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PSLE

PSLE Foundation Mathematics

<p>A curve $C$ has equation $y=\frac{2}{{{x}^{2}}-{{k}^{2}}}$ where&nbsp;$k\ge 2$. It is given that&nbsp;$C$ has only one stationary point.</p>
<ol class="i-list"><li data-list-item-id="item-1760439942706-1m9zyiuw3" data-list-item-path="item-1760439942706-1m9zyiuw3"><div class="list-item-paragraph">Find $\frac{\mathrm{d}y}{\mathrm{d}x}$. Hence find the coordinates of the stationary point in terms of $k$.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="item-1760439944386-u9p7qij1o" data-list-item-path="item-1760439944386-u9p7qij1o"><div class="list-item-paragraph">Sketch the graph of $C$, stating the equations of any asymptotes. coordinates of any points of intersection with the axes and the coordinates of the stationary point.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-1760439945173-d4tyuqve2" data-list-item-path="item-1760439945173-d4tyuqve2"><div class="list-item-paragraph">By adding a suitable graph on the same diagram in part <strong>(ii)</strong>, find the range of values of $h$, in terms of $k$, where&nbsp;$h$ is a positive real constant and $h\ne k$ such that&nbsp;${{\left( hx \right)}^{2}}+{{\left( \frac{2k}{{{x}^{2}}-{{k}^{2}}} \right)}^{2}}={{\left( kh \right)}^{2}}$ has at least one real root.<span class="inline-marks">[3]</span></div></li></ol>

A curve $C$ has equation $y=\frac{2}{{{x}^{2}}-{{k}^{2}}}$ where $k\ge 2$. It is given that $C$ has only one stationary point.

Sketching Rational Functions

Parameter Conditions for Graphs and Equations

Asymptotes and Discontinuities

Chain Rule

Stationary Points and Points of Inflection

Stationary Points and Their Nature

2020 TMJC Promo Q4

Techniques of Differentiation

Graphing Techniques

Applications of Differentiation

<p><strong>Do not use a calculator for this question</strong></p>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">Give that $y={{\mathrm{e}}^{\frac{2}{x+1}}}$, where $x\ne -1$, show that</span></div><div class="list-item-paragraph" style="text-align: center">${{(x+1)}^{2}}\frac{{{\mathrm{d}}^{2}}y}{\mathrm{d}{{x}^{2}}}+2(x+2)\frac{\mathrm{d}y}{\mathrm{d}x}=0$.</div><div class="list-item-paragraph">By further differentiation of this result, or otherwise, find the Maclaurin series of $y$, up to and including the term in ${{x}^{3}}$.</div><div class="list-item-paragraph">Leave the coefficients in exact form.</div><p style="text-align: right;">[7]</p></li><li><div class="list-item-content"><span class="list-item-text">Using appropriate expansions from the List of Formulae (MF26), find the expansion of ${{(4-x)}^{-1}}\cos \,\,3x$, up to and including the term in ${{x}^{3}}$.</span></div><div class="list-item-paragraph">Give the coefficients as exact fractions in their simplest form.</div><p style="text-align: right;">[5]</p></li></ol>

**Do not use a calculator for this question**

Give that $y={{\mathrm{e}}^{\frac{2}{x+1}}}$, where $x\ne -1$, show that

Using appropriate expansions from the List of Formulae (MF26), find the expansion of ${{(4-x)}^{-1}}\cos \,\,3x$, up to and including the term in ${{x}^{3}}$.

Maclaurin Series Using Standard Series

Maclaurin Series from Differential Equations

Maclaurin Series by Successive Differentiation

Higher Derivatives by Implicit Differentiation

Second and Higher Derivatives

Derivatives of Exponential Functions

Binomial Expansion as a Maclaurin Series

Products and Quotients of Power Series

2020 TMJC Promo Q6

Maclaurin Series and Power Series

<p>Find the derivatives of the following functions with respect to $x$:</p>
<ol class="i-list"><li><div class="list-item-paragraph">$y = \ln \left( \frac{\tan^{3}(1 - e^{x})}{\ln x^{3}} \right)$</div></li><li><div class="list-item-paragraph">$y = \ln \left( \frac{x^{5}}{\sin^{2}x} \right)$</div></li><li><div class="list-item-paragraph">$y = \ln \left( \frac{\ln x^{3}}{\tan x} \right)$</div></li></ol>

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

Derivatives of Trigonometric Functions

Derivatives of Natural Logarithmic Functions

<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 $x = x_1$.

Area Bounded by Curves

2025 JPJC P1 Q1

Applications of Integration

<p>It is given that $y={{e}^{{{\tan }^{-1}}4x}}$.</p>
<ol class="abc-list"><li><div class="list-item-paragraph">Show that $\left( 1+16{{x}^{2}} \right)\frac{{{\mathrm{d}}^{2}}y}{\mathrm{d}{{x}^{2}}}+k\frac{\mathrm{d}y}{\mathrm{d}x}\left( 8x-1 \right)=0$, where $k$ is a constant to be found.</div><p style="text-align: right;">[3]</p></li><li><div class="list-item-paragraph">By further differentiation of the result in part <strong>(a)</strong>, find the Maclaurin series for $y$ up to and including the term in ${{x}^{3}}$.</div><p style="text-align: right;">[3]</p></li><li><div class="list-item-paragraph">The first three non-zero terms of the Maclaurin expansion of ${{e}^{{{\tan }^{-1}}4x}}$ are equal to the first three non-zero terms of the series expansion of $\ln \left( \frac{a}{1-bx} \right)$, where $a$ and $b$ are constants. Using standard series from the List of Formulae (MF27), find the exact values of $a$ and $b$.</div><p style="text-align: right;">[2]</p></li></ol>

Show that $(1+16x^2)\frac{d^2y}{dx^2}+k\frac{dy}{dx}(8x-1)=0$, where $k$ is a constant to be found.

By further differentiation of the result in part (a), find the Maclaurin series for $y$ up to and including the term in $x^3$.

The first three non-zero terms of the Maclaurin expansion of $e^{\tan^{-1}4x}$ are equal to the first three non-zero terms of the series expansion of $\ln\left(\frac{a}{1-bx}\right)$, where $a$ and $b$ are constants. Using standard series from the List of Formulae (MF27), find the exact values of $a$ and $b$.

Finding Parameters from Power Series Coefficients

Derivatives of Inverse Trigonometric Functions

2025 DHS Promo Q6

<p>It is given that&nbsp;$y=\sqrt{1+\ln (1+x)}.$</p>
<ol class="i-list"><li data-list-item-id="item-1758949333462-cddz9iim8" data-list-item-path="item-1758949333462-cddz9iim8"><div class="list-item-paragraph">By differentiating successively, find the Maclaurin’s series for&nbsp;$y,$ up to and including the term in ${{x}^{2}}.$<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-1758949334189-rq45dru97" data-list-item-path="item-1758949334189-rq45dru97"><div class="list-item-paragraph">By substituting $x=\frac{1}{4},$approximate $\ln \left( \frac{4}{5} \right)$, leaving your answer in exact form.<span class="inline-marks">[3]</span></div></li></ol>

By differentiating successively, find the Maclaurin’s series for $y,$ up to and including the term in ${{x}^{2}}.$

Approximations Using Power Series

VJC P1 Q2

<p>Given that&nbsp;$y={{({{\sin }^{-1}}x)}^{2}}$, show that</p>
<p style="text-align: center">\[\left( 1-{{x}^{2}} \right){{\left( \frac{\mathrm{d}y}{\mathrm{d}x} \right)}^{2}}=4y\]</p>
<p style="text-align: center">and $\left( 1-{{x}^{2}} \right)\frac{{{\mathrm{d}}^{\text{2}}}y}{\mathrm{d}{{x}^{2}}}-x\frac{\mathrm{d}y}{\mathrm{d}x}=2$.<span class="inline-marks">[3]</span></p>
<p>By further differentiation of these results, find the Maclaurin series of $y$ up to including the term in ${{x}^{4}}$.<span class="inline-marks">[3]</span></p>
<p>The series found above is to be used to estimate a value for&nbsp;${{\pi }^{2}}$. Explain with the aid of a sketch why using $x=\frac{1}{2}$ in the series gives a better approximation than $x=1$, and hence find an approximation to ${{\pi }^{2}}$, giving your answer in fraction form. <span class="inline-marks">[4]</span></p>

Given that $y={{({{\sin }^{-1}}x)}^{2}}$, show that

\[\left( 1-{{x}^{2}} \right){{\left( \frac{\mathrm{d}y}{\mathrm{d}x} \right)}^{2}}=4y\]

and $\left( 1-{{x}^{2}} \right)\frac{{{\mathrm{d}}^{\text{2}}}y}{\mathrm{d}{{x}^{2}}}-x\frac{\mathrm{d}y}{\mathrm{d}x}=2$.

By further differentiation of these results, find the Maclaurin series of $y$ up to including the term in ${{x}^{4}}$.

The series found above is to be used to estimate a value for ${{\pi }^{2}}$. Explain with the aid of a sketch why using $x=\frac{1}{2}$ in the series gives a better approximation than $x=1$, and hence find an approximation to ${{\pi }^{2}}$, giving your answer in fraction form. 

Accuracy of Power Series Approximations

2013 ACJC P1 Q12

<p>A curve $C$ has parametric equations</p>
<p style="text-align: center">$x={{\sin }^{2}}\theta $, $y=2\sin \theta -{{\sin }^{3}}\theta $,</p>
<p>where $-\frac{\pi }{2}\le \theta \le \frac{\pi }{2}$.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Find the value(s) of the parameter $\theta$ at the point(s) where the curve $C$ is stationary.</div><p style="text-align: right;">[4]</p></li><li><div class="list-item-paragraph">Show that the equation of the tangent to the curve at the point where the gradient is $\frac{5}{4}$ is given by $y=\frac{5}{4}x+\frac{9}{16}$.</div><p style="text-align: right;">[4]</p></li><li><div class="list-item-paragraph">Determine whether the tangent $y=\frac{5}{4}x+\frac{9}{16}$ meets the curve $C$ again.</div><p style="text-align: right;">[3]</p></li></ol>

$x={{\sin }^{2}}\theta $, $y=2\sin \theta -{{\sin }^{3}}\theta $,

where $-\frac{\pi }{2}\le \theta \le \frac{\pi }{2}$.

Tangents and Normals to Parametric Curves

Differentiation of Parametric Equations

2016 MI P1 Q11

<p>Given that $y={{\mathrm{e}}^{{{\sin }^{-1}}\left( 2x \right)}}$, show that $\left( 1-4{{x}^{2}} \right)\frac{{{\mathrm{d}}^{2}}y}{\mathrm{d}{{x}^{2}}}-4x\frac{\mathrm{d}y}{\mathrm{d}x}=4y$.</p>
<p>By further differentiating this result, find the Maclaurin series for $y$ in ascending powers of $x$, up to and including the term in ${{x}^{3}}$.</p>
<p>Hence, find an approximation value of ${{\mathrm{e}}^{\frac{\pi }{2}}}$, by substituting a suitable value of $x$ in the Maclaurin series $y$.</p>
<p>Suggest one way to improve the accuracy of the approximated value obtained.</p>

Given that $y={{\mathrm{e}}^{{{\sin }^{-1}}\left( 2x \right)}}$, show that $\left( 1-4{{x}^{2}} \right)\frac{{{\mathrm{d}}^{2}}y}{\mathrm{d}{{x}^{2}}}-4x\frac{\mathrm{d}y}{\mathrm{d}x}=4y$.

By further differentiating this result, find the Maclaurin series for $y$ in ascending powers of $x$, up to and including the term in ${{x}^{3}}$.

Hence, find an approximation value of ${{\mathrm{e}}^{\frac{\pi }{2}}}$, by substituting a suitable value of $x$ in the Maclaurin series $y$.

Suggest one way to improve the accuracy of the approximated value obtained.

2006 NJC Promo Q5

<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">Differentiate ${{\sin }^{-1}}\left( {{\mathrm{e}}^{2x}} \right)$ with respect to $x$.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">Hence find $\displaystyle\int{\frac{2{{\mathrm{e}}^{2x}}+{{\mathrm{e}}^{4x}}}{\sqrt{1-{{\mathrm{e}}^{4x}}}}}\,\mathrm{d}x$.</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 ${{\sin }^{n-2}}\theta {{\cos }^{2}}\theta ={{\sin }^{n-2}}\theta -{{\sin }^{n}}\theta $, where $n\in {{\mathbb{Z}}^{+}}$, $n\ge 2$.</div><p style="text-align: right;">[1]</p></li><li><div class="list-item-paragraph">By considering ${{\sin }^{n}}\theta ={{\sin }^{n-1}}\theta \sin \theta $, show that</div><div class="list-item-paragraph" style="text-align: center">$\displaystyle\int_{0}^{\frac{\pi }{2}}{{{\sin }^{n}}\theta \,\,\mathrm{d}\theta =\frac{n-1}{n}\displaystyle\int_{0}^{\frac{\pi }{2}}{{{\sin }^{n-2}}\theta \,\,\mathrm{d}\theta }}$,</div><div class="list-item-paragraph">where $n\in {{\mathbb{Z}}^{+}}$, $n\ge 2$.</div><p style="text-align: right;">[4]</p></li><li><div class="list-item-paragraph">Hence find the exact value of $\displaystyle\int_{0}^{\frac{\pi }{2}}{{{\sin }^{3}}\theta \,\,\mathrm{d}\theta }$.</div><p style="text-align: right;">[2]</p></li></ol></li></ol>

Reduction Formulae for Integrals

Integration by Parts

Integration by the Reverse Chain Rule

Integrable Forms of Trigonometric Functions

Integration as an Inverse Process of Differentiation

Product, Quotient and Chain Rules

2023 MI P1 Q10

Techniques of Integration

<p>Given that&nbsp;$y={{({{\sin }^{-1}}x)}^{2}}$, show that&nbsp;$(1-{{x}^{2}}){{\left(\frac{\mathrm{d}y}{\mathrm{d}x} \right)}^{2}}=4y$ and $(1-{{x}^{2}})\frac{{{\mathrm{d}}^{2}}y}{\mathrm{d}{{x}^{2}}}-x\frac{\mathrm{d}y}{\mathrm{d}x}=2$.</p>

Given that $y={{({{\sin }^{-1}}x)}^{2}}$, show that $(1-{{x}^{2}}){{\left(\frac{\mathrm{d}y}{\mathrm{d}x} \right)}^{2}}=4y$ and $(1-{{x}^{2}})\frac{{{\mathrm{d}}^{2}}y}{\mathrm{d}{{x}^{2}}}-x\frac{\mathrm{d}y}{\mathrm{d}x}=2$.

<p>A curve has parametric equation $x=2t-1$ , $y=\frac{1}{{{t}^{2}}+1}$.</p>
<ol class="abc-list"><li><div class="list-item-paragraph">Prove that the equation of the tangent at the point with parameter $t$ is $(t^{2} + 1)^{2}y + tx = 3t^{2} - t + 1$.</div></li><li><div class="list-item-paragraph">The tangent at the point where&nbsp;$t = 3$ meets the curve again at the point where&nbsp;$t = q$. Find the value of $q$ and hence find the coordinates of the point of intersection.</div></li></ol>

A curve has parametric equation $x=2t-1$ , $y=\frac{1}{{{t}^{2}}+1}$.

<p>It is given that $\mathrm{f}\left( x \right)=\ln \left( \sin x+\cos x \right)$ for $-\frac{\pi }{4} < x\le \frac{\pi }{4}$.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Show that $\mathrm{f}\,\mathrm{ }\!\!'\!\!\mathrm{ }\!\!'\!\!\mathrm{ }\left( x \right)+{{\left[ \mathrm{{f}'}(x) \right]}^{2}}+1=0$.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">By further differentiation of the result in part <strong>(i)</strong>, find the Maclaurin series for $\mathrm{f}\left( x \right)$ up to and including the term in ${{x}^{3}}$.</div><p style="text-align: right;">[4]</p></li><li><div class="list-item-paragraph">Using the expansion obtained in part <strong>(ii)</strong> and the standard series from the List of Formulae (MF26), find the series expansion of $\mathrm{g}\left( x \right)=\ln \left( 1+\tan x \right)$ for $-\frac{\pi }{4} < x\le \frac{\pi }{4}$ as far as the term in ${{x}^{3}}$.</div><p style="text-align: right;">[4]</p></li></ol>

It is given that $\mathrm{f}\left( x \right)=\ln \left( \sin x+\cos x \right)$ for $-\frac{\pi }{4} < x\le \frac{\pi }{4}$.

Substitution and Composition of Power Series

2018 AJC P1 Q5

<p>Differentiate $\tan^3 x$ with respect to $x$ and express the derivative in terms of $\sec x$ only. Hence, evaluate $\displaystyle\int_{0}^{\frac{\pi}{4}} \sec^4 x \, \mathrm{d}x$.</p>

Differentiate $\tan^3 x$ with respect to $x$ and express the derivative in terms of $\sec x$ only. Hence, evaluate $\displaystyle\displaystyle \displaystyle\int_{0}^{\frac{\pi}{4}} \sec^4 x \, \mathrm{d}x$.

Definite Integration as the Reverse of Differentiation

Integration

Differentiation

<p>Given that&nbsp;$y=(3x+1)\sqrt{6x-1}$, express $\frac{\mathrm{d}y}{\mathrm{d}x}$ in the form $\frac{kx}{\sqrt{6x - 1}}$, where&nbsp;$k$ is a constant. </p>
<p>Hence, find $\displaystyle\displaystyle \displaystyle\int \frac{9x}{\sqrt{6x - 1}} \, \mathrm{d}x.$</p>

Given that $y=(3x+1)\sqrt{6x-1}$, express $\frac{\mathrm{d}y}{\mathrm{d}x}$ in the form $\frac{kx}{\sqrt{6x - 1}}$, where $k$ is a constant. 

Hence, find $\displaystyle\displaystyle \displaystyle\int \frac{9x}{\sqrt{6x - 1}} \, \mathrm{d}x.$

Product Rule

<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">Express $\frac{27x^2-6x+9}{(4x-1)(x^2+3)}$ in partial fractions.</span></div></li><li><div class="list-item-content"><span class="list-item-text">Differentiate $\ln(x^2+3)$ with respect of $x$.</span></div></li><li><div class="list-item-paragraph">Using the results of <strong>(a)</strong> and <strong>(b)</strong>, find $\displaystyle\int \frac{27x^2-6x+9}{(4x-1)(x^2+3)} \, \mathrm{d}x$.</div></li></ol>

Express $\frac{27x^2-6x+9}{(4x-1)(x^2+3)}$ in partial fractions.

Differentiate $\ln(x^2+3)$ with respect of $x$.

Using the results of **(a)** and **(b)**, find $\displaystyle\displaystyle \displaystyle\int \frac{27x^2-6x+9}{(4x-1)(x^2+3)} \, \mathrm{d}x$.

Integration Using Partial Fractions

Partial Fractions

Polynomials, Cubic Equations and Partial Fractions

<p>Find $\frac{\mathrm{d}}{\mathrm{d}x}\left( \frac{\cos^5 2x}{10} \right)$ and hence, find $\displaystyle\displaystyle \displaystyle\int (x - \sin 2x \cos^4 2x) \, \mathrm{d}x$.</p>

Find $\frac{\mathrm{d}}{\mathrm{d}x}\left( \frac{\cos^5 2x}{10} \right)$ and hence, find $\displaystyle\displaystyle \displaystyle\int (x - \sin 2x \cos^4 2x) \, \mathrm{d}x$.

<p>Show that $\frac{\mathrm{d}}{\mathrm{d}x} \left[ (2x-1)\sqrt{x+3} \right] = \frac{6x+11}{2\sqrt{x+3}}$. Hence, find $\displaystyle\int \frac{6x+11}{\sqrt{x+3}} \, \mathrm{d}x$.</p>

Show that $\frac{\mathrm{d}}{\mathrm{d}x} \left[ (2x-1)\sqrt{x+3} \right] = \frac{6x+11}{2\sqrt{x+3}}$. Hence, find $\displaystyle\displaystyle \displaystyle\int \frac{6x+11}{\sqrt{x+3}} \, \mathrm{d}x$.

<p>Differentiate $\frac{2}{2-x^2}$ with respect to $x$. Hence find $\displaystyle\displaystyle \displaystyle\int \frac{x}{(2-x^2)^2} \, \mathrm{d}x.$</p>

Differentiate $\frac{2}{2-x^2}$ with respect to $x$. Hence find $\displaystyle\displaystyle \displaystyle\int \frac{x}{(2-x^2)^2} \, \mathrm{d}x.$

<p>Given that $\frac{6+5x-8x^2}{(2x-1)(x^2+3)} = \frac{A}{2x-1} + \frac{Bx+C}{x^2+3}$, where $A$, $B$ and $C$ are constants</p>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">find the values of $A$, $B$ and $C$,</span></div></li><li><div class="list-item-content"><span class="list-item-text">differentiate $\ln(x^2 + 3)$ with respect to $x$.</span></div></li><li><div class="list-item-paragraph">Using the results from parts <strong>(a)</strong> and <strong>(b)</strong>, find $\displaystyle\int \frac{6+5x-8x^2}{(2x-1)(x^2+3)} \, \mathrm{d}x$.</div></li></ol>

Given that $\frac{6+5x-8x^2}{(2x-1)(x^2+3)} = \frac{A}{2x-1} + \frac{Bx+C}{x^2+3}$, where $A$, $B$ and $C$ are constants

find the values of $A$, $B$ and $C$,

differentiate $\ln(x^2 + 3)$ with respect to $x$.

Using the results from parts **(a)** and **(b)**, find $\displaystyle\displaystyle \displaystyle\int \frac{6+5x-8x^2}{(2x-1)(x^2+3)} \, \mathrm{d}x$.

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