A Math

A Math (IGCSE)

E Math

E Math (IGCSE)

H1 Math

H2 Math

IB AA HL Math

IB AA SL Math

IB AI HL Math

IB AI SL Math

IP Math

N(A) A Math

N(A) E Math

PSLE

PSLE Foundation Mathematics

<p>It is given that&nbsp;$y=\mathrm{f}(x)$ is such that \[\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{\mathrm{d}y}{\mathrm{d}x}\left(2\tan x+p\sec^2x\right),\] where&nbsp;$p$ is a constant. The Maclaurin series for&nbsp;$\mathrm{f}(x)$ is given by \[y=1+x+\frac12x^2+qx^3+\cdots,\] where&nbsp;$q$ is a constant.</p>
<ol class="abc-list"><li data-list-item-id="tjc-q05-item01" data-list-item-path="tjc-q05-item01"><div class="list-item-paragraph">State the values of $\mathrm{f}(0)$ and $\mathrm{f}'(0)$.<span class="inline-marks">[1]</span></div></li><li data-list-item-id="tjc-q05-item02" data-list-item-path="tjc-q05-item02"><div class="list-item-paragraph">Find the values of $p$ and $q$.<span class="inline-marks">[5]</span></div></li><li data-list-item-id="tjc-q05-item03" data-list-item-path="tjc-q05-item03"><div class="list-item-paragraph">Using the given series for&nbsp;$\mathrm{f}(x)$ and standard series from the List of Formulae (MF27), find the first three non-zero terms of the series expansion of $\sqrt{\mathrm{f}(x)}$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="tjc-q05-item04" data-list-item-path="tjc-q05-item04"><div class="list-item-paragraph">The given Maclaurin series for&nbsp;$\mathrm{f}(x)$ is used to estimate $\displaystyle \int_0^{0.2}\mathrm{f}(x)\,\mathrm{d}x$ and $\displaystyle \int_{0.2}^{0.4}\mathrm{f}(x)\,\mathrm{d}x$. Without any further computations, determine which estimate has a smaller percentage error. Give an explanation for your choice.<span class="inline-marks">[1]</span></div></li></ol>

It is given that $y=\mathrm{f}(x)$ is such that \[\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{\mathrm{d}y}{\mathrm{d}x}\left(2\tan x+p\sec^2x\right),\] where $p$ is a constant. The Maclaurin series for $\mathrm{f}(x)$ is given by \[y=1+x+\frac12x^2+qx^3+\cdots,\] where $q$ is a constant.

State the values of $\mathrm{f}(0)$ and $\mathrm{f}'(0)$.

Using the given series for $\mathrm{f}(x)$ and standard series from the List of Formulae (MF27), find the first three non-zero terms of the series expansion of $\sqrt{\mathrm{f}(x)}$.

The given Maclaurin series for $\mathrm{f}(x)$ is used to estimate $\displaystyle \int_0^{0.2}\mathrm{f}(x)\,\mathrm{d}x$ and $\displaystyle \int_{0.2}^{0.4}\mathrm{f}(x)\,\mathrm{d}x$. Without any further computations, determine which estimate has a smaller percentage error. Give an explanation for your choice.

Finding Parameters from Power Series Coefficients

Maclaurin Series from Differential Equations

Substitution and Composition of Power Series

Derivatives of Reciprocal Trigonometric Functions

Accuracy of Power Series Approximations

Higher Derivatives by Implicit Differentiation

2026 TJC P1 Q5

Maclaurin Series and Power Series

Techniques of Differentiation

<p>It is given that&nbsp;$y = \sin^{-1} px$, where&nbsp;$p$ is a constant.</p>
<ol class="abc-list"><li data-list-item-id="imported-item-block-1-item-0" data-list-item-path="imported-item-block-1-item-0"><div class="list-item-paragraph">Show that&nbsp;$\frac{\mathrm{d}y}{\mathrm{d}x} = p \sec y$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="imported-item-block-1-item-1" data-list-item-path="imported-item-block-1-item-1"><div class="list-item-paragraph"><p>By further differentiation of the result in part <strong>(a)</strong> find, in terms of $p$, the Maclaurin series for&nbsp;$y$ up to and including the term in $x^3$.</p><span class="inline-marks">[5]</span></div></li><li data-list-item-id="imported-item-block-1-item-2" data-list-item-path="imported-item-block-1-item-2"><div class="list-item-paragraph"><p>Use your answer to part <strong>(b)</strong> to find the series expansion of $\frac{1}{\sqrt{1 - 9x^2}}$, up to and including the term in $x^2$.</p><span class="inline-marks">[3]</span></div></li></ol>

It is given that $y = \sin^{-1} px$, where $p$ is a constant.

Show that $\frac{\mathrm{d}y}{\mathrm{d}x} = p \sec y$.

Derivatives of Inverse Trigonometric Functions

Maclaurin Series by Successive Differentiation

Problems involving Calculus

Deriving Series by Differentiation and Integration

N2025 P1 Q9

<p>Mrs Koh wants to hang a decoration on the vertical wall above her bookshelf. She needs a ladder to climb up.</p>
<p>The rectangle $ABCD$ is the side-view of the bookshelf and $HK$ is the side-view of the ladder where&nbsp;$AB = 24\text{ cm}$ and $BC = 192\text{ cm}$ (see <strong>Figure 1</strong>). The ladder touches the wall at $H$, the edge of the top of the bookshelf at $B$ and the floor at $K$.</p>
<div class="graph-block" data-graph-config="eyJheGlzIjp0cnVlLCJncmlkIjpmYWxzZSwid2lkdGgiOjMwMCwiaGVpZ2h0IjoyNTAsImVsZW1lbnRzIjpbeyJpZCI6IjFmMjkwODUxLTY3N2ItNDUzYS05ZDc2LTFhYjlmMTA1ODcyMSIsInR5cGUiOiJzZWdtZW50IiwiY29sb3IiOiIjMDAwMDAwIiwicG9pbnQxIjpbLTMuMiwtMi44XSwicG9pbnQyIjpbLTMuMTMsMi41NV0sInN0cm9rZVdpZHRoIjoxLjV9LHsiaWQiOiI5MjAyNjViOC1iM2UwLTRkZTUtODcxOS0zZWRmN2E5MTdlYmMiLCJ0eXBlIjoic2VnbWVudCIsImNvbG9yIjoiIzAwMDAwMCIsInBvaW50MSI6Wy0zLjIsLTIuOF0sInBvaW50MiI6WzMuMiwtMi44XSwic3Ryb2tlV2lkdGgiOjEuNX0seyJpZCI6ImI1Y2FkZWE1LTBmNzYtNDVlMC1iNWVjLWE5OGY5MzNlMjc3NiIsInR5cGUiOiJzZWdtZW50IiwiY29sb3IiOiIjMDAwMDAwIiwicG9pbnQxIjpbLTMuMTUsMC44XSwicG9pbnQyIjpbLTIsMC44XSwic3Ryb2tlV2lkdGgiOjEuNX0seyJpZCI6ImIxNjhlMTI2LWI0OGItNDBkZC1hZWI3LTc1NTYxYThiYWIzMCIsInR5cGUiOiJzZWdtZW50IiwiY29sb3IiOiIjMDAwMDAwIiwicG9pbnQxIjpbLTIsLTIuOF0sInBvaW50MiI6Wy0yLDAuOF0sInN0cm9rZVdpZHRoIjoxLjV9LHsiaWQiOiIxOTM0Njk4ZC0yNDRlLTRkZmQtOWQwYy05OWFkMzM4Y2E1MDMiLCJ0byI6LTIsImZyb20iOi0zLjE1LCJ0eXBlIjoiaW50ZWdyYWwiLCJ4TWF4IjotMiwieE1pbiI6LTMuMTUsImNvbG9yIjoiIzAwMDAwMCIsInhNYXgyIjozLjIsInhNaW4yIjotMy4yLCJmaWxsQ29sb3IiOiIjZjlmOGYxIiwiZXhwcmVzc2lvbiI6IjAgKiB4ICsgMC44IiwiZXhwcmVzc2lvbjIiOiIwICogeCAtIDIuOCIsImZpbGxPcGFjaXR5IjowLjQsInN0cm9rZVdpZHRoIjoxLjV9LHsiaWQiOiI3ZDQ1ODIzMS03YTA3LTQ2NWQtYWUwOC1mYmZhM2Y1YTNmYzQiLCJ0eXBlIjoic2VnbWVudCIsImNvbG9yIjoiIzAwMDAwMCIsInBvaW50MSI6WzEuNiwtMi44XSwicG9pbnQyIjpbLTMuMTQsMl0sInN0cm9rZVdpZHRoIjozfSx7ImlkIjoiMTY3YzI4ODEtMDdmYy00NTQ4LWI1NzgtZDA3ZTE5ODc1NzU3IiwidHlwZSI6ImFyYyIsImNvbG9yIjoiIzAwMDAwMCIsImNlbnRlciI6WzEuNiwtMi44XSwicG9pbnQxIjpbMC45LC0yLjA5XSwicG9pbnQyIjpbLTIuOCwtMi44XSwic3Ryb2tlV2lkdGgiOjEuNX0seyJ4IjotMy41NiwieSI6Mi4wNCwiaWQiOiI4ZTk1NmFiNy1hYjZmLTRmNTMtYTc0NS1jNTdiNTk2ZjJlY2EiLCJ0eXBlIjoidGV4dCIsImNvbG9yIjoiIzAwMDAwMCIsImxhYmVsIjoiXFwoSFxcKSJ9LHsieCI6LTMuNjIsInkiOjAuOCwiaWQiOiI1NzdkMDg4MC03Y2YyLTRhMjItOTExMy02NWQ5NGNhMTdkY2IiLCJ0eXBlIjoidGV4dCIsImNvbG9yIjoiIzAwMDAwMCIsImxhYmVsIjoiXFwoQVxcKSJ9LHsieCI6LTMuNjIsInkiOi0zLjExLCJpZCI6IjQxMTcxNGRjLTRhODAtNGI2Yi04NmExLWMyNGE1MmRmMDIzZCIsInR5cGUiOiJ0ZXh0IiwiY29sb3IiOiIjMDAwMDAwIiwibGFiZWwiOiJcXChEXFwpIn0seyJ4IjotMS42MywieSI6MC44NCwiaWQiOiI0YzE1ODQ1Mi05N2QwLTRlOWYtOGI0NC01ODFiYmNkMjFlNWIiLCJ0eXBlIjoidGV4dCIsImNvbG9yIjoiIzAwMDAwMCIsImxhYmVsIjoiXFwoQlxcKSJ9LHsieCI6LTIsInkiOi0zLjExLCJpZCI6IjFiYmQ0NGQ5LTk3OGItNGZkMC04Zjc0LTcxN2RkMjQ1Mzg0OCIsInR5cGUiOiJ0ZXh0IiwiY29sb3IiOiIjMDAwMDAwIiwibGFiZWwiOiJcXChDXFwpIn0seyJ4IjoxLjYsInkiOi0zLjExLCJpZCI6IjRiMWQwZmQ2LTEwYzgtNDc1Mi04NzFhLTNmNzY3YjcwYzM2MiIsInR5cGUiOiJ0ZXh0IiwiY29sb3IiOiIjMDAwMDAwIiwibGFiZWwiOiJcXChLXFwpIn0seyJ4IjowLjI4LCJ5IjotMi4zLCJpZCI6IjI4MWM1ZGQzLTg2YTctNDI0YS05MGY3LWNmMjE4YTI5ZmE0MyIsInR5cGUiOiJ0ZXh0IiwiY29sb3IiOiIjMDAwMDAwIiwibGFiZWwiOiJcXChcXHRoZXRhXFwpIn0seyJ4IjotMi42LCJ5IjotMS44LCJpZCI6ImE4MmJkNGIwLWYwNmQtNDdmNi04MjBhLTc0NmU4NDgwZDlmYSIsInR5cGUiOiJ0ZXh0IiwiY29sb3IiOiIjMDAwMDAwIiwibGFiZWwiOiJCb29rc2hlbGYiLCJ0ZXh0Um90YXRpb24iOjkwfV0sInNob3dYQXhpcyI6ZmFsc2UsInNob3dZQXhpcyI6ZmFsc2UsInhBeGlzTGFiZWwiOiJcXCh4XFwpIiwieUF4aXNMYWJlbCI6IlxcKHlcXCkiLCJib3VuZGluZ0JveCI6Wy01LDUsNSwtNV0sIm9yaWdpbkxhYmVsIjoiXFwoT1xcKSIsImdyaWRTZXR0aW5ncyI6eyJtYWpvckNvbG9yIjoiIzk0YTNiOCIsIm1ham9yU3RlcFgiOjEsIm1ham9yU3RlcFkiOjEsIm1pbm9yQ29sb3IiOiIjY2JkNWUxIiwibWFqb3JPcGFjaXR5IjowLjU1LCJtaW5vck9wYWNpdHkiOjAuMzUsIm1pbm9yRWxlbWVudHNYIjo0LCJtaW5vckVsZW1lbnRzWSI6NCwibWFqb3JTdHJva2VXaWR0aCI6MC44LCJtaW5vclN0cm9rZVdpZHRoIjowLjV9LCJzaG93QXhpc0xhYmVscyI6ZmFsc2UsInNob3dOYXZpZ2F0aW9uIjpmYWxzZSwia2VlcEFzcGVjdFJhdGlvIjp0cnVlLCJ4QXhpc0xhYmVsT2Zmc2V0IjpbMiwtMTFdLCJ5QXhpc0xhYmVsT2Zmc2V0IjpbLTE0LDVdLCJvcmlnaW5MYWJlbE9mZnNldCI6Wy01LC0yXX0=" data-width="300" data-height="250" style="width:300px;height:250px;max-width:100%;display:block;margin:0 auto;"></div>
<ol class="abc-list"><li><div class="list-item-paragraph">Given that&nbsp;$\angle HKD = \theta$, show that the length, $L\text{ cm}$ of the ladder is given by \[L = \frac{24}{\cos \theta} + \frac{192}{\sin \theta}.\]<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">Use differentiation to find the exact value of the shortest length of the ladder as&nbsp;$\theta$ varies.<br>[You do not need to verify that this length of the ladder is the shortest.]<span class="inline-marks">[5]</span></div></li></ol>

Mrs Koh wants to hang a decoration on the vertical wall above her bookshelf. She needs a ladder to climb up.

The rectangle $ABCD$ is the side-view of the bookshelf and $HK$ is the side-view of the ladder where $AB = 24\text{ cm}$ and $BC = 192\text{ cm}$ (see **Figure 1**). The ladder touches the wall at $H$, the edge of the top of the bookshelf at $B$ and the floor at $K$.

Problems Involving Maximisation and Minimisation

Applications of Differentiation

<ol class="abc-list"><li data-list-item-id="dss-p2-q4-q-0-content-items-0" data-list-item-path="dss-p2-q4-q-0-content-items-0"><div class="list-item-paragraph">Show that&nbsp;$\dfrac{d}{dx}(\cot x)=-\cosec^2 x$.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="dss-p2-q4-q-0-content-items-1" data-list-item-path="dss-p2-q4-q-0-content-items-1"><div class="list-item-paragraph">Hence find the value of $\displaystyle \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} -\cot^2 x\, dx$.<span class="inline-marks">[4]</span></div></li></ol>

Show that $\dfrac{d}{dx}(\cot x)=-\cosec^2 x$.

Hence find the value of $\displaystyle \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} -\cot^2 x\, dx$.

Definite Integration as the Reverse of Differentiation

Integrable Forms of Trigonometric Functions

2025 DSS PRELIMS P2 Q4

Integration

Differentiation

<p>Differentiate with respect to $x$ of the following expressions leaving your answers in terms of $x$:</p>
<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">$\frac{\sec (\ln x)}{x}$</div></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">$\frac{e^{-2x}}{5^x}$</div></li><li data-list-item-id="item-c" data-list-item-path="item-c"><div class="list-item-paragraph">$\ln \sqrt{e^{1+\sec x}}$</div></li><li data-list-item-id="item-d" data-list-item-path="item-d"><div class="list-item-paragraph">$x^\pi$</div></li><li data-list-item-id="item-e" data-list-item-path="item-e"><div class="list-item-paragraph">$\pi^x$</div></li><li data-list-item-id="item-f" data-list-item-path="item-f"><div class="list-item-paragraph">$2^{\ln x}$</div></li><li data-list-item-id="item-g" data-list-item-path="item-g"><div class="list-item-paragraph">$5^{\cos x}$</div></li><li data-list-item-id="item-h" data-list-item-path="item-h"><div class="list-item-paragraph">$x^x$</div></li><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">$(\ln x)^x$</div></li></ol>

Differentiate with respect to $x$ of the following expressions leaving your answers in terms of $x$:

Derivatives of Trigonometric Functions

Derivatives of Natural Logarithmic Functions

Logarithmic Differentiation

Quotient Rule

Derivatives of Exponential Functions

Power Rule and Derivatives of Algebraic Functions

Chain Rule

<ol class="abc-list"><li data-list-item-id="item-1772432944155-jtc5x5fqi" data-list-item-path="item-1772432944155-jtc5x5fqi"><div class="list-item-paragraph">Show that&nbsp;$\frac{\mathrm{d}}{\mathrm{d}x}\left(\cot{\frac{x}{2}}\right)=-\frac{1}{2}\textbf{cosec}^2\frac{x}{2}$</div></li><li data-list-item-id="item-1772433178033-egd5ipjmr" data-list-item-path="item-1772433178033-egd5ipjmr"><div class="list-item-paragraph">The diagram shows an isosceles triangle $ABC$ with $AC = BC$ and $AB = 8$ cm. The vertex $C$ is moving towards $AB$ in the direction perpendicular to $AB$ at a constant speed of $1.5$ cm/s. Let angle $ACB = \theta$. Find the rate of change of $\theta$ when triangle $ABC$ is equilateral.</div><div class="graph-block" data-graph-config="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" data-width="283" data-height="176" data-proportional style="width:283px;height:176px;max-width:100%;display:flow-root;margin:0 auto;--graph-width:283px;--graph-aspect:283 / 176;"></div></li></ol>

Connected Rates of Change

<p>Find the derivative of each of the following with respect to $x$.</p>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">$y = 4 \cos x \sin 3x$</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">$y = \sin^3(2x^2 - \mathrm{e})$</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">$y = \mathrm{cosec}^3(\pi x^4 + 1) - 3x^5$</div></li><li data-list-item-id="item-iv" data-list-item-path="item-iv"><div class="list-item-paragraph">$y = \sin \left( \mathrm{e}x + \frac{\pi}{4} \right) + \cos(2x^3 - 5x)$</div></li></ol>

Find the derivative of each of the following with respect to $x$.

$y = \mathrm{cosec}^3(\pi x^4 + 1) - 3x^5$

$y = \sin \left( \mathrm{e}x + \frac{\pi}{4} \right) + \cos(2x^3 - 5x)$

Product Rule

<p>Given that&nbsp;$x=\ln \left( \cot \frac{1}{2}\theta  \right)-\cos \theta $, $y=-\sin \theta $ , show that&nbsp;$\frac{\mathrm{d}y}{\mathrm{d}x}=\tan \theta $ .</p>

Given that $x=\ln \left( \cot \frac{1}{2}\theta  \right)-\cos \theta $, $y=-\sin \theta $ , show that $\frac{\mathrm{d}y}{\mathrm{d}x}=\tan \theta $ .

Differentiation of Parametric Equations

<p>Given the following relations between $x$ and $y$, find $\frac{\mathrm{d}y}{\mathrm{d}x}$ in terms of $x$ and $y$.</p>
<ol class="abc-list"><li><div class="list-item-paragraph">${{y}^{x}}=y+x$ </div></li><li><div class="list-item-paragraph">$y={{x}^{\text{sec}\hspace{0.5em}x}}$ </div></li><li><div class="list-item-paragraph">$y={{x}^{{{x}^{2}}}}$ </div></li></ol>

Given the following relations between $x$ and $y$, find $\frac{\mathrm{d}y}{\mathrm{d}x}$ in terms of $x$ and $y$.

Implicit Differentiation

<p>Find the derivative of each of the following:</p>
<ol class="abc-list"><li><div class="list-item-paragraph">$y = \sin^3(ex^2-\pi)$</div></li><li><div class="list-item-paragraph">$y = \mathrm{cosec}^3(\pi x^4+1)$</div></li><li><div class="list-item-paragraph">$\mathrm{f}(x) = 4x^2 \tan x$</div></li></ol>

Find the derivative of each of the following:

<p>Find the derivative of each of the following:</p>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">$y = \sin^3(2x^2 - \pi)$</span></div></li><li><div class="list-item-content"><span class="list-item-text">$y = \mathrm{cosec}^3(\pi x^4 + 1) - 3x^5$</span></div></li><li><div class="list-item-content"><span class="list-item-text">$\mathrm{f}(x) = \sin\left(ex + \frac{\pi}{4}\right) + \cos(2x^3 - 5x)$</span></div></li></ol>

$\mathrm{f}(x) = \sin\left(ex + \frac{\pi}{4}\right) + \cos(2x^3 - 5x)$

<p>Find $\frac{\mathrm{d}}{\mathrm{d}x}(\sec x)$ and hence find $\displaystyle\displaystyle \displaystyle\int \tan x \sec x \, \mathrm{d}x.$</p>

Find $\frac{\mathrm{d}}{\mathrm{d}x}(\sec x)$ and hence find $\displaystyle\displaystyle \displaystyle\int \tan x \sec x \, \mathrm{d}x.$

Integration as an Inverse Process of Differentiation

<p>A curve has parametric equations given by $x = 2\ln(\sec \theta)$ and $y = 2\theta - \tan \theta$.<br>Show that&nbsp;$\frac{\mathrm{d}y}{\mathrm{d}x} = \cot 2\theta$.</p>

A curve has parametric equations given by $x = 2\ln(\sec \theta)$ and $y = 2\theta - \tan \theta$.
Show that $\frac{\mathrm{d}y}{\mathrm{d}x} = \cot 2\theta$.

<p>Find the derivative of the following functions with respect to $x$:</p>
<ol class="i-list"><li><div class="list-item-paragraph">$\mathrm{f}(x) = e^{\csc x} \cdot 2^{1 - 3x^{2}}$</div></li><li><div class="list-item-paragraph">$y = 3^{\csc 3x}$</div></li><li><div class="list-item-paragraph">$y = \frac{\sin^{2}x}{5^{\cos x}}$</div></li></ol>

Find the derivative of the following functions with respect to $x$:

<p>The parametric equations of the curve $C$ are</p>
<p style="text-align: center">$x=\tan \theta -1$, $y=2\sec \theta $, where&nbsp;$-\frac{\pi }{2}&lt;\theta &lt;\frac{\pi }{2}$.</p>
<ol class="abc-list"><li data-list-item-id="item-1760961275279-njpvri7j9" data-list-item-path="item-1760961275279-njpvri7j9"><div class="list-item-paragraph">Find $\frac{\mathrm{d}y}{\mathrm{d}x}$ in terms of $\theta $, leaving your answer in a single trigonometric function.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-1760961275936-uic3edge1" data-list-item-path="item-1760961275936-uic3edge1"><div class="list-item-paragraph">Find the equation of the tangent to $C$ at the point where it cuts the $y$-axis, leaving your answer in exact form.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-1760961276913-61m5k7kii" data-list-item-path="item-1760961276913-61m5k7kii"><div class="list-item-paragraph">Find the Cartesian equation of $C$.<span class="inline-marks">[1]</span></div></li><li data-list-item-id="item-1760961277949-yh1yw236v" data-list-item-path="item-1760961277949-yh1yw236v"><div class="list-item-paragraph">Find the equations of the asymptotes and state the coordinates of any turning point(s).<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-1760961279831-43bl3h9nr" data-list-item-path="item-1760961279831-43bl3h9nr"><div class="list-item-paragraph">The region bounded by the tangent found in part <strong>(b)</strong>, the curve $C$ and the line $x=-1$ is rotated completely about the $x$-axis. Calculate the volume obtained.<span class="inline-marks">[3]</span></div></li></ol>

The parametric equations of the curve $C$ are

$x=\tan \theta -1$, $y=2\sec \theta $, where $-\frac{\pi }{2}<\theta <\frac{\pi }{2}$.

Tangents and Normals to Parametric Curves

Volume of Revolution about the x-axis

Asymptotes and Discontinuities

Graphs of Conic Sections

Converting Parametric Equations to Cartesian Form

2024 HCI Promo Q12

Graphing Techniques

Applications of Integration

<p>It is given that&nbsp;$y=\tan 3x$.</p>
<ol class="abc-list"><li data-list-item-id="item-1760850665985-g2h1fy0me" data-list-item-path="item-1760850665985-g2h1fy0me"><div class="list-item-paragraph">Show that&nbsp;$\frac{{{\mathrm{d}}^{2}}y}{\mathrm{d}{{x}^{2}}}=Ay\frac{\mathrm{d}y}{\mathrm{d}x}$, where&nbsp;$A$ is a constant to be determined.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-1760850666517-ggj0teyxq" data-list-item-path="item-1760850666517-ggj0teyxq"><div class="list-item-paragraph">Hence show that&nbsp;$\frac{{{\mathrm{d}}^{4}}y}{\mathrm{d}{{x}^{4}}}=B\frac{\mathrm{d}y}{\mathrm{d}x}\frac{{{\mathrm{d}}^{2}}y}{\mathrm{d}{{x}^{2}}}+Cy\frac{{{\mathrm{d}}^{3}}y}{\mathrm{d}{{x}^{3}}}$, where&nbsp;$B$ and $C$ are constants to be determined.<span class="inline-marks">[3]</span></div></li></ol>

Second and Higher Derivatives

2024 ASRJC Promo Q2

<p>Differentiate the following expressions with respect to $x$.</p>
<ol class="abc-list"><li><div class="list-item-paragraph">$\frac{4{{x}^{3}}+2{{x}^{2}}+1}{x}$</div></li><li><div class="list-item-paragraph">$\mathrm{cose}{{\mathrm{c}}^{3}}\sqrt{x}$</div></li><li><div class="list-item-paragraph">${{\cos }^{2}}\left( x{}^\circ \right)+\tan \left( x{}^\circ \right)$</div></li><li><div class="list-item-paragraph">$\ln \sqrt{\frac{x-a}{x+a}}$ , where $0 < a < x$</div></li><li><div class="list-item-paragraph">$\lg \left( {{x}^{2}} \right)$ , where $x<0$</div></li><li><div class="list-item-paragraph">${{e}^{{{\sin }^{2}}x}}\left( {{e}^{-{{x}^{2}}}}+1 \right)$</div></li></ol>

Differentiate the following expressions with respect to $x$.

Derivatives of Logarithmic Functions (Non-Natural Base)

Product, Quotient and Chain Rules

Basics of Differentiation Techniques Q1

<p>An architect designs and constructs a hump such that when a square wheel of dimensions 2 cm by 2 cm is rolled over the hump, the height of its centre above the horizontal ground (represented by the $x$-axis) is always constant. To test his invention, he places a square wheel against the hump, as shown in the diagram below. The points $R$, $S$ and $Q$ represent the centre of the square, the vertex of the square that is initially on the hump (at the origin $O$), and another vertex of the square which has a common side as $S$, respectively.</p>
<figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/converted/1760256329216-hwqcu63lcde.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvY29udmVydGVkLzE3NjAyNTYzMjkyMTYtaHdxY3U2M2xjZGUucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5Mjc4OTM1NiwiZXhwIjoxNzkyNzg5NjU2fQ.7wi0aDwQiNwqn6vyqPpVFQqLfc-o8OxU2a1qKTjcEP4" width="400" class="aligncenter" /></figure>
<p>The architect begins pushing the square wheel over the hump in the clockwise direction. He discovers that the centre of the square is always vertically above the variable point of contact between the square and the hump, $P$, as shown in the diagram below. It is further given that the angle $\theta$ is the angle between the side $QS$ of the square and the horizontal line passing through $S$ , measured in the positive $x$-direction. ($\theta$ is taken to be positive if the vertex $Q$ lies above the horizontal line passing through $S$, and negative if otherwise.)</p>
<figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/converted/1760256344668-ewta89bvx9r.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvY29udmVydGVkLzE3NjAyNTYzNDQ2NjgtZXd0YTg5YnZ4OXIucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5Mjc4OTM1NiwiZXhwIjoxNzkyNzg5NjU2fQ.Xi00J_X0TaJwlvtz3RuWLxUa469IikZLPRd3Bezdmxg" width="400" class="aligncenter" /></figure>
<ol class="i-list"><li><div class="list-item-paragraph">By considering the length of $RP$ or otherwise, show that the $y$-coordinate of $P$ can be expressed as $\sqrt{2}-\sec \theta $, for $-\frac{\pi }{4}\le \theta \le \frac{\pi }{4}$.</div><p style="text-align: right;">[3]</p></li><li><div class="list-item-paragraph">State the gradient of $QS$ in terms of $\theta$ for  $-\frac{\pi }{4}\le \theta \le \frac{\pi }{4}$. Deduce that $\frac{\mathrm{d}x}{\mathrm{d}\theta }=-\sec \theta $, where $x$ is the $x$-coordinate of $P$.</div><p style="text-align: right;">[3]</p></li><li><div class="list-item-paragraph">Find the area bounded between the hump and the $x$-axes, correct to 4 decimal places.</div><p style="text-align: right;">[3]</p></li></ol>

An architect designs and constructs a hump such that when a square wheel of dimensions 2 cm by 2 cm is rolled over the hump, the height of its centre above the horizontal ground (represented by the $x$-axis) is always constant. To test his invention, he places a square wheel against the hump, as shown in the diagram below. The points $R$, $S$ and $Q$ represent the centre of the square, the vertex of the square that is initially on the hump (at the origin $O$), and another vertex of the square which has a common side as $S$, respectively.

The architect begins pushing the square wheel over the hump in the clockwise direction. He discovers that the centre of the square is always vertically above the variable point of contact between the square and the hump, $P$, as shown in the diagram below. It is further given that the angle $\theta$ is the angle between the side $QS$ of the square and the horizontal line passing through $S$ , measured in the positive $x$-direction. ($\theta$ is taken to be positive if the vertex $Q$ lies above the horizontal line passing through $S$, and negative if otherwise.)

Area Bounded by Curves

Parametric Equations

2017 NJC PROMO Q12

<p>A curve $C$ has parametric equations $x=\ln (\sec \theta )$,$y=\ln (\sin \theta )$, $0 < \theta < \frac{\pi }{2}$.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Sketch the curve $C$, stating the equation of any asymptotes.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">Find the equation of the tangent to the curve at the point where $\theta =\frac{\pi }{4}$, leaving your answer in the form of $y=ax+b$ where $a$ and $b$ are exact values.</div><p style="text-align: right;">[5]</p></li><li><div class="list-item-paragraph">$P$ is a point on the curve $C$ which moves along $C$ at a rate of $\frac{\mathrm{d}\theta }{\mathrm{d}t}=2$ units/s. The point $B$ has coordinates $(1,0)$ and $OBP$forms a triangle, where $O$ is the origin. Find the rate of change of the area of triangle $OBP$ when $\theta =\frac{\pi }{3}$.</div><p style="text-align: right;">[4]</p></li></ol>

A curve $C$ has parametric equations $x=\ln (\sec \theta )$,$y=\ln (\sin \theta )$, $0 < \theta < \frac{\pi }{2}$.

Sketching Parametric Curves

2020 SAJC Promo Q9

<p>A curve $C$ has parametric equations $x=\cos 2\theta $ and $y=\cos \mathrm{ec }\theta $, where $0 < \theta \le \frac{\pi }{2}$.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Show that $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{4}\cos \mathrm{e}{{\mathrm{c}}^{3}}\theta $. What can be said about $\frac{\mathrm{d}y}{\mathrm{d}x}$ as $\theta \to 0$?</div><p style="text-align: right;">[3]</p></li><li><div class="list-item-paragraph">Sketch $C$, stating the equation of the asymptote and coordinates of the point of intersection with the axis.</div><p style="text-align: right;">[3]</p></li><li><div class="list-item-paragraph">Find the equation of the normal to $C$ at the point where $\theta =\frac{\pi }{6}$.</div><p style="text-align: right;">[3]</p></li><li><div class="list-item-paragraph">Show that the point(s) of intersection of the normal found in part <strong>(iii)</strong> and the curve $C$ satisfy the equation ${{\sin }^{3}}\theta +\frac{7}{4}\sin \theta -1=0$. Hence using a non-calculator method, show that this normal does not meet the curve again.</div><p style="text-align: right;">[4]</p></li></ol>

A curve $C$ has parametric equations $x=\cos 2\theta $ and $y=\cos \mathrm{ec }\theta $, where $0 < \theta \le \frac{\pi }{2}$.

Math Question Bank