Cambridge Prelim MX1 Textbook Marathon/Q&A (13 Viewers)

InteGrand · Mar 5, 2016

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

si2136 said:
View attachment 32979

Is this correct?

$\noindent The answer is $y = |x-2| -1$. To write it as a piecewise function, it'd be $y = \begin{cases} -x+1, \quad x\leq 2 \\ x -3, \quad x>2 \end{cases} .$

ml125 · Mar 5, 2016

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

InteGrand said:
$\noindent The answer is $y = |x-2| -1$. To write it as a piecewise function, it'd be $y = \begin{cases} -x+1, \quad x\leq 2 \\ x -3, \quad x>2 \end{cases} .$

Ohh.. right, neglected the definitions of the x. For the -x+1 part, why would that be x≤2 rather than x<2? I mean, looking at the second part.

InteGrand · Mar 5, 2016

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

ml125 said:
Ohh.. right, neglected the definitions of the x. For the -x+1 part, why would that be x≤2 rather than x<2? I mean, looking at the second part.

$\noindent We can put the equality part on either branch and it'll still be correct, since if we sub. in $x =2$ into either branch's equation, we get the correct $y$-value of $-1$.$

ml125 · Mar 5, 2016

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

InteGrand said:
$\noindent We can put the equality part on either branch and it'll still be correct, since if we sub. in $x =2$ into either branch's equation, we get the correct $y$-value of $-1$.$

Oh yeah just realised haha thanks

drsabz101 · Mar 5, 2016

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

how do u solve for a and b?

Paradoxica · Mar 5, 2016

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

drsabz101 said:
how do u solve for a and b?
View attachment 32981

$\noindent From the question, we know $\frac{\text{d}x}{\text{d}t}=6\,\text{cm/s}\\\therefore \frac{\text{d}A}{\text{d}t}=12x\,\text{cm}^2\text{/s}\,,\,\,\frac{\text{d}l}{\text{d}t}=6 \sqrt{2}x \, \text{cm/s}\\$In the first part, $x=70\, \text{cm}$ so the rates are straightforward to find. In the second part, $x=100\, \text{cm}$ since lengths are positive and the figure is a square. The rest is straightforward.$

drsabz101 · Mar 5, 2016

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

thankyou. I got it

drsabz101 · Mar 5, 2016

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

this q

drsabz101 · Mar 5, 2016

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

I just realised I have been using the SA for a cylinder...

drsabz101 · Mar 5, 2016

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

and then i used the SA for a sphere and finally got it

Carrotsticks · Mar 5, 2016

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

http://i0.kym-cdn.com/photos/images/original/000/002/116/Longcat_War.jpg

davidgoes4wce · Mar 7, 2016

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

drsabz101 said:
this q View attachment 32982

Im not sure but in Exercise 7H Q7 (b) I get a different answer to the book

Could someone confirm the workings to the question are correct?

Drongoski · Mar 7, 2016

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

I got the book's answer: $\frac {32000 \pi}{3} cm^3$ but the working out is a bit longish. Maybe there is a much shorter solution.

Paradoxica · Mar 7, 2016

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

Drongoski said:
I got the book's answer: $\frac {32000 \pi}{3} cm^3$ but the working out is a bit longish. Maybe there is a much shorter solution.

$\noindent It can be shown that the radius increases at a rate of $\frac{1}{8\pi r}\,\text{cm/s}\\$Taking the derivative of the volume with respect to time, we have $\\ \frac{\text{d}V}{\text{d}t} = 4\pi r^2 \frac{\text{d}r}{\text{d}t} \Rightarrow 10 \text{cm}^3\text{/s} = \frac{4\pi r^2}{8\pi r}\text{cm}^3\text{/s} \Rightarrow r = 20$cm$ \\ $The $r^2$ term introduces the extra square units we need in order to make the rate of change cubic units.$

davidgoes4wce · Mar 7, 2016

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

Don't want to sound like an after-timer but I eventually did manage to get r=20 cm as well in the end. ( I had a big weekend on Saturday and Sunday. e.g Mardi Gras)

Paradoxica · Mar 7, 2016

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

How was that? I didn't attend because I had other matters to tend to.

Blitz_N7 · Mar 7, 2016

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

davidgoes4wce · Mar 7, 2016

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

$$ Cambridge CH 13E, Q12. Current research into Alzheimer's disease suggests that the rate of loss of percentage brain function is proportional to the percentage brain function already lost. That is, if L is the percentage brain function lost, then $ \frac{dL}{dt}=kL, $ for some constant $ k > 0.$

$$ (a) Two years ago a patient was initially diagnosed with Alzheimer's disease, with a 15 \% loss of brain function. This year the patient was diagnosed with 20\% loss of brain function. Show that $ L=15 e^{kt}, $ where k $= \frac{1}{2} log \frac{4}{3}$

$(b) The nearby care centre will admit patients to a 24-hour nursing care when a patient reaches 60 \% loss of brain function. In how many more years will that be? Answer to the nearest year.$

I got (a) right but got an answer of 9.64 years for part (b).

The answer in the back of the book says " 8 More years"

davidgoes4wce · Mar 9, 2016

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

davidgoes4wce said:
$$ Cambridge CH 13E, Q12. Current research into Alzheimer's disease suggests that the rate of loss of percentage brain function is proportional to the percentage brain function already lost. That is, if L is the percentage brain function lost, then $ \frac{dL}{dt}=kL, $ for some constant $ k > 0.$

$$ (a) Two years ago a patient was initially diagnosed with Alzheimer's disease, with a 15 \% loss of brain function. This year the patient was diagnosed with 20\% loss of brain function. Show that $ L=15 e^{kt}, $ where k $= \frac{1}{2} log \frac{4}{3}$

$(b) The nearby care centre will admit patients to a 24-hour nursing care when a patient reaches 60 \% loss of brain function. In how many more years will that be? Answer to the nearest year.$

I got (a) right but got an answer of 9.64 years for part (b).

The answer in the back of the book says " 8 More years"

My working out, I eventually figured the question was asking in how "many more years" until it takes for 60% loss. (2+8=10). Remember you have to round 9.64 years to the nearest year. Not the most well worded question but it comes down to reading the question in more closer detail.

davidgoes4wce · Mar 9, 2016

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

Cambridge Prelim MX1 Textbook Marathon/Q&A (13 Viewers)

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