Cambridge Prelim MX1 Textbook Marathon/Q&A (2 Viewers)

appleibeats · Apr 4, 2015

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

The answer to that question in the book says:

y / y + 1

I understand your answer, but is the answer in the book right??

appleibeats · Apr 4, 2015

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

Does anyone know how to answer question 21 from 6A:

a) if x = 2 ^ 1/3 + 4 ^ 1/3, show that x^3 = 6(1+x)

b) if x = 1/2 + 1/2root 5, show that x^2 + x^-2 / x - x^-1 = 3

c) show that pq^-1 - p^-1q / p^2q^-2 - p^-2q^2 = pq/p^2 + q^2

Thanks for your help.

appleibeats · Apr 4, 2015

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

Does anyone know how to answer question 21 from 6A:

a) if x = 2 ^ 1/3 + 4 ^ 1/3, show that x^3 = 6(1+x)

b) if x = 1/2 + 1/2root 5, show that x^2 + x^-2 / x - x^-1 = 3

c) show that pq^-1 - p^-1q / p^2q^-2 - p^-2q^2 = pq/p^2 + q^2

Thanks for your help.

Ambility · Apr 4, 2015

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

appleibeats said:
The answer to that question in the book says:

y / y + 1

I understand your answer, but is the answer in the book right??

Oh crap, I didn't simplify fully

that's awkward. I'll begin where I left off.

$\frac{y^2-y}{y^2-1}\\\text{Now, we factorise}\\=\frac{y(y-1)}{(y+1)(y-1)}\\\text{Cancel the }(y-1)\\=\frac{y}{y+1}$

WrittenLoveLetters · Apr 5, 2015

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

enigma_1 said:
it's a terrible book. Use Cambridge instead. Make Maths in focus extinct from your lives pls

ty

I do MIF for class work - even my teacher says its shit, so she gives us hardcore past papers, work from selective schools and uses Cambridge.

appleibeats · Apr 5, 2015

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

Thanks Ambility for answer that question.
Also wondering if you could help me with Q 5 from 6G:

The question is:

1/ (k^1/2 + (k + 1 ) ^1/2 (k^1/4 +( k +1 )^ 1/4 = ( k + 1) ^ 1/4 - k^1/4

I know its hard to read, best to look at the textbook. Also wondering how do you write your maths on BOS, so I don't have to write it like I do above.

Thanks.

Ambility · Apr 6, 2015

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

appleibeats said:
Thanks Ambility for answer that question.
Also wondering if you could help me with Q 5 from 6G:

The question is:

1/ (k^1/2 + (k + 1 ) ^1/2 (k^1/4 +( k +1 )^ 1/4 = ( k + 1) ^ 1/4 - k^1/4

I know its hard to read, best to look at the textbook. Also wondering how do you write your maths on BOS, so I don't have to write it like I do above.

Thanks.

Honestly, I've spent 40 minutes on this problem and I can't get at it. Either it's something simple that I've been overlooking, or it requires some method I haven't yet learnt.

To answer you question on writing maths, I use a thing called LaTeX. It allows you to type in a code between ["tex"] tags (without the quotes), and it will format it in mathematics. If you want to learn how to use it, I recommend looking at this guide and looking at how some people set their LaTeX out. If you click the "Reply with Quote" button on the bottom left of people's comments, you should be able to see their LaTeX.

Anyway, if anyone wants to work out his question, this is it:

$\frac{1}{(k^{\frac{1}{2}}+(k+1)^{\frac{1}{2}})(k^{\frac{1}{4}}+(k+1)^{\frac{1}{4}})}$

SpiralFlex · Apr 6, 2015

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

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

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

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

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

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

[HR][/HR]

$\textup{Now I have,}$

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

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

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

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

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

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

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

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

Ambility · Apr 6, 2015

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]

$\textup{Now I have,}$

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

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

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

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

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

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

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

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

How silly of me to overlook this.

SpiralFlex · Apr 6, 2015

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

^Not silly at all, we all overlook things.

yog · Apr 7, 2015

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

How would you rationalise the denominator of:

1/(cuberoot(2)-1)

SpiralFlex · Apr 7, 2015

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

yog said:
How would you rationalise the denominator of:

1/(cuberoot(2)-1)

$\frac{1}{\sqrt[3]{2} -1 }$

$\textup{Let's see if we can do it using our difference between two squares technique,}$

$\frac{1}{\sqrt[3]{2} -1 } *\frac{\sqrt[3]{2} + 1 }{\sqrt[3]{2} +1 }$

$\frac{\sqrt[3]{2} + 1 }{(\sqrt[3]{2} -1)(\sqrt[3]{2} +1) }$

$\frac{\sqrt[3]{2} + 1 }{{\color{Red} (\sqrt[3]{2})^2} - 1^2 }$

$\textup{We have a}\; {\color{Red} (\sqrt[3]{2})^2= 2^\frac{2}{3}}\; \textup{on the denominator, this doesn't work.}$

$\textup{So we must adopt another technique.}$

$\textup{Note that} \;(a-b)(a^2+ab+b^2) = a^3-b^3$

$\textup{So, if I multiple the top and bottom by}\; ((\sqrt[3]{2})^2 +\sqrt[3]{2}+1^2)\; \textup{it should get rid of the cube root.}$

$\frac{1}{\sqrt[3]{2} -1 }*\frac{((\sqrt[3]{2})^2 +\sqrt[3]{2}+1)}{((\sqrt[3]{2})^2 +\sqrt[3]{2}+1)}$

$=\frac{((\sqrt[3]{2})^2 +\sqrt[3]{2}+1)}{(\sqrt[3]{2})^3 -(1)^3 }$

$=(\sqrt[3]{2})^2 +\sqrt[3]{2}+1$

appleibeats · Apr 8, 2015

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

Question 18 a and c from 6A:

Simplify:

a) $\frac{6^n - 3^n}{2^(n+1) + 2}$

the denominator is meant to be 2^n+1 +2, can't get it to work for some reason.

answer is : $\frac{3^n}{2}$

c) $\frac{12^n - 18^n}{3^n - 2^n}$

answer is: -2^n3^n

appleibeats · Apr 8, 2015

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

Also Question 20F from 6A:

if a = 2^ 1/2 + 2^-1/2 and b = 2^1/2 - 2 ^-1/2

find:

a^3 + b^3

answer is 7 root 2

Also 21a)

If x = 2^1/3 + 4^ 1/3, show that x^3 = 6(1 + x )

InteGrand · Apr 8, 2015

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

appleibeats said:
Question 18 a and c from 6A:

Simplify:

a) $\frac{6^n - 3^n}{2^(n+1) + 2}$

the denominator is meant to be 2^n+1 +2, can't get it to work for some reason.

answer is : $\frac{3^n}{2}$

c) $\frac{12^n - 18^n}{3^n - 2^n}$

answer is: -2^n3^n

To write $2^{n+1}$ , the LaTeX code needs to be written like this: 2^{n+1} (put curly braces around the thing you want in the exponent).

Part a)

Based on what you say the answer is, I think you made a typo in the numerator, it should be $6^n + 3^n$ instead (OR, the denominator should be $2^{n+1} - 2$ . I'll assume the numerator is $6^n + 3^n$ .

Then the numerator can be written as:

$(2\times 3)^n + 3^n$ (as $6=2\times 3$ )

$=2^n \times 3^n + 3^n$ (using the index law $(ab)^n = a^n \times b^n$ )

$=3^n (2^n +1)$ (factorising).

The denominator $2^{n+1}+2$ can be written as:

$2\times 2^{n}+2$ (as $2^{n+1}=2^1 \times 2^n = 2\times 2^n$ , using the index law $a^{m+n}=a^m \times a^n$ )

$=2(2^n + 1)$ (factorising).

So the original fraction is $\frac{6^n + 3^n}{2^{n+1}+2}=\frac{3^n (2^n +1)}{2(2^n + 1)}=\frac{3^n}{2}$ (cancelling $2^n +1$ ).

Part c)

The idea is to rewrite the numerator in terms of $2^n$ and $3^n$ using index laws, and then hopefully we will be able to cancel something with the denominator (notice that 12 and 18 can both be written in terms of powers of 2 and 3).

The numerator $12^n - 18^n$ can be written as:

$\left(2^2 \times 3\right)^n - \left(2\times 3^2\right)^n$ (as $12=2^2 \times 3$ and $18 = 2\times 3^2$ )

$= \left(2^2 \right)^n \times 3^n - 2^n \times \left(3^2 \right)^n$ (using the index law $(ab)^n = a^n \times b^n$ )

$=2^{2n} \times 3^n - 2^n \times 3^{2n}$ (using the index law $\left(a^b \right)^n = a^{bn}$ )

$= 2^n 3^n (2^n - 3^n)$ (factorising by taking out a common factor of $2^n 3^n$ )

$=-2^n 3^n (3^n - 2^n)$ (as $2^n - 3^n = -(3^n - 2^n)$ ).

So the original fraction becomes $\frac{-2^n 3^n (3^n - 2^n)}{3^n - 2^n}=-2^n 3^n\text{ }(=-6^n)$ .

Feynman · Apr 8, 2015

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

appleibeats said:
Also Question 20F from 6A:

if a = 2^ 1/2 + 2^-1/2 and b = 2^1/2 - 2 ^-1/2

find:

a^3 + b^3

answer is 7 root 2

Also 21a)

If x = 2^1/3 + 4^ 1/3, show that x^3 = 6(1 + x )

First question:

Find the values of a, b, a squared, b squared and ab from previous parts of the question. Take the positive square roots of a squared and b squared to find a and b (because the question gives us positive values for a and b).

Factor to make calculating easier (sum of two cubes):

$a^3+b^3 = (a+b)(a^2-ab+b^2)$

Substitute the values:

$(\frac{3\sqrt{2}}{2} + \frac{\sqrt{2}}{2})(\frac{9}{2}-\frac{3}{2}+\frac{1}{2})$

which becomes

$(\frac{4\sqrt{2}}{2})(\frac{7}{2})$

Multiply:

$\frac{28\sqrt{2}}{4}$

$= 7\sqrt{2}$

Second question:

$x = 2^\frac{1}{3} + 4^\frac{1}{3}$

$Factor 4 to its primes, 2 times 2, or 2 squared. Then because we raise a power to a power, multiply the indices (this will help later):$

$x = 2^\frac{1}{3} + 2^\frac{2}{3}$

$$ Now $ x^3 = (2^\frac{1}{3} + 2^\frac{2}{3})^3$

$Use the fact that $ (A+B)^3 = A^3 + 3A^2B + 3AB^2 + B^3 $ (cube of a sum)$

$(2^\frac{1}{3} + 2^\frac{2}{3})^3 $ becomes $ 2 + (3*2^\frac{4}{3}) + (3*2^\frac{5}{3}) + 4$

$= 6 + (3*2^\frac{4}{3}) + (3*2^\frac{5}{3})$

$Change 6 to its prime factors:$

$(3*2) + (3*2^\frac{4}{3}) + (3*2^\frac{5}{3})$

$We can take out a common factor of 3^1:$

$3(2+2^\frac{4}{3} + 2^\frac{5}{3}) $ and we can take out a common factor of 2 to the power of 1 $$

$3*2(1+2^\frac{1}{3}+2^\frac{2}{3})$

$= 6(1+2^\frac{1}{3}+2^\frac{2}{3})$

$Which can be changed back to the original x:$

$6(1+2^\frac{1}{3} + 4^\frac{1}{3}) $ which is equal to 6(1+x) as x = 2^\frac{1}{3} + 4^\frac{1}{3}.$

appleibeats · Apr 9, 2015

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

Question 18b and c from 6B:

$Let S = \frac{1}{2}({2^x + 2^{-x}}) and D = \frac{1}{2}({2^x - 2^{-x}})$

a) ALREADY ACHIEVED

Simplify SD, S + D, S - D, S^2 - D^2

b) Rewrite the formulae for S and D as quadratic equations in $2^{\text{x}}$ . Hence express x in terms of S, and in terms of D, in the case where x > 1

c) $$Show that x$ = \frac{1}{2} log_{\text{2}}\frac{1 + y }{1 - y} ,$ where y $= DS^{\text{-1}}.$

Feynman · Apr 28, 2015

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

dead thread?

Ambility · Apr 29, 2015

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

Feynman said:
dead thread?

I guess no one has any more questions...

shexz · May 11, 2015

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

appleibeats said:
Hi, not sure how to go about question 16 from chapter 4J. The question is

the side of a triangle are n2 + n + 1 , 2n + 1 and n2 - 1, where n>1. Find the largest angle of the triangle.

( n2 being --> n squared )

Thanks.

use cosine rule

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