3U Maths Question Thread

[RuiAce]

Hi, I had a question about finiding limits when graphing functions. I have learnt many methods for example, the polynomial divsion method, finding the limit at infinity ect.. I was wondering if you could give me a simple guide on how to find limits for all power functions, eg Horizontal and oblique asymptotes, not vertical because thats simple. Also, in terms of graphing, how do you know if it crosses the horizontal asymptote, because sometimes it does. Thank you.

Horizontal asymptotes: 2 cases
1) Degree of numerator is less than degree of denominator. When this happens, if you just do the usual method of taking limits to infinity, you'll find the asymptote is y=0.
2) Degree of numerator is equal to degree of denominator. Pretty much you'll be limiting something like \( \frac{ax^2 + ...}{dx^2 + ...} \), and again if you do the usual method of limits to infinity, you'll find it to be \( y = \frac{a}{d} \), i.e. the quotient of the leading coefficients.

Oblique asymptotes: Occurs only when the degree of the numerator is exactly one greater than the degree of the denominator. This is when you do polynomial long division. Once you do polynomial long division, you get \( y = ax + b + \frac{\text{something}}{\text{something}}\), and you can show that the oblique asymptote is \( y = ax + b\).

Also, as far as 3U goes, you'll never encounter stranger scenarios with asymptotes - oblique asymptotes are the weirdest.

(Also, I don't know what you mean when you say "limits for all power functions". Horizontal and oblique asymptotes are computed with the aid of limits, but I fail to see how they're the same thing.)
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More often than not, you don't. You can work around this issue by locating the coordinates of any stationary points, and then using
- x-intercepts
- vertical asymptotes
- stationary points
make a deduction about whether or not it crosses the horizontal asymptote. But it's very hard to say what will happen if you do not have these ingredients.

[Jane20]

Hi can anyone please help me with this question ? It looks kinda weird to me

Thanks a lot !!

[greenhouse12]

HI guys! Does anyone know where to find exercises and questions sorted by topic for the 3U maths exam? I really would like them sorted by topic as I want to do extra practice for upcoming tasks, but obviously these tasks only test certain topics at a time If this has already been posted somewhere, please link me and thanks!

Have a good day!

[RuiAce]

Hi can anyone please help me with this question ? It looks kinda weird to me

Thanks a lot !!

But of course, using the formula \( \binom{N}{K} = \frac{N!}{K!(N-K)!} \) we see that \( \binom{2n}{n} = \frac{(2n)!}{(n!)^2} \) so therefore C

[Calley123]

Hey,
Please help. Questions on inverse functions attached below!
Answers
13b) pi/6 + 1-(square root 3/6)
15) square root 2 inverse tan x
Thanks

[RuiAce]

Hey,
Please help. Questions on inverse functions attached below!
Answers
13b) pi/6 + 1-(square root 3/6)
15) square root 2 inverse tan x
Thanks

\begin{align*}\frac{d}{dx} \left( \frac{v^2}{2}\right)&= \frac{1}{1+x^2}\\ \frac{v^2}{2} &= \tan^{-1}x + C\\ v^2 &+ 2\tan^{-1}x + 2C\end{align*}

Q13 is missing

[arii]

Solve sin2x + cos2x = sin²x +1 for 0≤x≤2pi

[RuiAce]

Solve sin2x + cos2x = sin²x +1 for 0≤x≤2pi

We leave those three solutions aside and try to find the remainder
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[clovvy]

We leave those three solutions aside and try to find the remainder
___________________________________________________
[ over the domain }\boxed{0\leq x \leq 2\pi}\\ \text{is equivalent to solving }\sqrt{13} \cos \left(x + \tan^{-1} \frac32 \right) = 0\\ \text{ over the adjusted domain }\boxed{\tan^{-1}\frac32 \leq x + \tan^{-1}\frac32 \leq 2\pi + \tan^{-1}\frac32}[/tex]

How do I convert 2cosx+3sinx=0 to [tex]sqrt{13} \cos \left(x + \tan^{-1} \frac32 \right) = 0//[tex]

[RuiAce]

How do I convert 2cosx+3sinx=0 to

Like I said, auxiliary angle method.

Here a=2 and b=3.

[clovvy]

Like I said, auxiliary angle method.

Here a=2 and b=3.

Ah right yeah I forgot about that yeah I got it.. just 3U further trigs

[arii]

Had a bit of trouble with this question..

[RuiAce]

Had a bit of trouble with this question..

\begin{align*}&\quad\frac{\sin A + \sin (A+B) + \sin (A+2B)}{\cos A + \cos (A+B) + \cos (A+2B)}\\ &= \frac{\sin \big( (A+B) - B \big) + \sin (A+B) + \sin \big( (A+B) - B \big)}{\cos \big( (A+B) - B \big) + \cos (A+B) + \cos \big( (A+B) + B \big)}\\ &= \frac{[\sin (A+B) \cos B - \cos (A+B) \sin B] + \sin (A+B) + [\sin (A+B) \cos B + \cos (A+B) \sin B]}{[\cos (A+B) \cos B + \sin (A+B) \sin B] + \cos (A+B) + [\cos (A+B) \cos B - \sin (A+B) \sin B]}\\ &= \frac{2\sin(A+B)\cos B + \sin (A+B)}{2\cos(A+B)\cos B + \cos (A+B)}\\ &= \frac{\sin (A+B) [2\cos B + 1]}{\cos (A+B) [2\cos B + 1]}\\ &= \tan(A+B)\end{align*}

[Calley123]

\begin{align*}\frac{d}{dx} \left( \frac{v^2}{2}\right)&= \frac{1}{1+x^2}\\ \frac{v^2}{2} &= \tan^{-1}x + C\\ v^2 &+ 2\tan^{-1}x + 2C\end{align*}

Q13 is missing

Thank  you!!

[Calley123]

Please help!
Questions are from integration by subsitution