2023 HCI P1 Q4
Without the use of a calculator, show that
$\int_{-2}^{2}{\frac{\left| x-2 \right|}{{{x}^{2}}-2x+4}\text{d}x=p\ln 3+q\pi }$,
where $p$ and $q$ are exact real constants to be determined.
[6]
2024 DHS P1 Q1
For $a>4$, show that $\int_{0}^{a}{\left| 2{{x}^{2}}-3x-2 \right|\,\,\text{d}x}=\frac{1}{k}\left( 4{{a}^{3}}-9{{a}^{2}}-12a+56 \right)$ where $k$ is an integer to be found.
[4]
2022 MI P1 Q1
On the same axes, sketch the graphs of $y=2+{{\text{e}}^{-{{x}^{2}}}}$ and $y=\left| \ln \left( x-1 \right) \right|$. Label clearly the equations of any asymptotes and any points of intersection with the axes.
[4]
Hence solve the inequality $2+{{\text{e}}^{-{{x}^{2}}}}\le \left| \ln \left( x-1 \right) \right|$.
[2]
Answer:
$1 < x\le 1.10$ or $x\ge 8.39$ ($3$ s.f.)
2023 EJC P1 Q9
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Using standard series, find the Maclaurin series expansion of $\frac{1}{\sqrt{1-{{a}^{2}}{{x}^{2}}}}$ up to and including the term in ${{x}^{4}}$, where $a$ is a positive constant. Find also the range of values of $x$, in terms of $a$, for which the expansion is valid.
[5]
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Hence, by using a suitable value for $a$, show that ${{\cos }^{-1}}x\approx \frac{\pi }{2}-x-\frac{1}{6}{{x}^{3}}-\frac{3}{40}{{x}^{5}}$.
[3]
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Using the result in part (b), estimate $\int_{0}^{\frac{1}{2}}{{{\cos }^{-1}}x\,\,\text{d}x}$ to $~5$ significant figures.
[2]
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Comparing your estimate in (c) with the actual value of $\int_{0}^{\frac{1}{2}}{{{\cos }^{-1}}x\,\,\text{d}x}$, comment on the accuracy of your estimate and suggest how it can be further improved.
[2]
Answer:
(a) $\left| x \right|<\frac{1}{a}$ or $-\frac{1}{a} < x < \frac{1}{a}$ since $a$ is positive. (b) ${{\cos }^{-1}}x=\frac{\pi }{2}-\frac{{{x}^{3}}}{6}-\frac{3}{40}{{x}^{5}}+…$ (c) $0.65760$ ($5$ s.f.)
(d) Estimate in (c) is accurate to $4$s.f. but not to $5$ s.f. To improve estimate, we can include higher-order terms in the Maclaurin series expansion of ${{\cos }^{-1}}x$.
2023 TMJC P2 Q2
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Logarithmic functions are useful as measurement units for phenomena with large scale of values, such as decibels (for sound) or Richter scale (for earthquakes). Logarithmic functions make the measurement units smaller and easier to work with. An example of a logarithmic function is given as
$y=\ln \left( 1+2x+3{{x}^{2}} \right)$.
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By using differentiation, find the Maclaurin series for $y$, up to and including the term in ${{x}^{2}}$.
[4]
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By using the standard series in MF26, verify that the answer obtained in part (i) is correct.
[2]
-
-
Find the series expansion of $\frac{x}{\sqrt{4+x}}$, up to and including the term in ${{x}^{3}}$.
[4]
Answer:
(a)(i) $2x+{{x}^{2}}+…$ (up to ${{x}^{2}}$) (ii) $2x+{{x}^{2}}+…$(up to ${{x}^{2}}$) (verified)
(b) $\frac{1}{2}x-\frac{1}{16}{{x}^{2}}+\frac{3}{256}{{x}^{3}}$ (up to ${{x}^{3}}$)
2023 SAJC P2 Q2
In triangle $PQR$, $\angle PRQ=\frac{2\pi }{3}$, $\angle RPQ=\frac{\pi }{6}+\theta $, and $PR=a$ units, where $a$ is a positive real constant.
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Show that $PQ=\frac{\sqrt{3}a}{\cos \theta -\sqrt{3}\sin \theta }$.
[4]
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Given that $\theta $ is sufficiently small, show that $PQ\approx \sqrt{3}a\left( 1+b\theta +c{{\theta }^{2}} \right)$, where $b$ and $c$ are real constants.
[3]
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Given that $\theta >0$, find the range of values of $\theta $ such that the percentage error of the approximation in (ii) is less than $5\%$.
[2]
Answer:
(ii) $PQ\approx \sqrt{3}a\left( 1+\sqrt{3}\theta +\frac{7}{2}{{\theta }^{2}} \right)$ (iii) $\text{f}\left( \theta \right)<0.05$
2023 TJC P1 Q3
The function $\text{g}$ is defined by
$\text{g}\left( x \right)=\frac{x}{\sqrt{4-x}}$, $x\in \mathbb{R}$, $0<x<4$.
It is given that $\text{f}\left( x \right)=ax+b{{x}^{2}}+c{{x}^{3}}$ is the binomial expansion of $\text{g}\left( x \right)=\frac{x}{\sqrt{4-x}}$ in ascending powers of $x$ up to and including terms in ${{x}^{3}}$.
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Find the exact value of $a$, $b$ and $c$.
[4]
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Find the range of values of $x$ such that the percentage error in using $\text{f}\left( x \right)$ to approximate $\text{g}\left( x \right)$ is less than $4\%$.
[2]
Answer:
(a) $a=\frac{1}{2}$, $b=\frac{1}{16}$ and $c=\frac{3}{256}$ (b) $0 < x < 1.87$ ($3$ s.f.)
2020 TJC P2 Q3
It is given that $\text{f}(x)=\sin \left[ \ln \left( 1+x \right) \right]$, $x\in \mathbb{R}$, $x>-1$.
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Show that ${{(1+x)}^{2}}\text{f}”(x)+(1+x)\,\text{f}'(x)+\text{f}(x)=0$.
[3]
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By further differentiation of the result in (i), find the first three non-zero terms of the Maclaurin series for $\text{f}(x)$.
[3]
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Verify the correctness of the series found in part (ii) using the standard series from the List of Formulae (MF26).
[2]
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Use the series in part (ii) to approximate the value of $\int\limits_{0}^{2}{\text{f}(x)}\,\text{d}x$.
Use your calculator to find an accurate value of $\int\limits_{0}^{2}{\text{f}(x)}\,\text{d}x$. Why is the approximated value not very good?[3]
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Using the series obtained in part (ii), deduce the Maclaurin series for $\cos \left[ \ln (1+x) \right]$ in ascending powers of $x$, up to and including the term in ${{x}^{2}}$.
[2]
Answer:
(ii) $x-\frac{1}{2}{{x}^{2}}+\frac{1}{6}{{x}^{3}}+…$ (iii) the series found in (ii) is correct till the terms in ${{x}^{3}}$ (iv) For $1\le x\le 2$, the terms in higher powers of $x$ is not small and so are not negligible. Hence the approximation is not so good. (v) $1-\frac{1}{2}{{x}^{2}}+…$
2023 ASRJC P1 Q10
Glucose is a type of sugar in our blood and our body uses it for energy. Having low sugar in the blood for long periods of time can cause health problems if it is not treated.
At 10:00 am, a patient is given glucose via an intravenous drip at a constant rate of $r$ units of glucose per hour. It is known that the rate at which the glucose in the blood stream is absorbed by the body, is proportional to the amount of glucose present in the blood stream. At the start of the treatment, the patient has ${{x}_{0}}$units of glucose in his blood stream.
It is given that $x$ denotes the amount of glucose present in the blood stream of the patient at time $t$ hours after he starts the treatment.
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Form a differential equation in $x$ and show that $x=\left( {{x}_{0}}-\frac{r}{k} \right){{\text{e}}^{-kt}}+\frac{r}{k}$ where $k$ is a positive constant.
[4]
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State the theoretical limiting value of $x$ if the patient is given the treatment over a long period.
[1]
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Sketch the graph of $x$ for $t\ge 0$.
[2]
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At what time (to the nearest minute), would the glucose in his blood stream first exceed $90\%$ of the value found in part (b)?
[5]
It is now given that $\frac{r}{k}=2{{x}_{0}}$.
The patient has $10\%$ more glucose in his blood stream after $\frac{1}{2}$ hour of treatment.
Answer:
(b) $2{{x}_{0}}$ (d) $0.21072$
2016 RI P1 Q11 [Modified]
The line ${{l}_{1}}$ passes through the point $A$, whose position vector is $-\mathbf{i}+2\mathbf{j}$, and is parallel to the vector $\mathbf{i}+\mathbf{k}$. The line ${{l}_{2}}$ passes through the point $B$, whose position vector is $\mathbf{i}+\mathbf{j}+3\mathbf{k}$, and is parallel to the vector $\mathbf{j}+\mathbf{k}$.
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Show that the lines ${{l}_{1}}$ and ${{l}_{2}}$ are skew.
[2]
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Find the position vector of the point $N$ on ${{l}_{2}}$ such that $AN$ is perpendicular to ${{l}_{2}}$.
[3]
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Find a vector equation for $\prod $ in the form $\mathbf{r}=\mathbf{u}+a\mathbf{v}+\beta \mathbf{w}$, where $\mathbf{v}$ and $\mathbf{w}$ are perpendicular vectors.
[3]
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The point $X$ varies in such a way that the mid-point of $AX$ is always in $\prod $. Show that $X$ lies on a plane parallel to $\prod $.
[4]
The plane $\prod $ contains ${{l}_{2}}$ and is perpendicular to $AN$.
Answer:
(ii) $\overrightarrow{ON}=\left( \begin{matrix}
1 \\
0 \\
2 \\
\end{matrix} \right)$ (iii) $\prod :\mathbf{r}=\left( \begin{matrix}
1 \\
1 \\
3 \\
\end{matrix} \right)+\alpha \left( \begin{matrix}
0 \\
1 \\
1 \\
\end{matrix} \right)+\beta \left( \begin{matrix}
2 \\
1 \\
-1 \\
\end{matrix} \right)$, $\alpha $, $\beta \in \mathbb{R}$.