# Recent content by fan96

1. ### Simple 2U Trig Help

-\pi and \pi are typically both used to describe the angle of the negative x-axis (i.e. they're both really the same angle). 0 \le x \le \pi are the 1st and 2nd quadrants. -\pi \le x \le 0 are the 3rd and 4th quadrants. So -\pi \le x \le \pi describes all four quadrants.
2. ### help for proofs

Where did you get this question? You probably wouldn't be able to get this result from just e^n \ge {(n+1)^n}/{n!} as \frac{(n+1)^n}{n!} \stackrel{?}{\ge}1+n^2 is not actually true, even taking n \ge 0. You'd have to at least start a bit further back. I doubt this is the intended answer...
3. ### Mistakes with trivial arithmetic

Usually I'd expect a mistake like this to only incur a 1/2 mark penalty. If it was really really minor (e.g. you wrote the wrong units) then you might even get away for free.
4. ### Hard enrichment question from the Cambridge textbook!

In more concrete words, you are asked to show n = \cos^{-1} \frac 35, given the conditions a \cos \alpha = 1, a \cos (n+\alpha) = 5, a \cos (2n+\alpha) = 5. So, try to solve for n . You have three equations with as many variables, so intuitively this should be solvable. Bonus points -...
5. ### Mechanics question

Actually, the simplest way is to just get the answer from the back of the textbook, ask someone else who knows how to do it, or to search it up online. And this is what we would do if we only cared about the number we get at the end. So what is the point of even learning calculus and...
6. ### Binomial Help

\binom{n}{2} \left(\frac{9}{10}\right)^{n-2} < \binom{n}{3}\left(\frac{9}{10}\right)^{n-3} \frac 1{10} Divide by (9/10)^{n-3}: \binom{n}{2} \left(\frac{9}{10}\right) < \binom{n}{3} \frac 1{10} Now use \binom{n}{k} = \frac{n!}{(n-k)! \, k!} so \frac{n!}{(n-2)! \, 2!}...
7. ### "Describe the motion..." mechanics question

Your reasoning is correct. Another way to look at it is the following: If O is any point that is not \pi/2 , then the particle will start off with a positive velocity, which means its displacement will increase. But when its displacement increases, its velocity ( v = \cos^2x ) gets smaller...
8. ### HSC Maths Ext2 Questions + Answers

When we discuss real-valued functions it's often convenient or more practical for us to take "continuous" to mean "continuous on \mathbb R". Most people accept this because whatever meaning we take is usually clear to the reader from context. In this case it is (probably) not, so you should be...
9. ### Maths neatness

If your fraction stack is growing out of control, you can just get rid of the fraction entirely - write \frac{\frac{1}{x}+e^x}{1+\frac{1}{\sin^{-1} x}} as \left( \frac{1}{x}+e^x \right) \left( 1+\frac{1}{\sin^{-1} x} \right) ^{-1}. Another useful trick is e^{\frac{1}{x} + \sin x} \to...
10. ### Challenging (?) Proof Question

If one wanted to actually find examples of M , here is an approach that works for M \in \mathbb R . I'll use the hyperbolic function \cosh and its inverse, which aren't in the MX2 syllabus but their definitions are quite simple to understand: \cosh x = \frac 12 \left(e^x+e^{-x}\right) and...
11. ### Challenging (?) Proof Question

Is the condition M \notin \mathbb Z really necessary? The only case of this I can see is 1 + 1/1 = 2 \in \mathbb Z, so it would be simpler to leave it unmentioned.
12. ### Perms & Coms

If the queues are distinct, then we should have 17280 = \underbrace{\left( \binom{8}{4} \times 4! \times 4!\right)}_{\text{no restrictions}} - \underbrace{\left(\binom{6}{3} \times 4!\ \times 4!\right)\times 2}_{\text{Sean and Liam in diff. queues}}, or 17280 = \left( \binom{6}{4} \times 4...
13. ### Tricky projectile motion question, Thanks

For both balls we have \begin{cases} y_1(t) &= -\frac{9.8}{2}t^2+100\sin(\pi/6)t+h \\ x_1(t) &=100\cos(\pi/6)t\end{cases} \begin{cases} y_2(t) &= -\frac{9.8}{2}t^2+h \\ x_2(t) &=100t\end{cases} Solve y_1(t) = 0 and y_2(t) = 0 to get the time of flight for each ball (call these t_1 and...
14. ### Vectors

Recall that for two vectors \bold x, \bold y we have \cos \theta = \frac{\bold x \cdot \bold y}{|\bold x||\bold y|}, where \theta is the angle between \bold x and \bold y .
15. ### 4 types of relations

Unfortunately it's not that simple. You can't "break up" the absolute value function like that. If y = |x-3| , then y = \pm(x - 3). You can try graphing these to visualise the effect of the absolute value function.