Edexcel A-Level Mathematics AS Fp1 3 Numerical Solution of Equations Questions

Numerical-solution questions reward method control: evaluate the interval, substitute into the required iteration formula and round only at the requested accuracy.

Exam points

Prove an interval contains a root by evaluating f(a), f(b) and citing continuity.
Use one linear-interpolation step from the stated interval, then round the root.
Apply Newton-Raphson from x0 using f(x0) and f′(x0), then report x1 accurately.

Covered syllabus points

FP1.3.1 - Equations of the form f(x) = 0 f(x) will involve only

[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

[Maximum number: 5]

\mathrm{f}(x)=x^{3}-5 \sqrt{x}-4 x+7 \quad x \geqslant 0

The equation f(x)=0 has a root $\alpha$ in the interval [0.25,1]

$\mathrm{f}(x)=x^3-5\sqrt{x}-4x+7$

(a)

Use linear interpolation once on the interval [0.25,1] to determine an approximation to $\alpha$ , giving your answer to 3 decimal places.

The equation f(x)=0 has another root $\beta$ in the interval [1.5, 2.5]

[ 3 ]

$f(0.25)=3.515625,\quad f(1)=-1$ $\alpha\approx 0.25+\frac{3.515625}{3.515625+1}(1-0.25)=0.834$

(b)

Hence, using $x_{0}=1.75$ as a first approximation to $\beta$ , apply the Newton-Raphson process once to f(x) to determine a second approximation to $\beta$ , giving your answer to 3 decimal places.

[ 2 ]

$x_1=1.75-\frac{1.75^3-5\sqrt{1.75}-4(1.75)+7}{3(1.75)^2-\frac{5}{2\sqrt{1.75}}-4}$ $x_1=2.13057185\ldots$ $\beta\approx 2.131$

[Maximum number: 9]

f(x)=1-\frac{1}{8 x^{4}}+\frac{2}{7 \sqrt{x^{7}}} \quad x>0

The equation f(x)=0 has a single root, $\alpha$ , that lies in the interval [0.15,0.25]

(a)

Explain why 0.25 cannot be used as an initial approximation for $\alpha$ in the Newton-Raphson process.

[ 1 ]

Since $\mathrm{f}^{\prime}(0.25)=512-512=0$ the process cannot be carried out as it would require division by zero.

(b)

Taking 0.15 as a first approximation to $\alpha$ apply the Newton-Raphson process once to f(x) to obtain a second approximation to $\alpha$ Give your answer to 3 decimal places.

[ 5 ]

$\alpha=0.15-\frac{\mathrm{f}(0.15)}{\mathrm{f}^{\prime}(0.15)}=0.15-\frac{-27.332 \ldots}{1484.137 \ldots}=\ldots$

M1

=0.168 to 3 d.p.

A1cso

(5)

(c)

Use linear interpolation once on the interval [0.15, 0.25] to find another approximation to $\alpha$ Give your answer to 3 decimal places.

[ 3 ]

$\frac{f(0.25)-f(0.15)}{0.25-0.15}=\frac{f(0.15)-0}{0.15-\alpha}$
or equivalent interpolation strategy.
$\alpha=0.15-0.1\times\frac{f(0.15)}{f(0.25)-f(0.15)}$
or equivalent.
$=0.15-0.1\times\frac{-27.332}{5.571-(-27.332)}=0.23306\ldots$
awrt 0.233 (3 d.p.)
M1 correct interpolation strategy;
M1 proceeds to find $\alpha$ ;
A1 awrt 0.233.

All question bank results loaded

Covered syllabus points

Question 1

Question (a)

Question (i)

Question (ii)

Question (b)

Question (i)

Question 2

Question (a)

Question (b)

Question 3

Question (a)

Question (b)

Question (c)