| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Session | Specimen |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton-Raphson method |
| Type | Show root in interval |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question testing standard Newton-Raphson application. Part (a) requires simple substitution to show sign change, part (b) is direct formula application with given values, and part (c) uses basic calculus to show f'(x) > 0 for monotonicity. All parts are routine textbook exercises requiring no problem-solving insight. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05p Proof involving trig: functions and identities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(3.5) = -4.8\), \(f(4) = (+)3.1\) | M1 | Attempts \(f(x)\) at both \(x = 3.5\) and \(x = 4\) with at least one correct to 1 significant figure |
| Change of sign and function continuous in interval \([3.5, 4] \Rightarrow\) Root | A1* | \(f(3.5)\) and \(f(4)\) correct to 1 sig figure with correct reason and conclusion; reason could be change of sign, or \(f(3.5) \times f(4) < 0\); conclusion could be 'Hence root' or 'Therefore root in interval' |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts \(x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} \Rightarrow x_1 = 4 - \frac{3.099}{16.67}\) | M1 | Evidence of Newton-Raphson application |
| \(x_1 = 3.81\) | A1 | Correct answer only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Sketches both \(y = \ln(2x-5)\) and \(y = 30 - 2x^2\) on same axes | M1 | Valid attempt at showing only one root |
| States \(y = \ln(2x-5)\) meets \(y = 30 - 2x^2\) in just one place, therefore \(y = \ln(2x-5) = 30 - 2x\) has just one root \(\Rightarrow f(x) = 0\) has just one root | A1 | Correct conclusion. Alternatives: showing \(f(x) = \ln(2x-5) + 2x^2 - 30\) has no turning points, or sketching graph of \(f(x) = \ln(2x-5) + 2x^2 - 30\) |
# Question 8:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(3.5) = -4.8$, $f(4) = (+)3.1$ | M1 | Attempts $f(x)$ at both $x = 3.5$ and $x = 4$ with at least one correct to 1 significant figure |
| Change of sign and function continuous in interval $[3.5, 4] \Rightarrow$ Root | A1* | $f(3.5)$ and $f(4)$ correct to 1 sig figure with correct reason and conclusion; reason could be change of sign, or $f(3.5) \times f(4) < 0$; conclusion could be 'Hence root' or 'Therefore root in interval' |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts $x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} \Rightarrow x_1 = 4 - \frac{3.099}{16.67}$ | M1 | Evidence of Newton-Raphson application |
| $x_1 = 3.81$ | A1 | Correct answer only |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Sketches both $y = \ln(2x-5)$ and $y = 30 - 2x^2$ on same axes | M1 | Valid attempt at showing only one root |
| States $y = \ln(2x-5)$ meets $y = 30 - 2x^2$ in just one place, therefore $y = \ln(2x-5) = 30 - 2x$ has just one root $\Rightarrow f(x) = 0$ has just one root | A1 | Correct conclusion. Alternatives: showing $f(x) = \ln(2x-5) + 2x^2 - 30$ has no turning points, or sketching graph of $f(x) = \ln(2x-5) + 2x^2 - 30$ |
---8.
$$f ( x ) = \ln ( 2 x - 5 ) + 2 x ^ { 2 } - 30 , \quad x > 2.5$$
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathrm { f } ( x ) = 0$ has a root $\alpha$ in the interval [3.5,4]
A student takes 4 as the first approximation to $\alpha$.\\
Given $\mathrm { f } ( 4 ) = 3.099$ and $\mathrm { f } ^ { \prime } ( 4 ) = 16.67$ to 4 significant figures,
\item apply the Newton-Raphson procedure once to obtain a second approximation for $\alpha$, giving your answer to 3 significant figures.
\item Show that $\alpha$ is the only root of $\mathrm { f } ( x ) = 0$
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 1 Q8 [6]}}