
|---|---|
| SP: \(2x - \frac{1}{2x\sqrt{4 + \ln x}} = 0\) | M1 |
| let \(f(x) = 2x - \frac{1}{2x\sqrt{4 + \ln x}}, \quad f(0.3) = -0.40, f(0.4) = 0.088\) | M1 |
| sign change, \(f(x)\) continuous \(\therefore\) root | A1 |
| Answer | Marks |
|---|---|
| \(2x - \frac{1}{2x\sqrt{4 + \ln x}} = 0 \Rightarrow 2x = \frac{1}{2x\sqrt{4 + \ln x}}\) | M1 |
| \(x^2 = \frac{1}{4x\sqrt{4 + \ln x}} = \frac{1}{4}(4 + \ln x)^{-\frac{1}{2}}\) | M1 |
| \(x = \frac{1}{2}(4 + \ln x)^{-\frac{1}{2}} = \frac{1}{2}(4 + \ln x)^{-\frac{1}{2}}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(x_1 = 0.38151, x_2 = 0.37877, x_3 = 0.37900, x_4 = 0.37898\) (5dp) | M1 A2 | (13) |
**(a)**
$\frac{dy}{dx} = 2x - \frac{1}{2}(4 + \ln x)^{-\frac{1}{2}} \times \frac{1}{x} = 2x - \frac{1}{2x\sqrt{4 + \ln x}}$ | M1 A1 |
$x = 1, y = -1, \text{ grad } = \frac{7}{4}$ | A1 |
$\therefore y + 1 = \frac{7}{4}(x - 1)$ | M1 |
$4y + 4 = 7x - 7$ |
$7x - 4y = 11$ | A1 |
**(b)**
SP: $2x - \frac{1}{2x\sqrt{4 + \ln x}} = 0$ | M1 |
let $f(x) = 2x - \frac{1}{2x\sqrt{4 + \ln x}}, \quad f(0.3) = -0.40, f(0.4) = 0.088$ | M1 |
sign change, $f(x)$ continuous $\therefore$ root | A1 |
**(c)**
$2x - \frac{1}{2x\sqrt{4 + \ln x}} = 0 \Rightarrow 2x = \frac{1}{2x\sqrt{4 + \ln x}}$ | M1 |
$x^2 = \frac{1}{4x\sqrt{4 + \ln x}} = \frac{1}{4}(4 + \ln x)^{-\frac{1}{2}}$ | M1 |
$x = \frac{1}{2}(4 + \ln x)^{-\frac{1}{2}} = \frac{1}{2}(4 + \ln x)^{-\frac{1}{2}}$ | A1 |
**(d)**
$x_1 = 0.38151, x_2 = 0.37877, x_3 = 0.37900, x_4 = 0.37898$ (5dp) | M1 A2 | (13)
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**Total: (75)**8. A curve has the equation $y = x ^ { 2 } - \sqrt { 4 + \ln x }$.
\begin{enumerate}[label=(\alph*)]
\item Show that the tangent to the curve at the point where $x = 1$ has the equation
$$7 x - 4 y = 11$$
The curve has a stationary point with $x$-coordinate $\alpha$.
\item Show that $0.3 < \alpha < 0.4$
\item Show that $\alpha$ is a solution of the equation
$$x = \frac { 1 } { 2 } ( 4 + \ln x ) ^ { - \frac { 1 } { 4 } }$$
\item Use the iteration formula
$$x _ { n + 1 } = \frac { 1 } { 2 } \left( 4 + \ln x _ { n } \right) ^ { - \frac { 1 } { 4 } }$$
with $x _ { 0 } = 0.35$, to find $x _ { 1 } , x _ { 2 } , x _ { 3 }$ and $x _ { 4 }$, giving your answers to 5 decimal places.
END
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 Q8 [13]}}