HELP -->A few questions. (1 Viewer)

[haboozin]

Question 1
Find the values of K such that the equation x/(1-x²) = kx has three distinct real roots

normally these sort of questions r pretty easy except i cant find graphically any values of K that give 3 distinct roots.. I can find values of K with 2 distinct roots.
answered


Question 2
Use the substitution u = 4Pi - x to evaluate Int (from 5Pi/2 to 3Pi/2) x/(2 + cosX) dx

good formatting would be appretiated

Question 3
The point P represents the complex number z such that |z - (SQRT(3) + i)| = 1. find the set of possible values of |z| and argz..

I dont really understand this question
wouldnt |z| just be 2... and arg z could be anything ???
the locus would be a circle centre (SQRT(3), 1) radius 1..??
answered

Question 4
In an argand diagram the point P, Q, R represent the complex numbers z1, z2 and z2 + i(z2 - z1) respectively
(not z2 is not 2 * Z)

i Show that PQR is a right angled traigle....


ok umm is this because the its times by i and when u times by i its the same as moving 90 degrees.. ?
answered

Question 5 <-- NEW Question

P(C@, C/@) and Q(-C@,-C/@), Where @>0 and C>0, are two points on the rectangular hyperbola xy=C². THe circle with centre P and radius PQ cuts the hyperbola again at points A(Ca,C/a), B(Cb,C/b) and C(Cy,C/y). CP produced meets AB at D . MCN is tangent to the circle at C.

i. Show that the circle cuts the hyperbola at points (Ct,C/t) where t satisfies the equation

t⁴ - 2t³ - 3t²(@² + 1/@²) - 2t/@ + 1 =0

Hense deduce that aby@ = -1


ok i get very close but i have C's in it... :| i doono how to get rid of it.. Shafqat this ones for u..

 

[yrtherenonames]

1. Just using GraphCalc it looks like k>1 and k<0 would do it...I'm sure someone else will prove it. Is k an integer?

2.Well cos(4pi-u)=cosu, So it works out to be integral 1.5pi-2.5pi of
(4pi-u)/(2+cosu)du

Now 4pi/2+cosu you can do with some t-formula trickery...I'm not sure about a quick way to do u/(2+cosu) but apparently the ultimate value is 7.595...


3. That DOES look like a strange question...but surely it's a circle of radius 1?

4. I think you're on the right track...I think that very question is in a Grammar Past Trial paper, I'll have a look tomorrow.

 

[haboozin]

1. Just using GraphCalc it looks like k>1 and k<0 would do it...I'm sure someone else will prove it. Is k an integer?

2.Well cos(4pi-u)=cosu, So it works out to be integral 1.5pi-2.5pi of
(4pi-u)/(2+cosu)du

Now 4pi/2+cosu you can do with some t-formula trickery...I'm not sure about a quick way to do u/(2+cosu) but apparently the ultimate value is 7.595...


3. That DOES look like a strange question...

4. I think you're on the right track...I think that very question is in a Grammar Past Trial paper, I'll have a look tomorrow.

1. mmmm i prob drew my graph wrong ... silly...

2. but with T results why r u gonna do wiht the u?????/

3. indeed.
4. its catholic trial... 2003 fucking arnolds

 

[FinalFantasy]

1. mmmm i prob drew my graph wrong ... silly...

2. but with T results why r u gonna do wiht the u?????/

3. indeed.
4. its catholic trial... 2003 fucking arnolds

2.)after u use the given substitution the u on top should be gone

 

[haboozin]

i should say in Q2 the first part was

use substitution t = tan x/2to evaluate (same integrals and values) 1/(2 + cosx) dx


and i thinkkk that equaled 0 or smething im not sure though i forgot....

 

[haboozin]

2.)after u use the given substitution the u on top should be gone

what??? how/??

 

[FinalFantasy]

i should say in Q2 the first part was

use substitution t = tan x/2to evaluate (same integrals and values) 1/(2 + cosx) dx


and i thinkkk that equaled 0 or smething im not sure though i forgot....

if u evaluated int.1/(2 + cosx) in the first part
den in the second part u should get I=int.2pi\(2+cosx) or something similar
den u can just apply it

 

[FinalFantasy]

what??? how/??

ok
Question 2
Use the substitution u = 4Pi - x to evaluate Int (from 5Pi/2 to 3Pi/2) x/(2 + cosX) dx

let I=int. x\(2+cosx) dx from 5pi\2 to 3pi\2

let u=4pi-x
x=4pi-u
dx=-du

I=-int. (4pi-u)\(2+cos (4pi-u) ) du from 3pi\2 to 5pi\2
=-int. (4pi-u)\(2+cos u) du
=-int. 4pi\(2+cos u)du +int. u\(2+cos u) du from 3pi\2 to 5pi\2
=-int. 4pi\(2+cos u) du from 3pi\2 to 5pi\2 - int. u\(2+cos u) du from 5pi\2 to 3pi\2
but int. u\(2+cos u) du from 5pi\2 to 3pi\2 is equivalent to int. x\(2+cos x) dx with same terminals
.: 2I=-int. 4pi\(2+cos u) du from 3pi\2 to 5pi\2
I=-int. 2pi\(2+cos u) du from 3pi\2 to 5pi\2


now juz use wat u got from ur previous q. and times it by 2pi

 

[yrtherenonames]

Sounds good to me, but yeesh just doing t-formula would've bin easier

 

[thunderdax]

With question 3, draw the circle on an argand diagram and use the angle between the x axis and the tangents to that circle from the origin to find the limits of argz. To find modz, draw a line going through the origin and the center of the circle and find the distance between where it touches the circle and the origin. This gives you the min and max values of modz. I would do this question but its impossible to do without a picture and I'm not talented enough with computers to do that kind of thing.

 

[KFunk]

Question 3
The point P represents the complex number z such that |z - (SQRT(3) + i)| = 1. find the set of possible values of |z| and argz..

I dont really understand this question
wouldnt |z| just be 2... and arg z could be anything ???
the locus would be a circle centre (SQRT(3), 1) radius 1..??
?

That's pretty much it, let z = x + iy

|z - (&radic;3 + i)| = 1

|(x-&radic;3) + (y-1)i| = 1

&radic;[(x-&radic;3)² + (y-1)²] = 1

(x-&radic;3)² + (y-1)² = 1 , I'll write the next bit in a sec

--> Like thunderdax was saying with the circle, you should draw the two tangents from the origin to the outer rims of the circle. Then draw a line from the origin to the centre of the circle and lines from the centre perpendicular to where the tangents touch the circle. You then have two triangles which you can prove similar using (RHS), since they share the same hypotenuse, they are both right angled and they have equal sides (the radii). You can then use trig to find all the angles involved to find your max and min args.

 

[KFunk]

Question 1
Find the values of K such that the equation x/(1-x²) = kx has three distinct real roots

normally these sort of questions r pretty easy except i cant find graphically any values of K that give 3 distinct roots.. I can find values of K with 2 distinct roots.

For number one, consider the basic shape of a cubic polynomial and think of the local minima and local maxima that a cubic ussually has. When the local maxima touches the x-axis (to make a double root) you end up with < 3 distinct roots. When the local minima touches the axis (likewise) you end up with < 3 distinct roots. If you find the y value of the maxima/minima in terms of k, you can then let y=0 to find the values of k that create a double root (and hence the appropriate values will lie in between:

y = kx³ - kx +x (1)
y' = 3kx<sup.2</sup> - k + 1

At max/min x = &plusmn;&radic;[(k-1)/3k] (2)

sub (2) into (1) to find y values of max/min (using positive (2) first)

y = [(k-1)/3k].&radic;[(k-1)/3k] - k.&radic;[(k-1)/3k] + &radic;[(k-1)/3k]

y = -2k/3.&radic;[(k-1)/3k] + 2/3.&radic;[(k-1)/3k] (let y=0)

k&radic;[(k-1)/3k] = &radic;[(k-1)/3k]

k=1 -----> k &ne; 1

Actually.. just look at it and say when k=1 y=x³ , k=0 y=x (< 3 roots) k < 0 , k>1 . I think someone asked before whether k was an integer, 'cause while it's easy to prove that k &ne; 1 and k &ne; 0 , I'm not sure how to prove that it doesn't lie between those values... hmph

 

[Slidey]

Find the values of K such that the equation x/(1-x^2) = kx has three distinct real roots

normally these sort of questions r pretty easy except i cant find graphically any values of K that give 3 distinct roots.. I can find values of K with 2 distinct roots.

x/(1-x^2)=kx
x=kx(1-x^2)
x=kx-kx^3
0=(k-1)x-kx^3
0=x(k-1-kx^2) (now you already have a root at x=0, so that's one distinct root for you)
so three distinct real zeroes when
(k-1-kx^2)=0 has 2 distinct roots (i.e. discriminant >0)
kx^2-k+1=0
Discriminant:
4(k-1)k>0
Now think of that little parabola in your head and look at where it is greater than zero:
k<0, k>1

 

[ngai]

Question 3
In an argand diagram the point P, Q, R represent the complex numbers z1, z2 and z2 + i(z2 - z1) respectively
(not z2 is not 2 * Z)

i Show that PQR is a right angled traigle....

PR = r - p = z2 + i(z2 - z1) - z2 = i(z2 - z1)
PQ = q - p = z2 - z1
so PR = iPQ
so PR perpendicular to PQ
so angle at P is 90 degrees

 

[FinalFantasy]

hey Kfunk what happend to ur bear?

 

[haboozin]

PR = r - p = z2 + i(z2 - z1) - z2 = i(z2 - z1)
PQ = q - p = z2 - z1
so PR = iPQ
so PR perpendicular to PQ
so angle at P is 90 degrees

silly... sounds so easy.
Why didnt i do that.

 

[haboozin]

<b>NEW QUESTION LOOK AT ORIGINAL POST</b>

 

[who_loves_maths]

Originally Posted by haboozin
Question 5
P(C@, C/@) and Q(-C@,-C/@), Where @>0 and C>0, are two points on the rectangular hyperbola xy=C2. THe circle with centre P and radius PQ cuts the hyperbola again at points A(Ca,C/a), B(Cb,C/b) and C(Cy,C/y). CP produced meets AB at D . MCN is tangent to the circle at C.

i. Show that the circle cuts the hyperbola at points (Ct,C/t) where t satisfies the equation

t4 - 2t3 - 3t2(@2 + 1/@2) - 2t/@ + 1 =0

Hense deduce that aby@ = -1


ok i get very close but i have C's in it... :| i doono how to get rid of it.. Shafqat this ones for u..

i think you have c's in it haboozin because you just made some small silly mistakes in the factorisation:

The equation of the circle is: (x - c@)^2 + (y - c/@)^2 = (2c@)^2 + (2c/@)^2

the co-ordinates of the points of intersection between this circle and the hyperbola again is given by (ct, c/t) since the points lie on the hyperbola.
so, x= ct and y =c/t satifies the equation of the circle:

ie. (ct - c@)^2 + (c/t - c/@)^2 = (2c@)^2 + (2c/@)^2 ;
now, take the 'c' out of each bracket and then cancel:

c^2(t -@)^2 + c^2(1/t -1/@)^2 = c^2(4@^2 +4/@^2)

so the 'c^2' cancels ---> (t -@)^2 + (1/t -1/@)^2 = 4@^2 + 4/@^2
(this is where you get rid of the C's)

...

next, multiply each side by "t^2" to get:

t^4 -(2@)t^3 + (@^2)t^2 + 1 + (1/@^2)t^2 -(2/@)t = (4@^2)t^2 + (4/@^2)t^2 ;

which simplifies to: t^4 - 2@t^3 - 3t^2(@^2 + 1/@^2) - 2t/@ + 1 =0

now, the solutions to 't' are: t = {a, b, y, @} , which is info. already given to us.

hence, taking the product of the roots from the equation: aby@ = -1.

hope this helps

 

[who_loves_maths]

Originally Posted by haboozin
i. Show that the circle cuts the hyperbola at points (Ct,C/t) where t satisfies the equation

t4 - 2t3 - 3t2(@2 + 1/@2) - 2t/@ + 1 =0

also, just adding to my last post, i think you have the term (in bold above) in that equation wrong: i think it should be 2@t^3, and not just 2t^3.

 

[Slidey]

Use the substitution u = 4Pi - x to evaluate Int (from 5Pi/2 to 3Pi/2) x/(2 + cosX) dx
u=4pi-x
du=-dx
u(5pi/2)=3pi/2
u(3pi/2)=5pi/2

Int(x/[2+cosx])dx from 5pi/2 to 3pi/2
=
-Int([4pi-u]/[2+cosu])du from 3pi/2 to 5pi/2
=
Int([4pi-u]/[2+cosu])du from 5pi/2 to 3pi/2
=
Int(4pi/[2+cosu])du from 5pi/2 to 3pi/2 - Int(u/[2+cosu])du from 5pi/2 to 3pi/2

Recall Int(f(x))dx from a to b = Int(f(u))du from a to b, thus:

Int(x/[2+cosx])dx from 5pi/2 to 3pi/2 = Int(4pi/[2+cosu])du from 5pi/2 to 3pi/2 - Int(x/[2+cosx])du from 5pi/2 to 3pi/2

2 * Int(x/[2+cosx])dx from 5pi/2 to 3pi/2 = Int(4pi/[2+cosu])du from 5pi/2 to 3pi/2

Int(4pi/[2+cosu])du from 5pi/2 to 3pi/2
t=tan(u/2)
du= 2dt/(1+t^2)
cosu=(1-t^2)/(1+t^2)

Int(4pi/[2+cosu])du = Int(8pi/(3+t^2))dt = 8pi/sqrt(3) arctan(t/sqrt(3)) = 8pi/sqrt(3) arctan(tan(u/2)/sqrt(3))

So taking into account the division by two earlier, and the limits:
4pi/sqrt(3) [arctan(tan(u/2)/sqrt(3))] from 5pi/2 to 3pi/2
= 4pi/sqrt(3)*[pi/6+pi/6]
=sqrt3.pi^2/27

I think that was too long. I think I made a mistake.