<figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/public/question-images/converted/1760870710522-6pgexioletm.png" width="400" class="aligncenter" /></figure>
<p>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 $2$.</p>
<p>A sequence of real numbers ${{x}_{1}}$, ${{x}_{2}}$, ${{x}_{3}}$, ... satisfies the recurrence relation</p>
<p style="text-align: center">${{x}_{n+1}}={{x}_{n}}^{2}-2$</p>
<p>for $n\ge 1$.</p>
<ol class="abc-list"><li><div class="list-item-paragraph">It is given that as $n\to \infty $, ${{x}_{n}}\to L$. Show algebraically that $L$ can take values $-1$ or $2$.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">Hence or otherwise, determine the behaviour of the sequence when ${{x}_{1}}=-1$.</div><p style="text-align: right;">[1]</p></li><li><div class="list-item-paragraph">By considering ${{x}_{n+1}}-{{x}_{n}}$, prove that ${{x}_{n+1}}<{{x}_{n}}$ if $-1<{{x}_{n}} < 2$.</div><p style="text-align: right;">[2]</p></li></ol>

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

It is given that as $n\to \infty $, ${{x}_{n}}\to L$. Show algebraically that $L$ can take values $-1$ or $2$.

Hence or otherwise, determine the behaviour of the sequence when ${{x}_{1}}=-1$.

By considering ${{x}_{n+1}}-{{x}_{n}}$, prove that ${{x}_{n+1}}<{{x}_{n}}$ if $-1<{{x}_{n}} < 2$.

2025 TMJC P1 Q5

2025 TMJC P1 Q5

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