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

#### bookss12

##### New Member

Hi ! could someone lease help with these questions:

9
10c
11a

thanks heaps!

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#### iso1234

##### New Member

Does anyone have the answer (and preferably the logic) to 7B Q15? Thanks

#### Raiza_98

##### New Member

Hi I was just wondering where would the best and most useful content be within the book?? Like the 1st excercises are too much a waste of time, which chapters cover the most worth-wile course content??

#### aroon

##### New Member

Hi I was just wondering where would the best and most useful content be within the book?? Like the 1st excercises are too much a waste of time, which chapters cover the most worth-wile course content??
Do the series and sequences (you don't really need a teacher to understand it), try the calculus (chapter 7) and the trig in chapter 4

#### vickyera

##### New Member

Has anyone done 8I Q7
a) (p+q+x)(pq+px+qx)-pqx=(p+q)(p+x)(q+x) and the rest b,c,d
do they want me to expand everything or is there an easier way?
thankyou

#### davidgoes4wce

##### Well-Known Member

EX 9A Q 12, I've seen the first stage in writing the answer as :

$\sqrt{x^2+y^2}+y=2$

Part of the reason I've seen is that 'P must above the x-axis' $\implies |y|=y$

I'm not sure why they add the 'y' to it still? It has got me thinking for the last couple of hours. Is there a better way to understand this visually?

#### InteGrand

##### Well-Known Member

EX 9A Q 12, I've seen the first stage in writing the answer as :

$\sqrt{x^2+y^2}+y=2$

Part of the reason I've seen is that 'P must above the x-axis' $\implies |y|=y$

I'm not sure why they add the 'y' to it still? It has got me thinking for the last couple of hours. Is there a better way to understand this visually?
What is the question (don't have a copy of the textbook nearby)?

#### boredofstudiesuser1

##### Active Member

What is the question (don't have a copy of the textbook nearby)?
I think it might be this one:

Find the locus of a point P(x, y) which lies above the x-axis so that the sum of its distances to the origin and the x-axis is 2.

#### InteGrand

##### Well-Known Member

I think it might be this one:

Find the locus of a point P(x, y) which lies above the x-axis so that the sum of its distances to the origin and the x-axis is 2.
$\noindent In that case, the distance of P(x,y) to the origin is \sqrt{x^{2} + y^{2}} and the distance to the x-axis is y (since P is to lie above the x-axis). We want \emph{the sum} of these to be 2, so \sqrt{x^{2} + y^{2}}+y = 2.$

#### a_t

##### New Member

Need help with question 3(allparts) of 10E please

#### Drongoski

##### Well-Known Member

Need help with question 3(allparts) of 10E please
Should try to include the actual question. Not everyone willing to help you has the book. I have the older 3U Yr 11 - the Chapters and exercise numberings may have changed.

#### Pakka

##### New Member
Help with Question 11 Exercise 2J
For those without the book: Determine how the curve y = x^3 - x must be transformed in order to obtain the graph of y = x^3 - 3x. {hint: only stretchings are involved}
Full worked solutions would appreciated.
Cheers

#### InteGrand

##### Well-Known Member
Help with Question 11 Exercise 2J
For those without the book: Determine how the curve y = x^3 - x must be transformed in order to obtain the graph of y = x^3 - 3x. {hint: only stretchings are involved}
Full worked solutions would appreciated.
Cheers
$\noindent Stretch the original graph vertically by a factor of b and horizontally by a factor of a^{-1}, i.e. the graph becomes y = b\left((ax)^{3} -ax \right) = ba^{3}x^{3} -ba x. Comparing this to x^{3} -3x, we want ba^{3} = 1 and ba = 3. Dividing these equations shows a^{2} = \frac{1}{3}\Rightarrow a = \frac{1}{\sqrt{3}}. Then ba = 3 \Rightarrow b = \frac{3}{a} = 3\sqrt{3}. (We can check that these satisfy those two equations.) So the answer is to stretch the original graph vertically by a factor of b = 3\sqrt{3} and horizontally by a factor of a^{-1} = \sqrt{3}.$

Thank you

#### Thatstudentm9

##### Member
guys I'm in need of some help with [

]simplify
a)1+tan^2(90degrees- θ )
prove
b)cot θ cos θ/cot θ +cos θ=cos θ /1+sin θ

#### Drongoski

##### Well-Known Member
$1 + tan^2 (90^{\circ}-\theta) = 1 + cot^2 \theta = cosec^2 \theta \\ \\ \\ \frac {cot\theta cos \theta}{cot \theta + cos \theta} = \frac {\frac {cos \theta}{sin \theta} \cdot cos \theta}{\frac {cos \theta}{sin \theta} + cos \theta} = \frac {cos \theta \cdot cos \theta}{cos \theta + cos \theta \cdot sin \theta} = \frac { cos \theta \cdot cos \theta}{cos \theta (1 + sin \theta)} = \frac {cos \theta}{1+ sin \theta}$

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#### Drongoski

##### Well-Known Member
guys I'm in need of some help with [

]simplify
a)1+tan^2(90degrees- θ )
prove
b)cot θ cos θ/(cot θ +cos θ)=cos θ /(1+sin θ)
Should be amended as shown; otherwise it shows you don't know your algebra or you are being careless and/or sloppy.

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#### Mahan1

##### Member

$\textrm{Base case} , n=1, \textrm{then} \ \varnothing \ \textrm{and the whole set are the only subsets and the statement follows.}$

$\textrm{Suppose it is true for} \ n=k: \ \textrm{the set A with k elements has } 2^{k} \ \textrm{subsets}$

$\textrm{where } A = \{a_{1},a_{2},\dots, a_{k} \}$

$\textrm{add one element, } a_{k+1} \ \textrm{to A}$

$\textrm{let's call the new set B } = A \cup \{a_{k+1} \} \{a_{1}, \dots, a_{k+1} \}$

$\textrm{We note that all the subsets of B either contain } a_{k+1} \ \textrm{or not}$

$\textrm{number of subsets do not contain }a_{k+1} \ \ texture{is equal to the number of subsets of A} = 2^{k}$

$\textrm{number of subsets containing }a_{k+1} \ \textrm{if we add }a_{k+1} \ \textrm{to each subset of A} \ \textrm{we obtain}$

$\textrm{all the subsets containing } a_{k+1} \ \textrm{which is equal to } 2^{k}$

$\textrm{since the subsets of B either contain }a_{k+1} \textrm{or not then the total number of subsets of B is }$

$2^{k} + 2^{k} = 2^{k+1}$