Mock Exam Set 1 - Paper 3

[Maximum mark: 24]

This question asks you to investigate some properties of hexagonal numbers.

Hexagonal numbers can be represented by dots as shown below where $h_n$ denotes the $n$th hexagonal number, $n\in \mathbb{N}$.

Note that $6$ points are required to create the regular hexagon $h_2$ with side of length $1$, while $15$ points are required to create the next hexagon $h_3$ with side of length $2$, and so on.

1. Write down the value of $h_5$.[1]
   

2. By examining the pattern, show that $h_{n+1} = h_{n}+4n+1$, $n\in \mathbb{N}$. [3]
   

3. By expressing $h_n$ as a series, show that $h_n = 2n^2-n$, $n\in \mathbb{N}$.[3]
   

4. Hence, determine whether $2016$ is a hexagonal number.[3]
   

5. Find the least hexagonal number which is greater than $80\hspace{0.10em}000$.[5]
   

6. Consider the statement:

   $45$ is the only hexagonal number which is divisible by $9$.

   Show that this statement is false.[2]
   

Matt claims that given $h_1 = 1$ and $h_{n+1} = h_n + 4n + 1$, $n \in \mathbb{N}$, then

7. Show, by mathematical induction, that Matt's claim is true for all $n\in \mathbb{N}$.[7]
   

[Maximum mark: 31]

This question ask you to investigate the relationship between the number of sides and the area of an enclosure with a given perimeter.

A farmer wants to create an enclosure for his chickens, so he has purchased $28$ meters of chicken coop wire mesh.

1. Initially the farmer considers making a rectangular enclosure.

   1. Complete the following table to show all the possible rectangular enclosures with sides of at least $4\,$m he can make with the $28\,$m of mesh. The sides of the enclosure are always a whole number of metres.[3]
      

      

   2. What is the name of the shape that gives the maximum area?[1]
      

The farmer wonders what the area will be if instead of a rectangular enclosure he uses an equilateral triangular enclosure.

2. Show that the area of the triangular enclosure will be $\dfrac{196\hspace{0.05em}\sqrt 3}{9}$.[3]
   

Next, the farmer considers what the area will be if the enclosure has the form of a regular pentagon.

The following diagram shows a regular pentagon.

Let O be the centre of the regular pentagon. The pentagon is divided into five congruent isosceles triangles and angle $\textrm{A}\widehat{\textrm{O}}\textrm{B}\rule[-0mm]{0pt}{4mm}$ is equal to $\theta$ radians.

3. 1. Express $\theta$ in terms of $\pi$.

   2. Show that the length of OA is $\dfrac{14}{5} \operatorname{cosec}\left(\dfrac{\pi}{5}\right)$ m.

   3. Show that the area of the regular pentagon is $\dfrac{196}{5} \cot\left(\dfrac{\pi}{5}\right)$ m$^2$.[6]
      

Now, the farmer considers the case of a regular hexagon.

4. Using the method in part (c), show that the area of the regular hexagon is [5]
   
   $$\begin{aligned} \\ \dfrac{196}{6} \cot\left(\dfrac{\pi}{6}\right)\hspace{0.15em}\text{m}^2.\end{aligned}$$

The farmer notices that the hexagonal enclosure has a larger area than the pentagonal enclosure. He considers now the general case of an $n$-sided regular polygon. Let $A_n$ be the area of the $n$-sided regular polygon with perimeter of $28\,$m.

5. Show that $A_n=\dfrac{196}{n}\cot \left(\dfrac{\pi}{n}\right).$[5]
   

6. Hence, find the area of an enclosure that is a regular 14-sided polygon with a perimeter of $28$ m. Give your answer correct to one decimal place.[2]
   

7. 1. Evaluate $\displaystyle\lim_{n \to \infty}A_n.$

   2. Interpret the meaning of the result of part (g) (i). [6]