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PSLE

PSLE Foundation Mathematics

<p>A particle moves back and forth in a horizontal line between two end points. A fixed origin $O$ is the midpoint between these end points. Its displacement $x$ metres from $O$ at time $t$ seconds is $x=4\sin(2t+\frac\pi3)$, where&nbsp;$0\leq t\leq\pi$.</p>
<ol class="abc-list"><li data-list-item-id="p3-tzc-q02-item01" data-list-item-path="p3-tzc-q02-item01"><div class="list-item-paragraph">Determine</div><ol class="i-list"><li data-list-item-id="p3-tzc-q02-sub01-item01" data-list-item-path="p3-tzc-q02-item01/p3-tzc-q02-sub01-item01"><div class="list-item-paragraph">the amplitude;<span class="inline-marks">[1]</span></div></li><li data-list-item-id="p3-tzc-q02-sub01-item02" data-list-item-path="p3-tzc-q02-item01/p3-tzc-q02-sub01-item02"><div class="list-item-paragraph">the initial displacement;<span class="inline-marks">[2]</span></div></li><li data-list-item-id="p3-tzc-q02-sub01-item03" data-list-item-path="p3-tzc-q02-item01/p3-tzc-q02-sub01-item03"><div class="list-item-paragraph">the value of $t$ when the particle first passes through $O$.<span class="inline-marks">[2]</span></div></li></ol></li><li data-list-item-id="p3-tzc-q02-item02" data-list-item-path="p3-tzc-q02-item02"><div class="list-item-paragraph">The acceleration $a$ at displacement $s$ satisfies $a=-n^2s$, where&nbsp;$n&gt;0$.</div><div class="list-item-paragraph">The displacement is $s=A\sin(nt+b)$, where&nbsp;$t\geq0$, $A,n&gt;0$ and $-\pi\leq b\leq\pi$.</div><div class="list-item-paragraph">By finding expressions for&nbsp;$\frac{\mathrm{d}s}{\mathrm{d}t}$ and $\frac{\mathrm{d}^2s}{\mathrm{d}t^2}$, verify that&nbsp;$s=A\sin(nt+b)$ satisfies $a=-n^2s$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="p3-tzc-q02-item03" data-list-item-path="p3-tzc-q02-item03"><ol class="i-list"><li data-list-item-id="p3-tzc-q02-sub03-item01" data-list-item-path="p3-tzc-q02-item03/p3-tzc-q02-sub03-item01"><div class="list-item-paragraph">Use the chain rule to show that&nbsp;$a=v\frac{\mathrm{d}v}{\mathrm{d}s}$, where&nbsp;$v$ is velocity.<span class="inline-marks">[1]</span></div></li><li data-list-item-id="p3-tzc-q02-sub03-item02" data-list-item-path="p3-tzc-q02-item03/p3-tzc-q02-sub03-item02"><div class="list-item-paragraph">By solving $v\frac{\mathrm{d}v}{\mathrm{d}s}=-n^2s$, show that&nbsp;$v^2=n^2(A^2-s^2)$.<span class="inline-marks">[5]</span></div></li><li data-list-item-id="p3-tzc-q02-sub03-item03" data-list-item-path="p3-tzc-q02-item03/p3-tzc-q02-sub03-item03"><div class="list-item-paragraph">Hence, or otherwise, find the particle’s maximum speed.<span class="inline-marks">[2]</span></div></li></ol></li><li data-list-item-id="p3-tzc-q02-item04" data-list-item-path="p3-tzc-q02-item04"><div class="list-item-paragraph">The continuous random variable $S$ denotes the particle’s displacement $s$ from $O$ at time $t$.</div><div class="list-item-paragraph">\[f(s)=\begin{cases}\frac1{\pi\sqrt{A^2-s^2}},&-A&lt;s&lt;A,\\0,&\text{otherwise}.\end{cases}\]</div><div class="list-item-paragraph">Show that&nbsp;$P(-\frac A2\leq S\leq\frac A2)=\frac13$.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="p3-tzc-q02-item05" data-list-item-path="p3-tzc-q02-item05"><div class="list-item-paragraph">For $-A&lt;s&lt;A$, $f(s)$ can be expressed as $\frac{k}{|v(s)|}$, where $k&gt;0$ and $v(s)$ is the particle’s velocity at displacement $s$ from $O$.</div><div class="list-item-paragraph">Find the value of $k$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="p3-tzc-q02-item06" data-list-item-path="p3-tzc-q02-item06"><ol class="i-list"><li data-list-item-id="p3-tzc-q02-sub06-item01" data-list-item-path="p3-tzc-q02-item06/p3-tzc-q02-sub06-item01"><div class="list-item-paragraph">Determine $E(S)$, justifying your answer.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="p3-tzc-q02-sub06-item02" data-list-item-path="p3-tzc-q02-item06/p3-tzc-q02-sub06-item02"><div class="list-item-paragraph">Interpret the result in the context of the particle’s motion.<span class="inline-marks">[1]</span></div></li></ol></li></ol>

A particle moves back and forth in a horizontal line between two end points. A fixed origin $O$ is the midpoint between these end points. Its displacement $x$ metres from $O$ at time $t$ seconds is $x=4\sin(2t+\frac\pi3)$, where $0\leq t\leq\pi$.

Solving Differential Equations by Separation of Variables

Given the Displacement Function

Derivatives of Trigonometric Functions

Probability Density Functions

Expectation and Variance of Continuous Random Variables

Second and Higher Derivatives

Product, Quotient and Chain Rules

Probabilities by Integrating a Density Function

IB AA HL May 2026 P3 TZC Q2

Techniques of Differentiation

Continuous Random Variables

Differential Equations

Kinematics

<p>Let&nbsp;$\mathrm{f}(x)=(x^2+2x)\mathrm{e}^x$.</p>
<ol class="abc-list"><li data-list-item-id="quadraticexp-part-a" data-list-item-path="quadraticexp-part-a"><div class="list-item-paragraph">Using the definition $\mathrm{g}'(x)=\lim_{h\to0}\dfrac{\mathrm{g}(x+h)-\mathrm{g}(x)}{h}$, show from first principles that the derivative of $\mathrm{g}(x)=x^2+2x$ is $2x+2$.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="quadraticexp-part-b" data-list-item-path="quadraticexp-part-b"><div class="list-item-paragraph">Prove by induction that, for every positive integer $n$, the $n^{\mathrm{th}}$ derivative of $\mathrm{f}$ is $\mathrm{f}^{(n)}(x)=\mathrm{e}^x[x^2+2(n+1)x+n(n+1)]$.<span class="inline-marks">[9]</span></div></li></ol>

Let $\mathrm{f}(x)=(x^2+2x)\mathrm{e}^x$.

Using the definition $\mathrm{g}'(x)=\lim_{h\to0}\dfrac{\mathrm{g}(x+h)-\mathrm{g}(x)}{h}$, show from first principles that the derivative of $\mathrm{g}(x)=x^2+2x$ is $2x+2$.

Prove by induction that, for every positive integer $n$, the $n^{\mathrm{th}}$ derivative of $\mathrm{f}$ is $\mathrm{f}^{(n)}(x)=\mathrm{e}^x[x^2+2(n+1)x+n(n+1)]$.

Proof by Mathematical Induction

Differentiation by First Principles

Derivatives of Exponential Functions

Quadratic exponential derivative pattern

Mathematical Induction

<ol class="abc-list"><li data-list-item-id="ib2012mhp1tz2q13-part-a" data-list-item-path="ib2012mhp1tz2q13-part-a"><div class="list-item-paragraph">Using the definition of a derivative as&nbsp;$\mathrm{f}'(x)=\lim_{h\to0}\left(\frac{\mathrm{f}(x+h)-\mathrm{f}(x)}{h}\right)$, show that the derivative of $\frac{1}{2x+1}$ is $\frac{-2}{(2x+1)^2}$.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="ib2012mhp1tz2q13-part-b" data-list-item-path="ib2012mhp1tz2q13-part-b"><div class="list-item-paragraph">Prove by induction that the $n^{\mathrm{th}}$ derivative of $(2x+1)^{-1}$ is $(-1)^n\frac{2^n n!}{(2x+1)^{n+1}}$.<span class="inline-marks">[9]</span></div></li></ol>

Using the definition of a derivative as $\mathrm{f}'(x)=\lim_{h\to0}\left(\frac{\mathrm{f}(x+h)-\mathrm{f}(x)}{h}\right)$, show that the derivative of $\frac{1}{2x+1}$ is $\frac{-2}{(2x+1)^2}$.

Prove by induction that the $n^{\mathrm{th}}$ derivative of $(2x+1)^{-1}$ is $(-1)^n\frac{2^n n!}{(2x+1)^{n+1}}$.

IB Mathematics HL May 2012 P1 TZ2 Q13

<p>A particle moves back and forth in a horizontal line between two end points. A fixed origin $O$ is the midpoint between these end points. Its displacement $x$ metres from $O$ at time $t$ seconds is $x=4\sin(2t+\frac\pi3)$, where&nbsp;$0\leq t\leq\pi$.</p>
<ol class="abc-list"><li data-list-item-id="h2-motion-3" data-list-item-path="h2-motion-3"><div class="list-item-paragraph">Determine</div><ol class="i-list"><li data-list-item-id="h2-motion-4" data-list-item-path="h2-motion-3/h2-motion-4"><div class="list-item-paragraph">the amplitude;<span class="inline-marks">[1]</span></div></li><li data-list-item-id="h2-motion-6" data-list-item-path="h2-motion-3/h2-motion-6"><div class="list-item-paragraph">the initial displacement;<span class="inline-marks">[2]</span></div></li><li data-list-item-id="h2-motion-8" data-list-item-path="h2-motion-3/h2-motion-8"><div class="list-item-paragraph">the value of $t$ when the particle first passes through $O$.<span class="inline-marks">[2]</span></div></li></ol></li><li data-list-item-id="h2-motion-11" data-list-item-path="h2-motion-11"><div class="list-item-paragraph">More generally, a particle moves along a straight line with displacement $s=A\sin(nt+b)$ metres at time $t$ seconds, where&nbsp;$t\geq0$, $A,n&gt;0$ and $-\pi\leq b\leq\pi$.</div><div class="list-item-paragraph">Its velocity and acceleration are $v=\frac{\mathrm{d}s}{\mathrm{d}t}$ and $a=\frac{\mathrm{d}v}{\mathrm{d}t}$, respectively.</div><div class="list-item-paragraph">Find expressions for&nbsp;$\frac{\mathrm{d}s}{\mathrm{d}t}$ and $\frac{\mathrm{d}^2s}{\mathrm{d}t^2}$, and hence show that&nbsp;$a=-n^2s$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="h2-motion-15" data-list-item-path="h2-motion-15"><ol class="i-list"><li data-list-item-id="h2-motion-16" data-list-item-path="h2-motion-15/h2-motion-16"><div class="list-item-paragraph">Use the chain rule to show that&nbsp;$a=v\frac{\mathrm{d}v}{\mathrm{d}s}$, where&nbsp;$v$ is velocity.<span class="inline-marks">[1]</span></div></li><li data-list-item-id="h2-motion-18" data-list-item-path="h2-motion-15/h2-motion-18"><div class="list-item-paragraph">The particle is instantaneously at rest when&nbsp;$s=A$. By separating the variables in $v\frac{\mathrm{d}v}{\mathrm{d}s}=-n^2s$, show that&nbsp;$v^2=n^2(A^2-s^2)$.</div></li><li data-list-item-id="h2-motion-20" data-list-item-path="h2-motion-15/h2-motion-20"><div class="list-item-paragraph">Hence, or otherwise, find the particle’s maximum speed.<span class="inline-marks">[2]</span></div></li></ol></li></ol>

Verifying Solutions of Differential Equations

Chain Rule

IB AA HL May 2026 P3 TZC Q2 (H2 adaptation)

<p>A curve is given by the equation $y = \frac{3x + 10}{\sqrt{x + 1}}$.</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">Find $\frac{\mathrm{d}y}{\mathrm{d}x}$ and hence find the $x$-coordinate of the turning point of the curve.<span class="inline-marks">[3]</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">Use calculus to find the value of the second derivative at the turning point and state the nature of this turning point.<span class="inline-marks">[3]</span></div></li></ol>

A curve is given by the equation $y = \frac{3x + 10}{\sqrt{x + 1}}$.

Find $\frac{\mathrm{d}y}{\mathrm{d}x}$ and hence find the $x$-coordinate of the turning point of the curve.

Use calculus to find the value of the second derivative at the turning point and state the nature of this turning point.

Quotient Rule

Applications of Differentiation

Stationary Points and Their Nature

N2025 P1 Q4

<p>Given that</p>
<p style="text-align: center">\[y = A \cos [\ln(x+1)] + B \sin [\ln(x+1)],\]</p>
<p>where&nbsp;$A$ and $B$ are constants with $x &gt; -1$.</p>
<p>Show that&nbsp;$(x+1)^2 \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + (x+1) \frac{\mathrm{d}y}{\mathrm{d}x} + y = 0$.<span class="inline-marks">[5]</span></p>

\[y = A \cos [\ln(x+1)] + B \sin [\ln(x+1)],\]

where $A$ and $B$ are constants with $x > -1$.

Show that $(x+1)^2 \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + (x+1) \frac{\mathrm{d}y}{\mathrm{d}x} + y = 0$.

Derivatives of Natural Logarithmic Functions

2018 AJC C1BT1 Q3

<p>A curve is such that&nbsp;$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{9}{(2x-1)^2}$, where&nbsp;$x \neq \frac{1}{2}$. The point $(2, 12)$ lies on the curve.</p>
<ol class="i-list"><li data-list-item-id="item-1778561612436-sp6wzqj4x"><div class="list-item-paragraph">Find the equation of the curve.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-1778561619148-fviss5i9e"><div class="list-item-paragraph">Find the value of $\frac{\mathrm{d}^2y}{\mathrm{d}x^2}$ when&nbsp;$x = 0$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-1778561616412-w3ixzyy43"><div class="list-item-paragraph">A student concludes from <strong>(ii)</strong> that since the value of $\frac{\mathrm{d}^2y}{\mathrm{d}x^2} &gt; 0$ at $x = 0$, hence there is a minimum point at $x = 0$. State with reason and mathematical working whether the above conclusion is correct.<span class="inline-marks">[2]</span></div></li></ol>

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

Finding the Equation of the Curve

AMKSS 4AM 2022 Prelims P1 Q12

Integration

Differentiation

<p>Using the techniques of differentiation, or otherwise, answer the following questions:</p>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">Given that&nbsp;$y = \frac{x^2}{\mathrm{e}^x}$ and $\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + 2\frac{\mathrm{d}y}{\mathrm{d}x} = m \mathrm{e}^{-x} - y$, find the value of $m$.</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">Given that&nbsp;$y = a \mathrm{e}^{2x} + b \mathrm{e}^{-x}$ and $\frac{\mathrm{d}y}{\mathrm{d}x} + 4y = \mathrm{e}^{2x} - \mathrm{e}^{-x}$, find the values of $a$ and $b$.</div></li></ol>

Using the techniques of differentiation, or otherwise, answer the following questions:

Given that $y = \frac{x^2}{\mathrm{e}^x}$ and $\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + 2\frac{\mathrm{d}y}{\mathrm{d}x} = m \mathrm{e}^{-x} - y$, find the value of $m$.

Given that $y = a \mathrm{e}^{2x} + b \mathrm{e}^{-x}$ and $\frac{\mathrm{d}y}{\mathrm{d}x} + 4y = \mathrm{e}^{2x} - \mathrm{e}^{-x}$, find the values of $a$ and $b$.

<p>A curve has the equation $y = -5x^3 + 24x^2 - 7x - 5$.</p>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">Find the value of $x$ when the maximum gradient of the curve occurs.</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">Hence find the maximum gradient of the curve.</div></li></ol>

A curve has the equation $y = -5x^3 + 24x^2 - 7x - 5$.

Find the value of $x$ when the maximum gradient of the curve occurs.

Hence find the maximum gradient of the curve.

Derivative as the Gradient of a Curve

Problems Involving Maximisation and Minimisation

<p>A curve has the equation $y = 2x^3 - 9x^2 + 5x + 3$.</p>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">Find the value of $x$ when the minimum gradient of the curve occurs.</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">Hence find the minimum gradient of the curve.</div></li></ol>

A curve has the equation $y = 2x^3 - 9x^2 + 5x + 3$.

Find the value of $x$ when the minimum gradient of the curve occurs.

Hence find the minimum gradient of the curve.

<p>A curve has the equation $y = \frac{3x - 2}{x^2 + 4}$.</p>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">Obtain expressions for&nbsp;$\frac{\mathrm{d}y}{\mathrm{d}x}$ and $\frac{\mathrm{d}^2y}{\mathrm{d}x^2}$ in the simplest form.</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">Hence find, respectively, the range of values of $x$ for which $y$ is increasing and decreasing with respect to $x$.</div></li></ol>

A curve has the equation $y = \frac{3x - 2}{x^2 + 4}$.

Obtain expressions for $\frac{\mathrm{d}y}{\mathrm{d}x}$ and $\frac{\mathrm{d}^2y}{\mathrm{d}x^2}$ in the simplest form.

Hence find, respectively, the range of values of $x$ for which $y$ is increasing and decreasing with respect to $x$.

Increasing and Decreasing Functions

<p>A curve has the equation $y = \frac{2x^2}{x + 2}$.</p>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">Obtain expressions for&nbsp;$\frac{\mathrm{d}y}{\mathrm{d}x}$ and $\frac{\mathrm{d}^2y}{\mathrm{d}x^2}$ in the simplest form.</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">Hence find, respectively, the range of values of $x$ for which $y$ is increasing and decreasing with respect to $x$.</div></li></ol>

A curve has the equation $y = \frac{2x^2}{x + 2}$.

<p>The point $P$ travels along the curve $C$ with equation $y = x \sin^{-1} x$, $-1 &lt; x &lt; 1$. Let the gradient of the curve $C$ at the point $P$ be $m$.</p>
<p>If the $x$-coordinate of $P$ is increasing at the rate of $9$ units per second when&nbsp;$x = \frac{1}{2}$, find the exact value of the rate at which $m$ is changing at this instant.<span class="inline-marks">[4]</span></p>

The point $P$ travels along the curve $C$ with equation $y = x \sin^{-1} x$, $-1 < x < 1$. Let the gradient of the curve $C$ at the point $P$ be $m$.

If the $x$-coordinate of $P$ is increasing at the rate of $9$ units per second when $x = \frac{1}{2}$, find the exact value of the rate at which $m$ is changing at this instant.

Derivatives of Inverse Trigonometric Functions

Connected Rates of Change

Product Rule

2025 RI P1 Q2

<p>It is given that&nbsp;$y = \ln \left[ \sin \left( x + \frac{\pi}{4} \right) \right]$, where&nbsp;$-\frac{\pi}{4} &lt; x &lt; \frac{3\pi}{4}$.</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;$\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + \left( \frac{\mathrm{d}y}{\mathrm{d}x} \right)^2 + 1 = 0$. Hence find the first four non-zero terms of the Maclaurin expansion of $y$, leaving your answer in exact form.<span class="inline-marks">[6]</span></div></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph"><p>Verify the result obtained in part <strong>(a)</strong> using standard series from the List of Formulae (MF27).</p><span class="inline-marks">[5]</span></div></li></ol>

It is given that $y = \ln \left[ \sin \left( x + \frac{\pi}{4} \right) \right]$, where $-\frac{\pi}{4} < x < \frac{3\pi}{4}$.

Show that $\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + \left( \frac{\mathrm{d}y}{\mathrm{d}x} \right)^2 + 1 = 0$. Hence find the first four non-zero terms of the Maclaurin expansion of $y$, leaving your answer in exact form.

Maclaurin Series from Differential Equations

Substitution and Composition of Power Series

Higher Derivatives by Implicit Differentiation

2025 HCI P2 Q2

Maclaurin Series and Power Series

<p>The figure shows a conical container with slant height $3$ cm and semi-vertical angle $\theta$. The container has a volume of $V$ cm$^3$ and $\theta$ varies so that&nbsp;$0 &lt; \theta &lt; \frac{\pi}{2}$.</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/1768026580490-vm9d7iw29js.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgwMjY1ODA0OTAtdm05ZDdpdzI5anMucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5Mjc4OTMzMCwiZXhwIjoxNzkyNzg5NjMwfQ.KUQoWi_TTKaygbGY84XyzrF3sImEdCW2DpsVMjJLbYg" style="max-width: min(200px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<p></p>
<ol class="abc-list"><li><div class="list-item-paragraph">Show that&nbsp;$V = 9\pi \sin^2 \theta \cos \theta$.</div></li><li><div class="list-item-paragraph">Show that&nbsp;$\frac{\mathrm{d}V}{\mathrm{d}\theta} = 9\pi [2 \sin \theta - 3 \sin^3 \theta]$.</div></li><li><div class="list-item-paragraph">Find $\frac{\mathrm{d}^2 V}{\mathrm{d}\theta^2}$.</div></li><li><div class="list-item-paragraph">Determine the values of $\sin \theta$ and $\cos \theta$ for which $V$ has a stationary value, leaving your answers in surd form.</div></li><li><div class="list-item-paragraph">Find the maximum value of $V$ in terms of $\pi$.</div></li></ol>

The figure shows a conical container with slant height $3$ cm and semi-vertical angle $\theta$. The container has a volume of $V$ cm$^3$ and $\theta$ varies so that $0 < \theta < \frac{\pi}{2}$.

<p>Find the minimum gradient of the curve $y = 2x^3 + 6x^2 + 4x + 5$ and the value of $x$ when the minimum gradient occurs.</p>

Find the minimum gradient of the curve $y = 2x^3 + 6x^2 + 4x + 5$ and the value of $x$ when the minimum gradient occurs.

<p>Find the maximum gradient of the curve $y = -5x^3 + 24x^2 - 7x - 5$ and the value of $x$ when the maximum gradient occurs.</p>

Find the maximum gradient of the curve $y = -5x^3 + 24x^2 - 7x - 5$ and the value of $x$ when the maximum gradient occurs.

<p>Find the minimum gradient of the curve $y=2x^3-9x^2+5x+3$ and the value of $x$ when the minimum gradient occurs.<br><br></p>

Find the minimum gradient of the curve $y=2x^3-9x^2+5x+3$ and the value of $x$ when the minimum gradient occurs.



Minimum/Maximum Gradient of a Curve

<p>A curve has the equation $\mathrm{f}(x)=\frac{3x^2}{2x-1}$,</p>
<ol class="abc-list"><li data-list-item-id="item-1768016701411-aqcotticl" data-list-item-path="item-1768016701411-aqcotticl"><div class="list-item-paragraph">Obtain expressions for&nbsp;$\mathrm{f}'(x)$ and $\mathrm{f}''(x)$. Simplify your answers.</div></li><li data-list-item-id="item-1768016739139-s0u9mak9n" data-list-item-path="item-1768016739139-s0u9mak9n"><div class="list-item-paragraph">Hence find the range of values of $x$ for which $\mathrm{f}(x)$ is</div><ol class="i-list"><li data-list-item-id="subitem-1768016767400-ozcz2e5g3" data-list-item-path="item-1768016739139-s0u9mak9n/subitem-1768016767400-ozcz2e5g3"><div class="list-item-paragraph">an increasing function and</div></li><li data-list-item-id="subitem-1768016779890-h70btzpba" data-list-item-path="item-1768016739139-s0u9mak9n/subitem-1768016779890-h70btzpba"><div class="list-item-paragraph">a decreasing function.</div></li></ol></li></ol>

A curve has the equation $\mathrm{f}(x)=\frac{3x^2}{2x-1}$,

<p>A curve has the equation $\mathrm{f}(x)=\frac{2x^2}{(x+2)}$,</p>
<ol class="abc-list"><li data-list-item-id="item-1768016413995-ztpwgy8jp" data-list-item-path="item-1768016413995-ztpwgy8jp"><div class="list-item-paragraph">Obtain expressions for&nbsp;$\mathrm{f}'(x)$ and $\mathrm{f}''(x)$. Simplify your answers.</div></li><li data-list-item-id="item-1768016449360-x65x7rct3" data-list-item-path="item-1768016449360-x65x7rct3"><div class="list-item-paragraph">Hence find the range of values of $x$ for which $\mathrm{f}(x)$ is</div><ol class="i-list"><li data-list-item-id="subitem-1768016481486-ic04v4i0s" data-list-item-path="item-1768016449360-x65x7rct3/subitem-1768016481486-ic04v4i0s"><div class="list-item-paragraph">an increasing function and</div></li><li data-list-item-id="subitem-1768016492267-t5eq2xhk6" data-list-item-path="item-1768016449360-x65x7rct3/subitem-1768016492267-t5eq2xhk6"><div class="list-item-paragraph">a decreasing function.</div></li></ol></li></ol>

A curve has the equation $\mathrm{f}(x)=\frac{2x^2}{(x+2)}$,

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