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><strong>Do not use a calculator in answering this question.</strong></p>
<ol class="abc-list"><li data-list-item-id="ri-p2-q4-2-content-items-1" data-list-item-path="ri-p2-q4-2-content-items-1"><div class="list-item-paragraph">The curve $y=\mathrm{f}(x)$ passes through the origin and has gradient given by</div><div class="list-item-paragraph">\[\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{x+y}{1+xy}.\]</div><ol class="i-list"><li data-list-item-id="ri-p2-q4-2-content-items-1-subItems-1" data-list-item-path="ri-p2-q4-2-content-items-1/ri-p2-q4-2-content-items-1-subItems-1"><div class="list-item-paragraph">Show that&nbsp;$\displaystyle(1+xy)\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}+x\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2+(y-1)\dfrac{\mathrm{d}y}{\mathrm{d}x}=1$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="ri-p2-q4-2-content-items-1-subItems-2" data-list-item-path="ri-p2-q4-2-content-items-1/ri-p2-q4-2-content-items-1-subItems-2"><div class="list-item-paragraph">By further differentiation of the result in part <strong>(a)(i)</strong>, show that</div><div class="list-item-paragraph">\[(1+xy)\frac{\mathrm{d}^3y}{\mathrm{d}x^3}+\mathrm{p}(x)\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)\left(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\right)+\mathrm{q}(y)\frac{\mathrm{d}^2y}{\mathrm{d}x^2}+r\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2=0,\]</div><div class="list-item-paragraph">where the functions $\mathrm{p}(x)$, $\mathrm{q}(y)$ and the constant $r$ are to be determined.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="ri-p2-q4-2-content-items-1-subItems-3" data-list-item-path="ri-p2-q4-2-content-items-1/ri-p2-q4-2-content-items-1-subItems-3"><div class="list-item-paragraph">Hence find the Maclaurin series for&nbsp;$y$ up to and including the term in $x^3$.<span class="inline-marks">[2]</span></div></li></ol></li><li data-list-item-id="ri-p2-q4-2-content-items-2" data-list-item-path="ri-p2-q4-2-content-items-2"><div class="list-item-paragraph">The equation</div><div class="list-item-paragraph">\[2\sqrt{2}\cos\left(x-\frac\pi4\right)+3\sin2x-3=0\]</div><div class="list-item-paragraph">has a root $\theta$, which is close to zero.</div><ol class="i-list"><li data-list-item-id="ri-p2-q4-2-content-items-2-subItems-1" data-list-item-path="ri-p2-q4-2-content-items-2/ri-p2-q4-2-content-items-2-subItems-1"><div class="list-item-paragraph">Show that&nbsp;$\theta^2+a\theta+b\approx0$, where&nbsp;$a$ and $b$ are integers to be determined.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="ri-p2-q4-2-content-items-2-subItems-2" data-list-item-path="ri-p2-q4-2-content-items-2/ri-p2-q4-2-content-items-2-subItems-2"><div class="list-item-paragraph">Hence find an approximation for&nbsp;$\theta$ in surd form.<span class="inline-marks">[2]</span></div></li></ol></li></ol>

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

Small Angle Approximations

Approximations Using Power Series

Maclaurin Series from Differential Equations

Higher Derivatives by Implicit Differentiation

2026 RI P2 Q4

Techniques of Differentiation

Maclaurin Series and Power Series

<p>It is given that&nbsp;$\mathrm{f}(\theta) = a\theta^2 + b\theta + c$, where&nbsp;$a$, $b$ and $c$ are real constants. The curve with equation $y = \mathrm{f}(\theta)$ passes through the point with coordinates $(0, 1)$ and has a turning point at $\left( -\frac{\sqrt{3}}{3}, \frac{1}{2} \right)$.</p>
<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">Find the exact values of $a$, $b$ and $c$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">In a triangle $ABC$, $AB = 1$, $AC = \sqrt{3}$ and angle $BAC = \theta + \frac{\pi}{6}$ radians. Given that&nbsp;$\theta$ is sufficiently small for&nbsp;$\theta^3$ and higher powers of $\theta$ to be neglected, show that&nbsp;$BC^2 \approx \mathrm{f}(\theta)$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-c" data-list-item-path="item-c"><div class="list-item-paragraph">Hence, show that&nbsp;$BC \approx 1 + \frac{\sqrt{3}}{2}\theta + \frac{3}{8}\theta^2$.<span class="inline-marks">[3]</span></div></li></ol>

It is given that $\mathrm{f}(\theta) = a\theta^2 + b\theta + c$, where $a$, $b$ and $c$ are real constants. The curve with equation $y = \mathrm{f}(\theta)$ passes through the point with coordinates $(0, 1)$ and has a turning point at $\left( -\frac{\sqrt{3}}{3}, \frac{1}{2} \right)$.

Find the exact values of $a$, $b$ and $c$.

In a triangle $ABC$, $AB = 1$, $AC = \sqrt{3}$ and angle $BAC = \theta + \frac{\pi}{6}$ radians. Given that $\theta$ is sufficiently small for $\theta^3$ and higher powers of $\theta$ to be neglected, show that $BC^2 \approx \mathrm{f}(\theta)$.

Hence, show that $BC \approx 1 + \frac{\sqrt{3}}{2}\theta + \frac{3}{8}\theta^2$.

2025 TMJC P2 Q2

<p>In the triangle $\Delta ABC$, $AC = 1$, $\angle BAC = \frac{\pi}{3}$ radians and $\angle ABC = \left( \frac{\pi}{6} + \theta \right)$ radians.</p>
<ol class="abc-list"><li data-list-item-id="item-a-2" data-list-item-path="item-a-2"><div class="list-item-paragraph">Show that&nbsp;$BC = \frac{\sqrt{3}}{\cos \theta + \sqrt{3} \sin \theta}$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-b-2" data-list-item-path="item-b-2"><div class="list-item-paragraph"><p>Given that&nbsp;$\theta$ is sufficiently small such that&nbsp;$\theta^3$ and higher powers of $\theta$ may be neglected, show that\n\[BC \approx \sqrt{3} (1 + a\theta + b\theta^2)\] where $a$ and $b$ are constants to be determined.</p><span class="inline-marks">[3]</span></div></li></ol>

In the triangle $\Delta ABC$, $AC = 1$, $\angle BAC = \frac{\pi}{3}$ radians and $\angle ABC = \left( \frac{\pi}{6} + \theta \right)$ radians.

Show that $BC = \frac{\sqrt{3}}{\cos \theta + \sqrt{3} \sin \theta}$.

2025 TJC P1 Q4

<p>The diagram below shows $\Delta ABC$ and $\Delta AED$. $\Delta ABC$ is an equilateral triangle with each side measuring $2$ units and $\Delta AED$ is a right-angled triangle with $\angle ADE = \frac{\pi}{4} + x$. Points $C$ and $E$ both lie on the line $BD$.</p>
<div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1769241178125-iheacp12hg.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjkyNDExNzgxMjUtaWhlYWNwMTJoZy5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkzNTY0ODU1LCJleHAiOjE3OTM1NjUxNTV9.udmTuFpHVNKp8Gx8mNx0In30to-kHyuTBgBAXzCE77c" style="max-width: min(400px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<ol class="abc-list"><li data-list-item-id="item-a-geom" data-list-item-path="item-a-geom"><div class="list-item-paragraph">Show that&nbsp;$BD = 1 + \frac{\sqrt{3}}{\tan\left(\frac{\pi}{4} + x\right)}$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-b-geom" data-list-item-path="item-b-geom"><div class="list-item-paragraph">Given that&nbsp;$x$ is sufficiently small for&nbsp;$x^3$ and higher powers of $x$ to be neglected, show that&nbsp;$BD \approx a + bx + cx^2$ where&nbsp;$a$, $b$ and $c$ are exact constants to be determined.<span class="inline-marks">[4]</span></div></li></ol>

The diagram below shows $\Delta ABC$ and $\Delta AED$. $\Delta ABC$ is an equilateral triangle with each side measuring $2$ units and $\Delta AED$ is a right-angled triangle with $\angle ADE = \frac{\pi}{4} + x$. Points $C$ and $E$ both lie on the line $BD$.

Show that $BD = 1 + \frac{\sqrt{3}}{\tan\left(\frac{\pi}{4} + x\right)}$.

Given that $x$ is sufficiently small for $x^3$ and higher powers of $x$ to be neglected, show that $BD \approx a + bx + cx^2$ where $a$, $b$ and $c$ are exact constants to be determined.

2025 SAJC P2 Q3

<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">In a triangle $ABC$, $AB = 2$, angle $CAB = x$ radians and angle $CBA = \frac{\pi}{6}$ radians.</div><ol class="i-list"><li data-list-item-id="item-a-i" data-list-item-path="item-a/item-a-i"><div class="list-item-paragraph">Show that&nbsp;$AC = \frac{2}{\cos x + \sqrt{3} \sin x}$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-a-ii" data-list-item-path="item-a/item-a-ii"><div class="list-item-paragraph">Given that&nbsp;$x$ is a sufficiently small angle, show that</div><div class="list-item-paragraph" style="text-align: center">\[AC \approx a + bx + cx^2,\]</div><div class="list-item-paragraph">where&nbsp;$a$, $b$ and $c$ are constants to be determined.<span class="inline-marks">[3]</span></div></li></ol></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">It is given that&nbsp;$2xy + \ln y = \ln 3$. Show that</div><div class="list-item-paragraph" style="text-align: center">\[(2xy^2 + y)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 4y^2\frac{\mathrm{d}y}{\mathrm{d}x} - \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = 0.\]</div><div class="list-item-paragraph">Hence find the Maclaurin series for&nbsp;$y$, up to and including the term in $x^2$.<span class="inline-marks">[5]</span></div></li></ol>

Implicit Differentiation

Derivatives of Natural Logarithmic Functions

Maclaurin Series by Implicit Differentiation

2025 RI P2 Q2

<p>In the right-angled triangle $\Delta ABC$, angle $C$ is a right angle. $AB=13$ cm, $BC=5$ cm and $X$ is a point on $BC$ such that&nbsp;$\angle BAX$ is $\alpha$ radians.</p>
<div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1768709163495-j5jcg248bqk.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3Njg3MDkxNjM0OTUtajVqY2cyNDhicWsucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzU2NDg1NSwiZXhwIjoxNzkzNTY1MTU1fQ.JeL6Dz6T9B5xdc85XXsH8UXrNnBQ2480zMdgQP7tPXY" style="max-width: min(400px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">Find the exact value of $\cos \angle BAC$.<span class="inline-marks">[1]</span></div></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">By considering triangle $\Delta AXC$, or otherwise, show that&nbsp;$AX = \frac{156}{12 \cos \alpha + 5 \sin \alpha}$<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-c" data-list-item-path="item-c"><div class="list-item-paragraph"><p>Given that&nbsp;$\alpha$ is a sufficiently small angle, show that</p></div><div class="list-item-paragraph"><p style="text-align: center;"> $AX \approx 13 + p\alpha + q\alpha^2$, </p></div><div class="list-item-paragraph"><p>where $p$ and $q$ are constants to be determined exactly.</p><span class="inline-marks">[4]</span></div></li></ol>

In the right-angled triangle $\Delta ABC$, angle $C$ is a right angle. $AB=13$ cm, $BC=5$ cm and $X$ is a point on $BC$ such that $\angle BAX$ is $\alpha$ radians.

Find the exact value of $\cos \angle BAC$.

By considering triangle $\Delta AXC$, or otherwise, show that \[AX = \frac{156}{12 \cos \alpha + 5 \sin \alpha}\]

2025 EJC P1 Q3 

<ol class="abc-list"><li data-list-item-id="item-1761023884532-jinanf4nl" data-list-item-path="item-1761023884532-jinanf4nl"><div class="list-item-paragraph">Given that&nbsp;$\theta $ is small, show that&nbsp;$\sec \theta \approx 1+\frac{1}{2}{{\theta }^{2}}$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-1761023885515-6a30tn3dt" data-list-item-path="item-1761023885515-6a30tn3dt"><div class="list-item-paragraph">The diagram below shows a circle, centre $O$ and radius $r$, with points $A$ and $B$ on the circumference such that&nbsp;$\angle AOB=\theta $ radians, where&nbsp;$\theta $ is small. $AC$ is a tangent to the circle at $A$ and $OBC$ is a straight line.</div><div class="graph-block" data-graph-config="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" data-width="300" data-height="310" style="width:300px;height:310px;max-width:100%;display:block;margin:0 auto;"></div><ol class="i-list"><li data-list-item-id="subitem-1761023906822-tybkpbjf4" data-list-item-path="item-1761023885515-6a30tn3dt/subitem-1761023906822-tybkpbjf4"><div class="list-item-paragraph">Show that the length of chord $AB$ can be approximated by $r\theta $.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="subitem-1761023907415-v0x69j6sr" data-list-item-path="item-1761023885515-6a30tn3dt/subitem-1761023907415-v0x69j6sr"><div class="list-item-paragraph">Hence show that the perimeter of triangle $ABC$ can be approximated by $r\theta (a+b\theta )$, where&nbsp;$a$ and $b$ are constants to be determined.<span class="inline-marks">[4]</span></div></li></ol></li></ol>

2025 ASRJC P1 Q3

<p>The diagram below shows a circle with centre $C$ and radius $4$ cm. The line $AB$ is a tangent to the circle and angle $ACB=\left( \frac{\pi }{3}-\theta \right)$ radians.</p>
<div class="graph-block" data-graph-config="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" data-width="420" data-height="260" style="width:420px;height:260px;max-width:100%;display:block;margin:0 auto;"></div>
<p>Show that&nbsp;$A$, the area of the shaded region $R$, is given by $8\left[ \tan \left( \frac{\pi }{3}-\theta \right)-\left( \frac{\pi }{3}-\theta \right) \right]$.<span class="inline-marks">[2]</span></p>
<p>Given that&nbsp;$\theta $ is a sufficiently small angle, show that&nbsp;$A\approx \left( 8\sqrt{3}-\frac{8\pi }{3} \right)+a\theta +b{{\theta }^{2}}$, where&nbsp;$a$ and $b$ are constants to be determined.<span class="inline-marks">[4]</span></p>

The diagram below shows a circle with centre $C$ and radius $4$ cm. The line $AB$ is a tangent to the circle and angle $ACB=\left( \frac{\pi }{3}-\theta \right)$ radians.

Show that $A$, the area of the shaded region $R$, is given by $8\left[ \tan \left( \frac{\pi }{3}-\theta \right)-\left( \frac{\pi }{3}-\theta \right) \right]$.

Given that $\theta $ is a sufficiently small angle, show that $A\approx \left( 8\sqrt{3}-\frac{8\pi }{3} \right)+a\theta +b{{\theta }^{2}}$, where $a$ and $b$ are constants to be determined.

2016 VJC Promo Q2

<ol class="abc-list"><li data-list-item-id="item-1760748073985-6zkh4y147" data-list-item-path="item-1760748073985-6zkh4y147"><div class="list-item-paragraph">Find the series expansion for&nbsp;$\frac{\cos 2x}{{{\left( 1+\tan x \right)}^{2}}}$ up to and including the term in ${{x}^{2}}$, given that&nbsp;$x$ is a sufficiently small angle.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-1760748074735-fzd6yxb7p" data-list-item-path="item-1760748074735-fzd6yxb7p"><div class="list-item-paragraph">It is given that&nbsp;$\mathrm{f}(x)={{\left( 1-x \right)}^{\frac{1}{2}}}\ln \sqrt{1+x}$.</div><ol class="i-list"><li data-list-item-id="subitem-1760748088122-r4auxec5s" data-list-item-path="item-1760748074735-fzd6yxb7p/subitem-1760748088122-r4auxec5s"><div class="list-item-paragraph">Find the numerical value of $\mathrm{f}'(x)$ when&nbsp;$x=0.01$, leaving the answer in $5$ decimal places.<span class="inline-marks">[1]</span></div></li><li data-list-item-id="subitem-1760748095885-ni5ilr8hq" data-list-item-path="item-1760748074735-fzd6yxb7p/subitem-1760748095885-ni5ilr8hq"><div class="list-item-paragraph">Using the standard series from the List of Formulae (MF27), expand $\mathrm{f}\left( x \right)$ as far as the term in ${{x}^{3}}$. State the range of values of $x$ for which the expansion is valid.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="subitem-1760748103077-oufpzpbxf" data-list-item-path="item-1760748074735-fzd6yxb7p/subitem-1760748103077-oufpzpbxf"><div class="list-item-paragraph">Denote the answer in part <strong>(b)(ii)</strong> as&nbsp;$\mathrm{g}\left( x \right)$. Hence find $\mathrm{g }'\mathrm{ }\left( x \right)$ when&nbsp;$x=0.01$, leaving the answer in $5$ decimal places.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="subitem-1760748118528-hw1tpiuno" data-list-item-path="item-1760748074735-fzd6yxb7p/subitem-1760748118528-hw1tpiuno"><div class="list-item-paragraph">Comparing your answer in part <strong>(b)(i)</strong> against <strong>(b)(iii)</strong>, comment on the validity of this answer.<span class="inline-marks">[1]</span></div></li></ol></li></ol>

Maclaurin Series Using Standard Series

Series from Products, Substitution, Integration and Differentiation

Convergence and Validity of Power Series

Accuracy of Power Series Approximations

Binomial Expansion as a Maclaurin Series

Deriving Series by Differentiation and Integration

Products and Quotients of Power Series

2024 MI P2 Q5

<p>For a right-angled triangle $ABC$ with $BC=a$ and $AC=b$, $\measuredangle CAB=x$ and $\measuredangle ABC=\frac{\pi }{2}$ as shown below, a sector with radius $a$ is extended from the vertex $C$. This sector meets the side $AC$ at the point $Y$.</p>
<div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/converted/1760707177047-d4so0iwxo4a.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvY29udmVydGVkLzE3NjA3MDcxNzcwNDctZDRzbzBpd3hvNGEucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzU2NDg1NSwiZXhwIjoxNzkzNTY1MTU1fQ.xQRrEFWCVJtuHNRpkoZXyAgEIfCnmWjJK7IVM9dVo5s" style="max-width: min(500px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<ol class="abc-list"><li data-list-item-id="item-1760707176669-hpgk7x371" data-list-item-path="item-1760707176669-hpgk7x371"><div class="list-item-paragraph">Show that&nbsp;${{\left( \frac{AC}{AY} \right)}^{2}}=\frac{1}{{{\left( 1-\sin x \right)}^{2}}}$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-1760707177209-c5ixvvjxc" data-list-item-path="item-1760707177209-c5ixvvjxc"><div class="list-item-paragraph">Using appropriate expansions from the List of Formulae (MF26), find the Maclaurin series for&nbsp;$\mathrm{f}(x)={{\left( 1-\sin x \right)}^{-2}}$, up to and including the term in ${{x}^{3}}$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-1760707178147-p1fikf4k6" data-list-item-path="item-1760707178147-p1fikf4k6"><div class="list-item-paragraph">Deduce the series expansion for&nbsp;$\frac{2\cos x}{{{\left( 1-\sin x \right)}^{3}}}$, up to and including the term in ${{x}^{2}}$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-1760707178831-sau53jtmy" data-list-item-path="item-1760707178831-sau53jtmy"><div class="list-item-paragraph">Using your answer in part <strong>(b)</strong> or otherwise, show that&nbsp;$\frac{AC}{AY}\approx 1+x+{{x}^{2}}$, when&nbsp;$x$ is sufficiently small.<span class="inline-marks">[3]</span></div></li></ol>

For a right-angled triangle $ABC$ with $BC=a$ and $AC=b$, $\measuredangle CAB=x$ and $\measuredangle ABC=\frac{\pi }{2}$ as shown below, a sector with radius $a$ is extended from the vertex $C$. This sector meets the side $AC$ at the point $Y$.

Substitution and Composition of Power Series

2024 EJC P1 Q7

<figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/converted/1760706907296-yet4o3lzty.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvY29udmVydGVkLzE3NjA3MDY5MDcyOTYteWV0NG8zbHp0eS5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkzNTY0ODU1LCJleHAiOjE3OTM1NjUxNTV9.mC0fGRMNS7IEwAPIJNIW3evVxxcbwm4lpyjFQ-9ugZI" width="400" class="aligncenter" /></figure>
<p>In the triangle $ABC$, $AB=x$, $BC=y$, $AC=\frac{1}{6}$, angle $ABC=\frac{\pi }{6}$ radians and angle $ACB=\theta $ radians (see diagram).</p>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text"></span></div><ol class="i-list"><li><div class="list-item-paragraph">Show that $\frac{x}{y}=\frac{2\sin \theta }{\cos \theta +\sqrt{3}\sin \theta }$.</div><p style="text-align: right;">[3]</p></li><li><div class="list-item-paragraph">Given that $\theta $ is sufficiently small, express $\frac{x}{y}$ as a cubic polynomial in $\theta $.</div><p style="text-align: right;">[3]</p></li></ol></li><li><div class="list-item-content"><span class="list-item-text"></span></div><ol class="i-list"><li><div class="list-item-paragraph">Show that $\theta ={{\sin }^{-1}}\left( 3x \right)$.</div><p style="text-align: right;">[1]</p></li><li><div class="list-item-paragraph">Find the Maclaurin series for $\theta $, up to and including the term in ${{x}^{3}}$.</div><p style="text-align: right;">[5]</p></li></ol></li></ol>

In the triangle $ABC$, $AB=x$, $BC=y$, $AC=\frac{1}{6}$, angle $ABC=\frac{\pi }{6}$ radians and angle $ACB=\theta $ radians (see diagram).

Maclaurin Series by Successive Differentiation

2016 HCI P1 Q10

<div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1782814203109-chy79btud69.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3ODI4MTQyMDMxMDktY2h5NzlidHVkNjkucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzU2NDg1NSwiZXhwIjoxNzkzNTY1MTU1fQ.ur5kj8QDSIq9LGaZzaDAorKHiv6gI3hNibrbIRnplb0" style="max-width: min(250px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<p>In the diagram above, $OAB$ is an isosceles triangle where&nbsp;$OA=OB=r$ cm. It is given that the length $AB=2$ cm and angle $AOB=\frac{\pi }{3}-\theta $ radians.</p>
<ol class="i-list"><li data-list-item-id="item-1760706844247-c4bic633v" data-list-item-path="item-1760706844247-c4bic633v"><div class="list-item-paragraph">Using cosine rule, show that&nbsp;${{r}^{2}}=\frac{4}{2-\sqrt{3}\sin \theta -\cos \theta }$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-1760706845133-k5j9yw1lh" data-list-item-path="item-1760706845133-k5j9yw1lh"><div class="list-item-paragraph">Given that&nbsp;$\theta $ is a sufficiently small angle, show that</div><div class="list-item-paragraph" style="text-align: center">$r\approx 2+a\theta +b{{\theta }^{2}}$</div><div class="list-item-paragraph">where&nbsp;$a$ and $b$ are constants to be determined.<span class="inline-marks">[4]</span></div></li></ol>

In the diagram above, $OAB$ is an isosceles triangle where $OA=OB=r$ cm. It is given that the length $AB=2$ cm and angle $AOB=\frac{\pi }{3}-\theta $ radians.

2024 TMJC P1 Q4

<div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/converted/1760430160513-3fmegw259ta.jpg?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvY29udmVydGVkLzE3NjA0MzAxNjA1MTMtM2ZtZWd3MjU5dGEuanBnIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzU2NDg1NSwiZXhwIjoxNzkzNTY1MTU1fQ.bwJuiR7YPg-WwY8iEsmx0p9jglPErmmkZYKdsWnsVfk" style="max-width: min(220px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<ol class="abc-list" start="2"><li data-list-item-id="item-1760430167252-ya56ib33d" data-list-item-path="item-1760430167252-ya56ib33d"><div class="list-item-paragraph">An arc $PQ$ of a circle subtends an angle $\left( \pi -\theta \right)$ radians at the centre $O$. The length of chord $PQ$ is $L$ units and the area of the shaded segment $PRQ$ is $A$ square units.</div><div class="list-item-paragraph">Given that&nbsp;$\theta $ is small and $A=k{{L}^{2}}$, show that&nbsp;$2k{{\theta }^{2}}-2\theta +\pi -8k\approx 0$.<span class="inline-marks">[5]</span></div></li></ol>

2022 VJC Promo Q7

<p>In triangle $PQR$, $\angle PRQ=\frac{2\pi }{3}$, $\angle RPQ=\frac{\pi }{6}+\theta $, and $PR=a$ units, where&nbsp;$a$ is a positive real constant.</p>
<ol class="i-list"><li data-list-item-id="item-1760423747573-c4mvj1pjc" data-list-item-path="item-1760423747573-c4mvj1pjc"><div class="list-item-paragraph">Show that&nbsp;$PQ=\frac{\sqrt{3}a}{\cos \theta -\sqrt{3}\sin \theta }$.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="item-1760423748158-9fegtmk7c" data-list-item-path="item-1760423748158-9fegtmk7c"><div class="list-item-paragraph">Given that&nbsp;$\theta $ is sufficiently small, show that&nbsp;$PQ\approx \sqrt{3}a\left( 1+b\theta +c{{\theta }^{2}} \right)$, where&nbsp;$b$ and $c$ are real constants.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-1760423749013-a9nf00s5i" data-list-item-path="item-1760423749013-a9nf00s5i"><div class="list-item-paragraph">Given that&nbsp;$\theta &gt;0$, find the range of values of $\theta $ such that the percentage error of the approximation in <strong>(ii)</strong> is less than $5\%$.<span class="inline-marks">[2]</span></div></li></ol>

In triangle $PQR$, $\angle PRQ=\frac{2\pi }{3}$, $\angle RPQ=\frac{\pi }{6}+\theta $, and $PR=a$ units, where $a$ is a positive real constant.

2023 SAJC P2 Q2

<p>In geometric optics, the paraxial approximation is a small-angle approximation used in Gaussian optics and ray tracing of light through an optical system such as a lens.</p>
<p>In the diagram below, a light ray parallel to the horizontal axis is reflected at point $B$ on the circular lens centred at point $C$ and has radius $r$ cm. Let $\angle BCF=\theta $ radians. $FM$ is the perpendicular bisector of $CB$.</p>
<figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/converted/1760349826983-9hqilv7o5mv.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvY29udmVydGVkLzE3NjAzNDk4MjY5ODMtOWhxaWx2N281bXYucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzU2NDg1NSwiZXhwIjoxNzkzNTY1MTU1fQ.8Ab9A0gwWx08oDfGNQl3_nsqpJ0suIKO0-E12fhcovk" width="400" class="aligncenter" /></figure>
<ol class="i-list"><li><div class="list-item-paragraph">Show that $CF=\frac{r}{k\cos \theta }$, where $k$ is a real constant to be determined.</div><p style="text-align: right;">[1]</p></li><li><div class="list-item-paragraph">Hence find the series expansion for $CF$ if $\theta $ is sufficiently small for ${{\theta }^{3}}$ and terms in higher powers of $\theta $ to be neglected.</div><p style="text-align: right;">[2]</p></li></ol>
<p>Suppose that the source of the light ray is now repositioned such that $\angle BCF=\left( \theta +\frac{\pi }{6} \right)$ radians.</p>
<ol class="i-list" start="3"><li><div class="list-item-paragraph">Find the corresponding series expansion for $CF$, up to and including the term in ${{\theta }^{2}}$.</div><p style="text-align: right;">[4]</p></li></ol>

In geometric optics, the paraxial approximation is a small-angle approximation used in Gaussian optics and ray tracing of light through an optical system such as a lens.

In the diagram below, a light ray parallel to the horizontal axis is reflected at point $B$ on the circular lens centred at point $C$ and has radius $r$ cm. Let $\angle BCF=\theta $ radians. $FM$ is the perpendicular bisector of $CB$.

Suppose that the source of the light ray is now repositioned such that $\angle BCF=\left( \theta +\frac{\pi }{6} \right)$ radians.

2019 TJC P1 Q5

<div class="graph-block" data-graph-config="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" data-width="420" data-height="200" style="width:420px;height:200px;max-width:100%;display:block;margin:0 auto;"></div>
<p>In the triangle $ABC$, $AC=1$, angle $BAC=x$ radians and angle $ACB=\frac{1}{6}\pi $ radians (see diagram).</p>
<ol class="abc-list"><li data-list-item-id="item-1760349711336-rlv7rkmgn" data-list-item-path="item-1760349711336-rlv7rkmgn"><div class="list-item-paragraph">Show that&nbsp;$AB=\frac{1}{\cos x+\sqrt{3}\sin x}$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-1760349724409-q7gyr6zn1" data-list-item-path="item-1760349724409-q7gyr6zn1"><div class="list-item-paragraph">Given that&nbsp;$x$ is a sufficiently small angle, show that</div><div class="list-item-paragraph" style="text-align: center">$AB\approx 1+ax+b{{x}^{2}}$,</div><div class="list-item-paragraph">for constants $a$ and $b$ to be determined exactly.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-1760349743720-yntyhxtg9" data-list-item-path="item-1760349743720-yntyhxtg9"><div class="list-item-paragraph"><p>The first two terms in the Maclaurin series for&nbsp;$\ln \left( px+q \right)$ are equal to the first two terms in the series expansion in part <strong>(b)</strong>. Using standard series from the List of Formulae <strong>(MF26)</strong>, find the exact values of $p$ and $q$.</p><span class="inline-marks">[4]</span></div></li></ol>

In the triangle $ABC$, $AC=1$, angle $BAC=x$ radians and angle $ACB=\frac{1}{6}\pi $ radians (see diagram).

Finding Parameters from Power Series Coefficients

2023 YIJC P2 Q5

<p><strong>Maclaurin Series - Small Angle Approximation Q1</strong></p>
<figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/converted/1760342282490-1z9jaxf1tiy.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvY29udmVydGVkLzE3NjAzNDIyODI0OTAtMXo5amF4ZjF0aXkucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzU2NDg1NSwiZXhwIjoxNzkzNTY1MTU1fQ.hU-5FW56m7XofKl9D9kE_Qzt4ASlAYxn3OrxPbirsZU" width="400" class="aligncenter" /></figure>
<p>In the triangle $ABC$, angle $BAC=\frac{\pi }{3}$ radians, angle $ABC=\left( \frac{\pi }{3}+x \right)$ radians and angle $ACB=\left( \frac{\pi }{3}-x \right)$ radians, where $x$ is small. The lengths of $BC$, $AC$ and $AB$ are denoted by $a$, $b$ and $c$ respectively.</p>
<p>By using the sine formula, or otherwise, show that $b-c\approx \frac{2ax}{\sqrt{3}}$.</p>

**Maclaurin Series - Small Angle Approximation Q1**

In the triangle $ABC$, angle $BAC=\frac{\pi }{3}$ radians, angle $ABC=\left( \frac{\pi }{3}+x \right)$ radians and angle $ACB=\left( \frac{\pi }{3}-x \right)$ radians, where $x$ is small. The lengths of $BC$, $AC$ and $AB$ are denoted by $a$, $b$ and $c$ respectively.

By using the sine formula, or otherwise, show that $b-c\approx \frac{2ax}{\sqrt{3}}$.

Maclaurin Series Q1

<ol class="abc-list" start="2"><li data-list-item-id="item-1760244185411-ogvjhkpqm" data-list-item-path="item-1760244185411-ogvjhkpqm"><div class="graph-block" data-graph-config="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" data-width="420" data-height="140" style="width:420px;height:140px;max-width:100%;display:block;margin:0 auto;"></div><div class="list-item-paragraph"><p>In the triangle $ABC$, $AC=\sqrt{3}$, angle $BAC=\frac{2\pi }{3}$ radians and $ACB=\theta $ radians (see diagram).</p></div><ol class="i-list"><li data-list-item-id="subitem-1760244203380-gho3jgn8s" data-list-item-path="item-1760244185411-ogvjhkpqm/subitem-1760244203380-gho3jgn8s"><div class="list-item-paragraph">Show that&nbsp;$BC=\frac{3}{\sqrt{3}\cos \theta -\sin \theta }$ .<span class="inline-marks">[3]</span></div></li><li data-list-item-id="subitem-1760244204542-wr47l7i7o" data-list-item-path="item-1760244185411-ogvjhkpqm/subitem-1760244204542-wr47l7i7o"><div class="list-item-paragraph">Given that&nbsp;$\theta $ is a sufficiently small angle, show that&nbsp;$BC\approx \sqrt{3}+\theta +a{{\theta }^{2}}$, where&nbsp;$a$ is a constant to be determined.<span class="inline-marks">[4]</span></div></li></ol></li></ol>

2017 MI Promo Q11

<figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/converted/1760172794198-s239rqmk47l.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvY29udmVydGVkLzE3NjAxNzI3OTQxOTgtczIzOXJxbWs0N2wucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzU2NDg1NSwiZXhwIjoxNzkzNTY1MTU1fQ.ZxKFu-FDeFPYHSc6cp2kzm376utURPdsS2d-BMDhQSM" width="400" class="aligncenter" /></figure>
<p>A laser from a fixed point $O$ on a flat ground projects light beams to the top of two vertical structures $A$ and $B$ as shown above. To project the beam to the top of $A$, the laser makes an angle of elevation of $\frac{\pi }{6}$ radians. To project the beam to the top of $B$, the laser makes an angle of elevation of $\left( \frac{\pi }{6}+x \right)$ radians. The two structures $A$ and $B$ are of heights $h$ m and $\left( h+\sqrt{3}k \right)$m respectively and are $10$m and $\left( 10+k \right)$m away from $O$ respectively.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Show that the length of the straight beam from $O$ to $A$ is $\frac{20}{\sqrt{3}}$m.</div><p style="text-align: right;">[1]</p></li><li><div class="list-item-paragraph">Show that the length of $AB$ is $2k$m and that the angle of elevation of $B$ from $A$ is $\frac{\pi }{3}$ radians.</div><p style="text-align: right;">[3]</p></li><li><div class="list-item-paragraph">Hence, using the sine rule, show that $k=\frac{10\sin x}{\sqrt{3}\sin \left( \frac{\pi }{6}-x \right)}$.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">If $x$ is sufficiently small, show that $k\approx \frac{20}{\sqrt{3}}\left( x+a{{x}^{2}} \right)$, where $a$ is a constant to be determined.</div><p style="text-align: right;">[6]</p></li></ol>

A laser from a fixed point $O$ on a flat ground projects light beams to the top of two vertical structures $A$ and $B$ as shown above. To project the beam to the top of $A$, the laser makes an angle of elevation of $\frac{\pi }{6}$ radians. To project the beam to the top of $B$, the laser makes an angle of elevation of $\left( \frac{\pi }{6}+x \right)$ radians. The two structures $A$ and $B$ are of heights $h$ m and $\left( h+\sqrt{3}k \right)$m respectively and are $10$m and $\left( 10+k \right)$m away from $O$ respectively.

2017 JJC P1 Q10

<ol class="abc-list"><li><div class="list-item-paragraph">Given that&nbsp;$x$ is small enough for terms involving ${{x}^{4}}$ and above to be ignored, use the Maclaurin series for&nbsp;$\sin x$ and $\cos x$ to show that</div><div class="list-item-paragraph" style="text-align: center">$\tan x\approx x+\frac{1}{3}{{x}^{3}}$.<span class="inline-marks">[3]</span></div><div class="list-item-paragraph">A straight pole, $AB$, leans against a wall as shown in the diagram below. $AC$ is horizontal where&nbsp;$C$ is the base of the wall and $AC=2$ metres. $B$ is vertically above $C$ and angle $BAC={{85}^{\circ }}$.</div><div class="content-block list-item-image" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/converted/1760155508834-e2kr9cc98n.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvY29udmVydGVkLzE3NjAxNTU1MDg4MzQtZTJrcjljYzk4bi5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkzNTY0ODU1LCJleHAiOjE3OTM1NjUxNTV9.SHpyhz76eHwldEupW7JCriz3wDXrhJuqjczOTBe9EMM" alt="List item image" class="aligncenter custom-width" style="max-width: min(350px, 100%); height: auto;" /></div><div class="list-item-paragraph">Use the Maclaurin series for&nbsp;$\tan x$ to find an approximate value for&nbsp;$BC$, giving your answer correct to $5$ decimal places.<span class="inline-marks">[2]</span></div><div class="list-item-paragraph">Use a calculator to find the value of $BC$, giving your answer correct to $5$ decimal places, and hence show that the error in your approximation is about $0.0008\%$.<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">Given that&nbsp;$y=\tan \left( \frac{1}{x+a} \right)$, where&nbsp;$a$ is a positive constant, show that</div><div class="list-item-paragraph" style="text-align: center">${{\left( x+a \right)}^{2}}\frac{\mathrm{d}y}{\mathrm{d}x}+{{y}^{2}}+1=0$.<span class="inline-marks">[2]</span></div><div class="list-item-paragraph">Using $a=\frac{4}{\pi }$, find the Maclaurin series for&nbsp;$\tan \left( \frac{1}{x+a} \right)$, up to and including the term in ${{x}^{2}}$, giving all coefficients in exact form.<span class="inline-marks">[4]</span></div></li></ol>

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