<p>The diagram shows the graph of $y=\frac{1}{{{x}^{2}}+1}$ when $x&gt;0$.</p>
<figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/public/question-images/converted/1760099618699-646rzcyv5z.jpg" width="793" class="aligncenter" /></figure>
<ol class="i-list"><li><div class="list-item-paragraph">Evaluate $\int_{k}^{k+1}{\frac{1}{{{x}^{2}}+1}\mathrm{d}x}$ for $k&gt;0$, leaving your answer in terms of $k$.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">By considering appropriate rectangles on the interval $\left[ k,k+1 \right]$ for the curve $y=\frac{1}{{{x}^{2}}+1}$, show that</div><div class="list-item-paragraph" style="text-align: center">$\frac{1}{{{\left( k+1 \right)}^{2}}+1}&lt;{{\tan }^{-1}}\left( k+1 \right)-{{\tan }^{-1}}k&lt;\frac{1}{{{k}^{2}}+1}$ for $k\in {{\mathbb{Z}}^{+}}$.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">Use the identity $\tan \left( A-B \right)=\frac{\tan A-\tan B}{1+\tan A\tan B}$ to show that</div><div class="list-item-paragraph" style="text-align: center">${{\tan }^{-1}}x-{{\tan }^{-1}}y={{\tan }^{-1}}\frac{x-y}{1+xy}$, where $x&gt;y&gt;0$.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">By considering parts <strong>(ii)</strong> and <strong>(iii)</strong>, prove by the method of differences that</div><div class="list-item-paragraph" style="text-align: center">$\sum\limits_{k=1}^{n}{\frac{1}{{{\left( k+1 \right)}^{2}}+1}}&lt;{{\tan }^{-1}}\left( \frac{n}{n+2} \right)&lt;\sum\limits_{k=1}^{n}{\frac{1}{{{k}^{2}}+1}}$</div><p style="text-align: right;">[4]</p></li></ol>

The diagram shows the graph of $y=\frac{1}{{{x}^{2}}+1}$ when $x&gt;0$.

Evaluate $\displaystyle\displaystyle \int_{k}^{k+1}{\frac{1}{{{x}^{2}}+1}\mathrm{d}x}$ for $k&gt;0$, leaving your answer in terms of $k$.

By considering appropriate rectangles on the interval $\left[ k,k+1 \right]$ for the curve $y=\frac{1}{{{x}^{2}}+1}$, show that

$\frac{1}{{{\left( k+1 \right)}^{2}}+1}&lt;{{\tan }^{-1}}\left( k+1 \right)-{{\tan }^{-1}}k&lt;\frac{1}{{{k}^{2}}+1}$ for $k\in {{\mathbb{Z}}^{+}}$.

Use the identity $\tan \left( A-B \right)=\frac{\tan A-\tan B}{1+\tan A\tan B}$ to show that

${{\tan }^{-1}}x-{{\tan }^{-1}}y={{\tan }^{-1}}\frac{x-y}{1+xy}$, where $x&gt;y&gt;0$.

By considering parts <strong>(ii)</strong> and <strong>(iii)</strong>, prove by the method of differences that

$\sum\limits_{k=1}^{n}{\frac{1}{{{\left( k+1 \right)}^{2}}+1}}&lt;{{\tan }^{-1}}\left( \frac{n}{n+2} \right)&lt;\sum\limits_{k=1}^{n}{\frac{1}{{{k}^{2}}+1}}$

2020 EJC P1 Q10

2020 EJC P1 Q10

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