4U Maths Question Thread

[RuiAce]

Hey guys,
how do I recognise horizontal or vertical tangents at a stat point or an asymptote, thanks

Not sure what you mean by "tangents" at asymptotes.

[envisagator]

Hey, I'm having trouble recognising how to draw a guide graph or how it helps when drawing a graph. This is a question that I didnt quite understand: By investigating the behaviour of the curve y^2=x^2(x-1) at the neighbourhood of x=1 and fr large values of x sketch the curve. And is it possible to tackle this question without calculus??

[RuiAce]

Hey, I'm having trouble recognising how to draw a guide graph or how it helps when drawing a graph. This is a question that I didnt quite understand: By investigating the behaviour of the curve y^2=x^2(x-1) at the neighbourhood of x=1 and fr large values of x sketch the curve. And is it possible to tackle this question without calculus??

This should be unsurprising. What we are essentially sketching is \( y = \pm x \sqrt{x-1} \). For \(y = x\sqrt{x-1}\), \(y\to 1^+\). But for \(y = -x\sqrt{x-1}\), \(y\to 1^-\).


Intuitive explanation: It's very important to keep focusing on the point \( (1,0) \). We know that \(x \geq 1\). Suppose that as \(x \to 1^+\), \( \frac{dy}{dx}\) does not explode.

Edit: Earlier explanation wasn't sufficient. The most important criteria is that this graph needs to remain smooth. If we assume that \( \frac{dy}{dx}\) is finite, i.e. the curve is either increasing, decreasing or stationary, ultimately as we approach \(x\to 1^+\):
- The upper half of the graph, i.e. \( y = x\sqrt{x+1} \), approaches it at an angle or flat
- The lower half of the graph, i.e. \(y = -x\sqrt{x+1}\), does the same.

Either of these would lead to a corner or a cusp. These are sometimes feasible options, but they aren't here. This occurs more or less due to the fact we never introduce surds in our equation (until we manually take square roots), and all relevant functions \(y^2\), \(x^2\) and \(x+1\) are always differentiable everywhere along their respective domains: all real \(y\) and \(x \geq 1\).

So since we can't have a cusp or corner, everything above breaks, so our original assumption was wrong. \( \frac{dy}{dx} \) has to be infinite for the graph to smoothly bounce back off. (In fact, this is similar to just proving that \( \frac{dx}{dy} = 0\), but of course we wanted to avoid calculus here.)

Upon reflection, the intuitive method is a lot more catastrophic (at the 4U level) in contrast to just implicitly differentiating \(y^2 = x^3-x^2\).

[mxrylyn]

I am not sure how to integrate (x +2) / (x^2 -1)

And
1/ (3x^2 + 6x + 10)

Can I get the integral of 1/3x  and times 8t by the integral of 1/x and the integeral of 1/6x+10?

[RuiAce]

I am not sure how to integrate (x +2) / (x^2 -1)

And
1/ (3x^2 + 6x + 10)

Can I get the integral of 1/3x  and times 8t by the integral of 1/x and the integeral of 1/6x+10?

You need to complete the square for the second one: \( \int \frac{1}{3x^2+8x+10}dx = \int \frac{1}{3(x+1)^2+7}dx \)
\[ \int \frac{x+2}{x^2-1} dx = \int \frac{x}{x^2-1}dx + \int \frac{2}{x^2-1}dx\\ \text{The first integral goes to }\frac12 \ln |x^2-1|\\ \text{The second requires partial fractions.} \]
I don't understand what that third one is supposed to be.

[mxrylyn]

You need to complete the square for the second one: \( \int \frac{1}{3x^2+8x+10}dx = \int \frac{1}{3(x+1)^2+7}dx \)
\[ \int \frac{x+2}{x^2-1} dx = \int \frac{x}{x^2-1}dx + \int \frac{2}{x^2-1}dx\\ \text{The first integral goes to }\frac12 \ln |x^2-1|\\ \text{The second requires partial fractions.} \]
I don't understand what that third one is supposed to be.

Thank you!
Yes I messed up with both explaining and typing the 3rd one

[mxrylyn]

What method do I use to integrate

 1/(x(3-x))

(1-x^2)/(1+x^2)

And (sec^2x)/(9 - tan^2 x)

[RuiAce]

What method do I use to integrate

 1/(x(3-x))

(1-x^2)/(1+x^2)

And (sec^2x)/(9 - tan^2 x)

1. That is screaming partial fractions right there.
3. Looking at it again, after subbing \(u = \tan x\) you need partial fractions for this one as well.
\begin{align*}\int \frac{1-x^2}{1+x^2}\,dx &= -\int \frac{x^2-1}{x^2+1}\,dx\\ &= -\int \left( 1 - \frac{2}{x^2+1} \right)\,dx\\ &= -x +2\tan^{-1}x +C\end{align*}

[mxrylyn]

I am having trouble showing that 1/9-10sin^2 = sec^2 X / 9-tan^2 X

[RuiAce]

I am having trouble showing that 1/9-10sin^2 = sec^2 X / 9-tan^2 X

Hint: After dividing top and bottom by \(\cos^2 x\), you need to use a Pythagorean identity involving \(\sec^2 x\) on the bottom

[Caleb Campion]

Hi there, I was wondering if there's a chance my version has been misprinted or all of them are like that, but both my HSC Mathematics Extension 1 and 2 TOPIC TESTS books from ATARNotes have significant mistakes in them. The issue is, that there are so many of them. Many of the answers answer a variant of the actual question, with different values etc. Some of them, like Q3 in the Volumes part of the MX2 tests book, have printed the question differently to the answer, such as the question deals with a function with (e^3 . x) whereas the answer answers a question that uses (e^(3x)).

There are several spelling errors in my other atarnotes books also.

I love these books, and the topic tests are helpful, but with this many mistakes, its embarrasing to my self and my teacher (who bought them), and stops me from reccommending these products.

[RuiAce]

Hi there, I was wondering if there's a chance my version has been misprinted or all of them are like that, but both my HSC Mathematics Extension 1 and 2 TOPIC TESTS books from ATARNotes have significant mistakes in them. The issue is, that there are so many of them. Many of the answers answer a variant of the actual question, with different values etc. Some of them, like Q3 in the Volumes part of the MX2 tests book, have printed the question differently to the answer, such as the question deals with a function with (e^3 . x) whereas the answer answers a question that uses (e^(3x)).

There are several spelling errors in my other atarnotes books also.

I love these books, and the topic tests are helpful, but with this many mistakes, its embarrasing to my self and my teacher (who bought them), and stops me from reccommending these products.

Can you please compile the list of mistakes you have found and send them through to my inbox?

Whilst I can't comment on the MX2 TT's since I wasn't directly involved with them, a lot of things can easily get missed in either publication. Mistakes are things we all want to avoid, yet they still creep up without us noticing and it bugs all of us. I do not wish to disclose any information regarding my role in the MX1 TT's through online means - if any serious matters have come up I prefer to address them in person at the upcoming lecture series.

(Alternatively, can you just show me them at any one of the lectures?)


Edit: This list is not exhaustive (and is a bit lost in the forum as well), but here are some known typos in the MX2 publication.

[clovvy]

Having trouble with this integral

[RuiAce]

\begin{align*}\int_3^8 \frac{x}{x+1-\sqrt{x+1}}dx &= \int_2^3 \frac{u^2-1}{u^2 - u} 2u\,du\\ &= \int_2^3 \frac{2u(u-1)(u+1)}{u(u-1)}\,du\\ &= \int_2^3 (2u+2)\,du\\ &= [u^2+2u]_2^3\\ &= 7\end{align*}

[clovvy]

\begin{align*}\int_3^8 \frac{x}{x+1-\sqrt{x+1}}dx &= \int_2^3 \frac{u^2-1}{u^2 - u} 2u\,du\\ &= \int_2^3 \frac{2u(u-1)(u+1)}{u(u-1)}\,du\\ &= \int_2^3 (2u+2)\,du\\ &= [u^2+2u]_2^3\\ &= 7\end{align*}

Just implicit? God why did I forget implicit simply because this is integration haha.. thanks a lot tho