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: 4]

\mathrm{f}(x)=10-2 x-\frac{1}{2 \sqrt{x}}-\frac{1}{x^{3}} \quad x>0

(a)

Using $x_{0}=0.5$ as a first approximation to $\alpha$ , apply the Newton-Raphson procedure once to f(x) to find a second approximation to $\alpha$ , giving your answer to 3 decimal places.

The equation f(x)=0 has another root $\beta$ in the interval [4.8, 4.9]

[ 2 ]

$x_{1}=0.5-\frac{\mathrm{f}(0.5)}{\mathrm{f}^{\prime}(0.5)}=0.5-\frac{0.29289321 \ldots}{46.70710678 \ldots}$

Correct application of Newton-Raphson
=0.494

Correct value 3dp. A correct derivative
must have been used
(2)

(b)

Use linear interpolation once on the interval [4.8, 4.9] to find an approximation to $\beta$ , giving your answer to 3 decimal places.

[ 2 ]

$\frac{4.9-\beta}{|f(4.9)|}=\frac{\beta-4.8}{f(4.8)}$ $\beta=4.883$
ALT:
$\beta=\frac{4.8|f(4.9)|+4.9|f(4.8)|}{|f(4.8)|+|f(4.9)|}$ $\beta=\frac{4.8\times0.0344+4.9\times0.1627}{0.0344+0.1627}=4.883$

[Maximum number: 2]

f(x)=x^{3}-\frac{10 \sqrt{x}-4 x}{x^{2}} \quad x>0

Using $x_{0}=1.4$ as a first approximation to $\alpha$ , apply the Newton-Raphson procedure once to f(x) to calculate a second approximation to $\alpha$ , giving your answer to 3 decimal places.

$x_1=1.4-\frac{f(1.4)}{f'(1.4)}$ $=1.4-\frac{-0.43567\ldots}{10.30720\ldots}=1.442$

[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 ]

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

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Covered syllabus points

Question 1

Question (a)

Question (b)

Question 2

Question 3

Question (a)

Question (b)