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

#### Nacseo

##### Member
Yes but really difficult to spot is to use the 'reverse quotient rule,' whenever we are integrating a rational function it;s always a good idea to keep this trick in mind.

First note that $\bg_white 7x^6(1+x)^7-7x^7(1+x)^6=7x^6(1+x)^6(1+x-x)=7x^6(1+x)^6$
So rewriting the integral:

$\bg_white I=\frac{1}{7}\int{\frac{7x^6(1+x)^7-7x^7(1+x)^6}{(1+x)^{14}}}\ dx$

This is in the form $\bg_white \int{\frac{u'v-uv'}{v^2} dx$ where $\bg_white u=x^7 \quad \text{and} \quad v=(1+x)^7$

So $\bg_white I=\frac{x^7}{7(1+x)^7}+C$
Jeez my head hurt just by looking at these equations, I'm also gonna apply for Ext 1 Maths, but now when I see this I'm like, sh*t, SH*T.

#### Qeru

##### Well-Known Member
Yep I wanted to showcase the 'reverse product rule,'

$\bg_white \int\sin^{99}(x)\sin(101x)dx$

$\bg_white =\int\sin^{99}(x)\sin(100x+x)dx$

$\bg_white \int\sin^{99}(x) \sin(100x) \cos{x}+\sin^{100}(x)\cos(100x) dx$

$\bg_white \text{Now notice how:}\quad\frac{d}{dx}\sin^{100}(x)=100\sin^{99}(x)\cos{x} \quad\text{and}\quad\frac{d}{dx}\sin(100x)=100\cos(100x)$

SO our integral is in the form: $\bg_white \int uv'+vu' dx$ but off by a constant of $\bg_white \frac{1}{100}$

$\bg_white \therefore \int\sin^{99}(x)\sin(101x)dx=\frac{1}{100}\sin^{100}(x)\sin(100x)+C$

#### Qeru

##### Well-Known Member
Jeez my head hurt just by looking at these equations, I'm also gonna apply for Ext 1 Maths, but now when I see this I'm like, sh*t, SH*T.
This is extension 2 so don't worry haha.

#### Nacseo

##### Member
This is extension 2 so don't worry haha.
Oh thank god, I was abouta jump off a microwave for a sec

#### tito981

##### Active Member
Yep I wanted to showcase the 'reverse product rule,'

$\bg_white \int\sin^{99}(x)\sin(101x)dx$

$\bg_white =\int\sin^{99}(x)\sin(100x+x)dx$

$\bg_white \int\sin^{99}(x) \sin(100x) \cos{x}+\sin^{100}(x)\cos(100x) dx$

$\bg_white \text{Now notice how:}\quad\frac{d}{dx}\sin^{100}(x)=100\sin^{99}(x)\cos{x} \quad\text{and}\quad\frac{d}{dx}\sin(100x)=100\cos(100x)$

SO our integral is in the form: $\bg_white \int uv'+vu' dx$ but off by a constant of $\bg_white \frac{1}{100}$

$\bg_white \therefore \int\sin^{99}(x)\sin(101x)dx=\frac{1}{100}\sin^{100}(x)\sin(100x)+C$
ye when i watched the mit integration bee i felt it was an easy expansion + reverse pro rule, iirc neither of the contestants got it right.

#### Trebla

Reverse product/quotient rule is literally equivalent to integration by parts in the same way that reverse chain rule is equivalent to substitution.

#### Qeru

##### Well-Known Member
Yes you can also do IBP on the first integral from the third line: $\bg_white I=\int\sin^{99}(x) \sin(100x) \cos{x}dx+\int\sin^{100}(x)\cos(100x) dx$ to get

$\bg_white I=\frac{1}{100}\sin^{100}(x)\sin(100x)-\int\sin^{100}(x)\cos(100x) dx+\int\sin^{100}(x)\cos(100x) dx$
Which is the same thing.

#### Jojofelyx

##### Member
What do you guys reckon the hardest papers are from the last 10 years in order (for ext 2 and 1)?

#### idkkdi

##### Well-Known Member
What do you guys reckon the hardest papers are from the last 10 years in order (for ext 2 and 1)?
anything more than 10 years ago.

#### Drdusk

##### π
Moderator
What do you guys reckon the hardest papers are from the last 10 years in order (for ext 2 and 1)?
Last 10 years?? Maybe 2010? 2010 had a pretty nice final question.

#### vernburn

##### Active Member
Not really related but I remember reading that 1989 (especially the last question) is considered the hardest 4U paper ever written. As for the last 10 years I’d say 2010.

#### black.mamba

##### Active Member
What do you guys reckon the hardest papers are from the last 10 years in order (for ext 2 and 1)?
easiest ext 2 2020 lol

#### vernburn

##### Active Member
they really shouldnt have removed harder 3U.
Couldn’t agree more!

#### idkkdi

##### Well-Known Member
Couldn’t agree more!
maybe u would have state ranked with harder ext 1

#### idkkdi

##### Well-Known Member
Not really related but I remember reading that 1989 (especially the last question) is considered the hardest 4U paper ever written. As for the last 10 years I’d say 2010.
i also heard 1989 last q is very hard.

#### CM_Tutor

##### Moderator
Moderator
1989 last question was too hard, which was bad because that doesn't help in creating a ranking order, leaving aside how it makes the candidates feel.

Having removed 'harder 3u', the place with the opportunity for really challenging materials (IMO) is proof. The recent thread on second-order recursion illustrates one way the proof topic can be used to make challenging questions.

#### idkkdi

##### Well-Known Member
1989 last question was too hard, which was bad because that doesn't help in creating a ranking order, leaving aside how it makes the candidates feel.

Having removed 'harder 3u', the place with the opportunity for really challenging materials (IMO) is proof. The recent thread on second-order recursion illustrates one way the proof topic can be used to make challenging questions.
Mech is hard lol can’t do algebra