<p><strong>Do not use a graphing calculator in answering this question.</strong></p>
<ol class="i-list"><li><div class="list-item-paragraph">Let $y=\frac{2x}{4+{{x}^{2}}}-{{\tan }^{-1}}\frac{x}{2}$. Show that $\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{4{{x}^{2}}}{{{\left( 4+{{x}^{2}} \right)}^{2}}}$, and deduce that $y0$.</div><p style="text-align: right;">[3]</p></li><li><div class="list-item-paragraph">Hence show that if $\mathrm{f}\left( x \right)=\frac{1}{x}{{\tan }^{-1}}\frac{x}{2}$, where $x>0$, then $\mathrm{f}$ is a decreasing function.</div><p style="text-align: right;">[2]</p></li></ol>

**Do not use a graphing calculator in answering this question.**

Let $y=\frac{2x}{4+{{x}^{2}}}-{{\tan }^{-1}}\frac{x}{2}$. Show that $\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{4{{x}^{2}}}{{{\left( 4+{{x}^{2}} \right)}^{2}}}$, and deduce that $y0$.

Hence show that if $\mathrm{f}\left( x \right)=\frac{1}{x}{{\tan }^{-1}}\frac{x}{2}$, where $x>0$, then $\mathrm{f}$ is a decreasing function.

Functions

International Baccalaureate Higher Level Mathematics

2019 DHS J2 CT1 Q2

2019 DHS J2 CT1 Q2

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