| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton-Raphson method |
| Type | Inverse trigonometric or hyperbolic functions |
| Difficulty | Standard +0.8 This question combines multiple techniques (sign change, iterative methods with inverse trig functions, and Newton-Raphson for stationary points) requiring careful algebraic manipulation of sec x and its derivative. The rearrangement to form g(x) = 2/(x² + 5x) for the arccos iteration is non-trivial, and part (c) requires implicit differentiation of sec x tan x terms, making this substantially harder than routine C3 questions. |
| Spec | 1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives1.09a Sign change methods: locate roots1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(f(1) = 2.30, f(1.5) = -18.5\) | M1 | |
| sign change, \(f(x)\) continuous \(\therefore\) root | A1 | |
| (b) \(x^2 + 5x - 2\sec x = 0 \Rightarrow x^2 + 5x = \frac{2}{\cos x}\) | M1 | |
| \(\cos x = \frac{2}{x^2+5x}\) | M1 | |
| \(x = \arccos\frac{2}{x^2+5x}\) \(\therefore g(x) = \frac{2}{x^2+5x}\) | A1 | |
| \(x_1 = 1.3119, x_2 = 1.3269, x_3 = 1.3302, x_4 = 1.3310 = 1.331\) (3dp) | M1 A2 | |
| (c) \(f'(x) = 2x + 5 - 2\sec x \tan x\) | M1 | |
| SP: \(2x + 5 - 2\sec x \tan x = 0\) | M1 | |
| \(f'(1.05345) = 0.00046, f'(1.05355) = -0.0022\) | M1 | |
| sign change, \(f'(x)\) continuous \(\therefore\) root \(\therefore\) x-coord of P \(= 1.0535\) (5sf) | A1 | (11) |
**(a)** $f(1) = 2.30, f(1.5) = -18.5$ | M1 |
sign change, $f(x)$ continuous $\therefore$ root | A1 |
**(b)** $x^2 + 5x - 2\sec x = 0 \Rightarrow x^2 + 5x = \frac{2}{\cos x}$ | M1 |
$\cos x = \frac{2}{x^2+5x}$ | M1 |
$x = \arccos\frac{2}{x^2+5x}$ $\therefore g(x) = \frac{2}{x^2+5x}$ | A1 |
$x_1 = 1.3119, x_2 = 1.3269, x_3 = 1.3302, x_4 = 1.3310 = 1.331$ (3dp) | M1 A2 |
**(c)** $f'(x) = 2x + 5 - 2\sec x \tan x$ | M1 |
SP: $2x + 5 - 2\sec x \tan x = 0$ | M1 |
$f'(1.05345) = 0.00046, f'(1.05355) = -0.0022$ | M1 |
sign change, $f'(x)$ continuous $\therefore$ root $\therefore$ x-coord of P $= 1.0535$ (5sf) | A1 | (11)3.
$$f ( x ) = x ^ { 2 } + 5 x - 2 \sec x , \quad x \in \mathbb { R } , \quad - \frac { \pi } { 2 } < x < \frac { \pi } { 2 } .$$
\begin{enumerate}[label=(\alph*)]
\item Show that the equation $\mathrm { f } ( x ) = 0$ has a root in the interval [1,1.5].
A more accurate estimate of this root is to be found using iterations of the form
$$x _ { n + 1 } = \arccos \mathrm { g } \left( x _ { n } \right) .$$
\item Find a suitable form for $\mathrm { g } ( x )$ and use this formula with $x _ { 0 } = 1.25$ to find $x _ { 1 } , x _ { 2 } , x _ { 3 }$ and $x _ { 4 }$. Give the value of $x _ { 4 }$ to 3 decimal places.
The curve $y = \mathrm { f } ( x )$ has a stationary point at $P$.
\item Show that the $x$-coordinate of $P$ is 1.0535 correct to 5 significant figures.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 Q3 [11]}}