Author Topic: 3U Maths Question Thread (Read 1521027 times) Tweet Share

winstondarmawan · « **Reply #2385 on:** July 10, 2017, 10:44:47 am »

Quote from: anotherworld2b on July 10, 2017, 09:43:43 am

Differentiate it
You would use the quotient rule right ?

No, split the logs first using log laws. Quotient rule takes too long and is unecessary

RuiAce · « **Reply #2386 on:** July 10, 2017, 11:32:05 am »

Quote from: anotherworld2b on July 09, 2017, 07:16:14 pm

Could I have help with Q26 , Q27 please?

$\text{As suggested above, when you're differentiating stuff involving logarithms}\\ \text{it helps to apply log laws first.}$
$\text{For 27}\\ \begin{align*}\frac{d}{dx}\left(\ln \left[\frac{(x+2)^3}{x-2}\right]\right)&=\frac{d}{dx}(3\ln (x+2)-\ln (x-2))\\ &= \frac{3}{x+2}-\frac{1}{x-2}\end{align*}$

MisterNeo · « **Reply #2387 on:** July 10, 2017, 11:39:15 am »

Quote from: anotherworld2b on July 09, 2017, 07:16:14 pm

Could I have help with Q26 , Q27 please?

Differentiating these logs requires the use of log laws.
25. $\ln (x^2+5)^4=4\ln (x^2+5)$
$\frac{d}{dx}4\ln (x^2+5)=\frac{4}{x^2+5}*2x$
$=\frac{8x}{x^2+5}$
26. $\ln (\frac{x}{x^2-1})=\ln x-\ln(x^2-1)$
$\frac{d}{dx}\ln x-\ln(x^2-1)=\frac1x-\frac{2x}{x^2-1}$
$=-\frac{x^2+1}{x^3-x}$
That last line, you get them on the same denominator.

legorgo18 · « **Reply #2388 on:** July 10, 2017, 01:40:38 pm »

Hello, how do i do this

1) Solve for x and y in exact form if sin^-1 x - cos^-1 y= pi/12 and cos^-1x + sin^-1y = 5pi/12 (added them and got sin^-1 x = cos^-1 y then stuck)
2) Solve arcsin x - arccos x = arccos x/2 (i tried differentiating, rearranging and got degree 8 polynomial)

Ty

RuiAce · « **Reply #2389 on:** July 10, 2017, 02:53:59 pm »

Quote from: legorgo18 on July 10, 2017, 01:40:38 pm

Hello, how do i do this

1) Solve for x and y in exact form if sin^-1 x - cos^-1 y= pi/12 and cos^-1x + sin^-1y = 5pi/12 (added them and got sin^-1 x = cos^-1 y then stuck)
2) Solve arcsin x - arccos x = arccos x/2 (i tried differentiating, rearranging and got degree 8 polynomial)

Ty

Not sure how you added because that result is not true.
$\text{Adding:}\\ \begin{align*}\sin^{-1}x-\cos^{-1}y&=\frac{\pi}{12}\\ \cos^{-1}x+\sin^{-1}y&= \frac{5\pi}{12}\\ \implies \frac{\pi}{2}+\sin^{-1}y-\cos^{-1}y&=\frac{\pi}{2} \tag{1}\end{align*}\\ \text{having noted the identity }\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}$
$\text{Also, }\sin^{-1}y+\cos^{-1}y=\frac{\pi}{2}\\ \text{Adding this to equation (1) gives }2\sin^{-1}y=\frac{\pi}{2}\\ \text{The rest should be doable from here.}$
____________________________

Not sure why you bothered with calculus. That question is dodgy though. I need more time. Edit: I see where the degree 8 polynomial came from though. Time to find a work around.

legorgo18 · « **Reply #2390 on:** July 10, 2017, 03:09:15 pm »

Rui when you get pi/2 + arcsiny - arccosy = pi/2, if you cancel out the pi/2s then move the arccosy over, dont you get arcsiny = arccosy ?? oh i typed it wrong in the thing lmaoooooooooooooooooooooooo

RuiAce · « **Reply #2391 on:** July 10, 2017, 03:10:29 pm »

Ok, really dodgy question. The instant I saw the degree 8 polynomial (which for me ended up being a degree 6 polynomial) I said back up. I tried going backwards and just working with what I had and it turned out better.
$\textit{Lemma: } \cos(\arcsin x) = \sin (\arccos x) = \sqrt{1-x^2}\\ \text{Proof: Exercise. Requires a Pythagorean identity.}$
$\text{From }\arcsin x - \arccos x = \arccos \frac{x}2\\ \text{Applying 'cos' to both sides and expanding the compound angle gives}\\ x\sqrt{1-x^2}+x\sqrt{1-x^2}=\frac{x}{2} \implies \sqrt{1-x^2}=\frac{1}4\\ \text{noting that }x\neq 0\text{ by directly substituting into the original equation}$
$\text{This then solves to give }1-x^2=\frac{1}{16} \implies x= \pm \frac{\sqrt{15}}{4}\\ \text{and I'm sure you can reject the negative}\\ \text{by substituting back into the original equation}$

RuiAce · « **Reply #2392 on:** July 10, 2017, 03:11:37 pm »

Quote from: legorgo18 on July 10, 2017, 03:09:15 pm

Rui when you get pi/2 + arcsiny - arccosy = pi/2, if you cancel out the pi/2s then move the arccosy over, dont you get arcsiny = arccosy ?? oh i typed it wrong in the thing lmaoooooooooooooooooooooooo

Yeah. I was wondering if there was a typo. But I didn't assume it to be on the safe side

legorgo18 · « **Reply #2393 on:** July 10, 2017, 03:15:44 pm »

Omg all you had to do was cos both sides from the beginning... oh man i added 2 arccosx to both sides then cos both sides and squared etc, dont know what i was thinking...

RuiAce · « **Reply #2394 on:** July 10, 2017, 03:21:50 pm »

That was something along the lines of what I originally did. It was only when I saw how messy it was I backtracked a bit. I missed the "symmetry" you get from applying cos on the fly

legorgo18 · « **Reply #2395 on:** July 10, 2017, 03:28:24 pm »

Btw, how do you questions of finding f^-1(f(a)) if a is not in the domain of the inverse function. I was thinking of changing the domain but not sure how. Also got trials coming up like everyone should i be focusing 2u past papers more than 3 and what should i start from easy schools like tech or just grammar as 2u isnt that challenging?

RuiAce · « **Reply #2396 on:** July 10, 2017, 03:38:21 pm »

Quote from: legorgo18 on July 10, 2017, 03:28:24 pm

Btw, how do you questions of finding f^-1(f(a)) if a is not in the domain of the inverse function. I was thinking of changing the domain but not sure how. Also got trials coming up like everyone should i be focusing 2u past papers more than 3 and what should i start from easy schools like tech or just grammar as 2u isnt that challenging?

Sometimes you can do it via direct computation but depending on what f(x) is and the restrictions on a are, we can work around it accordingly.

limtou · « **Reply #2397 on:** July 10, 2017, 11:48:06 pm »

Hello! Please help with this question I'm very confused

RuiAce · « **Reply #2398 on:** July 10, 2017, 11:52:37 pm »

Quote from: limtou on July 10, 2017, 11:48:06 pm

Hello! Please help with this question I'm very confused

At first glance, this looks a question that's best approached by L'Hopital's rule, which is not permitted for the course. Please provide the source of the question.

limtou · « **Reply #2399 on:** July 10, 2017, 11:56:09 pm »

Quote from: RuiAce on July 10, 2017, 11:52:37 pm

At first glance, this looks a question that's best approached by L'Hopital's rule, which is not permitted for the course. Please provide the source of the question.

It was from a 4U James Ruse paper, but since it's harder 3U I posted it here.

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