MAT Prep Thread - 2nd November 2016

Prove that x+Square root(1-x) <= 5/4

(edited 7 years ago)

Reply 581

Zacken

22

Original post by Mystery.

yes that's the type I meant. Like this question :https://www.imperial.ac.uk/media/imperial-college/faculty-of-natural-sciences/department-of-mathematics/public/study/admissions/ug/mat-past-papers/Past-Test-and-Solutions-2010.pdf
Q3 I

No, that's really not what I'm talking about, this is what algebraic geometry is: https://www.dpmms.cam.ac.uk/study/II/AlgebraicGeometry/2015-2016/HW1.pdf

Original post by RuairiMorrissey

I don't fully understand how you got that last term?

Do you mean how I got

u\int_{-1}^1 3x^2 \, \mathrm{d}x = 2u

? If so: what's the value of

\int_{-1}^1 3x^2 \, \mathrm{d}x

.

Reply 582

anujsr

3

Original post by GrungeGirl

Can someone please explain to how to do this? It is from 2011, Q3.

Put the RHS under a common denominator i.e. 3b^2 -1 .

(edited 7 years ago)

Reply 583

Also How would you sketch the below? I did algebraic division and got 1-2/x^2-5x+6 and drew a parabola but when I put it into wolfram alpha it looks different

((x-1)(x-4))/((x-2)(x-3))

Original post by Zacken

No, that's really not what I'm talking about, this is what algebraic geometry is: https://www.dpmms.cam.ac.uk/study/II/AlgebraicGeometry/2015-2016/HW1.pdf

ohh looks like fun.

nevermind i got it

Edit: Also this;

Spoiler

Is this question correct, the mark scheme shows a different answer

Spoiler

(edited 7 years ago)

Reply 586

I think I am being really stupid here but I tried to do a graph because the standard way doesn't work

Could anyone explain q3 2013 pls. I dont understand the last two bits at all.
https://www.maths.ox.ac.uk/system/files/attachments/test13.pdf

stuck on part ii

Original post by Mystery.

stuck on part ii

Lad are these MAT questions? where are you finding them?

Reply 590

Original post by joodaa

Lad are these MAT questions? where are you finding them?

yeah man like 2000s

Reply 591

Original post by Mystery.

stuck on part ii

a^3x + a^-3x is a sum of cubes so equals (a^x + a^-x)(a^2x -1 + a^-2x)

a^2x + a^-2x = 4y^2-2 which is just squaring a^x + a^-x = 2y

so equals 2z=(2y)(4y^2-3) then just expand

Reply 592

Original post by 11234

Could anyone explain q3 2013 pls. I dont understand the last two bits at all.
https://www.maths.ox.ac.uk/system/files/attachments/test13.pdf

pls help im so confused with this one

Reply 593

Original post by 11234

a^3x + a^-3x is a sum of cubes so equals (a^x + a^-x)(a^2x -1 + a^-2x)

a^2x + a^-2x = 4y^2-2 which is just squaring a^x + a^-x = 2y

so equals 2z=(2y)(4y^2-3) then just expand

Thanks!
What's the way to spot a sum of cubes ?

Reply 594

Original post by Mystery.

Thanks!
What's the way to spot a sum of cubes ?

i just learnt the identities so x^3 + y^3 = (x+y)(x^2-xy+y^2)

x^3-y^3= (x-y)(x^2 + xy + y^2)

Reply 595

Original post by Mystery.

yeah man like 2000s

i found that the old ones are much easier than the new ones

Reply 596

Original post by 11234

i found that the old ones are much easier than the new ones

yeah they are.

Reply 597