# BoS Maths Trials 2017 (1 Viewer)

#### janzrn

##### New Member
aye lmao apparently people think

$\bg_white (1-\sqrt{h})^2 = 1-h$

rip there goes ur marks

also

a lot of people forgot that √k = 1-√h

their area function was (1-√h)² instead of (1-√h)⁴

there goes another mark

bless u janzen choi for being the one person who got my spiky volume correct <3

another kid almost got the answer correct but they forgot to double the result bc they were only finding the upper half volume...

rip in pieces

also a lot of u stoners thought the curved edges were like a square minus a circle????

can u even graphs plz
<3 back

#### Trebla

Updated the solutions to cover typos that were drawn to my attention.

Also, a particular point of feedback on one of the questions I marked. Far too many students actually did not attempt Q15b)iii) which should have been a free mark, especially if you're familiar with these types of questions. A surprising number of students actually wrote the limit was 1 despite the sum showing that it is clearly larger than 1.

On the flip side it was good to see a handful of students manage to get out Q15b)i) using a direct geometric series approach (which is often quite obscure to recognise) instead of the typical reduction formula approach which also would have worked.

Another point to make is Q11d), some were using inequalities with the complex numbers themselves rather than the modulus (which is a real number). You cannot create inequalities between complex numbers! You can only create inequalities describing something like their modulus or argument (which are real values).

Also, Q11a) gg so many people got owned by that integral...

Last edited:

#### seanieg89

##### Well-Known Member
I think the Q16h solution is still not quite correct.

Whilst it is true that

x-k\alpha =< x for each x, we cannot simply bound the product of factors of this form by the product of the x's, as some of these terms will in general be negative.

Another way of seeing this problem is that if we take absolute values and k\alpha happens to be close to n, then we see that in fact the early terms in this product are the largest, (but the inequality is still true because the later terms in the product are small).

##### -insert title here-
Also, Q11a) gg so many people got owned by that integral...
Oh come on, it wasn't THAT hard....

#### aycaramba

##### Member
Well, good luck for tomorrow everybody - let's smash it

##### Member
Well, good luck for tomorrow everybody - let's smash it
Yes, good luck indeed. I hope it's easy enough that we don't need it tho

#### Carrotsticks

##### Retired
I think the Q16h solution is still not quite correct.

Whilst it is true that

x-k\alpha =< x for each x, we cannot simply bound the product of factors of this form by the product of the x's, as some of these terms will in general be negative.

Another way of seeing this problem is that if we take absolute values and k\alpha happens to be close to n, then we see that in fact the early terms in this product are the largest, (but the inequality is still true because the later terms in the product are small).
Hey man, thanks for pointing this out. I've attached a more detailed solution. I got a bit lazy towards the end and also wanted to keep it within 2 pages, but I think this should do the trick. I was supposed to write the Q16 solutions on Sat night but I̶ ̶w̶e̶n̶t̶ ̶t̶o̶ ̶E̶l̶ ̶J̶a̶n̶n̶a̶h̶ ̶a̶n̶d̶ ̶c̶h̶i̶l̶l̶e̶d̶ ̶w̶i̶t̶h̶ ̶m̶a̶t̶e̶s̶ ̶t̶i̶l̶l̶ ̶1̶a̶m̶ ̶i̶n̶s̶t̶e̶a̶d̶ I didn't have time to do them, so this one's on me.

View attachment Correction.pdf

##### Member
Hey Guys,

Just wondering around what time you think the 3U solutions will be released?

Boom

#### Attachments

• 194.1 KB Views: 247
• 280.7 KB Views: 196

cheers