# Thread: Cambridge Prelim MX1 Textbook Marathon/Q&A

1. ## 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.

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

Originally Posted by appleibeats
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:

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

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

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

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

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

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

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

$\textup{Now I have,}$

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

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

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

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

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

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

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

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

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

Originally Posted by SpiralFlex
$\frac{1}{(\sqrt{k}+\sqrt{k+1})(\sqrt[4]{k} + \sqrt[4]{k+1})}$

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

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

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

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

$\textup{Now I have,}$

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

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

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

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

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

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

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

$=\sqrt[4]{k+1} - \sqrt[4]{k}$
How silly of me to overlook this.

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

^Not silly at all, we all overlook things.

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

How would you rationalise the denominator of:

1/(cuberoot(2)-1)

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

Originally Posted by yog
How would you rationalise the denominator of:

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

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

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

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

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

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

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

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

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

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

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

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

8. ## 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) $\frac{12^n - 18^n}{3^n - 2^n}$

9. ## 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

Also 21a)

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

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

Originally Posted by appleibeats
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) $\frac{12^n - 18^n}{3^n - 2^n}$

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 $\frac{-2^n 3^n (3^n - 2^n)}{3^n - 2^n}=-2^n 3^n\text{ }(=-6^n)$.

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

Originally Posted by appleibeats
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

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:

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

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

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

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

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

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

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

$Change 6 to its prime factors:$

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

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

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

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

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

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

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

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

Question 18b and c from 6B:

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

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

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

Originally Posted by Feynman
I guess no one has any more questions...

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

Originally Posted by appleibeats
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

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

Q 10 from 7H:

Sand being poured from a conveyor belt forms a cone with heigh h and semivertical angle 60 degrees. Show that the volume of the pile is V = pih^3 ( can do this), and differentiate with respect to t. (can do this)

a) Suppose that the sand is being poured at a constant rate of 0.3 m^3 /min , and let A be the area of the base. Find the rate at which the height is increasing:

i) when the height is 4 metres, ( can do this), ii) when the radius is 4 metres ( CANT DO THIS)

b) (CANT DO THIS) Show that dA/dt = 6pih dh/dt, and find the rate of increase o the base area at these times.

c) (CANT DO THIS) At what rate mist the sand be poured if it is required that the height increase at 8 cm/min, when the height is 4m?

ALSO

Q11)

An upturned cone of semivertical angle 45 degrees is being filled with water at a constnat rate of 20cm^3 /s. Find the rate at which the height, the area of the water surface, and the area of the cone wetted by the water, are increasing when the height is 50 cm.

Thanks.

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

Hi, worked out the previous question.

Have another one though.

Question 14 from 7I.

Find zeroes and discontinuities of:

a) y = cosx + sinx / cosx - sinx

b) y = cosx - sinx / cosx + sinx

Thanks

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

hey, does anyone know how to do question 17 and 20 from 2G?

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

Originally Posted by appleibeats
Hi, worked out the previous question.

Have another one though.

Question 14 from 7I.

Find zeroes and discontinuities of:

a) y = cosx + sinx / cosx - sinx

b) y = cosx - sinx / cosx + sinx

Thanks
Seeing as the problem doesn't give a domain, and the functions go on forever, we need to represent the answer as a general solution because there is going to be infinitely many possible solutions. These fractions will have zeros when the numerator is equal to zero ("zero divided by anything is zero") and will have discontinuities when the denominator is equal to zero ("dividing by zero is undefined"). With that in mind, let's work on 14a:

$\text{Where is the numerator equal to zero?}\\\cos x+\sin x=0\\\text{Divide by }\cos x\\\tan x=-1\\x=135^\circ, 315^\circ, 495^\circ, 675^\circ...\\x=135^\circ+180n^\circ, \text{where }n\in \mathbb{Z}\\\\\text{Where is the denominator equal to zero?}\\\cos x-\sin x=0\\\text{Divide by }\cos x\\\tan x-1=0\\\tan x=1\\x=45^\circ, 225^\circ, 405^\circ, 585^\circ...\\x=45^\circ+180n^\circ, \text{where }n\in \mathbb{Z}$

You can use the answers from 14a. to work out 14b. 14b's denominator is the same as 14a's numerator, and 14b's numerator is the same as 14a's denominator.

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

Thanks for that help.

Also I don't quite understand how to get the answers to the questions in 4 in 7J.

I know how to sketch them and their continuity, but am unsure about how to find where it is not differentiable.

eg, question 4a)

y = |x + 2 |

Also, question 7d from chapter 8A.

write down the general form of a monic quadratic for which one of the zeroes is x = 1. (I know how to do this.) Then find the equation of such a quadratic in which :

d) the curve passes through (3,9).

Thank you.

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

Originally Posted by appleibeats
Thanks for that help.

Also I don't quite understand how to get the answers to the questions in 4 in 7J.

I know how to sketch them and their continuity, but am unsure about how to find where it is not differentiable.

eg, question 4a)

y = |x + 2 |

Also, question 7d from chapter 8A.

write down the general form of a monic quadratic for which one of the zeroes is x = 1. (I know how to do this.) Then find the equation of such a quadratic in which :

d) the curve passes through (3,9).

Thank you.
I have a method to solve 8A 7d, but it's hardly ideal. I haven't studied that chapter so I recommend asking a teacher.

As for 7J 4a, the graph is not differentiable at a point where there is a sharp turn. This is because there are multiple tangent lines which can be drawn to the graph, multiple slopes of those tangent lines, multiples derivatives, no defined derivative. If you look at the graph of $y=|x+2|$, there is a point which takes a sudden turn at x=-2. So it's not differentiable at x=-2.

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

Originally Posted by appleibeats
Thanks for that help.

Also I don't quite understand how to get the answers to the questions in 4 in 7J.

I know how to sketch them and their continuity, but am unsure about how to find where it is not differentiable.

eg, question 4a)

y = |x + 2 |

Also, question 7d from chapter 8A.

write down the general form of a monic quadratic for which one of the zeroes is x = 1. (I know how to do this.) Then find the equation of such a quadratic in which :

d) the curve passes through (3,9).

Thank you.
Q 7d) from Chapter 8A:

$The general form of the monic quadratic is Q(x) = A(x-1)(x-a). If we want such a quadratic to go through (3,9), since we just need a particular quadratic, we can just let a be some convenient value like 0 (so Q(x) = Ax(x-1)), and then we solve for A by substituting in x=3 and Q(3)=9: 9 = A\times 3 \times (3-1) \Rightarrow 9 = 6A\Rightarrow A = \frac{3}{2}. So a suitable quadratic is Q(x)=\frac{3}{2}x(x-1).$

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

Originally Posted by InteGrand
Q 7d) from Chapter 8A:

$The general form of the monic quadratic is Q(x) = A(x-1)(x-a). If we want such a quadratic to go through (3,9), since we just need a particular quadratic, we can just let a be some convenient value like 0 (so Q(x) = Ax(x-1)), and then we solve for A by substituting in x=3 and Q(3)=9: 9 = A\times 3 \times (3-1) \Rightarrow 9 = 6A\Rightarrow A = \frac{3}{2}. So a suitable quadratic is Q(x)=\frac{3}{2}x(x-1).$
There does exist other suitable quadratics though, right? I came up with $y=-\frac{9}{4}(x-5)(x-1)$.

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

The answer in the text book for that question is y = 1/2 (2x + 3)(x - 1)

Also does anyone know how to do Question 23 from 8 F.

A piece of string of length l is bent to form the sector of a circle of radius r. Show that the area of the sector is maximised when r = 1/4l .

Also Question 27 from the same chapter.

A rectangle is inscribed in an isosceles triangle with one of the sides of the rectangle on the base of the triangle. Prove that the rectangle of greatest area occupies half the area of the triangle.

Thanks.

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

Originally Posted by Ambility
There does exist other suitable quadratics though, right? I came up with $y=-\frac{9}{4}(x-5)(x-1)$.
Yep there are infinitely many possible quadratics we could use.

The general form was $Q(x) = A(x-a)(x-1), A \neq 0$. In order for the point (3, 9) to lie on the curve, we just need to choose a and A (with $A\neq 0$) satisfying $9=A(3-a)(3-1)\Longleftrightarrow A = \frac{\frac{9}{2}}{3-a}, a\neq 3$, and there are clearly an infinite number of pairs $(a,A)$ satisfying this condition.

Page 2 of 61 First 12341252 ... Last

There are currently 1 users browsing this thread. (0 members and 1 guests)

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•