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Question 45

Show that the equation $x^3 + 5x = 1$ has a solution between $x = 0$ and $x = 1$ .

[2]

Show that the equation $x^3 + 5x = 1$ can be rearranged to give: $x=15−x35\displaystyle x = \frac{1}{5} - \frac{x^3}{5}$

[1]

Starting with $x_0 = 0$ , use the iteration formula $xn+1=15−xn35\displaystyle x_{n+1} = \frac{1}{5} - \frac{x_n^3}{5}$ twice to find an estimate for the solution to $x^3 + 5x = 1$

[3]

Markscheme

Iteration Questions

GCSE
Maths
Iteration

124 exam-style questions on OCR GCSE Maths Iteration. Each one has a worked solution and a mark scheme showing where the marks go.

Iteration

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Iteration

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Question 45

Show that the equation $x^3 + 5x = 1$ has a solution between $x = 0$ and $x = 1$ .

[2]

Show that the equation $x^3 + 5x = 1$ can be rearranged to give: $x=15−x35\displaystyle x = \frac{1}{5} - \frac{x^3}{5}$

[1]

Starting with $x_0 = 0$ , use the iteration formula $xn+1=15−xn35\displaystyle x_{n+1} = \frac{1}{5} - \frac{x_n^3}{5}$ twice to find an estimate for the solution to $x^3 + 5x = 1$

[3]

Markscheme

Iteration Questions

GCSE
Maths
Iteration

124 exam-style questions on OCR GCSE Maths Iteration. Each one has a worked solution and a mark scheme showing where the marks go.