Cambridge HSC MX1 Textbook Marathon/Q&A (4 Viewers)

davidgoes4wce · Apr 30, 2016

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$

davidgoes4wce · Apr 30, 2016

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

Drongoski · Apr 30, 2016

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

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

si2136 · May 4, 2016

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

http://prntscr.com/b0363w

How can you work backwards from factorial? Thanks.

InteGrand · May 4, 2016

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

si2136 said:
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 $r=3.$

si2136 · May 4, 2016

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

InteGrand said:
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?

InteGrand · May 4, 2016

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

si2136 said:
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.

si2136 · May 4, 2016

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.

InteGrand · May 4, 2016

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

si2136 said:
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, $y^\prime = \frac{\mathrm{d}}{\mathrm{d}x}\left(x \ln x\right)\cdot \mathrm{e}^{x\ln x}=\frac{\mathrm{d}}{\mathrm{d}x}\left(x \ln x\right)\cdot x^x \Rightarrow y^\prime = x^x \left(\ln x + 1\right)$.$

si2136 · May 4, 2016

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

InteGrand said:
$\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?

InteGrand · May 4, 2016

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

si2136 said:
Since it's implicit differentiating, do you need to include dy/dx? in the answer?

The y' is the dy/dx.

si2136 · May 4, 2016

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

InteGrand said:
The y' is the dy/dx.

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

Your answer wouldn't be 2g, but rather 2g dy/dx

InteGrand · May 4, 2016

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

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

Your answer wouldn't be 2g, but rather 2g dy/dx

$\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 $\frac{\mathrm{d}y}{\mathrm{d}x}$.$

WildestDreams · May 23, 2016

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

Can someone please help me with this question?? I'm really bad at reasoning with geometry!

si2136 · May 24, 2016

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

Similar Triangles. Circular Geo.

WildestDreams · May 27, 2016

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

si2136 said:
Similar Triangles. Circular Geo.

Oh haha its not circle geo, its only 2U geometry. I want to see how good reasoning would look like.

Drongoski · May 27, 2016

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

WildestDreams · May 29, 2016

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

Thank you so much!!

davidgoes4wce · Jun 7, 2016

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."

davidgoes4wce · Jun 8, 2016

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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