Author Topic: BEC'S methods questions (Read 106705 times) Tweet Share

bec · « **Reply #555 on:** November 04, 2008, 06:47:49 pm »

Really! I had it in my head that J was positive integers, so when you said that you were pretty much just "confirming" what I thought...good to know they're the same thing though, it simplifies things...

trinon · « **Reply #556 on:** November 04, 2008, 06:50:55 pm »

Quote from: bec on November 04, 2008, 05:57:53 pm

Thanks mao. I'll have a look now at that other thread.

With the J/Z thing (and yes, I did mean circular functions sorry!), would it also work if you used Z but not $\pm$ , as opposed to just + and using J?

And, sorry for the overload, but one more question...

4. Consider the graphs of y=kx and y=x^2 +bx +c^2 where b, c and k are all real numbers. Which of the following statements is false?

A. If k=b+2c, the graph of y=kx is a tangent to the graph of y=x^2 + bx + c^2.
B. If k=b-2c, the graph of y=kx touches the graph of y=x^2 + bx + c^2.
C. If b+2c<k<b-2c, the graph of y=kx does not intersect the graph of y = x^2 +bx +c^2.
D. If k>b+2c the graph of y=kx intersects the graph of y = x^2 +bx +c^2.
at two distinct points.
E. If k<b-2c the graph of y=kx intersects the graph of y = x^2 +bx +c^2.
at two distinct points.

Let $b = 1, c = 2$

$y=x^2+x+4, \frac{dy}{dx}=2x + 1$

Test A:

$k = b+2c=1+4=5$

Let the two graphs equal:

$5x=x^2+x+4$

$\implies 0=(x-2)^2$

$\implies x=2$

Sub back into original equation:

$y=5(2)=10, \therefore (2, 10)$

Find gradient:

$m_T=2(2)+1=5$

$y_T-10=5(x-2)$

$\implies y_T=5x=kx$

A is true.

Test B:

$y=x(1-4)=-3=x^2+x+4$

$\implies 0=x^2+4x+4$

$\implies 0=(x+2)^2$

$\implies x=-2$

B is true.

Test C:

Did you mean $b+2c>k>b-2c$ ?

$1+3 > k > 1-4$

$\implies 5 > k > -3$

Let $k=0$

$0=x^2+x+4$

$\Delta = 1 -16 = -15$

$\Delta < 0, \therefore no intercepts$

C is true.

Test D:

$k>1+4$

Let $k = 10$

$10x=x^2+x+4$

$\implies 0=x^2-9x+4$

$\Delta = 81-16$

$\Delta > 0, \therefore 2 intercepts$

D is true.

By process of elimination, E is incorrect. Do you want me to work that out for you aswell?

bec · « **Reply #557 on:** November 04, 2008, 06:55:41 pm »

Thanks Trinon, that must've taken you an age to write.

I often sub in values for variables but sometimes it doesn't prove anything - it just shows that an equation is true (or not) for one particular case. You've shown that this isn't an issue in this question since it only worked for E, but what if it didn't, and 2 of the possibilities "looked" like they were right?

Do you know what I mean?

Mao · « **Reply #558 on:** November 04, 2008, 07:21:13 pm »

E is also correct.

try graphing that with y=-4x, the first intersection is when x=-1.

I have a feeling the answer is C, which is when "c" is 0. I'll take a look at it tomorrow night.

trinon · « **Reply #559 on:** November 04, 2008, 07:47:20 pm »

Yeh, I know what you mean. I actually made that mistake myself with E. I found it to be one solution, maybe Mao was correct. What do the solutions say?

Oh and btw, for the first few tests I used it in the form of b and c, then I got tired of that and subbed in values.

dekoyl · « **Reply #560 on:** November 04, 2008, 08:22:58 pm »

Sorry to hi-jack your thread bec, but does one need to know the different "numerical fields" for methods? R and C I know but Z, N, Q(?), J(?) etc.. isn't stuck in my head yet.

Quickclaw · « **Reply #561 on:** November 04, 2008, 08:31:15 pm »

from what I saw in the study design they only mention them in cases like this:
• rules for derivatives of xn, (for n ∈ Q), ex, loge(x), sin(x), cos(x), tan(x), and linear combinations of
these functions (formal derivation not required);
so they don't have a point allocated just for it, but anyhow it might be a good idea to have an understanding of what they represent.

Mao · « **Reply #562 on:** November 04, 2008, 08:34:04 pm »

you don't need to know them =]

but for interest's sake [they may be wrong.. i have never properly studied these]

C is complex
R is real, subset of C
Q is rational, subset of R
Z/J is integers, subset of Q
N is natural numbers, subset of Z

bec · « **Reply #563 on:** November 05, 2008, 08:57:44 am »

I've had at look at the discussion in dekoyl's thread on that Kilbaha question but I still can't do it. I don't know if I'm finding the wrong area, or what I'm doing wrong...I've highlighted the area in orange on my graph.

Can anyone give me some pointers?
Thanks

Mao · « **Reply #564 on:** November 05, 2008, 10:56:04 am »

let A be half the area enclosed by $f^{-1}(x)$ and $f(x)$

$A=\int_0^p x-f^{-1}(x)\; dx = \int_0^p x - 2e^{x/3} + 2\; dx = \left[ \frac{x^2}{2}-6e^{x/3} + 2x\right]_0^p = \frac{p^2}{2}-6e^{p/3} + 2p + 6$

remembering that $f^{-1}(p)=2(e^{p/3}-1)=p$

$\implies A=\frac{p^2}{2}+2p - 3(2e^{p/3}-2) = \frac{p^2}{2}-p$

hence, $\int_0^p f(x)\; dx = A + \int_0^p x \; dx = \frac{p^2}{2}-p + \frac{p^2}{2} = p^2-p$

shown as required.

[as you see, i didn't like their method either]

bec · « **Reply #565 on:** November 05, 2008, 11:29:02 am »

Wow, I never would've thought of doing it that way. I understand what you've done, and why it worked (I think)...but why didn't my method work? What was wrong with it?

Edit: actually, I just realised I don't understand your working. I'm going out now but I'll ask later tonight...

And thank you, once again!

Mao · « **Reply #566 on:** November 05, 2008, 05:56:59 pm »

i don't exactly see what you did there, but from my understanding, $f(x)$ is a logarithmic function, and MM students are not required to know how to antidifferentiate them [it can be done via the method of integration by parts, something beyond the scope of Specialist Maths].

bec · « **Reply #567 on:** November 05, 2008, 07:50:08 pm »

I spoke to my teacher and she did it a different way, so I understand it now.
(She found the area of the square with corners (0,0) and (p,p) and subtracted $\int_0^p f^{-1}(x)$ )

bec · « **Reply #568 on:** November 05, 2008, 08:19:55 pm »

When a question asks to give the "proportion" of something, do you give it as a decimal or percentage?

Eg. "What proportion of old Germanic words, have a life greater than 300 years? Answer correct to 3dp."

(if I show this as a percentage, to 3dp, it has 5 significant figures - 22.313% - as opposed to the 3 it would have if I presented it as 0.223...)

Collin Li · « **Reply #569 on:** November 05, 2008, 08:25:25 pm »

I would go with 0.223

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