MaxWong

According to the table, there are 23 girls who like pizza.

P(pizza and girl) = \(\dfrac{\text{Number of girls who likes pizza}}{\text{Total number of people}} = \dfrac{23}{153}\).

Option D is correct.

Year 2019 is 19 years after Year 2000.

The current population is \(2e^{0.06(19)} \approx 6.253537 \text{ millions}\)

Approximately 6,253,537 people live in this city.

Obviously, the growth is an exponential growth, as the modelling function suggests.

A "strange" thing to say is that, year 2000 is 0 years after year 2000.

So we substitute t = 0, and we get:

Population living in the city in 2000 = \(2e^{0.06(0)} = 2(1) =2\text{ millions}\)

\(\text{Consider the range of }\tan^{-1} x.\\ \tan^{-1} x \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]\\ \text{That means we need to find a value } v\text{, which is between}-\dfrac{\pi}{2}\text{ and }\dfrac{\pi}{2}\text{, such that}\tan v = \tan\left(\dfrac{4\pi}{5}\right)\\ \)

\(\text{Solving the equation, }\\ v = \dfrac{4\pi}{5} + n\pi, \; n\in \mathbb Z\)

Only when n = -1, v is inside the required range.

The solution we need is \(v = \dfrac{4\pi}{5} - \pi = \dfrac{-\pi}{5}\).

Therefore \(\tan^{-1}\left(\tan\left(\dfrac{4\pi}{5}\right)\right) = \dfrac{-\pi}{5}\)

4(c)

\(\text{As shown above in 4(b), } \cos\left(2x-\dfrac{5\pi}{6}\right) = \sin\left(\dfrac{\pi}{3}\right)\implies x = n\pi+\dfrac{\pi}{2}\text{ or }x=n\pi+\dfrac{\pi}{3}\\ \text{As tan} x \text{ has a period of }\pi,\\ \tan x = \tan\left(\dfrac{\pi}{2}\right)\text{ or }\tan x = \tan\left(\dfrac{\pi}{3}\right)\\ \text{But considering the fact that}\tan x \text{ has an asymptote at } x = \dfrac{\pi}{2},\text{ the only finite value is }\tan\left(\dfrac{\pi}{3}\right).\\ \text{The required value is }\tan\left(\dfrac{\pi}{3}\right) = \sqrt3\)

4(b)

\(\cos\left(2x-\dfrac{5\pi}{6}\right) = \sin\left(\dfrac{\pi}{3}\right)\\ \cos\left(2x-\dfrac{5\pi}{6}\right) = \cos\left(\dfrac{\pi}{6}\right)\\ \text{As the cosine function is periodic with period }2\pi\text{, }\\ \cos\left(2x-\dfrac{5\pi}{6}\right) = \cos\left(\dfrac{\pi}{6}+2n\pi\right)\\ \text{As }\cos x = \cos\left(-x\right)\text{, }\\ \cos\left(2x-\dfrac{5\pi}{6}\right) = \cos\left(-\dfrac{\pi}{6}+2n\pi\right)\\ \text{Comparing the left hand side and the right hand side of the equation, }\\ 2x-\dfrac{5\pi}{6} = \dfrac{\pi}{6} + 2n\pi \text{ or } 2x-\dfrac{5\pi}{6} = -\dfrac{\pi}{6} + 2n\pi\\ 2x = \pi+2n\pi\text{ or }2x=\dfrac{2\pi}{3}+2n\pi\\ x = \dfrac{\pi}{2}+n\pi\text{ or }x=\dfrac{\pi}{3}+n\pi\)

This is trigonometry.

4(a)

\(\text{As }\sin x = \cos\left(\dfrac{\pi}{2}-x\right),\\ \sin \left(\dfrac{\pi}{3}\right) = \cos\left(\dfrac{\pi}{2}-\dfrac{\pi}3\right)=\cos\left(\pi\left(\dfrac{1}{2}-\dfrac{1}{3}\right)\right)=\cos\left(\pi\left(\dfrac{1}{6}\right)\right) =\cos\left(\dfrac{\pi}{6}\right)\\ \sin \left(\dfrac{\pi}{3}\right) = \cos\left(\dfrac{\pi}{k}\right)\\ \cos\left(\dfrac{\pi}{6}\right) = \cos\left(\dfrac{\pi}{k}\right)\\ \text{Comparing the left hand side and the right hand side of the equation,}\\ k =6\)

Please notice that \((a+b)^n \neq a^n+b^n\).

It is easy enough, even with that incorrect formula :)

The correct formula should be \(\displaystyle \sum^{n}_{i=1} i^2 = \dfrac{n(n+1)(2n+1)}{6}\).

Obviously, this limit is a Riemann integral.

\( S = \displaystyle{\lim_{n\rightarrow \infty} \sum_{i=1}^{n}\left(\left(\dfrac{2}{n}\right)\left(\dfrac{i}{n}\right)^2\right)\\ \;\;\;\!= 2\lim_{n\rightarrow \infty} \sum_{k=1}^{n}\left(\left(\dfrac{1}{n}\right)\left(\dfrac{k}{n}\right)^2\right)\\ \boxed{\text{Riemann integral: For any function f continuous on [0,1], } \lim_{n\rightarrow \infty}\sum^{n}_{k=1}\dfrac{1}{n}f\left(\dfrac{k}{n}\right) = \int^1_0f(x)dx }\\ S=2\int^1_0x^ 2dx\\\;\;\;\!=2\left(\dfrac{1}{3}\right)\\\;\;\;\!=\dfrac{2}{3} }\)