Eunoia Junior College H2 Math papers, with video solutions

A spherical container of radius \(5\) m is formed by rotating the following circle \(C\) about the \(y\)-axis.

The container has negligible thickness, and the circle \(C\) passes through the origin \(O\).

1. State a cartesian equation of \(C\).[1]

Initially the spherical container is completely filled with water. Two engineers are calculating the time needed for the container to be completely drained from a small circular hole at the bottom. The volume of water in the container at time \(t\) seconds is denoted by \(V\) m\(^{3}\).

[The volume of a sphere of radius \(r\) is \(\frac{4}{3}\pi {{r}^{3}}\).]

2. The first engineer proposes that the rate of change of \(V\) with respect to \(t\) is a constant \(k\).[1]
   1. Write down a differential equation relating \(V\), \(t\) and \(k\).[1]
   2. Determine, with justification, the sign of \(k\).[1]
   3. Find \(V\) in terms of \(t\) and \(k\), leaving your answer in exact form.[2]
3. The second engineer argues that the rate at which water flows out from the hole will be at its greatest in the beginning, and decreases as the depth of water in the container decreases. He suggests using Torricelli’s law, which says that
   \(\frac{\mathrm{d}V}{\mathrm{d}t}=-\alpha \sqrt{20h}\),
   where \(\alpha \) m\(^{2}\) is the area of the circular hole at the bottom of the container, and \(h\)m is the depth of water in the container at time \(t\) seconds. The radius of the hole at the bottom of the container is found to be constant at \(1\) cm.
   1. Show that \(\left( 10{{h}^{\frac{1}{2}}}-{{h}^{\frac{3}{2}}} \right)\frac{\mathrm{d}h}{\mathrm{d}t}=-\frac{\sqrt{5}}{5000}\).[4]
   2. Hence find the numerical value of \(t\) when the container is completely drained.[4]

The diagram shows the graph of $y=\frac{1}{{{x}^{2}}+1}$ when $x>0$.

1. Evaluate $\displaystyle\int_{k}^{k+1}{\frac{1}{{{x}^{2}}+1}\mathrm{d}x}$ for $k>0$, leaving your answer in terms of $k$.

   [2]

2. By considering appropriate rectangles on the interval $\left[ k,k+1 \right]$ for the curve $y=\frac{1}{{{x}^{2}}+1}$, show that
   $\frac{1}{{{\left( k+1 \right)}^{2}}+1}<{{\tan }^{-1}}\left( k+1 \right)-{{\tan }^{-1}}k<\frac{1}{{{k}^{2}}+1}$ for $k\in {{\mathbb{Z}}^{+}}$.

   [2]

3. Use the identity $\tan \left( A-B \right)=\frac{\tan A-\tan B}{1+\tan A\tan B}$ to show that
   ${{\tan }^{-1}}x-{{\tan }^{-1}}y={{\tan }^{-1}}\frac{x-y}{1+xy}$, where $x>y>0$.

   [2]

4. By considering parts (ii) and (iii), prove by the method of differences that
   $\sum\limits_{k=1}^{n}{\frac{1}{{{\left( k+1 \right)}^{2}}+1}}<{{\tan }^{-1}}\left( \frac{n}{n+2} \right)<\sum\limits_{k=1}^{n}{\frac{1}{{{k}^{2}}+1}}$

   [4]

Functions \(\mathrm{f}\) and \(\mathrm{g}\) are defined by

\(\mathrm{f}:x\mapsto {{x}^{3}}-7{{x}^{2}}-5x+11\), \(x\in \mathbb{R}\), \(x\ge k\),

\(\mathrm{g}:x\mapsto {{\left( x+1 \right)}^{2}}+2\),\(x\in \mathbb{R}\).

1. Let \(k=1\).
   1. Show that \(\mathrm{f}\) does not have an inverse.[2]
   2. Determine whether the composite function \(\mathrm{fg}\) exists.[2]
2. Find the value of \(k\) given that \({{\mathrm{f}}^{-1}}\) exists and that the domain of \({{\mathrm{f}}^{-1}}\) is \(x\in \mathbb{R}\), \(x\ge -24\).[2]
3. Let \(k=6\).
   1. Show algebraically that \(\mathrm{f}'(x)>0\) for all values of \(x\) in the domain of \(\mathrm{f}\).[2]
   2. Solve the equation \(\mathrm{g}{{\mathrm{f}}^{-1}}(x)=83\).[3]

State the derivative of \(\tan {{x}^{2}}\). Hence, or otherwise, find \(\displaystyle\displaystyle \displaystyle\int{{{x}^{3}}{{\sec }^{2}}{{x}^{2}}\mathrm{d}x}\)[4]

The parametric equations of curve $C$, is given as $x=at$, $y=a{{t}^{3}}$, where $a$ is a positive constant.

1. The point $P$ on the curve has parameter $p$ and the tangent to the curve at point $P$ cuts the $y$-axis at $S$ and the $x$-axis at $T$. The point $M$ is the midpoint of $ST$. Find a Cartesian equation of the curve traced by $M$ as $p$ varies.

   [5]

2. Find the exact area bounded by curve $C$, the line $x=0$, $x=3$ and $x$-axis, giving your answer in terms of $a$.

   [3]

A tank has a capacity of 100 litres. Initially, the tank contains 10 litres of water thoroughly mixed with 300 grams of salt. Salt water with a concentration of 5 g/litre is poured into the tank at a constant rate of 2 litres per minute, while the mixture flows out at a constant rate of 1 litre per minute.

Let \(S\) denote the amount of dissolved salt in the tank (in grams) at time \(t\) minutes after salt water is poured into the tank.

1. Show that \(\frac{\mathrm{d}S}{\mathrm{d}t}=10-\frac{S}{t+10}\), stating your assumption(s) clearly.[2]
2. By substituting \(Q=\left( t+10 \right)S\), solve the differential equation in (i), and find \(S\) in terms of \(t\).[4]
3. Hence, find the concentration of salt in the tank at the point just before it overflows.[2]

Once the volume of salt solution in the tank reaches 100 litres, the pouring stops, and the tank is allowed to drain off. The salt solution drains from the tank at a rate proportional to the volume of solution in the tank. Let \(V\) denote the volume of solution in the tank at time \(T\) after the tank starts to drain off.

4. If the tanks takes 10 minutes to drain half of its contents, find \(V\) in terms of  \(T\).[4]