Cambridge Prelim MX1 Textbook Marathon/Q&A (2 Viewers)

drsabz101 · Feb 25, 2016

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

What do we do with the x^2-1

InteGrand · Feb 25, 2016

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

drsabz101 said:
What do we do with the x^2-1

Get everything to one side (rearrange), so that we have a quadratic expression on one side and 0 on the other.

drsabz101 · Feb 25, 2016

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

Yeah, ended up solving it before. I understand now

davidgoes4wce · Feb 25, 2016

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

Again I tried to visualise it again, I used an example:

davidgoes4wce · Feb 25, 2016

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

drsabz101 said:
Can someone show how to get the solution for the second part of part B (finding M) and part c . Thankyou
View attachment 32945

$Q8a) \frac{dy}{dx}=a[(x-\beta)(1)+(x-\alpha)(1)]$
$=a[x-\beta+x-\alpha]$
$=a(2x-\alpha-\beta)$

$$ Part 8b) At x= $ \alpha$
$\frac{dy}{dx}=a(2 \alpha-\alpha-\beta)=a(\alpha-\beta)$

$$ At x = $ \beta$
$\frac{dy}{dx}=a(2\beta-\alpha-\beta)=a(\beta-\alpha)$

$The gradients have the same magnitude but are opposite in sign$

$$ This can be shown below: At x $=\beta, \frac{dy}{dx}=a[-(\alpha-\beta)]=-a(\alpha-\beta)$

$$ Where the tangents meet, we use the Point Gradient Formula in order to determine where they intersect but we should look to eliminate the x in order to get the y-value. $ y-y_{1}=m(x-x_{1})$

$At x=\alpha, m_{1}=\frac{dy}{dx}=a(\alpha-\beta), x-intercept (\alpha,0)$
$At x=\beta, m_{2}=\frac{dy}{dx}=a(\beta-\alpha), x-intercept (\beta, 0)$

$\frac{y-y_{1}}{m_{1}}+x_{1}=x \textcircled{1}$
$\frac{y-y_{2}}{m_{2}}+x_{2}=x \textcircled{2}$

$Equating \textcircled{1} and \textcircled{2} together$

$\frac{y-y_{1}}{m_1}+x_{1}=\frac{y-y_{2}}{m_2}+x_{2}$
$\frac{y-0}{a(\alpha-\beta)}+\alpha=\frac{y-0}{a(\beta-\alpha)}+\beta$
$\frac{y-0}{a(\alpha-\beta)}+\alpha=\frac{y-0}{-a(\alpha-\beta)}+\beta$
$\frac{y}{a(\alpha-\beta)}+\frac{y}{a(\alpha-\beta)}=\beta-\alpha$
$\frac{2y}{a(\alpha-\beta)}=\beta-\alpha$
$2y=-a(\alpha-\beta)^2$
$y=\frac{-a}{2} (\alpha-\beta)^2$
$$ The average of the x-intercepts is , what Integrand said, you simply add the two x-intercepts values and divide by 2, $ x=\frac{a+b}{2}$
$$ Hence, the coordinate of M is $ (\frac{a+b}{2}, \frac{-a}{2} (\alpha-\beta)^2 )$

$$ Part 8 (c) , the tangent is horizontal when $ \frac{dy}{dx}=0$

$Given \frac{dy}{dx}=a(2x-\alpha-\beta) from Part 8 (a)$

$0=a(2x-\alpha-\beta)$
$x=\frac{\alpha+\beta}{2}$
$At x=\frac{\alpha+\beta}{2}, $ by substituting into the original equation $$
$y=a[\frac{\alpha+\beta}{2}-\alpha] [\frac{\alpha+\beta}{2}-\beta]$
$y=a[(\frac{\beta}{2}-\frac{\alpha}{2})(\frac{\alpha}{2}-\frac{\beta}{2})]$

$=-a[(\frac{\alpha}{2}-\frac{\beta}{2})(\frac{\alpha}{2}-\frac{\beta}{2})]$
$=-a[\frac{\alpha}{2}-\frac{\beta}{2}]^2$
$=\frac{-a}{4}(\alpha-\beta)^2$
$$ Hence, the coordinate of V $(\frac{a+b}{2}, \frac{-a}{4}(\alpha-\beta)^2)$

drsabz101 · Feb 25, 2016

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

Thanks I really appreciate it ( I like visual representations)

davidgoes4wce · Feb 25, 2016

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

To prove that M is closer to the x-axis, in comparison to V, you could compare the ratios of each other, which in this case is half. (Or vice versa, you could divide V by M, and you will you get a ratio of 2.

davidgoes4wce · Feb 25, 2016

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

drsabz101 said:
Thanks I really appreciate it ( I like visual representations)

I think alot of the university students I tutor said the same thing. Like I said before, its a huge shame HSC does not allow graphing calculators. I thought learning was supposed to be enhanced? I guess you could always use a Desmos application in order to view it for free. (if you do a heap of questions, you can visualise these graphs WITHOUT a calculator or graphing facility)

I can understand both sides of the argument too....students may lose their touch of learning and rely on the calculators too much.

davidgoes4wce · Feb 25, 2016

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

Also felt like that question was more of an Extension question than a Development Question EX 7F, Q8

davidgoes4wce · Feb 26, 2016

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

drsabz101 said:
Can someone show the working out to this question , thanks

Solve the equation |x-2| -|x+3| = x^2 -1

$My first suggestion for this question is to again visualize it$

$I'd look for the zeroes for the functions in $ |x-2| $ and $ |x+3|, $ in this case x=2 and x=-3 are two possible zeroes$

$$ You will basically be looking at three major intervals on that graph, $ x < {-3}, -3 \leq x <{2} $ and $ x \geq {2}$

$$ From the combined Graph, one can see that the solution must lie between $ -3 \leq x < 2$

$$ How do we do this algebraically? We add up the functions -(x+3) and -(x-2). Remember these functions have to be negative when in the domain of -3 $ \leq x < 2$

$$ Solving algebraically , y=-(x-2)-(x+3)=-2x-1 $ \ \textcircled{1}$

$$ Equate this with the quadratic equation given, y=x^2-1 $ \ \textcircled{2}$

$$ By equating $ \textcircled{1} \ and \ \textcircled{2} , -2x-1=x^2-1 \implies 0=x^2+2x =x(x+2)$

$x=0 and x=-2. By substituting back into the original equation, we find that our coordinates are (0,-1) and (-2,3)$

drsabz101 · Feb 26, 2016

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

I have tried this question 4 times now, and am not getting it right! Can someone please show the working. I am using the formula y'=u'vw+uv'w+uvw'

InteGrand · Feb 26, 2016

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

drsabz101 said:
I have tried this question 4 times now, and am not getting it right! Can someone please show the working. I am using the formula y'=u'vw+uv'w+uvw'
View attachment 32948

Maybe post your working, since it could just be a silly mistake.

Paradoxica · Feb 26, 2016

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

drsabz101 said:
I have tried this question 4 times now, and am not getting it right! Can someone please show the working. I am using the formula y'=u'vw+uv'w+uvw'
View attachment 32948

Do you know how to do logarithmic differentiation? It's a lot easier.

InteGrand · Feb 26, 2016

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

If you don't know logarithmic differentiation, you can read about it on this old thread: http://community.boredofstudies.org/14/mathematics-extension-2/79552/interesting-possible-trick.html . (Or search it up online.)

Nailgun · Feb 26, 2016

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

InteGrand said:
If you don't know logarithmic differentiation, you can read about it on this old thread: http://community.boredofstudies.org/14/mathematics-extension-2/79552/interesting-possible-trick.html . (Or search it up online.)

Ooh didn't realize you could use it like that

drsabz101 · Feb 26, 2016

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

ill redo the question and post it in like 20 min

InteGrand · Feb 26, 2016

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

Nailgun said:
Ooh didn't realize you could use it like that

Yeah, it can be pretty useful for doing tedious derivatives of long products

. But it's rare that they'd make you do a differentiation involving a product of three (or more) functions in the HSC I think.

davidgoes4wce · Feb 26, 2016

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

drsabz101 said:
I have tried this question 4 times now, and am not getting it right! Can someone please show the working. I am using the formula y'=u'vw+uv'w+uvw'
View attachment 32948

$As one of the earlier posters mentioned, its a good method to take the natural logs of both sides.$

$ln \ y= ln \ x+ ln (x-2)^4 +ln (2x+1)^\frac{1}{2}$

$ln \ y= ln \ x + 4 \ ln (x-2) + \frac{1}{2} ln (2x+1)$

$\frac{dy}{dx} \frac{1}{y} = \frac{1}{x} + \frac{4(1)}{x-2} + \frac{1}{2} \ \frac{2}{2x+1}$

$$ In simplified terms $ , \frac{dy}{dx}= y [ \frac{1}{x}+\frac{4}{x-2}+\frac{1}{2x+1}]$

$\frac{dy}{dx}=x(x-2)^4 \ (2x+1)^{\frac{1}{2}} [\frac{1(2x+1)(x-2)+4x(2x+1)+1x(x-2)}{x(x-2)(2x+1)}]$

$$ Simplifying, $ \frac{x-2}{(2x+1)^{\frac{1}{2}}} \ (11x^2-x-2)$

$\frac{dy}{dx}=0, x=2 $ is the value when the tangent is horizontal. $$

Drongoski · Feb 27, 2016

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

davidgoes4wce said:
$As one of the earlier posters mentioned, its a good method to take the natural logs of both sides.$

$ln \ y= ln \ x+ ln (x-2)^4 +ln (2x+1)^\frac{1}{2}$

$ln \ y= ln \ x + 4 \ ln (x-2) + \frac{1}{2} ln (2x+1)$

$\frac{dy}{dx} \frac{1}{y} = \frac{1}{x} + \frac{4(1)}{x-2} + \frac{1}{2} \ \frac{2}{2x+1}$

$$ In simplified terms $ , \frac{dy}{dx}= y [ \frac{1}{x}+\frac{4}{x-2}+\frac{1}{2x+1}]$

$\frac{dy}{dx}=x(x-2)^4 \ (2x+1)^{\frac{1}{2}} [\frac{1(2x+1)(x-2)+4x(2x+1)+1x(x-2)}{x(x-2)(2x+1)}]$

$$ Simplifying, $ \frac{(x-2)^3}{(2x+1)^{\frac{1}{2}}} \ (11x^2-x-2)$

$\frac{dy}{dx}=0, x=2 $ is the value when the tangent is horizontal. $$

$(x-2)^3$ ??

davidgoes4wce · Feb 27, 2016

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

Drongoski said:
??

I took the natural logs of both sides. That is also the answer in the back of the book.

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