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<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">The diagram shows the graph of ${y = e^{2x} - kx}$, where ${k}$ is a positive real number.</span></div><div class="list-item-paragraph">The two roots of the equation ${e^{2x} - kx = 0}$ are denoted by ${\alpha}$ and ${\beta}$, where ${\alpha {{x}_{n}}}$ of ${{x}_{n}}\beta $.</div><div class="list-item-paragraph">It is given that there is a sequence of real numbers ${{x}_{1}}$ , ${{x}_{2}}$, ${{x}_{3}}$…  that satisfies the recurrence relation,</div><div class="list-item-paragraph" style="text-align: center">${{{x}_{n+1}}=\frac{1}{k}{{\mathrm{e}}^{2{{x}_{n}}}}}$, for ${n\ge 1}$.</div><div class="list-item-paragraph">By considering ${x_{n+1} - x_n}$, prove that ${{{x}_{n+1}}>{{x}_{n}}}$ of ${{x}_{n}}<\alpha $ or ${{x}_{n}}>\beta $.<span class="inline-marks">[2]</span></div></li></ol>

The diagram shows the graph of ${y = e^{2x} - kx}$, where ${k}$ is a positive real number.

The two roots of the equation ${e^{2x} - kx = 0}$ are denoted by ${\alpha}$ and ${\beta}$, where ${\alpha {{x}_{n}}}$ of ${{x}_{n}}\beta $.

It is given that there is a sequence of real numbers ${{x}_{1}}$ , ${{x}_{2}}$, ${{x}_{3}}$…  that satisfies the recurrence relation,

${{{x}_{n+1}}=\frac{1}{k}{{\mathrm{e}}^{2{{x}_{n}}}}}$, for ${n\ge 1}$.

By considering ${x_{n+1} - x_n}$, prove that ${{{x}_{n+1}}>{{x}_{n}}}$ of ${{x}_{n}}<\alpha $ or ${{x}_{n}}>\beta $.

2016 SRJC P1 Q3

2016 SRJC P1 Q3

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