MAT Prep Thread - 2nd November 2016

Guys, I am legitimately scared. I know what I need to know for the exam but the moment I look at the past papers everything looks much harder than it is. Do you have any cure for that?

Reply 625

Original post by denkata

Guys, I am legitimately scared. I know what I need to know for the exam but the moment I look at the past papers everything looks much harder than it is. Do you have any cure for that?

Keep doing them.

Reply 626

Original post by rtc16

How did you get the 2a/a-x ? i got 2x/a-x am i doing something stupidly wrong?

No, you're right.

Reply 627

11234

Original post by rtc16

How did you get the 2a/a-x ? i got 2x/a-x am i doing something stupidly wrong?

a+x/a-x = (-(a-x) + 2a)/a-x = -1 +2a/a-x

Reply 628

Hey Zacken just found out we're classmates - which college are you at?

Reply 629

Original post by GrungeGirl

How do you draw the graph for Q4(i) in 2013 paper? Screen Shot 2016-10-26 at 15.24.34.png

Thanks

When you get any rational function (of the general form f(x)=(a'x+b)/(cx+d)) you should consider only its axis intercepts and its asymptotes.

The vertical asymptote is x=-d/c, the horizontal asymptote is y=a'/c (as x approaches infinity b and d become insignificant) and the intercepts are (-b/a', 0) and (0, b/d) respectively. Obviously look out for cases such as d=0 or a'=0 where there is no intercept.

Then you only need to consider which "quadrants" (defined by the two asymptotes) the curve lies in. It will lie in diagonally opposite quadrants (1st and 3rd or 2nd and 4th). This could be done by substituting values. All rational functions have the same general shape (see y=1/x).

In this case, a'=1, b=a, c=-1, d=a. So the vertical asymptote is x=a, the horizontal asymptote is y=-1 (not required) and the axis intercepts are (0,1) and (-a, 0). The question only requires us to draw the graph in the "second quadrant", which the curve indeed lies in. This means that it also lies in "quadrant 4" (where it doesn't cross any axis).

(edited 7 years ago)

Reply 630

Original post by Insecure

Hey Zacken just found out we're classmates - which college are you at?

King's. You?

Original post by 11234

a+x/a-x = (-(a-x) + 2a)/a-x = -1 +2a/a-x

You wrote -1 + in your last post, hence his confusion.

Reply 631

Trinity bruh

(edited 7 years ago)

Reply 632

11234

Original post by Zacken

King's. You?

You wrote -1 + in your last post, hence his confusion.

sorry i got lazy....plus cant do latex too much effort

Reply 633

11234

Original post by Insecure

Trinity bruh

hi im applying trinity this year. Any tips? Do you happen to be on any IMO or olympiad team...cos they nearly scared me into not applying

Reply 634

Original post by 11234

hi im applying trinity this year. Any tips? Do you happen to be on any IMO or olympiad team...cos they nearly scared me into not applying

Hey, there's this huge misconception about IMO at Trinity. It's true that there are many IMO contestants here and that most IMO contestants apply to Trinity, but a bit less than half of us here (including me) have no experience in maths olympiads at all. I would, however, advise you not to apply to Trinity.

Tip would be just to work hard. Feel free to ask more questions.

Reply 635

RuairiMorrissey

5

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

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

.

It's more how you could just multiply them together like that.

Also, would the integral of f(-x) not be -u?

Reply 636

Original post by RuairiMorrissey

It's more how you could just multiply them together like that.

Read above, I've explained this three times now. The integral from -1 to 1 of f(x) dx or f(t) dt is just a random number. It's some constant u. It's not a function, it's not a variable. It's just a way of writing a number. It represents a constant. You can pull constants outside integrals.

Also, would the integral of f(-x) not be -u?

No. Draw a sketch. Reflect it in the y-axis. If you still don't believe me, use the substitution v = -x and see what the integral simplifies down to.

Reply 637