Mathematics HL/SL/Studies Help

[Guest 1529]

What are parameters and restrictions on IB homeworks?

I think none... it's simple help in homework. it's not even an IA... but what is it that is troubling you??

[hel]

Is there a missing picture?

yup.

Sorry about that...

[Chelsea21]

Hello

In the function p(x-q)(x-r) -> they give me this coordinates (-2,0), (0,-4) and (4,0)

How do I find the value of q, r and p ?

[Slovakov]

Hello

In the function p(x-q)(x-r) -> they give me this coordinates (-2,0), (0,-4) and (4,0)

How do I find the value of q, r and p ?

You havwe a function y=p(x-q)(x-r). So, just put the values of x and y from the given coordinates into the equation of the function so that it forms a system of equations, and solve it.

That is:

0=p(-2-q)(-2-r)

-4=p(0-q)(0-r)

0=p(4-q)(4-r)

[Chelsea21]

Is there any way of working with vectors, with the TI-84 calculator?

[genepeer]

Is there a missing picture?

yup.

Sorry about that...

From the given diagram, f(x) could be any function that passes through (-2,-8) and has a gradient of 3 at arbitrary x-value greater than ²/₃, a. Assuming f(x) passes through the origin and that the gradient at this point is zero (since it looks flat) could probably help but I won't even bother since it says "diagram not drawn to scale". The problem is either really hard or I need to use part (a) to solve it, thus making it 'less harder'.

What was (a)? If you want to upload any other images, please upload to imgur.com this time and not imageshack.us... long story

Edited May 1, 2011 by Gene-Peer

[Chronofox]

Am I allowed to ask a question from an old exam paper?

[Drake Glau]

Am I allowed to ask a question from an old exam paper?

Yes.

The only restriction on exams is a 24 hour time period after the exam is given for that session.

Edited May 1, 2011 by Drake Glau

[Chronofox]

Am I allowed to ask a question from an old exam paper?

Yes.

The only restriction on exams is a 24 hour time period after the exam is given for that session.

Okay. My question is:

f(x) = ln|x| + i[arg(x)] for 0 <= arg(x) <= 2π

And I have to prove that f(x₁x₂) = f(x₁) + f(x₂) is not always true.

Well, the first part of the questions asks you to solve for x = 1 - i and x = -1 + i.

f(1 - i) = ln|1 - i| + i[arctan(-1)]

= ln√2 + i(3 / 4)π or π ln√2 + i(7 / 4)π

f(i - 1) = ln|i - 1| + i[arctan (-1)]

= ln√2 + i(3 / 4)π or π ln√2 + i(7 / 4)π

Since this is paper 1, no calculators are allowed and thus I don't think the equation can be simplified any further.

So f(x₁) + f(x₂) is just 2[ln√2 + i(3 / 4)π or π ln√2 + i(7 / 4)π]

As for f(x₁x₂):

f(x₁x₂) = f(2i) = ln|2i| + i[arg(∞)]

=ln(2) + i(π / 2)

And ln(2) + i(π / 2) =/= ln√2 + i(3 / 4)π or π ln√2 + i(7 / 4)π

Is this an adequate enough proof? The unsimplified nature of the answers to each solution of f(x) makes me very doubtful.

[dessskris]

f(1 - i) = ln|1 - i| + i[arctan(-1)]

= ln√2 + i(3 / 4)π or π ln√2 + i(7 / 4)π

f(i - 1) = ln|i - 1| + i[arctan (-1)]

= ln√2 + i(3 / 4)π or π ln√2 + i(7 / 4)π

is x a complex number?

why is that π there?

f(1-i) = ln√2 + i7π/4

f(-1+i) = ln√2 + i3π/4

(1-i)(-1+i)=2i

f(2i) = ln2 + iπ/2

Want to prove:

f(2i) ≠ f(1-i) + (-1+i)

ln2 + iπ/2 ≠ ln√2 + i7π/4 + ln√2 + i3π/4

ln2 + iπ/2 ≠ 2ln√2 + i10π/4

ln2 + iπ/2 ≠ ln(√2²) + i5π/2

ln2 + iπ/2 ≠ ln2 + i5π/2

Proven..?

I am not sure though.. I've just finished learning complex

[timtamboy63]

Would really appreciate help with this, it's killing me:

Vectors btw,

[iber2011]

Would really appreciate help with this, it's killing me:

Vectors btw,

a)

AC=AO+OC= -p+r

BC=BO+OC= -q+r

for b...i'm not sure I think you have to use the dot product

for c...For OC and AB to be perpendicular their magnitudes should equal to zero when you multiply.

Sorry this is all I could help

[timtamboy63]

Yeah, a) was pretty easy, b) I can't work out, but I know that they must all equal 0.

Thanks anyway

[timtamboy63]

Here are a couple more I have no idea how to even start

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Help would be greatly appreciated

Edited May 1, 2011 by timtamboy63

[bomaha]

Here are a couple more I have no idea how to even start

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---

---

Help would be greatly appreciated

Sorry we didn't study vectors so I can't help, but in what books are these questions?

[timtamboy63]

Its the Haese and Harris Second Edition Chapter 14 Review Questions

They're pretty tough, our tests are usually cut and paste straight from there.

[genepeer]

Want to prove:

f(2i) ≠ f(1-i) + f(-1+i)

ln2 + iπ/2 ≠ ln√2 + i7π/4 + ln√2 + i3π/4

ln2 + iπ/2 ≠ 2ln√2 + i10π/4

ln2 + iπ/2 ≠ ln(√2²) + i5π/2

ln2 + iπ/2 ≠ ln2 + i5π/2

Proven..?

There's something our teacher liked to emphasize proving statements. Not sure if that important though. You can't start with the statement you want to prove and end with a true statement. It's supposed to be the other way round. So either write those lines in reverse or:

f(1-i) + f(-1+i) = ... = ln2 + i5π/2

f(2i) = ln2 + iπ/2

therefore, f(2i) ≠ f(1-i) + f(-1+i)

The reason he said the earlier method is wrong, let's say you want to (falsely) prove -2=2

-2=2

(-2)²=2²

4=4

in this case, the reverse won't hold... Just nitpicking.

[genepeer]

Would really appreciate help with this, it's killing me:

Vectors btw,

a)

AC=AO+OC= -p+r

BC=BO+OC= -q+r

b)

I'll use a full stop for the dot product symbol

OB (q) is perpendicular to AC (r-p), so:

OB.AC = 0

q.(r-p) = 0

q.r-q.p = 0

q.r = q.p

OA (p) is perpendicular to BC (r-q), so:

OA.BC = 0

p.(r-q) = 0

p.r-p.q = 0

p.r = p.q

Therefore, q.r = q.p = p.r

c)

AB = p-q

OC = r

AB.OC = (p-q).r

=p.r-q.r, since q.r = p.r

=0 therefore OC and AB are perpendicular.

S divides AB externally in the ratio 3:5 (img: S---A--B)

SA = 3/2*AB

=3/2*(b-a)

OS = OA + AS

= a - 3b/2 + 3a/2

= 5a/2 - 3b/2

CS = 5a/2 - 3b/2 - c

T divides CS internally in the ration 1:2 (img: C-T--S)

CT = 1/3*CS

= 5a/6 - b/2 - c/3

OT = OC + CT

= c + 5a/6 - b/2 - c/3

= 5a/6 - b/2 + 2c/3

Let's say a = i + rj + 2k and b = 2i + 2j - k

If a vector is parrallel to vector a and perpendicular to vector b, then a and b are perpendicular. Therefore,

a.b = 0

(1)(2) + ( r )(2) + (2)(-1) = 0

2r = 0

r = 0

Now a = i + 2k

If a unit vector, c, is parallel to a then

c = x a, where x is just a real number.

= x(i + 2k)

= xi + 2xk

and |c| = 1 since it's a unit vector.

x² + (2x)² = 1

5x² = 1

x = ±^√5/₅

Since they asked for one unit vector, we can choose x = ^√5/₅ and

c = ^√5/₅i + ^2√5/₅k

Edited May 1, 2011 by Gene-Peer

[hel]

Is there a missing picture?

yup.

Sorry about that...

From the given diagram, f(x) could be any function that passes through (-2,-8) and has a gradient of 3 at arbitrary x-value greater than ²/₃, a. Assuming f(x) passes through the origin and that the gradient at this point is zero (since it looks flat) could probably help but I won't even bother since it says "diagram not drawn to scale". The problem is either really hard or I need to use part (a) to solve it, thus making it 'less harder'.

What was (a)? If you want to upload any other images, please upload to imgur.com this time and not imageshack.us... long story

https://mr-t-tes.wikispaces.com/file/view/Nov+2010+Mathematics+SL+paper+1+QP.pdf Q10. There's the whole question

[genepeer]

https://mr-t-tes.wikispaces.com/file/view/Nov+2010+Mathematics+SL+paper+1+QP.pdf Q10. There's the whole question

You forgot to tell us f(x) = x³. Would have guessed it if it weren't for the "Not drawn to scale"

T = integral from -2 to k of ( f(x) - tangent )

= _-2∫ ^k x³ - (3x - 2) dx

= _-2∫ ^k x³ - 3x + 2 dx

= [ x⁴/4 - 3x²/2 + 2x ]_-2^k

= (k⁴/4-3k²/2 + 2k) - (4-6-4)

= k⁴/4 - 3k²/2 + 2k + 6

and T = 2k+4, therefore:

k⁴/4 - 3k²/2 + 2k + 6 = 2k + 4

k⁴/4 - 3k²/2 + 2 = 0

k⁴ - 6k² + 8 = 0

Edited May 1, 2011 by Gene-Peer