Another extra TMUA paper 1, from Zetta

[@absorbentcolt2966]

Just want to say thank you for your TMUA help videos. You really helped me get the mark I need. Thanks a lot

[@njb2032-r9w]

what did you get?

[@absorbentcolt2966]

@ 5.9 nothing crazy but over 5 so that’s all that matters

[@FakemanNotme]

Here is a version of the paper with the errors fixed: drive.google.com/file/d/1F5-vF-TGJhzFh7fx2KontIyFmUdWAE0C/view?usp=sharing

[@darkzonegaming9169]

8:15, for Q8. Let S = sqrt(100001)- sqrt(100000). notice how difference of 2 squares can help here. so S *(sqrt(100001)+sqrt(100000)) = 100001-100000=1. so S = 1/[sqrt(100001) + sqrt(100000)]. Lastly note that sqrt(100001) is approximately equal to sqrt(100000) and you get the ans

[@olahashemi8457]

What does S* mean?

[@darkzonegaming9169]

I meant multiplication

[@timilehinadenekan678]

@@darkzonegaming9169 is S REPRESSENTING BOTH SQRT(100001) AND SQRT(100000) OR JUST 1 OF THEM

[@johnwick54143]

i did the oct sitting and got a q very similar to q9

[@malignusvonbottershnike563]

If you were in a rush, I think Q13 can be done slightly quicker, provided you happen to know that the area of a rhombus is just half the product of the lengths of its diagonals (consider the rhombus lying inside a rectangle, with its vertices at the midpoints of the sides of that rectangle, and it just turns into summing the area of some triangles). I suppose you also have to recognise that it's a rhombus, which does require spotting the symmetry, and a nice diagram definitely helps with that.

[@rtwodrew2]

Nice, I had spotted that it was a rhombus, but that formula didn't cross my mind unfortunately

[@olahashemi8457]

Q12. Surely because it says given at least 2 die show a 3, doesn’t that mean you take into account whet A and C show 3’s and not B?

[@ywhatm8081]

For 11 was it just me or downloading the paper changed the -2ad to +2ad so it couldn’t be solved

[@haris7068]

for 15, u gotta use double angle formulas, which i dont think is in the spec. but then y = 2sin(arctan(x))cos(arctan(x)). let v = arctan(x), then u can have a right angle triangle with angle v, side x, 1 and sqrt(x^2+1). then we can turn v into arcsin and arccos which will lead to the answer D

[@Sh3ikhy]

Do you have a link for all papers from Zetta (like a google drive) instead of just one?

[@RajSandhu-gm8iz]

Hi, welcome back, nice to see some videos again. In Q8 you could let x=100000, then you have (√(x+1) - √(x)) /1, multiply top and bottom by the congugate √(x+1) + √(x), leads to 1/√(x+1) +√(x), for large vaues of x that is approximately 1/2√(x), hence B. Maybe that was intended solution??

[@nandoboiii5215]

Do you have any extra papers for the ESAT maths 1, maths 2 and physics?

[@timilehinadenekan678]

for question 4 how do you know the first term of the GP is x and not --6

[@sargeras410]

is this paper actually brutal or is it just my issue 🤐

[@michaelswahla4927]

Yeah I'm with you on this one I just hope i get above 10 marks on each paper.

[@LeBronJames-mt3fd]

@@michaelswahla4927 how come you do two papers im only doing tmua paper 2

[@metanoia-dd5oi]

What would 13/20 be on a paper like this? roughly

[@thomasamour3147]

Where are you getting these papers from? Thanks. Great channel

[@LeBronJames-mt3fd]

surely question 12 is a badly worded question, it is ambiguous

[@LeBronJames-mt3fd]

it should just be what is a probability of rolling a dice three times landing "three" given that it rolls a "three" twice