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STEP 2016 Solutions

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Reply 240
Original post by riquix
Are we allowed to post Step III solutions yet?


Yes

Edit: no
(edited 7 years ago)
Original post by riquix
Are we allowed to post Step III solutions yet?


I believe you have to wait 24 hours after the start time of the exam, i.e. you have to wait until 2pm
Reply 242
Original post by 1 8 13 20 42
I believe you have to wait 24 hours after the start time of the exam, i.e. you have to wait until 2pm


Ah right
Grade boundary predictions?
Original post by Shrek1234
Grade boundary predictions?


No discussion until 2pm (BST) :smile:

Posted from TSR Mobile
(edited 7 years ago)
2pm **** sake
saving this space for question 7

don't steel it :<
so grade boundary predictions?
Reply 248
STEP III: Question 4

Sum to N



Sum to infinity



Hyperbolic sum

(edited 7 years ago)
Original post by Zacken
STEP III: Question 4

Sum to N



Sum to infinity



Hyperbolic sum



Yes so, I did the right thing for last part but didnt simplify and ofcourse i used x=e^2y. Lose like 5?


Posted from TSR Mobile
I'll post a solution to 8 if you'd like?
Original post by student0042
I'll post a solution to 8 if you'd like?


Please do
Q2 is my *****. Had a beauty to that, it was eazy. Need to go pick up my little brother first.


Posted from TSR Mobile
EDIT:
someone else do q7, Latex ****ing trolled me hard
(edited 7 years ago)
Reply 254
Q12 is mine guys :wink:

Posted from TSR Mobile
Anyone got a copy of the paper yet?
Reply 256
Original post by Mathemagicien
2 late


😥😥 I cri
Original post by Shrek1234
Please do


Alright Q8. I don't have access to a paper, so I may have remembered the question incorrectly, but I got the same answers as yesterday, so...
STEP III 2016 Q8 1.jpg
bagsy q1

In=1(x2+2ax+b)ndx[br]Usingthesubstitution: I_n = \int^{\infty}_{-\infty}\frac{1}{(x^2 + 2ax + b)^n} dx[br] Using the substitution:

x+a=ba2tanu x + a = \sqrt{b-a^2} \tan u

dxdu=ba2sec2u \Rightarrow \frac{dx}{du} = \sqrt{b-a^2} \sec ^2 u

In=1((x+a)2+ba2)ndx \Rightarrow I_n = \int^{\infty}_{-\infty}\frac{1}{((x+a)^2 + b - a^2)^n} dx

In=π/2π/2ba2sec2u((ba2)tan2u+(ba2))ndu \Rightarrow I_n = \int^{\pi/2}_{-\pi/2}\frac{\sqrt{b-a^2}\sec ^2 u}{((b-a^2)\tan ^2 u + (b-a^2))^n } du

tan2u+1=sec2u \tan ^2 u + 1 = \sec ^2 u

In=π/2π/2ba2sec2u((ba2)sec2u)ndu I_n = \int^{\pi/2}_{-\pi/2}\frac{\sqrt{b-a^2}\sec ^2 u}{((b-a^2)\sec ^2 u)^n } du

In=π/2π/21(ba2)n12sec2n2udu I_n = \int^{\pi/2}_{-\pi/2}\frac{1}{(b-a^2)^{n-\frac{1}{2}} \sec ^{2n-2} u } du

i)Find I1 i) Find\ I_1

I1=π/2π/21(ba2)12du I_1 = \int^{\pi/2}_{-\pi/2}\frac{1}{(b-a^2)^{\frac{1}{2}}} du

I1=[x(ba2)12]π/2π/2 I_1 = \left[ \frac{x}{(b-a^2)^\frac{1}{2}} \right]_{-\pi/2}^{\pi/2}

I1=πba2 as required I_1 = \frac{\pi}{\sqrt{b-a^2}}\ as\ required

apparently my latex is ugly
(edited 7 years ago)
Original post by Mathemagicien
...


Q 12 was beautiful, probably the easiest STEP III stats question ever.

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