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

drsabz101 · Feb 13, 2016

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

Thanks heaps ! I'll ask my teacher or a friend about h, if no one knows

davidgoes4wce · Feb 13, 2016

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

appleibeats said:
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 $= DS^{\text{-1}}.$

18b and 18c remain unanswered on this thread. Im getting stuck on these two parts as well.

drsabz101 · Feb 13, 2016

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

Help with these 2 questions:

jathu123 · Feb 13, 2016

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

http://i67.tinypic.com/2wntk0n.jpg

Soz its sideways, 16h. Basically you find the the gradient of the line l which is 12, then use the point gradient forumla to find the equation (sub in the origin points). Since B is directly above H, the x coordinate of B will be 6. So sub in 6 into your line formula (y=12x) and you'll get 72. Now by inspection, you can conclude that BH is 36. Since the gradient of the line is 12, BH/AH=12
36/AH=12
Therefore AH=3

Edit: try right clicking on the link and saving it/downloading it, if you want a straight pic. The website makes it landscape for some reason haha

drsabz101 · Feb 13, 2016

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

Soz its sideways, 16h. Basically you find the the gradient of the line l which is 12, then use the point gradient forumla to find the equation (sub in the origin points). Since B is directly above H, the x coordinate of B will be 6. So sub in 6 into your line formula (y=12x) and you'll get 72. Now by inspection, you can conclude that BH is 36. Since the gradient of the line is 12, BH/AH=12
36/AH=12
Therefore AH=3

Thankyou

Paradoxica · Feb 13, 2016

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

davidgoes4wce said:
18b and 18c remain unanswered on this thread. Im getting stuck on these two parts as well.

$\noindent$ v=2^x, \frac{1}{v} = 2^{-x},\,x>1 \\$ This condition is only to make sure $v$ and $\frac{1}{v}$ are distinct.$ \\ S = \frac{1}{2} \left(v+\frac{1}{v}\right),\,D = \frac{1}{2} \left(v-\frac{1}{v}\right)\\v^2-2Sv+1=0,\,v^2-2Dv-1=0 \\ $Subtract the equations to solve simultaneously.$ \\ 2Sv-2Dv-2=0 \Leftrightarrow v = \frac{1}{S-D} = S+D\\x = \log_2{v} = \log_2(S+D)=-\log_2(S-D) \\ x+x = \log_2(S+D)-\log_2(S-D)=\log_2\left(\frac{S+D}{S-D}\right)\\$The rest is obvious.$

drsabz101 · Feb 13, 2016

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

Anyone know how to do that 2 questions I posted previously. I am honestly stuck

InteGrand · Feb 13, 2016

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

drsabz101 said:
Anyone know how to do that 2 questions I posted previously. I am honestly stuck

It's basically just standard calculus Q's about tangents. Like for the second one, differentiate the function to find the derivative function. This tells us the slope of the tangent at a point x. To find the tangent to the curve at a given point, we need to find its slope (by subbing in the x-value to the derivative function) and then use point-gradient form. To show that a given line is tangent at a given point, just show that the line passes through that point and that the function's derivative at that point has the same slope as the given line.

drsabz101 · Feb 13, 2016

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

Okay I'll do it now

drsabz101 · Feb 13, 2016

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

I got the second question

davidgoes4wce · Feb 13, 2016

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

drsabz101 said:
Help with these 2 questions:
View attachment 32904

Here is the graph visually:

davidgoes4wce · Feb 13, 2016

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

drsabz101 said:
Help with these 2 questions:
View attachment 32904

The 2nd part of that question:

Again here is the graph to support that answer:

davidgoes4wce · Feb 14, 2016

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

Paradoxica said:
$\noindent$ v=2^x, \frac{1}{v} = 2^{-x},\,x>1 \\$ This condition is only to make sure $v$ and $\frac{1}{v}$ are distinct.$ \\ S = \frac{1}{2} \left(v+\frac{1}{v}\right),\,D = \frac{1}{2} \left(v-\frac{1}{v}\right)\\v^2-2Sv+1=0,\,v^2-2Dv-1=0 \\ $Subtract the equations to solve simultaneously.$ \\ 2Sv-2Dv-2=0 \Leftrightarrow v = \frac{1}{S-D} = S+D\\x = \log_2{v} = \log_2(S+D)=-\log_2(S-D) \\ x+x = \log_2(S+D)-\log_2(S-D)=\log_2\left(\frac{S+D}{S-D}\right)\\$The rest is obvious.$

I'm also not 100% sure how you went from $\log_{2} (S+D)=-\log_{2} (S-D)$

$$My thinking is if we wrote$ \ \frac{1}{S-D} \ like \ (S-D)^{-1}$
$we would be able to write in terms of log because the indices is to a power of -1$

InteGrand · Feb 14, 2016

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

davidgoes4wce said:
I'm also not 100% sure how you went from $\log_{2} (S+D)=-\log_{2} (S-D)$

$$My thinking is if we wrote$ \ \frac{1}{S-D} \ like \ (S-D)^{-1}$
$we would be able to write in terms of log because the indices is to a power of -1$

$\noindent He used the fact that $S+D = \frac{1}{S-D}$, which is true because $(S+D)(S-D) = 1$ (this can be proved by using the definition of $S$ and $D$ in terms of $v$).$

davidgoes4wce · Feb 14, 2016

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

InteGrand said:
$\noindent He used the fact that $S+D = \frac{1}{S-D}$, which is true because $(S+D)(S-D) = 1$ (this can be proved by using the definition of $S$ and $D$ in terms of $v$).$

The answer seems to be a bit off from the text book:

$x=log_{2} (S+\sqrt{S^2-1})$

and

$x=log_{2} (D+\sqrt{D^2+1})$

Paradoxica · Feb 14, 2016

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

davidgoes4wce said:
The answer seems to be a bit off from the text book:

$x=log_{2} (S+\sqrt{S^2-1})$
and

$x=log_{2} (D+\sqrt{D^2+1})$

What Integrand said is sufficient to find the textbook answer.

appleibeats · Feb 14, 2016

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

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

Unsure how to answer part c). Do I use the base triangle?

davidgoes4wce · Feb 14, 2016

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

appleibeats said:
View attachment 32906

Unsure how to answer part c). Do I use the base triangle?

$My thinking with this question is to use the sine Rule in order to get$ $\angle PCD$

$Given that we know the length of PC=h,$ $PD=\frac{h}{\sqrt{3}}$ , $CD^2=\frac{h^2}{3}(4-\sqrt{6})$ , $\angle DPC=45\degree$

I keep getting an angle of $34\degree 36'$ doing a number of different ways including the Cosine rule.

I then subtract that answer from 360, which leaves me with 325 degrees. Just wondering am I on the right track? Is there a mistake in the answers ? I've seen a mistake from one of the Sadler, Pender, Shea and Ward textbook , so just want to confirm with others on here?

jathu123 · Feb 14, 2016

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

you got the correct angle, 34.6 degrees, but you don't subtract it from 360. If you look closely you actually do 34.6 + 180 + 90 = 305 (to the nearest degree)

http://imgur.com/oM1L49p

davidgoes4wce · Feb 14, 2016

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

jathu123 said:
you got the correct angle, 34.6 degrees, but you don't subtract it from 360. If you look closely you actually do 34.6 + 180 + 90 = 305 (to the nearest degree)

http://imgur.com/oM1L49p

Thank you, I can see it now.

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