A20Iteration

• M1 for substituting \(x = 2\) and \(x = 3\) into \(x^{3} + 2x^{2} - 30\)
• A1 for \(-14\) and \(15\) with a conclusion referring to the change of sign

Worked solution: \(x = 2\): \(2^3 + 2 \times 2^2 - 30 = -14\)
\(x = 3\): \(3^3 + 2 \times 3^2 - 30 = 15\)
One value is negative and one is positive. There is a change of sign (and the graph is continuous), so there is a solution between 2 and 3.

• M1 for substituting \(x = 1.4\) and \(x = 1.5\) into \(x^{3} + x^{2} - 5\)
• A1 for \(-0.296\) and \(0.625\) with a conclusion referring to the change of sign

Worked solution: \(x = 1.4\): \(1.4^3 + 1.4^2 - 5 = -0.296\)
\(x = 1.5\): \(1.5^3 + 1.5^2 - 5 = 0.625\)
One value is negative and one is positive. There is a change of sign (and the graph is continuous), so there is a solution between 1.4 and 1.5.

• C1 for stating that \(\frac{1}{x - 2}\) is never 0 (a fraction with numerator 1 cannot be 0), so there is no solution
• C1 for explaining that the sign change happens at \(x = 2\), where the expression is not defined (the graph has a break / asymptote), so a change of sign does not mean a solution here

Worked solution: A fraction with numerator 1 can never equal 0, so there is no solution at all.
The sign changes because the graph of \(y = \frac{1}{x - 2}\) jumps from negative to positive at \(x = 2\) (you cannot divide by 0). The change-of-sign method only works when the graph is continuous (has no break).

• M1 for substituting both values: \(1 - 3 - 1 = -3\) and \(8 - 6 - 1 = 1\)
• C1 for 'change of sign' with a conclusion that there is a solution between 1 and 2

Worked solution: \(1^3 - 3 \times 1 - 1 = -3 \lt 0\) and \(2^3 - 3 \times 2 - 1 = 1 \gt 0\).
A cubic graph has no breaks, so it must cross zero between \(x = 1\) and \(x = 2\).