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]

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]

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

1. Evaluate $\displaystyle\int\nolimits_{k}^{k+1}{\frac{1}{{{x}^{2}}+1}\mathrm{d}x}$ 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]

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]

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]

A teacher, Mr. Ku, suspects that the average time a student spends on his or her mobile phone per day is ${{\mu }_{0}}$ minutes. He selected a random sample of $97$ students in the school who own mobile phones and recorded the amount of time each student spent on his or her phone in a randomly selected day. The results are displayed in the table below.

1. Calculate unbiased estimates of the population mean and variance of the time a student spends on his or her mobile phone per day.

   [2]

The null hypothesis that the average time a student spends on his or her mobile phone per day is ${{\mu }_{0}}$ minutes is tested, at $5\%$ level of significance, against the alternative hypothesis that the average time a student spends on his or her mobile phone per day differs from ${{\mu }_{0}}$ minutes.

2. Determine the range of values of ${{\mu }_{0}}$ for which the null hypothesis is rejected.

   [5]

3. Explain, in the context of this question, the meaning of ‘at $5\%$ level of significance’.

   [1]

4. If the null hypothesis in (ii) is rejected at $5\%$ significance level, can we reject the null hypothesis at $1\%$ level of significance? Explain your answer.

   [1]