<p>A curve is such that&nbsp;\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{9}{(2x-1)^2}\), where&nbsp;\(x \neq \frac{1}{2}\). The point \((2, 12)\) lies on the curve.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Find the equation of the curve.<span class="inline-marks">[3]</span></div></li><li><div class="list-item-paragraph">Find the value of \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\) when&nbsp;\(x = 0\).<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">A student concludes from <strong>(ii)</strong> that since the value of \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} &gt; 0\) at \(x = 0\), hence there is a minimum point at \(x = 0\). State with reason and mathematical working whether the above conclusion is correct.<span class="inline-marks">[2]</span></div></li></ol>

A curve is such that \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{9}{(2x-1)^2}\), where \(x \neq \frac{1}{2}\). The point \((2, 12)\) lies on the curve.

Find the value of \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\) when \(x = 0\).

A student concludes from **(ii)** that since the value of \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} > 0\) at \(x = 0\), hence there is a minimum point at \(x = 0\). State with reason and mathematical working whether the above conclusion is correct.

AMKSS 4AM 2022 Prelims P1 Q12

AMKSS 4AM 2022 Prelims P1 Q12