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>A seating specialist was tasked to install seats for a performance arena. In the curved section of the arena, he installed $21$ seats in the front row facing the arena and for each subsequent row, there are $2$ more seats than the row before it. The seating capacity of this curved section is $1664$.</p>
<ol class="abc-list"><li data-list-item-id="ri-q12-l2-item01" data-list-item-path="ri-q12-l2-item01"><div class="list-item-paragraph">Find the total number of rows in this section of the arena, and the number of seats in the tenth row from the back of the section.<span class="inline-marks">[3]</span></div></li></ol>
<p>At a performance held at this arena, the cost of a ticket for a front row seat at this curved section is $151. The cost of a ticket for a seat in every subsequent row decreases by $2\%$ compared to the row before. The cost of tickets for seats in the same row are equal.</p>
<ol class="abc-list" start="2"><li data-list-item-id="ri-q12-l4-item01" data-list-item-path="ri-q12-l4-item01"><ol class="i-list"><li data-list-item-id="ri-q12-l4-item01-sub-item01" data-list-item-path="ri-q12-l4-item01/ri-q12-l4-item01-sub-item01"><div class="list-item-paragraph">Sam can only afford a ticket below $100 and he would like to sit as near to the front of the section as possible. Find the row nearest to the front for which he can purchase a ticket.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="ri-q12-l4-item01-sub-item02" data-list-item-path="ri-q12-l4-item01/ri-q12-l4-item01-sub-item02"><div class="list-item-paragraph">If&nbsp;$\mathrm F(r)$ is the total cost of tickets for the seats in the $r$th row from the front of the section, write down $\mathrm F(r)$ in terms of $r$.<span class="inline-marks">[1]</span></div></li><li data-list-item-id="ri-q12-l4-item01-sub-item03" data-list-item-path="ri-q12-l4-item01/ri-q12-l4-item01-sub-item03"><div class="list-item-paragraph">Find the total cost of tickets for all the seats in the first $n$ rows from the front, in the form of</div><div class="list-item-paragraph">$\displaystyle143450(1-A^n)+B\sum_{r=1}^{n}\left(rA^{r-1}\right),$</div><div class="list-item-paragraph">where&nbsp;$A$ and $B$ are constants to be determined.</div></li><li data-list-item-id="ri-q12-l4-item01-sub-item04" data-list-item-path="ri-q12-l4-item01/ri-q12-l4-item01-sub-item04"><div class="list-item-paragraph">Suppose a company can purchase tickets for this performance before public sales are open and they would like to purchase $600$ tickets in this curved section with a budget of $52 000. Determine whether this budget is sufficient for this purchase.<span class="inline-marks">[3]</span></div></li></ol></li></ol>

A seating specialist was tasked to install seats for a performance arena. In the curved section of the arena, he installed $21$ seats in the front row facing the arena and for each subsequent row, there are $2$ more seats than the row before it. The seating capacity of this curved section is $1664$.

Find the total number of rows in this section of the arena, and the number of seats in the tenth row from the back of the section.

At a performance held at this arena, the cost of a ticket for a front row seat at this curved section is $151. The cost of a ticket for a seat in every subsequent row decreases by $2\%$ compared to the row before. The cost of tickets for seats in the same row are equal.

Problems involving inequalities

Geometric Progression

Arithmetic Progression

Sigma Notation

2026 RI P1 Q12

Sequence and Series

<p>The diagram shows the parts of the graph of $y = \frac{2}{1+x}$ between $x = 1$ and $x = 2$. Prove that the total area of the four rectangles, each of equal width, as shown in the diagram below, may be expressed as&nbsp;$\sum_{r=1}^{4} \frac{2}{8+r}$.<span class="inline-marks">[2]</span></p>
<div class="graph-block" data-graph-config="eyJheGlzIjp0cnVlLCJncmlkIjpmYWxzZSwid2lkdGgiOjMwMCwiaGVpZ2h0IjoyMDAsImVsZW1lbnRzIjpbeyJpZCI6ImltcG9ydGVkLWZ1bmN0aW9uZ3JhcGgtYmxvY2stMS1jb250ZW50LWVsZW1lbnQtMCIsInR5cGUiOiJmdW5jdGlvbmdyYXBoIiwieE1heCI6Mi44LCJ4TWluIjowLCJjb2xvciI6IiMwMDAwMDAiLCJleHByZXNzaW9uIjoiMiAvICgxICsgeCkiLCJzdHJva2VXaWR0aCI6MS41fSx7ImlkIjoiaW1wb3J0ZWQtcG9seWdvbi1ibG9jay0xLWNvbnRlbnQtZWxlbWVudC0xIiwidHlwZSI6InBvbHlnb24iLCJjb2xvciI6IiMwMDAwMDAiLCJwb2ludHMiOltbMSwwXSxbMSwwLjldLFsxLjI1LDAuOV0sWzEuMjUsMF1dLCJmaWxsQ29sb3IiOiJub25lIiwic2hhcGVLaW5kIjoicmVjdGFuZ2xlIiwic3Ryb2tlV2lkdGgiOjEuNX0seyJpZCI6ImltcG9ydGVkLXBvbHlnb24tYmxvY2stMS1jb250ZW50LWVsZW1lbnQtMiIsInR5cGUiOiJwb2x5Z29uIiwiY29sb3IiOiIjMDAwMDAwIiwicG9pbnRzIjpbWzEuMjUsMF0sWzEuMjUsMC44XSxbMS41LDAuOF0sWzEuNSwwXV0sImZpbGxDb2xvciI6Im5vbmUiLCJzaGFwZUtpbmQiOiJyZWN0YW5nbGUiLCJzdHJva2VXaWR0aCI6MS41fSx7ImlkIjoiaW1wb3J0ZWQtcG9seWdvbi1ibG9jay0xLWNvbnRlbnQtZWxlbWVudC0zIiwidHlwZSI6InBvbHlnb24iLCJjb2xvciI6IiMwMDAwMDAiLCJwb2ludHMiOltbMS41LDBdLFsxLjUsMC43M10sWzEuNzUsMC43M10sWzEuNzUsMF1dLCJmaWxsQ29sb3IiOiJub25lIiwic2hhcGVLaW5kIjoicmVjdGFuZ2xlIiwic3Ryb2tlV2lkdGgiOjEuNX0seyJpZCI6ImltcG9ydGVkLXBvbHlnb24tYmxvY2stMS1jb250ZW50LWVsZW1lbnQtNCIsInR5cGUiOiJwb2x5Z29uIiwiY29sb3IiOiIjMDAwMDAwIiwicG9pbnRzIjpbWzEuNzUsMF0sWzEuNzUsMC42N10sWzIsMC42N10sWzIsMF1dLCJmaWxsQ29sb3IiOiJub25lIiwic2hhcGVLaW5kIjoicmVjdGFuZ2xlIiwic3Ryb2tlV2lkdGgiOjEuNX0seyJ4IjowLjk1LCJ5IjotMC4xNiwiaWQiOiJpbXBvcnRlZC10ZXh0LWJsb2NrLTEtY29udGVudC1lbGVtZW50LTUiLCJ0eXBlIjoidGV4dCIsImNvbG9yIjoiIzAwMDAwMCIsImxhYmVsIjoiXFwoMVxcKSJ9LHsieCI6MS45NywieSI6LTAuMTUsImlkIjoiaW1wb3J0ZWQtdGV4dC1ibG9jay0xLWNvbnRlbnQtZWxlbWVudC02IiwidHlwZSI6InRleHQiLCJjb2xvciI6IiMwMDAwMDAiLCJsYWJlbCI6IlxcKDJcXCkifSx7IngiOjEuNDIsInkiOjEuMzMsImlkIjoiaW1wb3J0ZWQtdGV4dC1ibG9jay0xLWNvbnRlbnQtZWxlbWVudC03IiwidHlwZSI6InRleHQiLCJjb2xvciI6IiMwMDAwMDAiLCJsYWJlbCI6IlxcKHkgPSBcXGZyYWN7Mn17MSt4fVxcKSJ9XSwic2hvd1hBeGlzIjp0cnVlLCJzaG93WUF4aXMiOnRydWUsInhBeGlzTGFiZWwiOiJcXCh4XFwpIiwieUF4aXNMYWJlbCI6IlxcKHlcXCkiLCJib3VuZGluZ0JveCI6Wy0wLjUsMi41LDMsLTAuNV0sIm9yaWdpbkxhYmVsIjoiXFwoT1xcKSIsImdyaWRTZXR0aW5ncyI6eyJtYWpvckNvbG9yIjoiIzk0YTNiOCIsIm1ham9yU3RlcFgiOjEsIm1ham9yU3RlcFkiOjEsIm1pbm9yQ29sb3IiOiIjY2JkNWUxIiwibWFqb3JPcGFjaXR5IjowLjQ1LCJtaW5vck9wYWNpdHkiOjAuNDUsIm1pbm9yRWxlbWVudHNYIjo0LCJtaW5vckVsZW1lbnRzWSI6NCwibWFqb3JTdHJva2VXaWR0aCI6MC44LCJtaW5vclN0cm9rZVdpZHRoIjowLjUsInNob3dNYWpvckdyaWRMaW5lcyI6dHJ1ZSwic2hvd01pbm9yR3JpZExpbmVzIjp0cnVlfSwic2hvd0F4aXNMYWJlbHMiOmZhbHNlLCJzaG93TmF2aWdhdGlvbiI6ZmFsc2UsImtlZXBBc3BlY3RSYXRpbyI6ZmFsc2UsInhUaWNrTGFiZWxFdmVyeSI6MSwieVRpY2tMYWJlbEV2ZXJ5IjoxLCJ4QXhpc0xhYmVsT2Zmc2V0IjpbMiwtMTFdLCJ5QXhpc0xhYmVsT2Zmc2V0IjpbLTE0LDVdLCJvcmlnaW5MYWJlbE9mZnNldCI6Wy01LC0yXX0=" data-width="300" data-height="200" style="width:300px;height:200px;max-width:100%;display:block;margin:0 auto;"></div>
<p>In general, when there are $n$ rectangles, each of equal width, under the curve between $x = 1$ and $x = 2$ (instead of just $4$), find an expression, in the form $\sum_{r=1}^{n} \mathrm{f}(r)$, for their total area.</p>
<p>Hence show that&nbsp;$\sum_{r=1}^{n} \frac{2}{2n+r} &lt; 2 \ln \left(\frac{3}{2}\right)$.</p>
<p>Deduce that&nbsp;$\sum_{r=1}^{n} \frac{2}{2n+r} &lt; 2 \ln \left(\frac{3}{2}\right) &lt; \sum_{r=1}^{n} \frac{2}{2n+r-1}$.</p>

The diagram shows the parts of the graph of $y = \frac{2}{1+x}$ between $x = 1$ and $x = 2$. Prove that the total area of the four rectangles, each of equal width, as shown in the diagram below, may be expressed as $\sum_{r=1}^{4} \frac{2}{8+r}$.

In general, when there are $n$ rectangles, each of equal width, under the curve between $x = 1$ and $x = 2$ (instead of just $4$), find an expression, in the form $\sum_{r=1}^{n} \mathrm{f}(r)$, for their total area.

Hence show that $\sum_{r=1}^{n} \frac{2}{2n+r} < 2 \ln \left(\frac{3}{2}\right)$.

Deduce that $\sum_{r=1}^{n} \frac{2}{2n+r} < 2 \ln \left(\frac{3}{2}\right) < \sum_{r=1}^{n} \frac{2}{2n+r-1}$.

Riemann Sums

PJC P1 Q9

Applications of Integration

<p>In this question, $a$ is a positive constant.</p>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">Explain, with the aid of a sketch, why $\frac{a}{n} \sum_{k=1}^{n} \ln \left(1 + \frac{ak}{n}\right) &gt; \int_{0}^{a} \ln(1+x) \, \mathrm{d}x$ for all positive integers $n$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">State a similar expression that is less than $\int_{0}^{a} \ln(1+x) \, \mathrm{d}x$.<span class="inline-marks">[1]</span></div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">Use your sketch in part <strong>(i)</strong> to explain why $\lim_{n \to \infty} \left[ \frac{a}{n} \sum_{k=1}^{n} \ln \left(1 + \frac{ak}{n}\right) \right] = \int_{0}^{a} \ln(1+x) \, \mathrm{d}x$ and find the value of this limit exactly in terms of $a$.<span class="inline-marks">[5]</span></div></li></ol>

In this question, $a$ is a positive constant.

Limit of an expression

Integration by Parts

Techniques of Integration

<p>The diagram below shows a sketch of the graph of $y = \sin^2 x$ for&nbsp;$0 \le x \le \frac{\pi}{2}$.</p>
<p>Rectangles each of width $\frac{\pi}{2n}$ are drawn under the curve for&nbsp;$0 \le x \le \frac{\pi}{2}$.</p>
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data-width="400" data-height="300" style="width:400px;height:300px;max-width:100%;display:block;margin:0 auto;"></div>
<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;$A$, the total area of all the rectangles, is given by $a \sum_{k=1}^{n-1} \sin^2 \frac{k\pi}{2n}$, where&nbsp;$a$ is to be determined.<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 exact value of $\displaystyle \lim_{n\to\infty} a \sum_{k=1}^{n-1} \sin^2 \frac{k\pi}{2n}$.<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">Hence, find the value of $\int_{0}^{1} \sin^{-1} \sqrt{y} \ \mathrm{d}y$.<span class="inline-marks">[2]</span></div></li></ol>

The diagram below shows a sketch of the graph of $y = \sin^2 x$ for $0 \le x \le \frac{\pi}{2}$.

Rectangles each of width $\frac{\pi}{2n}$ are drawn under the curve for $0 \le x \le \frac{\pi}{2}$.

Show that $A$, the total area of all the rectangles, is given by $a \sum_{k=1}^{n-1} \sin^2 \frac{k\pi}{2n}$, where $a$ is to be determined.

Integrating Powers of Sine and Cosine

Integrable Forms of Trigonometric Functions

Area Enclosed by a Curve and the y-axis

2024 YIJC P1 Q2

<p>The diagram shows the curve with equation $y = \frac{1}{x}$. <br>$n$ rectangles of equal width are drawn under the curve between $x=1$ and $x=2$.</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/1768709448742-jn6j120hppd.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3Njg3MDk0NDg3NDItam42ajEyMGhwcGQucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5Mjc2MzYxMCwiZXhwIjoxNzkyNzYzOTEwfQ.9E3AwJRymZcbTRN3KzcvvIUEguqiu710wGie-tReLXk" style="max-width: min(400px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<p>Let&nbsp;$S_n$ be the <strong>total</strong> area of the $n$ rectangles.</p>
<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">Let&nbsp;$n=2$. By considering $S_2$, show that&nbsp;$\frac{7}{12} &lt; \ln 2$. State your reasoning clearly.<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">By considering $S_n$ for a suitable value of $n$, find a rational number $q$ such that&nbsp;$\frac{7}{12} &lt; q &lt; \ln 2$.<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"><p>Express $S_n$ in the for</p></div><div class="list-item-paragraph"><p style="text-align: center;">$\sum_{r=1}^n \mathrm{g}(r)$, </p></div><div class="list-item-paragraph"><p>where the function $\mathrm{g}$ is to be determined.</p><span class="inline-marks">[1]</span></div></li><li data-list-item-id="item-d" data-list-item-path="item-d"><div class="list-item-paragraph">By considering another suitable set of $n$ rectangles, show with the aid of a diagram that \[\ln 2 &lt; S_n + \frac{1}{2n}\] where&nbsp;$n$ is any given positive integer.<span class="inline-marks">[3]</span></div></li></ol>

The diagram shows the curve with equation $y = \frac{1}{x}$. 
$n$ rectangles of equal width are drawn under the curve between $x=1$ and $x=2$.

Let $S_n$ be the **total** area of the $n$ rectangles.

Let $n=2$. By considering $S_2$, show that $\frac{7}{12} < \ln 2$. State your reasoning clearly.

By considering $S_n$ for a suitable value of $n$, find a rational number $q$ such that $\frac{7}{12} < q < \ln 2$.

Manipulating Summation Limits

2025 EJC P1 Q5 

<p>Write down each of the following series in summation notation</p>
<ol class="i-list"><li><div class="list-item-paragraph">$0!+1!+2!+3!+…+6!$</div></li><li><div class="list-item-paragraph">$1!+3!+5!+7!…$ to $n$ terms</div></li><li><div class="list-item-paragraph">$1\cdot 2\cdot 3+2\cdot 3\cdot 4+3\cdot 4\cdot 5+4\cdot 5\cdot 6+…$ to $n$ terms</div></li><li><div class="list-item-paragraph">$-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}+…$ to $n+1$ terms</div></li><li><div class="list-item-paragraph">$2\cdot 3\cdot 4-3\cdot 4\cdot 5+4\cdot 5\cdot 6-5\cdot 6\cdot 7+…$ to $n$ terms</div></li><li><div class="list-item-paragraph">$-\frac{3}{2}+\frac{5}{4}-\frac{7}{8}+\frac{9}{16}+…$ to $n+1$ terms</div></li><li><div class="list-item-paragraph">$2+6+18+54+…$ to $n$ terms</div></li><li><div class="list-item-paragraph">$-\frac{1}{2}+3-\frac{1}{4}+5-\frac{1}{6}+7-\frac{1}{8}…$ to $n$ terms</div></li></ol>

Write down each of the following series in summation notation

Writing the series in sigma notation

<ol class="abc-list"><li><div class="list-item-paragraph">Find $\sum\limits_{r=0}^{n}{\left[ \left( n+2 \right)r+n \right]}$, giving your answer in terms of $n$.</div><p style="text-align: right;">[3]</p></li></ol>
<p>[You may use the result $\sum\limits_{r=1}^{n}{{{r}^{3}}=\frac{1}{4}{{n}^{2}}{{\left( n+1 \right)}^{2}}}$ for the rest of this question.]</p>
<ol class="abc-list" start="2"><li><div class="list-item-paragraph">By writing down the first two and the last two terms in the series, find $\sum\limits_{r=1}^{n}{{{\left( r+2 \right)}^{3}}}$, giving your answer in terms of $n$.</div><p style="text-align: right;">[3]</p></li><li><div class="list-item-paragraph">Find ${{1}^{3}}-{{2}^{3}}+{{3}^{3}}-{{4}^{3}}+{{5}^{3}}-{{6}^{3}}+\ldots +{{\left( 2n-1 \right)}^{3}}-{{\left( 2n \right)}^{3}}$, giving your answer in terms of $n$.</div><p style="text-align: right;">[3]</p></li></ol>

[You may use the result $\sum\limits_{r=1}^{n}{{{r}^{3}}=\frac{1}{4}{{n}^{2}}{{\left( n+1 \right)}^{2}}}$ for the rest of this question.]

Sigma Notation By Substitution

Summing Polynomial Series

2024 RI Promo Q5

<p>A financial institution, Future Investments Inc., has introduced a new investment scheme. The scheme pays a compound interest of $5\%$ per annum at the end of the year, based on the amount in the account at the beginning of each year.</p>
<p>John and Sarah are both interested in this scheme.</p>
<ol class="abc-list"><li><div class="list-item-paragraph">John invests $\$x$ at the start of the first year and a further $\$x$ on the first day of each subsequent year. He chooses to leave the money in his account for the interest to accumulate.</div><ol class="i-list"><li><div class="list-item-paragraph">Write down the amount in John’s account, including the interest, at the end of the first year.<span class="inline-marks">[1]</span></div></li><li><div class="list-item-paragraph">Show that John will have a total of $21x\left( {{1.05}^{n}}-1 \right)$ in his account at the end of $n$ years.<span class="inline-marks">[3]</span></div></li><li><div class="list-item-paragraph">If John invests $\$10000$ at the start of every year, find the number of years for the total in his account to first exceed $\$500\,\,000$. Determine if this happens at the start or at the end of that year.<span class="inline-marks">[4]</span></div></li></ol></li><li><div class="list-item-paragraph">Sarah invests $\$6000$ at the start of the first year. On the first day of each subsequent year, she invests $\$400$ more than the amount invested at the start of the previous year.</div><ol class="i-list"><li><div class="list-item-paragraph">Explain why the amount in Sarah’s account at the end of $n$ years can be given by $\sum\limits_{r=1}^{n}{\left( a+b\left( r-1 \right) \right)}{{\left( 1.05 \right)}^{n-\left( r-1 \right)}}$, and determine the values of the constants $a$ and $b$.<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">Hence determine the number of complete years for the total amount in her account to first exceed $\$500\,\,000$.<span class="inline-marks">[2]</span></div></li></ol></li></ol>

A financial institution, Future Investments Inc., has introduced a new investment scheme. The scheme pays a compound interest of $5\%$ per annum at the end of the year, based on the amount in the account at the beginning of each year.

John and Sarah are both interested in this scheme.

Annuities and Amortization

2024 RI Promo Q11

<p>The diagram below shows a curve with equation $y=\sqrt{\left( 1-x \right)}$. Let&nbsp;$A$ be the area bounded by the curve and the axes. A total of $n$ rectangles, each of equal width, are constructed to approximate the value of $A$.</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/1760850560508-weww0d34ntd.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvY29udmVydGVkLzE3NjA4NTA1NjA1MDgtd2V3dzBkMzRudGQucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5Mjc2MzYxMCwiZXhwIjoxNzkyNzYzOTEwfQ.dEZj5P3f_Hxfe9L8R2Pz8FAP6vf6JCTmUUfSOblpso4" style="max-width: min(250px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<ol class="abc-list"><li data-list-item-id="item-1760850560032-jiyd8va04" data-list-item-path="item-1760850560032-jiyd8va04"><div class="list-item-paragraph">Show that the total area of $n$ rectangles is given by $\frac{1}{n\sqrt{n}}\sum\limits_{r=0}^{n-1}{\sqrt{n-r}}$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-1760850560609-zebhhcdsx" data-list-item-path="item-1760850560609-zebhhcdsx"><div class="list-item-paragraph">Without using a calculator, evaluate $\mathop{\lim}\limits_{n\to \infty }\,\frac{1}{n\sqrt{n}}\sum\limits_{r=0}^{n-1}{\sqrt{n-r}}$.<span class="inline-marks">[3]</span></div></li></ol>

The diagram below shows a curve with equation $y=\sqrt{\left( 1-x \right)}$. Let $A$ be the area bounded by the curve and the axes. A total of $n$ rectangles, each of equal width, are constructed to approximate the value of $A$.

2024 HCI Promo Q3

<ol class="i-list"><li data-list-item-id="item-1760753768264-g5h1mn28j" data-list-item-path="item-1760753768264-g5h1mn28j"><div class="content-block list-item-image" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/converted/1760753776557-mkbtrqaje6p.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvY29udmVydGVkLzE3NjA3NTM3NzY1NTctbWtidHJxYWplNnAucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5Mjc2MzYxMCwiZXhwIjoxNzkyNzYzOTEwfQ._qOZJ7ob5ZSZbdk1gKqZfywqjVfkCLkDxuaTWDOxzHc" alt="List item image" class="aligncenter custom-width" style="max-width: min(400px, 100%); height: auto;" /></div><div class="list-item-paragraph">The diagram shows the graph of $y={{x}^{2}}$ and $n$ rectangles, each with equal width, from $x=0$ to $x=1$. Show that the total area of the $n$ rectangles can be written as&nbsp;$\frac{1}{{{n}^{3}}}\sum\limits_{r=1}^{n}{{{r}^{2}}}$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-1760753768878-a449vnaaz" data-list-item-path="item-1760753768878-a449vnaaz"><div class="list-item-paragraph">Hence find ${\displaystyle \lim_{n\to\infty} \frac{1}{n^3} \sum_{r=1}^{n} r^2.}$<span class="inline-marks">[2]</span></div></li></ol>

2020 MI P2 Q1

<p>Figure 1 shows a sketch of the curve $y=\mathrm{f}\left( x \right)$ and $A$ is the region under the curve between $x=1$ and $x=3$. Yvonne and Irene use different ways to draw $4$ rectangles of equal width, $h$, to estimate the area of $A$.</p>
<p>Figure 2 shows $4$ rectangles drawn by Yvonne, with the curve intersecting each rectangle at the mid-point of its width. The $x$-coordinates of the mid-points are ${{m}_{1}}$, ${{m}_{2}}$, ${{m}_{3}}$ and ${{m}_{4}}$.</p>
<figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/converted/1760424606336-adgx6qij1ze.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvY29udmVydGVkLzE3NjA0MjQ2MDYzMzYtYWRneDZxaWoxemUucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5Mjc2MzYxMCwiZXhwIjoxNzkyNzYzOTEwfQ.edaImr4d7fwYbkcDXDhOPkVcaKbGS_BD-2epCp_Mv1w" width="400" class="aligncenter" /></figure>
<ol class="abc-list"><li><div class="list-item-paragraph">State the values of $h$, ${{m}_{1}}$, ${{m}_{2}}$, ${{m}_{3}}$ and ${{m}_{4}}$.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">The sum of area of rectangles using Yvonne’s method is denoted by $B$. Show that $B=h\sum\limits_{r=0}^{3}{\left( \mathrm{f}\left( a+rh \right) \right)}$, where the value of $a$ is to be determined.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">Irene finds that the sum of area of $4$ rectangles that she has drawn is $C=h\sum\limits_{r=0}^{3}{\left( \mathrm{f}\left( 1+rh \right) \right)}$. Draw these rectangles in Figure 1.</div><p style="text-align: right;">[1]</p></li></ol>
<p>You are now given $\mathrm{f}\left( x \right)=\frac{1}{x}+1$.</p>
<ol class="abc-list" start="4"><li><div class="list-item-paragraph">By finding the numerical values of $B$, $C$ and the actual area of region $A$, explain how these values and the rectangles drawn in Figures 1 and 2 show that Yvonne’s estimation of the area of $A$ is better than Irene’s.</div><p style="text-align: right;">[2]</p></li></ol>

Figure 1 shows a sketch of the curve $y=\mathrm{f}\left( x \right)$ and $A$ is the region under the curve between $x=1$ and $x=3$. Yvonne and Irene use different ways to draw $4$ rectangles of equal width, $h$, to estimate the area of $A$.

Figure 2 shows $4$ rectangles drawn by Yvonne, with the curve intersecting each rectangle at the mid-point of its width. The $x$-coordinates of the mid-points are ${{m}_{1}}$, ${{m}_{2}}$, ${{m}_{3}}$ and ${{m}_{4}}$.

You are now given $\mathrm{f}\left( x \right)=\frac{1}{x}+1$.

2022 YIJC P1 Q2

<p>Arthur uses a special type of clay to make a multi-layered structure in which each layer is in the shape of a cylinder. The base layer has a radius of $60$ cm and a height of $20$ cm. The radius of each layer above is $3$ cm less than the radius of the previous layer. The height of each layer remains the same.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Show that the maximum number of layers that a structure can have is $20$.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">Ribbon is pasted around the circumference of each layer of a structure. Find the total length of a ribbon required for a $20$-layered structure.</div><p style="text-align: right;">[3]</p></li><li><div class="list-item-paragraph">Show the total volume of a $20$-layered structure is given by</div><div class="list-item-paragraph" style="text-align: center">$p\pi {{\sum\limits_{n=1}^{20}{\left[ \mathrm{f}(n) \right]}}^{2}}$</div><div class="list-item-paragraph">where $p$ is a constant to be determined and the function $\mathrm{f}(n)$ is to be found.</div><p style="text-align: right;">[3]</p><div class="list-item-paragraph">Given that Arthur has $1500\,000$ cm$^{3}$ of clay, explain if Arthur is able to construct a $20$-layered structure.</div><p style="text-align: right;">[1]</p></li></ol>
<p>To make a multi-layered structure more appealing, the height of each layer is now varied such that the height of the layer above is $80\%$ of the height of the previous layer.</p>
<ol class="i-list" start="4"><li><div class="list-item-paragraph">Assuming that the height of the base layer is still $20$ cm, determine the maximum number of layers a structure can have if the height of the structure is to be at most $85$ cm.</div><p style="text-align: right;">[3]</p></li></ol>

Arthur uses a special type of clay to make a multi-layered structure in which each layer is in the shape of a cylinder. The base layer has a radius of $60$ cm and a height of $20$ cm. The radius of each layer above is $3$ cm less than the radius of the previous layer. The height of each layer remains the same.

To make a multi-layered structure more appealing, the height of each layer is now varied such that the height of the layer above is $80\%$ of the height of the previous layer.

2020 MI P1 Q11

<p>Two series ${{S}_{n}}$ and ${{T}_{n-1}}$ are given by</p>
<p style="text-align: center">${{S}_{n}}=\sum\limits_{r=1}^{n}{\left( \frac{r}{{{2}^{r-1}}} \right)}$ and ${{T}_{n-1}}=\sum\limits_{r=1}^{n-1}{\left( \frac{r}{{{2}^{r}}} \right)}$.</p>
<p>By listing the terms of ${{S}_{n}}$ and ${{T}_{n-1}}$, write ${{S}_{n}}-{{T}_{n-1}}$ as a 3 geometric series and find ${{S}_{n}}-{{T}_{n-1}}$, leaving your answer in terms of $n$.<span class="inline-marks">[3]</span></p>
<p>Hence find the value of ${{S}_{n}}-{{T}_{n-1}}$ as&nbsp;$n\to \infty $. <span class="inline-marks">[1]</span></p>

Two series ${{S}_{n}}$ and ${{T}_{n-1}}$ are given by

${{S}_{n}}=\sum\limits_{r=1}^{n}{\left( \frac{r}{{{2}^{r-1}}} \right)}$ and ${{T}_{n-1}}=\sum\limits_{r=1}^{n-1}{\left( \frac{r}{{{2}^{r}}} \right)}$.

By listing the terms of ${{S}_{n}}$ and ${{T}_{n-1}}$, write ${{S}_{n}}-{{T}_{n-1}}$ as a 3 geometric series and find ${{S}_{n}}-{{T}_{n-1}}$, leaving your answer in terms of $n$.

Hence find the value of ${{S}_{n}}-{{T}_{n-1}}$ as $n\to \infty $. 

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