These Ten-Year-Series (TYS) worked solutions with video explanations for 2011 A Level H2 Mathematics are suggested by Mr Gan. For any comments or suggestions please contact us at support@timganmath.edu.sg.
Without using a calculator, solve the inequality
$\frac{{{x}^{2}}+x+1}{{{x}^{2}}+x-2}<0$.
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The parametric equation of a curve are
$x={{t}^{2}}$, $y=\frac{2}{t}$
(i)
Find the equation of the tangent to the curve at the point $\left( {{p}^{2}},\frac{2}{p} \right)$, simplifying your answer.
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(i) Find the equation of the tangent to the curve at the point $\left( {{p}^{2}},\frac{2}{p} \right)$, simplifying your answer.
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(ii)
Hence find the coordinates of the points $Q$ and $R$ where this tangent meets the $x$- and $y$- axes respectively.
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(ii) Hence find the coordinates of the points $Q$ and $R$ where this tangent meets the $x$- and $y$- axes respectively.
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(iii)
Find a cartesian equation of the locus of (of the curve traced by) the mid-point of $QR$ as $p$ varies.
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(iii) Find a cartesian equation of the locus of (of the curve traced by) the mid-point of $QR$ as $p$ varies.
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(i)
Use the first three non-zero terms of the Maclaurin series for $\cos x$ to find the Maclaurin series for $\text{g}\left( x \right)$, where $\text{g}\left( x \right)={{\cos }^{6}}x$, up to and including the term in ${{x}^{4}}$.
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(i) Use the first three non-zero terms of the Maclaurin series for $\cos x$ to find the Maclaurin series for $\text{g}\left( x \right)$, where $\text{g}\left( x \right)={{\cos }^{6}}x$, up to and including the term in ${{x}^{4}}$.
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(ii)
(ii)(a) Use your answer to part (i) to give an approximation for $\int_{0}^{a}{\text{g}\left( x \right)}\text{ d}x$ in terms of $a$, and evaluate this approximation in the case where $a=\frac{1}{4}\pi $.
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(a) Use your answer to part (i) to give an approximation for $\int_{0}^{a}{\text{g}\left( x \right)}\text{ d}x$ in terms of $a$, and evaluate this approximation in the case where $a=\frac{1}{4}\pi $.
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(b) Use your calculator to find an accurate value for $\int_{0}^{\frac{1}{4}\pi }{\text{g}\left( x \right)}\text{ d}x$. Why is the approximation in part (ii)(a) not very good?
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(b) Use your calculator to find an accurate value for $\int_{0}^{\frac{1}{4}\pi }{\text{g}\left( x \right)}\text{ d}x$. Why is the approximation in part (ii)(a) not very good?
[2]
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Referred to the origin $O$, the points $A$ and $B$ are such that $\overrightarrow{OA}=\mathbf{a}$ and $\overrightarrow{OB}=\mathbf{b}$. The point $P$ on $OA$ is such that $OP:PA=1:2$, and the point $Q$ on $OB$ is such that $OQ:QB=3:2$. The mid-point of $PQ$ is $M$ (see diagram).
(i)
Find $\overrightarrow{OM}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ and show that the area of triangle $OPM$ can be written as $k\left| \mathbf{a}\times \mathbf{b} \right|$, where $k$ is a constant to be found.
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(i) Find $\overrightarrow{OM}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ and show that the area of triangle $OPM$ can be written as $k\left| \mathbf{a}\times \mathbf{b} \right|$, where $k$ is a constant to be found.
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(ii)
The vectors $\mathbf{a}$ and $\mathbf{b}$ are now given by
$\mathbf{a}=2p\mathbf{i}-6p\mathbf{j}+3p\mathbf{k}$ and $\mathbf{b}=\mathbf{i}+\mathbf{j}-2\mathbf{k}$,
where $p$ is a positive constant. Given that $\mathbf{a}$ is a unit vector,
(ii) The vectors $\mathbf{a}$ and $\mathbf{b}$ are now given by
$\mathbf{a}=2p\mathbf{i}-6p\mathbf{j}+3p\mathbf{k}$ and $\mathbf{b}=\mathbf{i}+\mathbf{j}-2\mathbf{k}$,
where $p$ is a positive constant. Given that $\mathbf{a}$ is a unit vector,
(a) find the exact value of $p$,
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(a) find the exact value of $p$,
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(b) give a geometrical interpretation of $\left| \mathbf{a}\cdot \mathbf{b} \right|$,
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(b) give a geometrical interpretation of $\left| \mathbf{a}\cdot \mathbf{b} \right|$,
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(c) evaluate $\mathbf{a}\times \mathbf{b}$.
[2]
(c) evaluate $\mathbf{a}\times \mathbf{b}$.
[2]
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Do not use a graphing calculator in answering this question.
(i)
The roots of the equation ${{z}^{2}}=-8\mathbf{i}$ are ${{z}_{1}}$ and ${{z}_{2}}$. Find ${{z}_{1}}$ and ${{z}_{2}}$ in cartesian form $x+\mathbf{i}y$, showing your working.
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(ii)
Hence, or otherwise, find in cartesian form the roots ${{w}_{1}}$ and ${{w}_{2}}$ of the equation
${{w}^{2}}+4w+\left( 4+2\mathbf{i} \right)=0$
[3]
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The continuous random variable $X$ has the distribution $N\left( \mu ,{{\sigma }^{2}} \right)$. It is known that $\text{P}\left( X<40.0 \right)=0.05$ and $\text{P}\left( X<70.0 \right)=0.975$. Calculate the values of $\mu $ and $\sigma $.
[4]
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A committee of $10$ people is chosen at random from a group consisting of $18$ women and $12$ men. The number of women on the committee is denoted by $R$.
(i)
Find the probability that $R=4$.
(i) Find the probability that $R=4$.
(ii)
The most probable number of women on the committee is denoted by $r$. By using the fact that $\text{P}\left( R=r \right)>\text{P}\left( R=r+1 \right)$, show that $r$ satisfies the inequality
$\left( r+1 \right)!\left( 17-r \right)!\left( 9-r \right)!\left( r+3 \right)!>r!\left( 18-r \right)!\left( 10-r \right)!\left( r+2 \right)!$
And use this inequality to find the value of $r$.
(ii) The most probable number of women on the committee is denoted by $r$. By using the fact that $\text{P}\left( R=r \right)>\text{P}\left( R=r+1 \right)$, show that $r$ satisfies the inequality
$\left( r+1 \right)!\left( 17-r \right)!\left( 9-r \right)!\left( r+3 \right)!>r!\left( 18-r \right)!\left( 10-r \right)!\left( r+2 \right)!$
And use this inequality to find the value of $r$.
(iii)
Use the calculator to find $\text{E}\left( R \right)$ and $\text{Var}\left( R \right)$.
(iii) Use the calculator to find $\text{E}\left( R \right)$ and $\text{Var}\left( R \right)$.
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