2025 TMJC P1 Q5

The diagram shows the graph of \(y={{x}^{2}}-x-2\). The two roots of the equation \({{x}^{2}}-x-2=0\) are \(-1\) and \(2\).

A sequence of real numbers \({{x}_{1}}\), \({{x}_{2}}\), \({{x}_{3}}\), ... satisfies the recurrence relation

\({{x}_{n+1}}={{x}_{n}}^{2}-2\)

for \(n\ge 1\).

1. It is given that as \(n\to \infty \), \({{x}_{n}}\to L\). Show algebraically that \(L\) can take values \(-1\) or \(2\).[2]
2. Hence or otherwise, determine the behaviour of the sequence when \({{x}_{1}}=-1\).[1]
3. By considering \({{x}_{n+1}}-{{x}_{n}}\), prove that \({{x}_{n+1}}<{{x}_{n}}\) if \(-1<{{x}_{n}} < 2\).[2]