# Maths-Ext 1/2 Marks Lost and Hardest Topics (1 Viewer)

#### SplashJuice

##### Active Member
Yall reckon Cambridge is the GOAT for maths ext 1 n 2? because our school uses Fitzpatrick and eh idk about that tbh
Yes Cambridge is the GOAT.

#### vernburn

##### Active Member
As for less structure, a question that was debated in my 4u class (and which we found two solutions for) on integration:

$\bg_white \int{\frac{x^6}{(1+x)^8}}\ dx$

See if you can find a way to find this indefinite integral. I can make this into a reasonable MX1 question with sufficient structure, but without structure it is definitely a challenge even at MX2 level.
I believe that this involves a very nice (and very sneaky!) trick, namely dividing the numerator and denominator by $\bg_white x^8$. I also found a less elegant method which involves the substitution $\bg_white x=\cos2\theta$.

#### quickoats

##### Well-Known Member
It is certainly true that questions can be made more difficult by giving less structure, but it is also true that structure can be less helpful.

On the latter, there are questions where the structure is meant to support you taking approach A when the approach that some students / people / teachers will be inclined towards instinctively is approach B. There was an MX2 question that I saw last year that I solved parts (a) and (b) and then found a proof / solution to part (d) which I used to get the result in part (c) because I didn't see the connection to go from (b) directly to (c) without establishing (d). It's better to get a solution than not (obviously) so the structure of the question doesn't necessarily have to be followed unless there are words like "hence" in the question... and on this, more able students need to recognise the unstated implication of the phrase "hence or otherwise" as it has two distinctly different potential meanings.

As for less structure, a question that was debated in my 4u class (and which we found two solutions for) on integration:

$\bg_white \int{\frac{x^6}{(1+x)^8}}\ dx$

See if you can find a way to find this indefinite integral. I can make this into a reasonable MX1 question with sufficient structure, but without structure it is definitely a challenge even at MX2 level.
Typical bulldozer approach would be let u = 1+x then do a binomial expansion of (1 - u)^6 on the top. Not the most efficient but gets the job done.

#### CM_Tutor

##### Moderator
Moderator
I believe that this involves a very nice (and very sneaky!) trick, namely dividing the numerator and denominator by $\bg_white x^8$. I also found a less elegant method which involves the substitution $\bg_white x=\cos2\theta$.
Yes, the substitution

$\bg_white u=1+\frac{1}{x}$

will get the result that

$\bg_white \int{\frac{x^6}{(1+x)^8}}\ dx = \frac{x^7}{7(1+x)^7} + C$

after demonstrating that

$\bg_white \frac{x^6}{(1+x)^8} = \frac{1}{x^2\left(1+\frac{1}{x}\right)^8}$

This would be one way to give the question structure:

(a)(i) Show that
$\bg_white \frac{x^6}{(1+x)^8} = \frac{1}{x^2\left(1+\frac{1}{x}\right)^8}$
(ii) Hence, find
$\bg_white \int{\frac{x^6}{(1+x)^8}}\ dx$
by using the substitution $\bg_white u=1+\frac{1}{x}$

I do know of a trig substitution that will work but it is not $\bg_white x=\cos{2\theta}$, though that might work too. It's nice to see another approach.

#### CM_Tutor

##### Moderator
Moderator
Typical bulldozer approach would be let u = 1+x then do a binomial expansion of (1 - u)^6 on the top. Not the most efficient but gets the job done.
Yes, though it is messy. Actually, it could be the basis for a binomial question...

1. Consider the integral
$\bg_white I=\int{\frac{x^6}{(1+x)^8}}\ dx$

(a) Using the substitution $\bg_white x=u-1$, find an expression for $\bg_white I$ as a series of terms in $\bg_white (x+1)$.

(b) Using a trig substitution (which would be given), show (for some constant $\bg_white C$) that

$\bg_white I=\frac{x^7}{7(1+x)^7}+C$

(c) Prove that the results from (a) and (b) are the same, or explain why one of the results is false.

#### vernburn

##### Active Member
I do know of a trig substitution that will work but it is not $\bg_white x=\cos{2\theta}$, though that might work too. It's nice to see another approach.
Here is how $\bg_white x=\cos2\theta$ sub works out:
\bg_white \begin{align*}\int\frac{x^6}{(1+x)^8}dx&=-2\int\frac{\cos^62\theta}{(1+\cos2\theta)^8}\sin2\theta d\theta\\
&=-2\int\frac{\cos^62\theta}{(2\cos^2\theta)^8}2\sin\theta\cos\theta d\theta\\
&=-\frac{1}{64}\int\frac{(2\cos^2\theta-1)^6}{\cos^{15}\theta}\sin\theta d\theta\\
&=-\frac{1}{64}\int\frac{64\cos^{12}\theta-192\cos^{10}\theta+240\cos^8\theta-160\cos^6\theta+60\cos^4\theta-12\cos^2\theta+1}{\cos^{15}\theta}\sin\theta d\theta\\
&=-\frac{1}{64}\int\left(\frac{64}{\cos^3\theta}-\frac{192}{\cos^5\theta}+\frac{240}{\cos^7\theta}-\frac{160}{\cos^9\theta}+\frac{60}{\cos^{11}\theta}-\frac{12}{\cos^{13}\theta}+\frac{1}{\cos^{15}\theta}\right)\sin\theta d\theta\end{align*}

Now I'm sure you can see how this works out from here - far from elegant but it still works!

As for another trig sub, maybe $\bg_white x=\tan^2\theta$:
\bg_white \begin{align*}\int\frac{x^6}{(1+x)^8}dx&=\int\frac{\tan^{12}\theta}{(1+\tan^2\theta)^8}2\tan\theta\sec^2\theta d\theta\\
&=2\int\frac{\tan^{13}\theta}{\sec^{14}\theta}d\theta\\
&=2\int\sin^{13}\theta\cos\theta d\theta\\
&=\frac17\sin^{14}\theta+C\\
&=\frac{x^7}{7(1+x)^7}+C\end{align*}

#### idkkdi

##### Well-Known Member
Here is how $\bg_white x=\cos2\theta$ sub works out:
\bg_white \begin{align*}\int\frac{x^6}{(1+x)^8}dx&=-2\int\frac{\cos^62\theta}{(1+\cos2\theta)^8}\sin2\theta d\theta\\
&=-2\int\frac{\cos^62\theta}{(2\cos^2\theta)^8}2\sin\theta\cos\theta d\theta\\
&=-\frac{1}{64}\int\frac{(2\cos^2\theta-1)^6}{\cos^{15}\theta}\sin\theta d\theta\\
&=-\frac{1}{64}\int\frac{64\cos^{12}\theta-192\cos^{10}\theta+240\cos^8\theta-160\cos^6\theta+60\cos^4\theta-12\cos^2\theta+1}{\cos^{15}\theta}\sin\theta d\theta\\
&=-\frac{1}{64}\int\left(\frac{64}{\cos^3\theta}-\frac{192}{\cos^5\theta}+\frac{240}{\cos^7\theta}-\frac{160}{\cos^9\theta}+\frac{60}{\cos^{11}\theta}-\frac{12}{\cos^{13}\theta}+\frac{1}{\cos^{15}\theta}\right)\sin\theta d\theta\end{align*}

Now I'm sure you can see how this works out from here - far from elegant but it still works!

As for another trig sub, maybe $\bg_white x=\tan^2\theta$:
\bg_white \begin{align*}\int\frac{x^6}{(1+x)^8}dx&=\int\frac{\tan^{12}\theta}{(1+\tan^2\theta)^8}2\tan\theta\sec^2\theta d\theta\\
&=2\int\frac{\tan^{13}\theta}{\sec^{14}\theta}d\theta\\
&=2\int\sin^{13}\theta\cos\theta d\theta\\
&=\frac17\sin^{14}\theta+C\\
&=\frac{x^7}{7(1+x)^7}+C\end{align*}
anyone bothered to finish off the sectan integrating in the cos2theta substitution lol.? btw, u would factor out sec tan , and yeet that out with a sec theta = u substitution.

any smarter way than doing that algebra in cos2theta solution?

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#### Jojofelyx

##### Active Member
lmao it looks like yall finished the course already damn. leads me to a question, do most band 6ers in ext 1 and 2 have the course finished by term 2 of year 12 or wot?

#### quickoats

##### Well-Known Member
lmao it looks like yall finished the course already damn. leads me to a question, do most band 6ers in ext 1 and 2 have the course finished by term 2 of year 12 or wot?
nope lol

#### quickoats

##### Well-Known Member
no gods in this thread pls
Seriously pace yourself and try to understand the concepts fully. This is a much better approach than rushing through content.

#### Jojofelyx

##### Active Member
Seriously pace yourself and try to understand the concepts fully. This is a much better approach than rushing through content.
idkkid seems to have it down pat tho and it seems to be term 2 of his yr 12 journey?

#### shashysha

##### Well-Known Member
lmao it looks like yall finished the course already damn. leads me to a question, do most band 6ers in ext 1 and 2 have the course finished by term 2 of year 12 or wot?
I finished the whole course like middle-end of the last term (few weeks after trials) so pretty late actually. School just not did cover mechanics until after trials and same for 3U projectile. Although they did throw some 2U questions in our 3U exam and the first CDF question I did ever was in that trial lol

Moderator

#### Drdusk

##### π
Moderator
Is the 100 atar orange too annoying for you?
I never said it was me....

#### idkkdi

##### Well-Known Member
idkkid seems to have it down pat tho and it seems to be term 2 of his yr 12 journey?
bruh it's a 300 page book. you could probably finish the entire book in 2 weeks if you're good at maths and skip over the trivial questions lol.

I have a feeling that @Qeru was done with 4u before 4u even started lmao

#### idkkdi

##### Well-Known Member
Seriously pace yourself and try to understand the concepts fully. This is a much better approach than rushing through content.
not much to understand tbh. cambridge book is crisp. also new syllabus has quite a bit less content than old it seems.

After that it's a bloody long ass algebra grind lol.

#### idkkdi

##### Well-Known Member
lmao it looks like yall finished the course already damn. leads me to a question, do most band 6ers in ext 1 and 2 have the course finished by term 2 of year 12 or wot?
no way. band 6 in maths ext 1 and ext 2 doesn't mean anything, it's way too common. like half the students in ext 2.
As for super early finisher like @Qeru, we're looking at maybe 1 in 30 odds.

#### Jojofelyx

##### Active Member
I finished the whole course like middle-end of the last term (few weeks after trials) so pretty late actually. School just not did cover mechanics until after trials and same for 3U projectile. Although they did throw some 2U questions in our 3U exam and the first CDF question I did ever was in that trial lol
Dont wanna seem intrusive, but did you manage a mid band 6?

#### Jojofelyx

##### Active Member
bruh it's a 300 page book. you could probably finish the entire book in 2 weeks if you're good at maths and skip over the trivial questions lol.

I have a feeling that @Qeru was done with 4u before 4u even started lmao
my Cambridge ext 1 is like 750 lmao :\\

no way. band 6 in maths ext 1 and ext 2 doesn't mean anything, it's way too common. like half the students in ext 2.
As for super early finisher like @Qeru, we're looking at maybe 1 in 30 odds.
But isn't getting a band 6 based of the difficulty, I read something on SMH that was about extension subjects and having higher % of kids who get band 6, mainly cuz its something they are good at and willing to invest time into it, that's why music extension got like 50% or something band 6s?idk could be wrong, feel free to correct me tho.

not much to understand tbh. cambridge book is crisp. also new syllabus has quite a bit less content than old it seems.

After that it's a bloody long ass algebra grind lol.
and wdym algebra grind? copious amounts of working out or wot?