4U Maths Question Thread

[RuiAce]

hello

can someone help pleaseee because Im not getting the right answer.

thank you

\begin{align*} \int \frac{1}{1-\sin x}\,dx &= \int \frac{1 + \sin x}{1 - \sin^2 x}\,dx\\ &= \int \frac{1 + \sin x}{\cos^2 x}\,dx\\ &= \int \frac{1}{\cos^2 x} + \frac{\sin x}{\cos^2 x}\,dx\\ &= \int \sec^2 x + \sec x \tan x \,dx\\ &= \sec x +\tan x + C \end{align*}

[3.14159265359]

\begin{align*} \int \frac{1}{1-\sin x}\,dx &= \int \frac{1 + \sin x}{1 - \sin^2 x}\,dx\\ &= \int \frac{1 + \sin x}{\cos^2 x}\,dx\\ &= \int \frac{1}{\cos^2 x} + \frac{\sin x}{\cos^2 x}\,dx\\ &= \int \sec^2 x + \sec x \tan x \,dx\\ &= \sec x +\tan x + C \end{align*}

thank you!!

[kaustubh.patel]

hello people, need some help please, Q10 from 2014 hsc

[clovvy]

hello people, need some help please, Q10 from 2014 hsc

So the answer is D, and pay attention at how I use u as a dummy variable for x...

[3.14159265359]

hello

I just wanted to ask,  up to what year/how far back should I go with the hsc past papers since the syllabus changed and the older papers are more difficult?

thank you

[RuiAce]

hello

I just wanted to ask,  up to what year/how far back should I go with the hsc past papers since the syllabus changed and the older papers are more difficult?

thank you

They do get harder the further back you go, however anything from 2001 onwards is still relevant, so as far as you can go back to there (factoring in that you also have other subjects to study for).

[3.14159265359]

sorry can someone help me with these questions please. with the volume question I'm confused because they did not specify which axis its rotated by so how am I supposed to know.
thank you

[RuiAce]

sorry can someone help me with these questions please. with the volume question I'm confused because they did not specify which axis its rotated by so how am I supposed to know.
thank you

Actually that is a fair call with the volumes question. That question is in theory not doable because they haven't specified the line about which the rotation is taken with respect to.

As this is a reasonably huge dump of strenuous questions I (more or less because I'll get weary ) won't go into complete depth with them. However it is worth mentioning that for the 3 variable AM-GM inequality you've mentioned, in the current HSC there would be hints that help you build up to that result, as opposed to being shoved straight into the deep end. This is one way of proving the AM-GM inequality for 3 variables given that you've already proven it for 4 variables, and to prove it for 4 variables you can work through my trial lecture handout solutions.

The angular velocity \( \omega\) is defined by \( \omega = \frac{\d\theta}{dt} \), i.e. the rate at which the angle the particle makes, at the origin in the positive \(x\)-axis, changes with respect to time. That derivation was examined in the 1981 paper (and from memory it should've been stated in my 4U notes book), but it essentially relies on you starting with \( x = r\cos \theta\), \(y = r\sin \theta\) and then using implicit differentiation to obtain results like \( \dot{x} = \frac{d\theta}{dt} \times -r\sin \theta = -r\omega \sin \theta\), etc.

That locus essentially describes the arc, with endpoints at \(z_1\) and \(z_2\), going in an anticlockwise direction from \(z_1\) to \(z_2\) and not including the points \(z_1\) and \(z_2\) themselves.

[3.14159265359]

The angular velocity \( \omega\) is defined by \( \omega = \frac{\d\theta}{dt} \), i.e. the rate at which the angle the particle makes, at the origin in the positive \(x\)-axis, changes with respect to time. That derivation was examined in the 1981 paper (and from memory it should've been stated in my 4U notes book), but it essentially relies on you starting with \( x = r\cos \theta\), \(y = r\sin \theta\) and then using implicit differentiation to obtain results like \( \dot{x} = \frac{d\theta}{dt} \times -r\sin \theta = -r\omega \sin \theta\), etc.

That locus essentially describes the arc, with endpoints at \(z_1\) and \(z_2\), going in an anticlockwise direction from \(z_1\) to \(z_2\) and not including the points \(z_1\) and \(z_2\) themselves.

the answers for the volumes question did it around the y axis.

1981 😵😵 does that mean its highly unlikely that they will ask this question, ever? and I don't understand what is the point of that question. how am I  supposed to know wtf they are talking about. I did not do that in school (coz I have a crappy teacher) so when I first saw it I was like wtffffff

for that locus that is what I did. but the answers had something weird 

[RuiAce]

the answers for the volumes question did it around the y axis.

1981 😵😵 does that mean its highly unlikely that they will ask this question, ever? and I don't understand what is the point of that question. how I sone supposed to know wtf they are talking about. I did not do that in school (coz I have a crappy teacher) so when I first saw it I was like wtffffff

for that locus that is what I did. but the answers had something weird 

Yeah ignore that volumes question. They didn't hint at all that you were meant to do that.

Yeah that mechanics question is somewhat famous (can be found via a google search), because it basically asked for the proof of all the circular motion formulae we've been taking for granted all our lives. I can't see it being examined again unless they help set it up for you (just like with that AM-GM inequality question), because the intuition behind thinking that up without any help whatsoever is a bit of a huge stretch.

Their diagram looks like it basically gave the proof for the locus. The reason why that locus gives the arc is because if you look at the diagram, using the exterior angle of a triangle we have \( \arg(z-z_1) = \beta + \arg(z-z_2)\). Using \( \arg z - \arg w = \arg \frac{z}{w}\), this rearranges to give the equation you started off with. The arc gets traced out because you can move \(z\) anywhere along that arc, and the same proof is still valid.

[3.14159265359]

Their diagram looks like it basically gave the proof for the locus. The reason why that locus gives the arc is because if you look at the diagram, using the exterior angle of a triangle we have \( \arg(z-z_1) = \beta + \arg(z-z_2)\). Using \( \arg z - \arg w = \arg \frac{z}{w}\), this rearranges to give the equation you started off with. The arc gets traced out because you can move \(z\) anywhere along that arc, and the same proof is still valid.

I don't understand what you mean
why is the "circle" midway of the quadrant and not touching the x axis like the normal ones like this (but ofc the angle wouldn't be 90, it would be beta)?

[RuiAce]

I don't understand what you mean
why is the "circle" midway of the quadrant and not touching the x axis like the normal ones like this (but ofc the angle wouldn't be 90, it would be beta)?

(Yep, the angle would be \(\beta\).)

In that particular example, you didn't just take an arbitrary choice of \(z_1\) and \(z_2\). For your original question, \(z_1\) and \(z_2\) could've been anywhere you wanted it to. But for your newer question, you had specifically let \(z_1 = 2\), which represents the point \( (2,0)\) on the Argand plane, and you've also specifically let \(z_2 = -2\), which represents the point \( (-2,0)\) on the Argand plane. Both of these points specifically lie on the \(x\)-axis.

So pretty much, that newer example isn't a "normal" one. It's just a special case.

[3.14159265359]

(Yep, the angle would be \(\beta\).)

In that particular example, you didn't just take an arbitrary choice of \(z_1\) and \(z_2\). For your original question, \(z_1\) and \(z_2\) could've been anywhere you wanted it to. But for your newer question, you had specifically let \(z_1 = 2\), which represents the point \( (2,0)\) on the Argand plane, and you've also specifically let \(z_2 = -2\), which represents the point \( (-2,0)\) on the Argand plane. Both of these points specifically lie on the \(x\)-axis.

So pretty much, that newer example isn't a "normal" one. It's just a special case.

does that mean if I drew the original question like the answer I just sent, I would get it right (but again not 90, beta)?

[UStoleMyBike]

Hey, I'm new here

If

what is


and why?

[RuiAce]

does that mean if I drew the original question like the answer I just sent, I would get it right (but again not 90, beta)?

In theory, I would give you the marks for it.

Although in practice, usually when you aren't told anything about what \(z_1\) and \(z_2\) are, I like to keep my answer 'as arbitrary as possible' in a sense. Placing \(z_1\) and \(z_2\) in places that almost look like you're just chucking them at random places in thin air feels less 'restricted' in a sense, as opposed to doing what that other one did. Copying what they did can unintentionally give the impression that \(z_1\) and \(z_2\) have to lie on the \(x\)-axis no matter what, when really they don't. So whilst it's technically not incorrect to do it that way, it probably isn't the best habit to go towards.

Especially since, potentially in the exam they could give you something like \( \arg \left( \frac{z-3+4i}{z+2-i} \right) = \frac\pi3\) instead.