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Original post by ren.kunche

2-sqrt(2)

2-sqrt(2)

Was that D?

Original post by AryanRG

I got D for the radii q idk what exact number but probs got it wrong I couldn’t rationalise the denomantor correctly. I got the square ending in 225 too. And for integration I got 4 which is also a guess

For the square number one was that a guess? I got that too

Original post by Anonymous

For the square number one was that a guess? I got that too

Was this the one with the number starting with 64? Kinda discounted the other ones based on their first, second, and last digits.

Also, for the floor function question, was anyone getting like 11/4 or some **** instead of a whole number? I remember the sum being 3/4 + (1/2)(4/3) + (1/4)(16/9) + (1/8)(64/27) but when i worked out the infinite series it kept on giving me 3/4 + 2 for some reason lmao

Original post by _tjola7

How did everyone find the MAT paper?

Icl I didn’t want to go to Oxford anyway 😭

Anyone remember their answers for the multiple choice?

yeah it was the one starting w 64. I got that too but just a guess. what was your final answer to that integration one?

was the integral one: (3/4)^log base 2 x..... between 0 and 1

Original post by Anonymous

was the integral one: (3/4)^log base 2 x..... between 0 and 1

I got 5 for that area 4 for 0 and 1 and 1 for 1 to 2 so total 5 idk

Original post by Anonymous

was the integral one: (3/4)^log base 2 x..... between 0 and 1

between 0 and 2

did you do q4? that **** was beautiful when it magically simplified at the end

Original post by Anonymous

between 0 and 2

did you do q4? that **** was beautiful when it magically simplified at the end

did you do q4? that **** was beautiful when it magically simplified at the end

Didn’t got time to finish the writing work but trig one is so good like the identity is so nice

Original post by Anonymous

between 0 and 2

did you do q4? that **** was beautiful when it magically simplified at the end

did you do q4? that **** was beautiful when it magically simplified at the end

yeah i think i got 12-15/15 in q4. i got d in terms of a as they wanted and got the minimum of f(t) as 3 i think. and i think a was root 2, and d is bigger or equal to root 27.

Original post by Theacty119

I got 5 for that area 4 for 0 and 1 and 1 for 1 to 2 so total 5 idk

when i put into wolfram alpha it doesn't give whole number so maybe i input it wrong? i adjusted the limits between 0 and 2 now asw.

Original post by Anonymous

yeah i think i got 12-15/15 in q4. i got d in terms of a as they wanted and got the minimum of f(t) as 3 i think. and i think a was root 2, and d is bigger or equal to root 27.

yeah f(t) was a minimum when t = 2 which was 3. ngl though i think i got d bigger or equal to sqrt8? dunno i think i just substituted the minimum f(t) into the d^2/3 equation. maybe we got different f(t)

I guessed the integral was 2 last second, which is probably wrong but it is what it is

nvm im dumb as **** lmao its defo sqrt27 i'm gonna pray i put that down

Original post by Anonymous

yeah f(t) was a minimum when t = 2 which was 3. ngl though i think i got d bigger or equal to sqrt8? dunno i think i just substituted the minimum f(t) into the d^2/3 equation. maybe we got different f(t)

I guessed the integral was 2 last second, which is probably wrong but it is what it is

I guessed the integral was 2 last second, which is probably wrong but it is what it is

i cant remember my f(t) but we shouldve got same if our t and minimum were same. mine had something to the power of a 1/4 in it I think. i think I subbed in a=root 2 into my f(t) equation (after converting the t back to a squared), and got d = 3 to the power of 3/2. i think one of us probably just made a silly mistake.

for the integral one I completely guessed - I made a triangle that was in the shape ish of the graph and estimated it to be around 4.25 but just put 4 but almost deffo got that one wrong.

Original post by Anonymous

nvm im dumb as **** lmao its defo sqrt27 i'm gonna pray i put that down

oh just saw this lmao

Original post by Anonymous

nvm im dumb as **** lmao its defo sqrt27 i'm gonna pray i put that down

what did you get for q3? i got the 9 marks for sure I think till the final cos function, and for that I just guessed x to be -10 degrees and got the whole function equalling 4 to the power of 8 which seems like the highest itt could be . really bad logic though and even if its right probs wont get working marks.

Anyone know how the Multiple choice for that how many number of root for cx^2 +ax+b works I got 2 real root cause it’s just guess work lmao

Original post by Theacty119

Anyone know how the Multiple choice for that how many number of root for cx^2 +ax+b works I got 2 real root cause it’s just guess work lmao

i got no real roots. i did the algebra and eventually got the discriminant equalling -63asquared or something which is below 0 since a isn't 0, so no real roots. what did you get M to equal in the mc question?

Original post by Anonymous

what did you get for q3? i got the 9 marks for sure I think till the final cos function, and for that I just guessed x to be -10 degrees and got the whole function equalling 4 to the power of 8 which seems like the highest itt could be . really bad logic though and even if its right probs wont get working marks.

pretty sure the key idea was just to utilise the fact that cos(90-x) = sin(x).

i dont remember the exact expression, but there were two brackets multiplied iirc, one was (1 - cos^2(2x-60)) and the other (3-cos(150-2x))

by applying the identity to the left bracket, you get (1-sin^2(150-2x)) = cos^2(150-2x) which leaves the expression in a form previous to the prior part, but with something to the power of 4 instead?

anyways i got an answer of 16^4 = 4^8 = 2^16. If you know the expression, I could explain it probably, but i'm pretty confident it was maximised when cos(150-2x) = -1.

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