AQA A Level Maths Paper 1, June/November 2020

[@e_bi0938]

wow! this paper stinks! fingers crossed my exam tomorrow isnt like this one. somehow managed to fit in all the topics i hate! anyway, thank you! i can appreciate that lots of time goes into these videos and they're tremendously helpful.

[@denizyuksel-b6y]

this guy just knows the paper like the back of his hand thanks soooooooooo muc h :)

[@airblade8287]

OMG, YES! I was waiting for this upload. Are you able to do Paper 2 and 3?

[@PaulMackayTuition]

These are now available!

[@geoffphillips8183]

great work, thnx! Extremely helpful!

[@Tofu2003]

THANK YOU SO MUCH!!!!!

[@karotic]

33:25 I’m not with you on why h=1? because you say that it’s derived from the fact that h goes up to 6 in part a but I’d say it actually goes up to 5 seeing as the R started from when x=1 in part a.

[@karotic]

ahh I have just thoight about it. 6-1/5 = 1

[@amritlohia8240]

Your reasoning in 4(b) is incorrect. The cases are 4-(2x-6)>2 *and* x>=3, giving 3

[@minnanasseri9578]

Mark scheme says otherwise

[@amritlohia8240]

@@minnanasseri9578 I'm talking about the actual mathematics, not what any "mark scheme" may say (they can equally be wrong or incomplete)...

[@e_bi0938]

@@amritlohia8240 i dont reckon ill do too well if i start arguing with the markschemes, lol.

[@alexmerz7805]

hi there, these videos are incredibly helpful! at the beginning of question 15 is there an alternative way to get from e^x=6-e^x/2 to ln4? I just would never have thought of your way. When I tried it I started by taking ln of both sides but then got lost!

[@PaulMackayTuition]

I don't think there is a simpler way around here. When you've got equations containing multiple terms with variables in the powers, you just always have to be on the lookout for a quadratic - you're looking for the case where the coefficient of x in one power is twice that of another (here it was 1 and 1/2). If you were solving something like 4^6x + 5(4^3x)+6=0, you'd make a substitution for 4^3x and tackle it in the same sort of way because the coefficients are 6 and 3 - again, doubled.

[@alexmerz7805]

@@PaulMackayTuition so useful, thanks so much. for your problem - if y=4^3x I got (1/4)y^2+5y+6 = 0

[@PaulMackayTuition]

@@alexmerz7805 you wouldn't need the 1/4 there

[@shizukusan8141]

Hi there, question for you: would you mind to tell me how did you het the second value of t from ex 8 b please?!

[@PaulMackayTuition]

In general, for sine, the principal solution of sin x = a is x = sin^-1 a. However, because of symmetry, there's always a secondary solution at (pi - sin^-1 a) too (and you can add multiples of 2pi to each of these to find all the other solutions, but that's not relevant to this question)

[@mrmemphis2055]

Hi sir for question 10 (b)(I) how did you get 91 for number of terms? 90 would make sense but I don’t understand where the + 1 came from

[@PaulMackayTuition]

Just don't forget that you need the value at both the limits. If you just take the difference, then you're finding how many terms there are AFTER the first term, and that first term is being forgotten about! Imagine instead that the sum went from r=2 to 5. That would mean substituting it r=2, r=3, r=4 and r=5, which is 4 terms (not 5-2=3 terms). n is always found from (upper limit - the lower limit + 1). Hope that's clearer now.

[@mrmemphis2055]

@@PaulMackayTuition Ahhh okay that makes perfect sense. Many thanks sir

[@dr.pepper912]

Legend.

[@alamppost9051]

Hi, this was a really helpful video. For 10 bii, why is u40 49b + c? Thank you

[@PaulMackayTuition]

From sigma notation, you can see that the first term corresponds to r=10, and increases from there, so the 2nd term would be when r=11, 3rd when r=12 and so on up to the 40th term where r=49 (r=10 then 39 more).

[@octreos1864]

why do we use radians for Q8 and not degrees?

[@PaulMackayTuition]

Because of the 2pi in the argument of the function, which is a full cycle in radians.

[@octreos1864]

@@PaulMackayTuition ohhhhhh im an idiot for not realising 😂 thank you

[@johnnyseddon4028]

im so confused on 8a, i thought u would put the minimum as t=0 and then hours=11.7 i dont understand why u have put sin() = -1 and then you have got -3.87, makes no sense haha, thanks for the video tho, 2 months till a levels and im doing everything i can do get thru 100s of past papers

[@PaulMackayTuition]

The sine ranges between -1 and 1 so the minimum value is -1 and occurs when the argument of sine (what's inside it) is equal to 3pi/2.