help wif questions (1 Viewer)

[shsshs]

Hi i cant seem to do the following.

1) Find the relationship between a and c for
3
x - 3ax + c to have 3 distinct real roots

2) Sketch l z + 1 l < l z l + 1


3) integrate:
.......-x
__e____
........-2x
1 + e


----------

edit

sorry i cant work these ones out now
(a)

_3____2
x + px + qx + r = 0

one root is half the sum of the other 2

prove
__3
2p = 9pq - 27r

(b) Prove by induction that
_n___n
x + y is divisble by x+ y for all odd n

ty

 

[darkliight]

Ouch. Now I know this is kinda off topic, but why don't we have a latex mod for this board?

Question 3) set u = e^-x then -du = e^-x giving you int -du/1+u^2 which is just -tan^-1(u) ( and we know what u is, so we get -tan^-1(e^-x) )

 

[Mountain.Dew]

Hi i cant seem to do the following.

1) Find the relationship between a and c for
3
x - 3ax + c to have 3 distinct real roots

2) Sketch l z + 1 l < l z l + 1

here goes my 2 cents...

1) f(x) = x³ - 3ax + c...

if they are to have 3 distinct real roots, consider the polynomial's derivative, and the 'general' graph of the polynomial

on the number plane, ideally we have a poly curve that cuts the x-axis at 3 distinct points. the 'tails' of the curve will go off into different directions IE if the left-side tail goes up, then the right-side tail goes down.

now, when considering the polynomial's derivative, focus on the turning pts. realise that there HAS to be two turning pts, one max and one min. also notice that comparing the y values of the TPs, they are different in sign. THIS IS A CONSEQUENCE OF HAVING A POLYNOMIAL WITH 3 DISTINCT ROOTS.

therefore, we consider when f'(x) = 0

f'(x) = 3x² -3a = 0 ==> x² = a, x = +/- sqrt(a).

now, using this information, and from the fact that y values of the TPs, are different in sign, THEN

f(sqrt(a)) * f(-sqrt(a)) < 0 => substitute values, and now u have ur relationship between a and c.

2) substitute z = x + iy ==> get a cartesian equation ==> graph it.

 

[SeDaTeD]

For q1. There exist 3 distinct real roots if there are two stationary points on either side of the x-axis. ie, they have different signs.
f'(x) = 3x^2 - 3a = 0 => x = +\-sqrt[a]
therefore, f(sqrt[a])f(-sqrt[a]) < 0 , since they have different signs.
(asqrt[a] - 3asqrt[a] +c)(-asqrt[a] + 3asqrt[a] + c) < 0
c^2 - (2sqrt[a])^2 < 0
c^2 < 4a

Edit: Mountain Dew beat me to it.

For q2, the entire complex plane except for the non-negative real numbers.

for all z, |z+1| - |z| =< ||z+1| - |z|| =< |z+1-z| = |1| = 1 (using reverse triangle inequality)
Equality only occurs when z and z+1 are scalar multiples of each other and this only happens on the x-axis.
Solving |x+1| - |x| = 1 yields x>= 0, so we exclude all points of the form x + 0i, where x => 0.
For any other point |z+1| - |z| < 1 holds.

Note: Reverse triangle inequality just follows from the triangle inequality |x+y| =< |x| + |y| by setting x-> x-y, rearranging, then doing it again for y-> y-x and combining the two.

 

[onebytwo]

another way is to sub z = x + iy as Mountain Dew suggests

thus mod ( x + iy +1) =< mod (x +iy) + 1
i.e sqrt ( (x+1)^2 + y^2) <= sqrt (x^2 + y^2) +1
squaring both sides, we get,
x^2 + 2x + 1 + y^2 <= x^2 +y^2 + 2sqrt(x^2 + y^2) +1
cancelling, we get,
x<= sqrt(x^2 + y^2)
squaring both sides,
x^2 <= x^2 +y^2
so, y^2>=0
i.e, y>= 0

 

[webby234]

Be a bit careful when squaring both sides - what if x is negative?

 

[SeDaTeD]

Squaring both sides of an inequality does not always preserve the inequality, eg.
-2 < -1
4 < 1 is false.

 

[icycloud]

SeDaTeD's method is good.

If you really wanted to do it with z = x + iy, then:

|z+1| < |z|+1
|z+1|² < (|z|+1)² {both sides are positive |z+1| >= 0 and |z| + 1 >= 0 so you can square both sides and the inequality will still hold}

(z+1)(z*+1) < |z|² + 2|z| + 1
zz* + z + z* + 1 < zz* + 2|z| + 1
2Re(z) < 2|z|

Thus, Re(z) < |z|

#

NB: z* means the conjugate of z.

---

At this point, you can 'reason' the answer out:

Ask yourself this question: When is the real part of a complex number less than its modulus?

The answer is the whole complex plane, except when x >= 0 and y = 0, in which case the real part is the same as the modulus. (i.e. the set of C, minus the set of Z*).

---

Or you could continue the algebra:

Now, suppose z = x + iy and therefore x and y are real numbers.
For all z we have:

x <= sqrt(x²) <= sqrt(x²+y²)

Because:

x <= |x| = sqrt(x²)

And, y² >= 0
Add x² to both sides: x² + y² >= x²
And we have the result sqrt(x² + y²) >= sqrt(x²)

Thus, since x = Re(z) and sqrt(x² + y²) = |z|, we have: Re(z) <= |Re(z)| <= |z|
Since this is true for all z, we have proven the initial inequality to be true for all z, but we must discount the case when Re(z) = |z| because the original inequality was Re(z) < |z| and not Re(z) <= |z|

i.e. x = sqrt(x²+y²)
since LHS >= 0, x >= 0
and squaring both sides, x² = x² + y², y² = 0, y = 0

So we discount the case x >= 0 and y = 0, which is the same result we got by 'reasoning it out' in the previous section.

---

Of course in the exam, you wouldn't have to explain it in as much detail as the above and the working would be at max 10 lines long.

 

[del pietro]

another way is to sub z = x + iy as Mountain Dew suggests

thus mod ( x + iy +1) =< mod (x +iy) + 1
i.e sqrt ( (x+1)^2 + y^2) <= sqrt (x^2 + y^2) +1
squaring both sides, we get,
x^2 + 2x + 1 + y^2 <= x^2 +y^2 + 2sqrt(x^2 + y^2) +1
cancelling, we get,
x<= sqrt(x^2 + y^2)
squaring both sides,
x^2 <= x^2 +y^2
so, y^2>=0
i.e, y>= 0

I might be wrong but... isnt (yi)^2 = -y^2 since i^2 = -1
therefore this working out would be incorrect and the answer would come down to y < 0 ???
pls correct me if im wrong...

 

[shsshs]

nope. the modulus is given by root(x^2 + y^2)

yeh so from wat icycloud sed.. re(z) < lzl

wat i did was..

x < root(x^2 + y^2)

so IF x > 0

2
y > 0 this is baiscally everything except the axis

and IF x < 0
2
y < 0 this is nothing since y is real.

is this correct?

 

[shsshs]

hey soz added 2 new q's

 

[Riviet]

If x>0, then we can sqare both sides as the equality still holds for when both sides are positive:

x² < x²+y²
y² > 0

We can see that this is true for all values of y, except when y=0, which is what icycloud said about the solution not including values on the real axis.

 

[ianc]

(b) Prove by induction that
_n___n
x + y is divisble by x+ y for all odd n

ty

I attached my working for this question, although I'm not sure if its 100% correct.

Could someone check to see if my method's ok? Thanks

 

[SeDaTeD]

Yep that looks good.

 

[Riviet]

Got stuck in a line, but I finally get it.