GCSE · Iteration

Showing a root exists — change of signExam-style questions

Why a sign change in a continuous function traps a root between two values, how to present the argument for full marks, and how the same idea justifies a root to a given accuracy.

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13 questions · GCSE (Higher) level · mark scheme and worked solution on each · no sign-up to try

DifficultyAll Level 3 Level 4 Level 5

1
Show that the equation $x^3 - 5x + 1 = 0$ has a solution between $x = 0$ and $x = 1$
Level 4 · 2 marksTry it
2
$f(x) = x^3 + x - 20$

Show that the equation $f(x) = 0$ has a solution between $x = 2.5$ and $x = 2.6$
Level 4 · 2 marksTry it
3
Show that the equation $x^3 + 2x - 30 = 0$ has a solution between $x = 2$ and $x = 3$
Level 4 · 3 marksTry it
4
$f(x) = x^3 - 4x - 9$

Show that the equation $f(x) = 0$ has a solution between $x = 2.7$ and $x = 2.8$
Level 4 · 3 marksTry it
5
$f(x) = x^2 - 6x + 2$

(a) Show that the equation $f(x) = 0$ has a solution between $x = 0$ and $x = 1$

(b) Show that the equation has a second solution between $x = 5$ and $x = 6$
Level 4 · 4 marksTry it