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Re: MX2 Integration Marathon Here's an actual challenging Q: (Well not for you guys ofcourse :P) \int_{-1}^{1}\frac{2x}{(x^2+2x+5)^2}dx

Re: MX2 Integration Marathon Hehe made an easy q, What is this to 2 dp: e^(\int_{-1.518213265}^{-pi/2+0.0005005}\frac{2}{2cos^2x-sin2x}dx)

You are sitting on the borderline 90, you can get 91.5 (If you believe you can get 85 in modern try hard to do that it is dragging u)

63-70, if you push hard you'll get in

Re: MX2 Integration Marathon This time let u=x^n+1, do some partial fractions, inside the log you can split into 1-1/u which when put under boundaries of 2 to infinity... 1/infinity=0 Hence = 1/n * [ln(1)-ln(1/2)] =1/n * ln(1/2)

Re: MX2 Integration Marathon By inspection if we expand the bottom the power of x will be 5/6. Hence let u=x^1/6 Then the integral is reduced to 6u^2/(1+u^2), factor out constants, +1-1 and you get a 6 as a constant and a tan inverse integration (of 1=pi/4)... =6*(1-pi/4) =6-2pi/3 Nice shifty...

Lol why so troll haha. I believe that is around 98 definitely (bio is dragging u down like crazy)

Depends where u are. I'm doing physics and I thought I would do well... Nek minnet trials got destroyed. Best weigh up, if u like calculations then phys if memorising is ur thing bio

Doing 4u is automatically studying for 3u as most trial papers contain harder 3 unit. HOWEVER, there are many things such as easy binomials and perms that are less tested which should be solid before doing majority 4u papers. Best strategy is to build 3u until 95%+ average then shift focus to 4u...

Depends on ur school exams... Usually ur school can determine and recommend you: For my school they wanted: 90%+ for 2u (I got 98%) - 2u papers were easy 80%+ for 3u (I got 85%)- 3u papers were hard Basically you should like maths and be good at it.

Re: HSC 2014 4U Marathon We know from diagram (triangle inside the circle): APxPQ=BPxPC Working backwards from what to prove: BPxPC=PQx(PB+PC) BPxPC=PQxBC If we equate what to prove and what we know: BC=AP? But BC=AB so, AB=AP which makes no sense. (Does not necessarily have to be true and...

Re: HSC 2014 4U Marathon Not too hard: Begin with expansions of: (ly-mx)^2>=0 (nx-lz)^2>=0 (ny-mz)^2>=0 Then add them together to obtain: l^2 * y^2 + l^2 * z^2 + m^2 * x^2 + m^2 * z^2 + n^2 * x^2 + n^2 * y^2 >= 2(lxmy+lxnz+mgnz) Add l^2 * x^2 + m^2 * y^2 + n^2 * z^2 to both sides and...

LOL sorry but I can't draw a diagram on this but you should be able to follow my steps. Denoting the points with the initials of the person I drew a flat triangle with Jacob on left, Tylah on the right and the building up north. By using trig on the 3D diagram (you can draw and h is the height...

Not exactly sure what you're doing but this is how i did it. Connect radii to the ends of the chord and connecting the centers together creating essentially 4 right angled triangles as you said it was. (Radii from center bisects the chord) which means if u find the chord is divided in half...

Re: MX2 Integration Marathon Very nice :) Thanks a lot!

Re: HSC 2014 4U Marathon i) From the condition given, A+B=C. Multiply everything by e^x. Ae^x+Be^x=Ce^x. Now the only way to reach the * form is to multiply throughout by the same constant which in this case means that a=b=c must hold true. ii) As a result of dropping that condition, A, B and...

Re: MX2 Integration Marathon So wait... Is it auxillary angle method? JKS haha Nice job mate :P Q:Prove that the derivative of arccot x=-1/(1+x^2) by considering y=arccot x. Hence integrate 1/(1+x^2)dx. *Idk maybe too easy lol General Q: How would one approach this integral: x^3/(2x-1).dx

Re: MX2 Integration Marathon Congrats on that other question HAHA :) Can you explain how you got to the second step? to the cosec^4

Re: MX2 Integration Marathon *If one sketches the function inside the absolute value it is clear that the graph from 0 to infinity is positive. Let the integral be I. So integrating by parts using u=e^-x and dv=sinx then further intigrating by parts again we end up with: I=[e^-x(cosx+sinx)]...

i) x is the distance from the origin up the slope. So as it is rolled up the distance will increase and as we know by logic, it will roll down again right? Now there are different ways of doing this: X Method: Look at the maximum of the parabola. The x-value will be the maximum x value of the...