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A Math (IGCSE)

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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>A curve $C$ has equation $y=x^{-3}\ln x$.</p>
<ol class="abc-list"><li data-list-item-id="p1q7-q-b2-i1" data-list-item-path="p1q7-q-b2-i1"><div class="list-item-paragraph">Show that&nbsp;$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1-3\ln x}{x^4}$ and hence find the coordinates of the turning point of $C$.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="p1q7-q-b2-i2" data-list-item-path="p1q7-q-b2-i2"><div class="list-item-paragraph">Find the exact area enclosed by $C$, the $x$-axis and the line $x=3$.<span class="inline-marks">[5]</span></div></li></ol>

A curve $C$ has equation $y=x^{-3}\ln x$.

Show that $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1-3\ln x}{x^4}$ and hence find the coordinates of the turning point of $C$.

Find the exact area enclosed by $C$, the $x$-axis and the line $x=3$.

Derivatives of Natural Logarithmic Functions

Product Rule

Area Bounded by Curves

Stationary Points and Their Nature

N2022 P1 Q7

Techniques of Differentiation

Applications of Integration

Applications of Differentiation

Techniques of Integration

<p>The curve $D$ has equation $y = \ln(1 + 2x^2)$.</p>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">Use a non-calculator method to find the equation of the tangent to $D$ at the point where&nbsp;$x = 2$, giving your answer in the form $ax + by = c + d \ln 3$ where&nbsp;$a$, $b$, $c$ and $d$ are integers.</span><span class="inline-marks">[3]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Sketch the graphs of $D$ and the tangent found in part (a) on a single diagram, stating the coordinates of the point(s) of intersection of the two graphs and the points where the tangent meets the axes.</span><span class="inline-marks">[3]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Find the area of the finite region bounded by the curve $D$, the $x$-axis and the tangent line. Give your answer correct to $3$ significant figures.</span><span class="inline-marks">[3]</span></div></li></ol>
<p>A point $P(x, y)$ moves along the curve $D$ in such a way that its $x$ coordinate is increasing at a constant rate of $3$ units per second. Find the rate of change of its $y$ coordinate at the instant when&nbsp;$x=1$.<span class="inline-marks">[2]</span></p>

The curve $D$ has equation $y = \ln(1 + 2x^2)$.

Use a non-calculator method to find the equation of the tangent to $D$ at the point where $x = 2$, giving your answer in the form $ax + by = c + d \ln 3$ where $a$, $b$, $c$ and $d$ are integers.

Sketch the graphs of $D$ and the tangent found in part (a) on a single diagram, stating the coordinates of the point(s) of intersection of the two graphs and the points where the tangent meets the axes.

Find the area of the finite region bounded by the curve $D$, the $x$-axis and the tangent line. Give your answer correct to $3$ significant figures.

A point $P(x, y)$ moves along the curve $D$ in such a way that its $x$ coordinate is increasing at a constant rate of $3$ units per second. Find the rate of change of its $y$ coordinate at the instant when $x=1$.

Tangents and Normals at a Point

Sketching Logarithmic Functions and Solving Equations Graphically

Connected Rates of Change

2024 SAJC H1 Q4

Graphing Techniques

<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><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">On a single diagram, sketch the graphs of $y = \frac{3x}{x - 1}$ and $y = 3x$, stating the equation(s) of any asymptote(s) and the coordinates of points of intersection.<span class="inline-marks">[4]</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">Hence find the area bounded by the curve $y = \frac{3x}{x - 1}$ and the lines $y = 3x$ and $y = 8$.<span class="inline-marks">[2]</span></div></li></ol></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">A curve has the equation $y = \sqrt{x^2 + 1}$ where&nbsp;$x$ is non-negative.</div><ol class="i-list"><li data-list-item-id="item-b-i" data-list-item-path="item-b/item-b-i"><div class="list-item-paragraph">With the aid of a sketch of the above curve, explain why the value of $S = \lim_{n \to \infty} \frac{1}{n} \left[ 1 + \sqrt{\frac{1}{n^2} + 1} + \sqrt{\frac{4}{n^2} + 1} + \dots + \sqrt{\frac{(n - 1)^2}{n^2} + 1} \right]$ is $\displaystyle \displaystyle\int_0^1 \sqrt{x^2 + 1} \, \mathrm{d}x$.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="item-b-ii" data-list-item-path="item-b/item-b-ii"><div class="list-item-paragraph">Hence find the value of $S$.<span class="inline-marks">[1]</span></div></li></ol></li></ol>

A curve has the equation $y = \sqrt{x^2 + 1}$ where $x$ is non-negative.

Sketching Rational Functions

Asymptotes and Discontinuities

Riemann Sums

Graphs of Conic Sections

2025 SAJC P1 Q7

<p>The curve $C$ has equation $y = \frac{1}{4x - x^2}$.</p>
<ol class="abc-list"><li data-list-item-id="item-a"><div class="list-item-content"><span class="list-item-text">Find $\frac{\mathrm{d}y}{\mathrm{d}x}$. Hence, find the $x$-coordinate, $x = x_1$, of the turning point on $C$ and determine its nature.</span><span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-b"><div class="list-item-content"><span class="list-item-text">Using calculus, find the exact area of the region between $C$, the $x$-axis and the lines with equations $x = 1$ and $x = x_1$.</span><span class="inline-marks">[3]</span></div></li></ol>

The curve $C$ has equation $y = \frac{1}{4x - x^2}$.

Find $\frac{\mathrm{d}y}{\mathrm{d}x}$. Hence, find the $x$-coordinate, $x = x_1$, of the turning point on $C$ and determine its nature.

Using calculus, find the exact area of the region between $C$, the $x$-axis and the lines with equations $x = 1$ and $x = x_1$.

Chain Rule

Stationary Points and Points of Inflection

2025 JPJC P1 Q1

<p>A curve $C$ is defined parametrically by</p>
<p style="text-align: center">\[x = a \tan t, \quad y = a \sec^2 t \sin t, \quad -\frac{\pi}{4} &lt; t &lt; \frac{\pi}{4},\]</p>
<p>where&nbsp;$a$ is a positive constant.</p>
<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">Show that&nbsp;$\frac{\tan t}{\sqrt{1 + \tan^2 t}} = \sin t$.<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 using part <strong>(a)</strong> or otherwise, find the cartesian equation of $C$ in the form $y = \mathrm{f}(x)$, simplifying your answer.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-c" data-list-item-path="item-c"><div class="list-item-paragraph">Show that&nbsp;$\mathrm{f}(-x) = -\mathrm{f}(x)$. Hence sketch $C$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-d" data-list-item-path="item-d"><div class="list-item-paragraph">Find the exact area of the region bounded by $C$, the $x$-axis and the lines $x = -A$ and $x = A$, where&nbsp;$0 &lt; A &lt; a$, leaving your answer in terms of $A$ and $a$.<span class="inline-marks">[3]</span></div></li></ol>

A curve $C$ is defined parametrically by

\[x = a \tan t, \quad y = a \sec^2 t \sin t, \quad -\frac{\pi}{4} < t < \frac{\pi}{4},\]

Show that $\frac{\tan t}{\sqrt{1 + \tan^2 t}} = \sin t$.

Show that $\mathrm{f}(-x) = -\mathrm{f}(x)$. Hence sketch $C$.

Find the exact area of the region bounded by $C$, the $x$-axis and the lines $x = -A$ and $x = A$, where $0 < A < a$, leaving your answer in terms of $A$ and $a$.

Odd and Even Functions

Sketching Parametric Curves

Converting Parametric Equations to Cartesian Form

2025 DHS P1 Q6

Functions

<ol class="abc-list"><li data-list-item-id="item-a-3" data-list-item-path="item-a-3"><ol class="i-list"><li data-list-item-id="item-a-3-i" data-list-item-path="item-a-3/item-a-3-i"><div class="list-item-paragraph"><p>Show that&nbsp;$\frac{3x^2}{(x+1)(3x^2+x+1)}$ can be expressed as<br></p></div><div class="list-item-paragraph"><p style="text-align: center;">$\frac{A}{x+1} + \frac{Bx+C}{3x^2+x+1}$,<br></p></div><div class="list-item-paragraph"><p>where&nbsp;$A$, $B$ and $C$ are constants to be determined.</p><span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-a-3-ii" data-list-item-path="item-a-3/item-a-3-ii"><div class="list-item-paragraph">Hence, find $\displaystyle \displaystyle\int \frac{3x^2}{(x+1)(3x^2+x+1)} \, \mathrm{d}x$.<span class="inline-marks">[4]</span></div></li></ol></li></ol>
<ol class="abc-list" start="2"><li data-list-item-id="item-b-3" data-list-item-path="item-b-3"><div class="list-item-paragraph">The region $R$ is bounded by the curve $y = \frac{3x^2}{(x+1)(3x^2+x+1)}$, the line $x=4$ and the $x$-axis.</div><ol class="i-list"><li data-list-item-id="item-b-3-i" data-list-item-path="item-b-3/item-b-3-i"><div class="list-item-paragraph">Sketch the graph of $y = \frac{3x^2}{(x+1)(3x^2+x+1)}$ for&nbsp;$x \ge 0$, giving the coordinates of any intercepts with the axes and the equations of any asymptotes. Shade the region $R$ on your sketch.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-b-3-ii" data-list-item-path="item-b-3/item-b-3-ii"><div class="list-item-paragraph">Find the exact area of $R$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-b-3-iii" data-list-item-path="item-b-3/item-b-3-iii"><div class="list-item-paragraph">Find the volume of the solid generated when&nbsp;$R$ is rotated through $2\pi$ radians about the $x$-axis.<span class="inline-marks">[2]</span></div></li></ol></li></ol>

Integration by Splitting the Numerator

Integration Using Partial Fractions

Volume of Revolution about the x-axis

Partial Fractions

Integration by Completing the Square

Standard Integrals with Quadratic Denominators

2025 CJC P1 Q11

<p>The curve $C$ is defined by the parametric equations $x=at^2$, $y=\frac{a}{t}$, $t \neq 0$, where&nbsp;$a$ is a positive constant.</p>
<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">Show that the equation of the tangent to the curve at the point $P\left(ap^2, \frac{a}{p}\right)$ is $2p^3y+x=3ap^2$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">Find the equations of the tangents to $C$ at the points where&nbsp;$x=a$. Find also the acute angle between these tangents.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-c" data-list-item-path="item-c"><ol class="i-list"><li data-list-item-id="item-c-i" data-list-item-path="item-c/item-c-i"><div class="list-item-paragraph">Show that&nbsp;$(q-p)^2(q+2p)=q^3-3p^2q+2p^3$.<span class="inline-marks">[1]</span></div></li><li data-list-item-id="item-c-ii" data-list-item-path="item-c/item-c-ii"><div class="list-item-paragraph">The tangent to the curve at $P$ cuts the curve again at $Q\left(aq^2, \frac{a}{q}\right)$. Find $q$ in terms of $p$.<span class="inline-marks">[3]</span></div></li></ol></li></ol>
<p>The following diagram shows the graph of $C$ for&nbsp;$a=1$, and the Cartesian equation of the curve $C$ is $xy^2=1$. The tangent to $C$ at the point $\left(p^2, \frac{1}{p}\right)$ cuts the curve again at the point $\left(4p^2, -\frac{1}{2p}\right)$.</p>
<div class="graph-block" data-graph-config="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" data-width="300" data-height="200" data-proportional style="width:300px;height:200px;max-width:100%;display:flow-root;margin:0 auto;--graph-width:300px;--graph-aspect:300 / 200;"></div>

<p>The region $S$ is bounded by the tangent to $C$ at the point $\left( p^2, \frac{1}{p} \right)$, the curve $C$ and the line</p>
<p>$x = p^2$</p>
<ol class="abc-list" start="4"><li data-list-item-id="item-d" data-list-item-path="item-d"><div class="list-item-paragraph">Find the area of $S$ in terms of $p$.<span class="inline-marks">[3]</span></div></li></ol>

The curve $C$ is defined by the parametric equations $x=at^2$, $y=\frac{a}{t}$, $t \neq 0$, where $a$ is a positive constant.

Show that the equation of the tangent to the curve at the point $P\left(ap^2, \frac{a}{p}\right)$ is $2p^3y+x=3ap^2$.

Find the equations of the tangents to $C$ at the points where $x=a$. Find also the acute angle between these tangents.

The following diagram shows the graph of $C$ for $a=1$, and the Cartesian equation of the curve $C$ is $xy^2=1$. The tangent to $C$ at the point $\left(p^2, \frac{1}{p}\right)$ cuts the curve again at the point $\left(4p^2, -\frac{1}{2p}\right)$.

The region $S$ is bounded by the tangent to $C$ at the point $\left( p^2, \frac{1}{p} \right)$, the curve $C$ and the line

Find the area of $S$ in terms of $p$.

Tangents and Normals to Parametric Curves

Differentiation of Parametric Equations

2025 ACJC P1 Q11

<ol class="abc-list"><li><div class="list-item-paragraph">Solve the following integral.</div><ol class="i-list"><li><div class="list-item-paragraph">$\displaystyle \displaystyle\int \frac{1-2x}{\sqrt{3+2x-x^2}} \, \mathrm{d}x$.<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">$\displaystyle \displaystyle\int x \ln(2-x^2) \, \mathrm{d}x$, where&nbsp;$-\sqrt{2} &lt; x &lt; \ \sqrt{2}$.<span class="inline-marks">[3]</span></div></li></ol></li><li><div class="list-item-paragraph">A function $\mathrm{f}$ is defined by $\mathrm{f}(x) = \mathrm{e}^x - 2$.</div><ol class="i-list"><li><div class="list-item-paragraph">Sketch the graph of $y = |\mathrm{f}(x)|$, indicating clearly the equation(s) of asymptote(s), if any.<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">Hence find $\displaystyle \displaystyle\int_0^{\ln 3} |\mathrm{f}(x)| \, \mathrm{d}x$, giving your answer in exact form.<span class="inline-marks">[3]</span></div></li></ol></li></ol>

Sketching Exponential Functions and Solving Equations Graphically

Integration by Parts

Modulus Functions

Graphs of Modulus Function

2025 ASRJC P2 Q5

Transformations of Graphs

<p>The diagram shows the curves $C_{1}: x^{2} + y^{2} = 4$ and $C_{2}: \frac{x^{2}}{9} + y^{2} = 1$. Find the area of the shaded regions.</p>
<div class="graph-block" data-graph-config="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" data-width="338" data-height="236" style="width:338px;height:236px;max-width:100%;display:block;margin:0 auto;"></div>

The diagram shows the curves $C_{1}: x^{2} + y^{2} = 4$ and $C_{2}: \frac{x^{2}}{9} + y^{2} = 1$. Find the area of the shaded regions.

<p>Find the area of the loop enclosed by $y^{2} = x(2 - x)^{2}$.</p>

Find the area of the loop enclosed by $y^{2} = x(2 - x)^{2}$.

<p>Find the area enclosed by the curves $\mathrm{g}(x) = -x^{3} + 2x^{2} + x - 3$ and $\mathrm{f}(x) = e^{\frac{x}{3}} - 3$.</p>

Find the area enclosed by the curves $\mathrm{g}(x) = -x^{3} + 2x^{2} + x - 3$ and $\mathrm{f}(x) = e^{\frac{x}{3}} - 3$.

<p>Find the area enclosed by the curves $\mathrm{g}(x) = x^{3} + 2x^{2} - 2$ and $\mathrm{f}(x) = x - 1$.</p>

Find the area enclosed by the curves $\mathrm{g}(x) = x^{3} + 2x^{2} - 2$ and $\mathrm{f}(x) = x - 1$.

<p>A curve $C$ has parametric equations</p>
<p style="text-align: center">$x=a\left( \frac{4}{\pi }t+\sin 2t \right)+2a$, </p>
<p style="text-align: center">$y=a\cos t$,</p>
<p>where $-\frac{\pi }{2}\le t\le \frac{\pi }{2}$ and $a$ is a positive constant.</p>
<ol class="abc-list"></ol>
<ol class="i-list"><li><div class="list-item-paragraph">Sketch $C$. State clearly the coordinates of any points where $C$ meets the $x$-axis.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">Show that the area enclosed by $C$ and the $x$-axis is given by</div><div class="list-item-paragraph" style="text-align: center">${{a}^{2}}\displaystyle\int_{{{\theta }_{1}}}^{{{\theta }_{2}}}{\left( \frac{4}{\pi }\cos t+2\cos t\cos 2t \right)\mathrm{d}t}$</div><div class="list-item-paragraph">where ${{\theta }_{1}}$ and ${{\theta }_{2}}$ are values to be stated.</div><p style="text-align: right;">[3]</p></li><li><div class="list-item-paragraph">Hence find, in terms of $a$, the exact area enclosed by $C$ and the $x$-axis.</div><p style="text-align: right;">[3]</p></li></ol>

$x=a\left( \frac{4}{\pi }t+\sin 2t \right)+2a$, 

where $-\frac{\pi }{2}\le t\le \frac{\pi }{2}$ and $a$ is a positive constant.

Integrable Forms of Trigonometric Functions

Trigonometric Integration Using Product-to-Sum Formulae

2021 HCI P2 Q1

<ol class="i-list"><li><div class="list-item-paragraph">Given that $k$ is a positive constant, sketch the curve $E$ with equation $y=\left| {{x}^{2}}\left( x-k \right) \right|$, labelling the coordinates of the points where the curve cuts the axes.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">Hence, find, in terms of $k$, the area of the region(s) bounded by $E$ and the $x$-axis for $0\le x\le 2k$.</div><p style="text-align: right;">[3]</p></li></ol>

Integrals involving Modulus Functions

2024 NJC Promo Q3

<figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/converted/1760838133629-rtrhy1228n9.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvY29udmVydGVkLzE3NjA4MzgxMzM2MjktcnRyaHkxMjI4bjkucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5Mzg1MzgwNywiZXhwIjoxNzkzODU0MTA3fQ.qSGWLvbA2mVI1C5n0EJ_Vkquxsc_R0WJSOHTSirqT0U" width="400" class="aligncenter" /></figure>
<p>The figure above shows an ellipse with parametric equations</p>
<p style="text-align: center">${x = 2\cos\theta, y = 6\sin\left(\theta + \frac{\pi}{3}\right), -\pi \leq \theta < \pi}$</p>
<p>The curve meets the coordinate axes at points ${A}$, ${B}$, ${C}$ and ${D}$.</p>
<ol class="abc-list"><li><div class="list-item-paragraph">Find the coordinates of the points $A$, $B$, $C$ and $D$.</div></li></ol>
<p>The straight line $l$ is the tangent to the ellipse at the point $E$.</p>
<ol class="abc-list" start="2"><li><div class="list-item-paragraph">State the equation of $l$, given that $l$ is parallel to the $y$-axis.</div></li><li><div class="list-item-paragraph">Find the value of $\theta $ at the point $E$.</div></li></ol>
<p>The finite region bounded by the ellipse and the $x$-axis for which $y\ge 0$ is shown shaded in the figure above.</p>
<ol class="abc-list" start="4"><li><div class="list-item-paragraph">Show that the area of this region is given by the integral</div><div class="list-item-paragraph" style="text-align: center">$\displaystyle\int_{-\frac{\pi }{3}}^{\frac{2\pi }{3}}{3-3\cos 2\theta +3\sqrt{3}\sin 2\theta \,\mathrm{d}\theta }$.</div></li><li><div class="list-item-paragraph">Hence find the area of the shaded region.</div></li></ol>

The figure above shows an ellipse with parametric equations

${x = 2\cos\theta, y = 6\sin\left(\theta + \frac{\pi}{3}\right), -\pi \leq \theta < \pi}$

The curve meets the coordinate axes at points ${A}$, ${B}$, ${C}$ and ${D}$.

The straight line $l$ is the tangent to the ellipse at the point $E$.

The finite region bounded by the ellipse and the $x$-axis for which $y\ge 0$ is shown shaded in the figure above.

Tangents and Normals Parallel to the Axes

Parametric Equations

Parametric Equations Q2

<figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/converted/1760837813404-tu6u9e9jqj9.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvY29udmVydGVkLzE3NjA4Mzc4MTM0MDQtdHU2dTllOWpxajkucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5Mzg1MzgwNywiZXhwIjoxNzkzODU0MTA3fQ.Etk5lU2QsHWrqB7gg1_Ax1HTcWthrnkrNzxY_YwyflY" width="400" class="aligncenter" /></figure>
<p>The figure above shows a curve known as a re-entrant cycloid, with parametric equations</p>
<p style="text-align: center">$x=\theta -4\sin \theta $, $y=1-2\cos \theta $, $0\le \theta \le 2\pi $.</p>
<p>The curve crosses the $x$-axis at the points $P$ and $Q$.</p>
<ol class="abc-list"><li><div class="list-item-paragraph">Find the value of $\theta $ at the points $P$ and $Q$.</div></li><li><div class="list-item-paragraph">Show that the area of the finite region bounded by the curve and the $x$-axis, shown shaded in the figure above, is given by the integral</div><div class="list-item-paragraph" style="text-align: center">$\displaystyle\int_{{{\theta }_{1}}}^{{{\theta }_{2}}}{1-6\cos \theta +8{{\cos }^{2}}\theta \,\mathrm{d}\theta }$,</div><div class="list-item-paragraph">where ${{\theta }_{1}}$ and ${{\theta }_{2}}$ must be stated.</div></li><li><div class="list-item-paragraph">Find an exact value for the above integral.</div></li></ol>

The figure above shows a curve known as a re-entrant cycloid, with parametric equations

$x=\theta -4\sin \theta $, $y=1-2\cos \theta $, $0\le \theta \le 2\pi $.

The curve crosses the $x$-axis at the points $P$ and $Q$.

Integrating Powers of Sine and Cosine

Parametric Equations Q1

<p>A curve $C$ has parametric equations $x=1-\sin \theta $, $y=\theta +\cos \theta $, where $-\pi \le \theta \le \frac{\pi }{2}$.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Sketch the graph of $C$, stating the exact coordinates of the end points.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">Find $\frac{\mathrm{d}y}{\mathrm{d}x}$ in terms of $\theta $. What can be said about the tangent to $C$ as $\theta \to -\frac{\pi }{2}$?</div><p style="text-align: right;">[3]</p></li><li><div class="list-item-paragraph">A point $P$ on $C$ has parameter $p$, where $0 < p < \frac{\pi }{2}$. Show that the normal to $C$ at $P$ crosses the $y$-axis at point $Q$ with coordinates $\left( 0,p \right)$.</div><p style="text-align: right;">[3]</p></li><li><div class="list-item-paragraph">Show that the area of region bounded by $C$, the normal to $C$ at point $P$ and the $y$-axis is given by $a\pi +bp+c\cos p$, where $a$, $b$ and $c$ are to be determined.</div><p style="text-align: right;">[6]</p></li></ol>

A curve $C$ has parametric equations $x=1-\sin \theta $, $y=\theta +\cos \theta $, where $-\pi \le \theta \le \frac{\pi }{2}$.

2019 CJC P2 Q5

<p>The diagram shows the region ${R}$, which is bounded by the curve ${(y-2)^{2}=2x-3}$ and the line ${x=3.5}$.</p>
<div class="content-block content-block--columns"><div class="question-columns question-columns--2" data-layout="2"><div class="question-column" style="flex: 1; align-self: flex-start;"><p>Find the</p><ol class="i-list"><li data-list-item-id="item-1790411500021-19e5lk948" data-list-item-path="item-1790411500021-19e5lk948"><div class="list-item-paragraph"><p>area of region ${R}$.</p></div></li><li data-list-item-id="item-1790411500793-i7g0t6tg3" data-list-item-path="item-1790411500793-i7g0t6tg3"><div class="list-item-paragraph"><p>volume when&nbsp;${R}$ is rotated through four right angles about the ${x}$-axis.</p></div></li><li data-list-item-id="item-1790411501589-99mij59u7" data-list-item-path="item-1790411501589-99mij59u7"><div class="list-item-paragraph"><p>volume when&nbsp;${R}$ is rotated through ${2\pi}$ about the ${y}$-axis.</p></div></li></ol></div><div class="question-column" style="flex: 1; align-self: flex-start;"><div class="graph-block" data-graph-config="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" data-width="256" data-height="223" data-proportional style="width:256px;height:223px;max-width:100%;display:flow-root;margin:0 auto;--graph-width:256px;--graph-aspect:256 / 223;"></div></div></div></div>

The diagram shows the region ${R}$, which is bounded by the curve ${(y-2)^{2}=2x-3}$ and the line ${x=3.5}$.

Volume of Revolution about the y-axis

Volume of Revolution: Roots of Equations

<p>A curve has parametric equations $x={{\cos }^{2}}t$, $y={{\sin }^{3}}t$, for&nbsp;$0\le t\le \frac{\pi }{2}$.</p>
<ol class="i-list"><li data-list-item-id="item-1760750966244-f2jgujso1" data-list-item-path="item-1760750966244-f2jgujso1"><div class="list-item-paragraph">Sketch the curve.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-1760750966769-1jb7oaukh" data-list-item-path="item-1760750966769-1jb7oaukh"><div class="list-item-paragraph">The tangent to the curve at the point $\left( {{\cos }^{2}}\theta ,\,\,{{\sin }^{3}}\theta \right)$ where&nbsp;$0&lt;\theta &lt;\frac{\pi }{2}$, meets the $x$- and $y$-axes at $Q$ and $R$ respectively. The origin is denoted by $O$. Show that the area of triangle $OQR$ is $\frac{1}{12}\sin \theta {{\left( 3{{\cos }^{2}}\theta +2{{\sin }^{2}}\theta \right)}^{2}}$.<span class="inline-marks">[6]</span></div></li><li data-list-item-id="item-1760750967894-8wj87h6ye" data-list-item-path="item-1760750967894-8wj87h6ye"><div class="list-item-paragraph">Show that the area under the curve for&nbsp;$0\le t\le \frac{\pi }{2}$ is $2 \int_{0}^{\frac{\pi }{2}}{\cos t{{\sin }^{4}}\mathrm{t}\,\,\mathrm{d}t}$, and use the substitution $\sin t=u$ to find this area.<span class="inline-marks">[5]</span></div></li></ol>

A curve has parametric equations $x={{\cos }^{2}}t$, $y={{\sin }^{3}}t$, for $0\le t\le \frac{\pi }{2}$.

Derivatives of Trigonometric Functions

Integration by Substitution

N2007 P1 Q11

<p>A curve $C$ is given by the parametric equations</p>
<p style="text-align: center">$x=2+2\sin \theta $, $y=2\cos \theta +\sin 2\theta $, for $-\pi <\theta \le \pi $.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Sketch the curve, indicating clearly the coordinates of the axial intercepts.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">Find the exact area bounded by the curve.</div><p style="text-align: right;">[5]</p></li><li><div class="list-item-paragraph">Verify that $y=x\cos \theta $.</div><div class="list-item-paragraph">Deduce that the Cartesian equation of the curve $C$ is</div><div class="list-item-paragraph" style="text-align: center">$4{{y}^{2}}=4{{x}^{3}}-{{x}^{4}}$.</div><p style="text-align: right;">[3]</p></li><li><div class="list-item-paragraph">Find the volume of the solid of revolution formed when the curve $C$ is rotated $\pi $ radians about the $x$-axis.</div><p style="text-align: right;">[2]</p></li></ol>

A curve $C$ is given by the parametric equations

$x=2+2\sin \theta $, $y=2\cos \theta +\sin 2\theta $, for $-\pi <\theta \le \pi $.

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