4u Mathematics Marathon V 1.0 (1 Viewer)

[pLuvia]

x³-9x+9=0
(a-1)(b-1)(c-1)
=(ab-a-b+1)(c-1)
=abc-ac-bc+c-ab+a+b-1
=abc-(ac+ab+bc)+(a+b+c)-1
=-9-(-9)+0-1
=-1
As req'd

Have no quesiton at this point

 

[EmmR]

Why is it.. whenever I visit this thread, somebody is thinking of a question.

It seems so take you people longer to think of a question then to answer it.

Has the humanity advanced so far that our curiosity has been completely destroyed?, or is it that the people over in 4u English are the ones posing the questions.. and we (mathematicians) merely solve them with numbers...

Eg.
what is the meaning of life?
42

 

[SoulSearcher]

I'm not currently thinking of a question

 

[onebytwo]

integrate ln(lnx)

i saw this question in some year old post, but they didnt really come to a conclusion, i got x(ln(lnx)) + 1/(lnx)^2 + c, by using IBP twice

anyway, can anyone please try it

 

[Sober]

integrate ln(lnx)

i saw this question in some year old post, but they didnt really come to a conclusion, i got x(ln(lnx)) + 1/(lnx)^2 + c, by using IBP twice

anyway, can anyone please try it

Your first term is correct but the second term should be -li(x) where li is the logarithmic integral which is beyond the scope of the 4u course.

 

[Riviet]

Next Question:

i) Show that the area bounded by a concave down parabola and the x-axis with x-intercepts a, -a and y-intercept b is 4ab/3 units².

hint:

Spoiler

Use the form y=k(x-a)(x+a) and the given info to find k and hence integrate to show the required area.

ii) The base of a solid is the region bounded by y=cosx and y=-cosx for -pi/2 < x < pi/2. Parabolic slices are taken perpendicular to the x-axis with height e^x. Find the volume of the solid formed.

hint:

Spoiler

A diagram is definitely helpful for any volumes question. Use part i) to help find a general expression for the volume of a thin slice and sum these slices up to integrate and hence find the volume.

UPDATE: hints included.

 

[STx]

i) let y=(x-a)(x+a)=x²-a²

when, y=b => x=0 .'. b=-a²

Now, A= _-a∫^a x²-a² dx
= [ x³/3-a².x] {^-a-->^a}
= [{4.a.-a²}/3]
= [4ab/3] (Since b=-a²) #

ii)

Considering half the parabolic slice:
b= e^x and y=a=cos(x)

=> A(x)=2ab/3=[2e^x.cos(x)]/3

Volume*4 as im integrating half the parabolic slice
.'. ∆V= (8/3) ₀∫^π/2 e^x.cos(x) dx
let u=cos(x), u'=-sin(x) and v'=e^x, v=e^x
.'. ∆V= (8/3).[e^x.cos(x)]₀^π/2 + (8/3) ₀∫^π/2 e^x.sin(x) dx

let ∆V₁=(8/3) ₀∫^π/2 e^x.sin(x) dx
let u=sin(x), u'=cos(x) and v'=e^x, v=e^x
.'. ∆V₁= (8/3).[e^x.sin(x)]₀^π/2 - (8/3) ₀∫^π/2 e^x.cos(x) dx
= (8/3).[e^x.sin(x)]₀^π/2 - ∆V

.'. ∆V= (4/3).[e^π/2-1] u³ #

 

[Riviet]

ii) Is ∆V=4ae^x/3 ?, this part doesnt seem correct..
Then...integrating...i get [4a/3]{e^π/2-e^-π/2} u³

Note that "a" is not common to each parabolic slice.

 

[STx]

^is it correct now?

 

[Riviet]

Yep, except that you forgot to double the area (don't know why you found half a parabolic slice in the first place ). Post up a new question when you're ready.

 

[STx]

Yep, except that you forgot to double the area (don't know why you found half a parabolic slice in the first place ). Post up a new question when you're ready.

I found the area for half the parabolic slice then integrated from 0-->pi/2 then multiplied the volume by 4 due to the symmetry that would occur, to find the total volume.

Next Question

Let R(x₀,y₀), P(x₁,y₁) and Q(x₂,y₂) be points on the circle x²+y²+2gx+2fy+c=0

i) If d is the distance between the points R and P, show that
-d²/2 = x₀x₁+y₀x₁ + g(x₀+x₁) + f(y₀+y₀) + c

ii) Suppose that Q is also distance d from R. Explain why the equation of the chord PQ is

-d²/2 = x₀x+y₀y + g(x₀+x) + f(y₀+y) + c

 

[Riviet]

- Bump -

Anyone know how to do (ii)?

 

[haque]

I've done it but i can't seem to be able to upload my scan-ive had this trouble for a long time actually. i'll type it

d^2=(x0-x1)^2 +(y0-y1)^2 and
x0^2+y0^2+2gx0+2fy0+c=0 since R is on circle and we can say a similar thing for P(x1,y1)
thus x0^2+y0^2+x1^2+y1^2=-2g(x0+x1)-2f(y0+y1)-2c-2x0x1-2y0y1 thus
-d^2/2=g(x1+x0)+f(y0+y1) +x0x1+y0y1 +c

Now since Q is also distance d from R its coordinates will satisfy this equation
-d^2/2=g(x2+x0)+f(y0+y2) +x0x2+y0y2 +c and since this is an equation of a straight line in the form
-d^2/2=g(x+x0)+f(y0+y) +x0x+y0y +c which P and Q satisfies, it must be equation of the chord. QED

 

[haque]

whuile on the topic of conics -AB and CD are perpendicular chords passing through same focus S on ellipse with semimajor axis a and semiminor axis b. show that 1/AS.SB +1/CS.SD is constant.

 

[Riviet]

Ah that makes sense, it reminds me of a similar argument for the chord of contact proof.

 

[Riviet]

Bump!

Next Question:

Given that 0 < h < 1, prove by induction that if n is an integer, n > 1, then (1-h)ⁿ < 1/(1+nh).

 

[haque]

By the way Riviet i already posted aquestion in the meanwhile

let n=1 consider 1-h -1/1+h=-h^2/1+h<=0 therefore true for n=1. assume true for n=k so (1-h)^k<=1/1+kh consider n=k+1 so

(1-h)^k+1=(1-h)*(1-h)^k<=1/1+kh *1/1+h=1/1+kh+h+kh^2<=1/1+kh+h=1/1+(k+1)h therefore true for n=k+1 and bla bla so on(i left out the inductive assumption and all that). should i post a different question or u guys want the old one to stand?

 

[Riviet]

Ah sorry haque, didn't notice the question that you posted up previously, let's let your one stand.

P.S Just a reminder that you said you would do the question posted by Yip in the 3u marathon.

 

[haque]

Yea i know i did the induction bit and gave up on part 3 i think-i'll try again after i've done a few past hsc's and trials-I know where Yip got that question and its not 3U

 

[onebytwo]

whuile on the topic of conics -AB and CD are perpendicular chords passing through same focus S on ellipse with semimajor axis a and semiminor axis b. show that 1/AS.SB +1/CS.SD is constant.

*bump*

Spoiler

drawing a perpendicular chords on an ellipse. let the point P lie on AS such that angle ASX=@ and PS=r and Q is the point on the x-axis directly below the point P so that angle PQS is right. so then QS=x-ae
so from the triangle PQS cos@=(x-ae)/r, so x=rcos@+ae
also y/r = sin@, so y=rsin@
putting these and x and y values into the equation of the ellipse and on simplifying you get (a^2sin^2@+b^2cos^2@)r^2 + (2aeb^2cos@)r +a^2b^2(e^2-1) = 0
the roots of the equationa are AS and SB, so find 1/(AS.SB)

then put P' on DS so angle P'SQ'=90-@ and do the same there to find 1/(CS.SD)
then 1/(AS.SB) + 1/(CS.SD) = (a^2+b^2)/[a^2b^2(e^2-1)], which is independant of x, y and @

the same question i asked my maths teacher a few months ago, took him ages to get.

EDIT: next question:
prove that (secx+cosecx)/(tanx-cotx)=(tanx+cotx)/(secx-cosecx)