Edexcel IAL Maths FP1.3.1 Numerical Methods

[Maximum number: 6]

(a)

$\mathrm f(x)=x-4-\cos(5\sqrt{x}),\qquad x>0.$

[ 4 ]

$\mathrm{f}(x)=x-4-\cos (5 \sqrt{x}) \quad x>0$

(i)

Show that the equation f(x)=0 has a root $\alpha$ in the interval [2.5, 3.5]

[ 2 ]

$\mathrm{f}(2.5)=-1.44 \ldots, \mathrm{f}(3.5)=0.497 \ldots$
Sign change (negative, positive) and f(x) is continuous therefore (a root) $\alpha$
is between x=2.5 and x=3.5
(2)

(ii)

Use linear interpolation once on the interval [2.5, 3.5] to find an approximation to $\alpha$ , giving your answer to 2 decimal places.

[ 2 ]

E.g. $\frac{\alpha-2.5}{|\mathrm{f}(2.5)|}=\frac{3.5-\alpha}{\mathrm{f}(3.5)} \Rightarrow \alpha=\ldots$ or $\frac{\alpha-2.5}{0-\mathrm{f}(2.5)}=\frac{3.5-2.5}{\mathrm{f}(3.5)-\mathrm{f}(2.5)} \Rightarrow \alpha=\ldots$ $\alpha=$ awrt 3.24
(2)

(b)

$\mathrm g(x)=\frac{1}{10}x^2-\frac{1}{2x^2}+x-11,\qquad x>0.$

[ 2 ]

$\mathrm{g}(x)=\frac{1}{10} x^{2}-\frac{1}{2 x^{2}}+x-11 \quad x>0$

(i)

Using $x_{0}=6$ as a first approximation to $\beta$ , apply the Newton-Raphson procedure once to g(x) to find a second approximation to $\beta$ , giving your answer to 3 decimal places.

[ 2 ]

Newton-Raphson:
$x_1=6-\frac{\mathrm g(6)}{\mathrm g'(6)} =6-\frac{-1.41388\ldots}{2.20462\ldots}.$
M1

$x_1=6.641.$
A1cao

Edexcel A-Level Mathematics AS Fp1 3 1 Equations of the Form F X 0 F X Will Involve Only Questions

Exam points

Edexcel A-Level Mathematics AS Fp1 3 1 Equations of the Form F X 0 F X Will Involve Only Questions question 1

Question (a)

Question (i)

Question (ii)

Question (b)

Question (i)