Standard +0.3 This is a straightforward application of the Newton-Raphson method with a simple function f(x) = e^(-x) - x. While it involves an exponential function, the derivative is elementary (f'(x) = -e^(-x) - 1), and the iteration is routine calculation from a given starting value. Slightly above average difficulty only because it's Further Maths content and requires careful arithmetic, but the method is mechanical with no conceptual challenges.
2 Use the Newton-Raphson method to find the root of the equation \(\mathrm { e } ^ { - x } = x\) which is close to \(x = 0.5\). Give the root correct to 3 decimal places.
For correct denominator (in 2nd case must include \(1/4\))
\(= 1/2 [\frac{1}{2} \tan^{-1} \frac{2x-1}{2}]_1^{3/2}\) OR \(1/4[\tan^{-1}(x-1/2)]_1^{3/2}\)
M1
For integration to \(k\tan^{-1}(ax+b)\) or \(k\ln(\frac{ax+b-c}{ax+b+c})\)
\(= 1/4(\tan^{-1} 1 - \tan^{-1} 0) = 1/16 \pi\)
M1
For substituting limits in any \(\tan^{-1}\) expression
A1
For correct value
$\int_1^{3/2} \frac{1}{(2x-1)^2 + 4} dx$ OR $1/4 \int_1^{3/2} \frac{1}{(x-1/2)^2 + 1} dx$ | B1 | For correct denominator (in 2nd case must include $1/4$)
$= 1/2 [\frac{1}{2} \tan^{-1} \frac{2x-1}{2}]_1^{3/2}$ OR $1/4[\tan^{-1}(x-1/2)]_1^{3/2}$ | M1 | For integration to $k\tan^{-1}(ax+b)$ or $k\ln(\frac{ax+b-c}{ax+b+c})$
$= 1/4(\tan^{-1} 1 - \tan^{-1} 0) = 1/16 \pi$ | M1 | For substituting limits in any $\tan^{-1}$ expression
| A1 | For correct value
2 Use the Newton-Raphson method to find the root of the equation $\mathrm { e } ^ { - x } = x$ which is close to $x = 0.5$. Give the root correct to 3 decimal places.
\hfill \mbox{\textit{OCR FP2 Q2 [5]}}