Cambridge Prelim MX1 Textbook Marathon/Q&A (1 Viewer)

appleibeats · Feb 13, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Screen Shot 2016-02-13 at 11.29.24 am.png

For 19 part c)

I got the correct anwer for i) sinx - root3 cosx = 2sin(x +5/3 pi)

Now for part ii)

I get the answers - pi/2 and pi/6

however the answer says pi/6 and 3pi/2

Could someone explain part ii)

InteGrand · Feb 13, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
View attachment 32891

For 19 part c)

I got the correct anwer for i) sinx - root3 cosx = 2sin(x +5/3 pi)

Now for part ii)

I get the answers - pi/2 and pi/6

however the answer says pi/6 and 3pi/2

Could someone explain part ii)

$\noindent For part (ii), they said they want the answers in the range $0\leq x <2\pi$. So you need to convert your answer of $-\frac{\pi}{2}$ into the equivalent angle in the desired range, which is $\frac{3\pi}{2}$ (found by adding $2\pi$ to it).$

appleibeats · Feb 13, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

So in this question part ii), is it necessary to change the domain??
For example,

sinx - root3cosx = -1 for 0 < x < 2pi

Now

2sin(x + 5pi/3) = -1 ; does the domain change here: ie. does it become 0< x + 5pi/3 < 2pi or does it stay the same

InteGrand · Feb 13, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
So in this question part ii), is it necessary to change the domain??
For example,

sinx - root3cosx = -1 for 0 < x < 2pi

Now

2sin(x + 5pi/3) = -1 ; does the domain change here: ie. does it become 0< x + 5pi/3 < 2pi or does it stay the same

$\noindent Since $0\leq x <2\pi$, we have $\frac{5\pi}{3}\leq x +\frac{5\pi}{3} < 2\pi+\frac{5\pi}{3}$.$

appleibeats · Feb 13, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Does that mean there should be more than 2 answers, ie. 3 answers.

InteGrand · Feb 13, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
Does that mean there should be more than 2 answers, ie. 3 answers.

No

appleibeats · Feb 13, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Screen Shot 2016-02-13 at 1.11.26 pm.png

I get for a) PC = h and PD = h/root3

I am unsure on how to get part b)

I though it would use the base triangle, but we have no angles to use.

jathu123 · Feb 13, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Since its in a NE - E direction, the angle DPE will be 45. Then use cosine rule

http://imgur.com/loQCMDy

hedgehog_7 · Feb 13, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

proof by induction: n(n+1)(n+2) is divisible by 3 for all integers greater than or equal to 1

prove that 7^n+3n(7^n)-1 is divisible by 9 for all positive numbers n

prove that 5^n > n^5

davidgoes4wce · Feb 13, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

SpiralFlex said:
$\frac{1}{(\sqrt{k}+\sqrt{k+1})(\sqrt[4]{k} + \sqrt[4]{k+1})}$

$\textup{I notice that if I multiply}\; (\sqrt[4]{k} - \sqrt[4]{k+1}) \;\textup{on the denominator I will get a difference between two squares so it may be useful in simplifying the expression.}$

$\textup{But I can't just multiply the denominator by}\; (\sqrt[4]{k} - \sqrt[4]{k+1}) \;\textup{for nothing. There is no such thing as a free lunch.}$

$\textup{So what I'll do is I will multiply the numerator and denominator by}$

$\; (\sqrt[4]{k} - \sqrt[4]{k+1}). \;\textup{This is okay since I'm essentially multiplying the fraction by 1.}$

[HR][/HR]

$\textup{Now I have,}$

$\frac{1*(\sqrt[4]{k} - \sqrt[4]{k+1})}{(\sqrt{k}+\sqrt{k+1})({\color{Blue} \sqrt[4]{k} + \sqrt[4]{k+1})* (\sqrt[4]{k} - \sqrt[4]{k+1})}}$

$\textup{The term that looks like we can simplify is highlighted in blue, we can use the difference between two squares,}$

$=\frac{(\sqrt[4]{k} - \sqrt[4]{k+1})}{{\color{Red} (\sqrt{k}+\sqrt{k+1})(\sqrt{k} - \sqrt{k+1})}}$

$\textup{The highlighted term in red looks like something we can simplify again, oh it's the difference between two squares again! Happy accident!}$

$=\frac{(\sqrt[4]{k} - \sqrt[4]{k+1})}{k-(k+1)}$

$=\frac{(\sqrt[4]{k} - \sqrt[4]{k+1})}{-1}$

$=\frac{ -( \sqrt[4]{k+1} - \sqrt[4]{k})}{-1}$

$=\sqrt[4]{k+1} - \sqrt[4]{k}$

anyone can do part 5b do this question?

$\sum^{255}_{k=1} \frac{1}{(\sqrt{k}+\sqrt{k+1})(\sqrt[4]{k}+\sqrt[4]{k+1})}$

davidgoes4wce · Feb 13, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

hedgehog_7 said:
proof by induction: n(n+1)(n+2) is divisible by 3 for all integers greater than or equal to 1

prove that 7^n+3n(7^n)-1 is divisible by 9 for all positive numbers n

prove that 5^n > n^5

Answer to the 1st part of your question:

Theorem: If n is a positive integer, then n(n+1)(n+2) is divisible by 3

Proof by induction:
n=1

1*2*3 = 6, which is divisible by 6.

Assume that for n = k, k(k+1)(k+2) is divisible by 6

We need to show that, based on that assumption, (k+1)(k+2)(k+3) is also
divisible by 6.

(k+1)(k+2)(k+3) = (k+1)(k+2)k + (k+1)(k+2)3 = k(k+1)(k+2) + 3(k+1)(k+2).

By induction hypothesis, the first term is divisible by 6,
and the second term 3(k+1)(k+2) is divisible by 6 because it contains
a factor 3 and one of the two consecutive integers k+1 or k+2 is
even and thus is divisible by 2. Thus it is divisible by both 3 and
2, which means it is divisible by 6. The theorem is proved since
the sum of two multiples of 6 is also a multiple of 6.

Paradoxica · Feb 13, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

davidgoes4wce said:
anyone can do part 5b do this question?

$\sum^{255}_{k=1} \frac{1}{(\sqrt{k}+\sqrt{k+1})(\sqrt[4]{k}+\sqrt[4]{k+1})}$

$$\noindent SUM $ = \sum_{k=1}^{255} \left(\sqrt[4]{k+1}-\sqrt[4]{k}\right) = \left(\sum_{k=2}^{256} \sqrt[4]{k}\right) - \left(\sum_{k=1}^{255} \sqrt[4]{k}\right) \\ $SUM $ = \sqrt[4]{256}+ \left(\sum_{k=2}^{255} \sqrt[4]{k}\right) -\left(\sum_{k=2}^{255} \sqrt[4]{k}\right) -\sqrt[4]{1}= 4-1 = 3$

davidgoes4wce · Feb 13, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Nice pardoxica, I didn't see that trick coming

davidgoes4wce · Feb 13, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Paradoxica said:
$$\noindent SUM $ = \sum_{k=1}^{255} \sqrt[4]{k+1}-\sqrt[4]{k} = \sum_{k=2}^{256} \sqrt[4]{k} - \sum_{k=1}^{255} \sqrt[4]{k} \\ $SUM $ = \sqrt[4]{256}+ \sum_{k=2}^{255} \sqrt[4]{k} - \sqrt[4]{1} -\sum_{k=2}^{255} \sqrt[4]{k} = 4-1 = 3$

One thing though, in the last line should that be a '256' in the superscript for the 2nd term from the left?

davidgoes4wce · Feb 13, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Actually I don't get how the terms on the right are different from the left...................

Paradoxica · Feb 13, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

davidgoes4wce said:
One thing though, in the last line should that be a '256' in the superscript for the 2nd term from the left?

Yes. Shifting the variable across by one allows us to directly compare the terms, but shifting the index also changes the limits of summation, as indicated by 256. And besides, it's supposed to have a nice answer, is it not? The perfect 8th power isn't there for no reason.

davidgoes4wce · Feb 13, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Paradoxica said:
Yes. Shifting the variable across by one allows us to directly compare the terms, but shifting the index also changes the limits of summation, as indicated by 256. And besides, it's supposed to have a nice answer, is it not? The perfect 8th power isn't there for no reason.

Thanks I get it now.

drsabz101 · Feb 13, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

I need help differentiating 1,2,3 and then for 4 I am not quite sure about parts b,c,d,e,f,g,h. Thankyou

davidgoes4wce · Feb 13, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Is the first question supposed to be y=x^6+x^5+x^4+x^3+x^2+x+1?

drsabz101 · Feb 13, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

This s q4.
Part 1 of question 4:

Part 2:

Part 3:

Make a Difference – Donate to Bored of Studies!

Cambridge Prelim MX1 Textbook Marathon/Q&A (1 Viewer)

Member

Well-Known Member

Member

Well-Known Member

Member

Well-Known Member

Member

Active Member

Member

Well-Known Member

Well-Known Member

-insert title here-

Well-Known Member

Well-Known Member

Well-Known Member

-insert title here-

Well-Known Member

Member

Attachments

Well-Known Member

Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)