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

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

$Cambridge Ex 3A Q 13 (c)$

$Two engines, Thomas and Henry, move on closer parallel tracks. They start at the origin, and are together again at time t=e-1. Thomas' displacement-time equation, in units of metres and minutes, is x=300 log (t+1), and Henry's is x=kt, for some constant k.$

$I understand everything including the exact values but my question is how to convert it to a non-exact value form$

$300 \ log (e-1)-\frac{300(e-2)}{e-1} in which case they get 37 metres. I was a bit unsure how they converted t=e-2 to 43''$

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

37 metres and 43'' I dont quite get.

I get a value of -410.36 instead of 37 metres

With the time part , I went (e-2) x 60 (remembering these units were initially in minutes)= which gave me approximately 8 seconds

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

$Part (c) of the question asked : Use calculus to find the maximum distance between Henry and Thomas during the first e-1 minutes, and the time when it occurs (in exact form, and then correct to the nearest meter or the nearest second$

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

^^^ Don't worry I sorted it out , with my scientific calculator I needed to type 'e1' for it to read it as 'e'

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

$Cambridge Exercise 3C Q 4 (b)$

$The answer was a bit unclear so I wanted to clear it up here$

$A stone is thrown downwards from the top of a 120 metre building, with an initial speed of 25 m/s. Take g =10m/s^{2}, and take upwards as positive, so that \ddot{x}=-10$

$Rework part (a) with the origin at the top of the building, and downwards positive$

$I set the initial conditions as t=0, v=25 and x=-120.$

$\ddot{x}=10,\ \dot{x}=10t+25,\ x=5t^{2}+25t-120$

$Could someone confirm if those 3 equations are right?$

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

$The answer says this : x=5t^2+25t then it says Put 5t^2+25t=120$

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

I get it now, the initial conditions set , if you want to treat the top of the building as the origin, t=0, v=25, g=10, x=0.

All good

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

Yea - I noticed in your 1st post: x = -120 instead of 0.

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

http://prntscr.com/b0363w

How can you work backwards from factorial? Thanks.

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

Originally Posted by si2136
http://prntscr.com/b0363w

How can you work backwards from factorial? Thanks.
Inspection.

$\noindent We want \frac{8!}{\left(8-r\right)!}=336. So \left(8-r\right)! = \frac{8!}{336}=120. So 8-r = 5, so r=3.$

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

Originally Posted by InteGrand
Inspection.

$\noindent We want \frac{8!}{\left(8-r\right)!}=336. So \left(8-r\right)! = \frac{8!}{336}=120. So 8-r = 5, so r=3.$
Is there another way without inspection? And is there a technique used in inspection in which you can notice it straight away?

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

Originally Posted by si2136
Is there another way without inspection? And is there a technique used in inspection in which you can notice it straight away?
If you know your early factorials, it's easy to recognise that 5! = 120. You'll probably recognise early factorials if you do lots of perms and combs Q's.

In the above method, the only inspection comes about when at the stage (8-r)! = 120.

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

A curve is given y = x^x where x > 0. Find the gradient function of the curve y = x^x.

How do I manage to differentiate it using Logarithmic Functions? Thanks.

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

Originally Posted by si2136
A curve is given y = x^x where x > 0. Find the gradient function of the curve y = x^x.

How do I manage to differentiate it using Logarithmic Functions? Thanks.
$\noindent Since y=x^{x}, taking logs of both sides yields \ln y = x\ln x. Differentiating both sides w.r.t. x yields \frac{y^\prime}{y}=\ln x + 1. Hence y^\prime = y \left(\ln x + 1\right) = x^x \left(\ln x + 1\right). This technique is known as \textsl{logarithmic differentiation}.$

$\noindent An alternative way of finding the derivative is to rewrite it with base \mathrm{e}, i.e. y=x^x=\left(\mathrm{e}^{\ln x}\right)^x = \mathrm{e}^{x\ln x}. So by the chain rule, y^\prime = \frac{\mathrm{d}}{\mathrm{d}x}\left(x \ln x\right)\cdot \mathrm{e}^{x\ln x}=\frac{\mathrm{d}}{\mathrm{d}x}\left(x \ln x\right)\cdot x^x \Rightarrow y^\prime = x^x \left(\ln x + 1\right).$

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

Originally Posted by InteGrand
$\noindent Since y=x^{x}, taking logs of both sides yields \ln y = x\ln x. Differentiating both sides w.r.t. x yields \frac{y^\prime}{y}=\ln x + 1. Hence y^\prime = y \left(\ln x + 1\right) = x^x \left(\ln x + 1\right). This technique is known as \textsl{logarithmic differentiation}.$

$\noindent An alternative way of finding the derivative is to rewrite it with base \mathrm{e}, i.e. y=x^x=\left(\mathrm{e}^{\ln x}\right)^x = \mathrm{e}^{x\ln x}. So by the chain rule, y^\prime = \frac{\mathrm{d}}{\mathrm{d}x}\left(x \ln x\right)\cdot \mathrm{e}^{x\ln x}=\frac{\mathrm{d}}{\mathrm{d}x}\left(x \ln x\right)\cdot x^x \Rightarrow y^\prime = x^x \left(\ln x + 1\right).$
Since it's implicit differentiating, do you need to include dy/dx? in the answer?

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

Originally Posted by si2136
Since it's implicit differentiating, do you need to include dy/dx? in the answer?
The y' is the dy/dx.

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

Originally Posted by InteGrand
The y' is the dy/dx.
Yes, but I mean for example you're differentiating f(x) = g^2

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

Originally Posted by si2136
Yes, but I mean for example you're differentiating f(x) = g^2

$\noindent I did include \frac{\mathrm{d}y}{\mathrm{d}x} when differentiating the \ln y with respect to x. I wrote \frac{y^\prime}{y} for that, which is the same thing as writing \frac{1}{y} \frac{\mathrm{d}y}{\mathrm{d}x}. In other words, I just wrote y^\prime instead of \frac{\mathrm{d}y}{\mathrm{d}x}.$

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

Screen Shot 2016-05-23 at 8.43.39 PM.png
Screen Shot 2016-05-23 at 8.44.33 PM.png

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

Similar Triangles. Circular Geo.

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

Originally Posted by si2136
Similar Triangles. Circular Geo.
Oh haha its not circle geo, its only 2U geometry. I want to see how good reasoning would look like.

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

a)

In triangles DEG and DFC:

- angle EDG = angle FDC (same angle)

- DE/DF = DG/DC = 1/2

.: triangles DEG and DFC are similar. (pair of sides proportional and included angles equal)

.: angle DGE = angle DCF (corresp angles of similar triangles)

But these are corresponding angles

.: EG // FC (and .: AC)

b)

angle ABC (= angle EBG) = Angle ACB (AB = AC)

But angle DGE (= angle BGE) = angle ACB

.: for triangle EBG, angle EBG = angle BGE

.: EB = EG

Now EG/FC = DE/DF = 1/2

.: FC = 2 x EG = 2 x EB

QED

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

Thank you so much!!

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

$Cambridge Year 12 Ex 5D Q 5 (c)$

$Use the formula for ^nC_2 to show that ^nC_2+^{n+1}C_2=n^2, and verify the result on the third column of the Pascal Triangle.$

$This was my working out$

The part I wasn't sure about in the question was it said "verify the result on the third column of the Pascal triangle."

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

Bump

"verify the result on the third column of the Pascal triangle." Does anyone have any idea on this? Or is my answer shown above good enough ?

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