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

davidgoes4wce · Feb 22, 2016

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

$With Part 7 (i) here is my screenshot$

davidgoes4wce · Feb 23, 2016

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

$$ Just want to clarify a question. In the Cambridge text they have a Q8 Exercise 6A, they say 'write these expressions using fractional and negative indices:'$

$$ Would it not be right for instance in Q 8a , to write the answer as $ \frac{1}{7^{\frac{-1}{3}}}$

davidgoes4wce · Feb 24, 2016

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

$$ Exercise 14 A , Question 17 (a). Is there a better way to solve for $ tan x =\sqrt{2}-1$

$$ Personally, I did the inverse of tan $ , x=tan^{-1} [\sqrt{2}-1], $ this in turn gives x $=\frac{\pi}{8} $ and x $=\frac{9\pi}{8}$

$$ With tan in this case being a positive number, the solution for tan has to lie in the Quadrants from $ [0, \frac{\pi}{2}] $ and $ [\frac{3\pi}{2}, 2 {\pi}]$

Nailgun · Feb 24, 2016

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

davidgoes4wce said:
I honestly didn't know until you, leehuan and Carrotsticks corrected me that you could have a log with base e, without the subscript being written. (Having doing 15 years of maths).

Always thought that a log without a base 'e' was to the base 10.

As I understand, in the world of maths y = logx denotes the natural logarithm by default. in every other discipline (i.e. engineering) logx means base 10

davidgoes4wce · Feb 24, 2016

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

Nailgun said:
As I understand, in the world of maths y = logx denotes the natural logarithm by default. in every other discipline (i.e. engineering) logx means base 10

$So I guess that was the big surprise for me. Even in Chemistry we use pH=-log[H+], physics when looking at the Decibels reading L $ =10 log [\frac{I}{I_{0}}] . $ I do gather that the markers will accept either log or ln as an acceptable answer.$

integral95 · Feb 24, 2016

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

How I was taught was

$ln(x) = log_e(x) \\ \\ log(x) = log_{10}(x)$

integral95 · Feb 24, 2016

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

davidgoes4wce said:
$$ Exercise 14 A , Question 17 (a). Is there a better way to solve for $ tan x =\sqrt{2}-1$

$$ Personally, I did the inverse of tan $ , x=tan^{-1} [\sqrt{2}-1], $ this in turn gives x $=\frac{\pi}{8} $ and x $=\frac{9\pi}{8}$

$$ With tan in this case being a positive number, the solution for tan has to lie in the Quadrants from $ [0, \frac{\pi}{2}] $ and $ [\frac{3\pi}{2}, 2 {\pi}]$

How in the world does taking the inverse directly yield that answer?

One way of obtaining that solution is to consider the double angle formula with tan pi/4

$tan(\frac{\pi}{4}) = \frac{2tan(\frac{\pi}{8})}{1-tan^{2}(\frac{\pi}{8})} = 1 \\ \\ \\ \Rightarrow tan^{2}(\frac{\pi}{8})+2tan(\frac{\pi}{8})-1 = 0$

Use the quadratic formula and only take the positive solution

Shuuya · Feb 24, 2016

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

integral95 said:
How in the world does taking the inverse directly yield that answer?

One way of obtaining that solution is to consider the double angle formula with tan pi/4

$tan(\frac{\pi}{4}) = \frac{2tan(\frac{\pi}{8})}{1-tan^{2}(\frac{\pi}{8})} = 1 \\ \\ \\ \Rightarrow tan^{2}(\frac{\pi}{8})+2tan(\frac{\pi}{8})-1 = 0$

Use the quadratic formula and only take the positive solution

Taking the inverse does directly get you the related angle pi/8 though, and then he used that angle to find the solutions in the appropriate quadrants... Isn't that how you would normally do it?

davidgoes4wce · Feb 24, 2016

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

integral95 said:
How in the world does taking the inverse directly yield that answer?

One way of obtaining that solution is to consider the double angle formula with tan pi/4

$tan(\frac{\pi}{4}) = \frac{2tan(\frac{\pi}{8})}{1-tan^{2}(\frac{\pi}{8})} = 1 \\ \\ \\ \Rightarrow tan^{2}(\frac{\pi}{8})+2tan(\frac{\pi}{8})-1 = 0$

Use the quadratic formula and only take the positive solution

$$ The answer in the back of the book was x $ =\frac{\pi}{8} and \frac{9\pi}{8}, $ for $ 0 \leq x \leq 2{\pi}$

davidgoes4wce · Feb 24, 2016

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

Draw the 4 quadrants, 'ASTC- All Stations to Central' and you will see your solutions. If tan is a positive value, it must lie in the 1st and the 3rd quadrants.

Im a 'visualizer' , I've also drawn a diagram to represent my solution.

integral95 · Feb 24, 2016

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

Sorry my question is, how would you know that taking the inverse tan of $\sqrt{2} -1$ give you an exact value of pi/8 (i.e the principle value) without using a calculator? seeing that this isn't one of the common trig identities you've would learn.

davidgoes4wce · Feb 24, 2016

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

integral95 said:
Sorry my question is, how would you know that taking the inverse tan of $\sqrt{2} -1$ give you an exact value of pi/8 (i.e the principle value) without using a calculator? seeing that this isn't one of the common trig identities you've would learn.

The question was from a textbook and I think HSC and Prelim courses have an assumption where all calculators are allowed .There is no way you could look at that first up and determine the exact value (unless you have memorised that )

$\sqrt{2}-1 $ is not an exact value that we are expected to remember $$

integral95 · Feb 24, 2016

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

davidgoes4wce said:
The question was from a textbook and I think HSC and Prelim courses have an assumption where all calculators are allowed .There is no way you could look at that first up and determine the exact value (unless you have memorised that )

$\sqrt{2}-1 $ is not an exact value that we are expected to remember $$

and the formal proof behind that identity is based on my solution, no I'm saying that using a calculator (the ones approved by the board of studies) to take the inverse tan would just give some numbers rounded to like 5 decimal places rather than the exact value of pi/8, so it wouldn't really work.

Shuuya · Feb 24, 2016

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

integral95 said:
and the formal proof behind that identity is based on my solution, no I'm saying that using a calculator (the ones approved by the board of studies) to take the inverse tan would just give some numbers rounded to like 5 decimal places rather than the exact value of pi/8, so it wouldn't really work.

You can convert the long decimal into an exact radian value by dividing pi by the answer.

drsabz101 · Feb 24, 2016

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

Can someone show how to get the solution for the second part of part B (finding M) and part c . Thankyou

InteGrand · Feb 24, 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

$\noindent One word: symmetry. :$\mathcal{P}$$

InteGrand · Feb 24, 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

$\noindent The tedious and expected way to do it is to first find the equations of the tangents at the $x$-intercepts, i.e. at the points where $x=\alpha$ and $x=\beta$ (finding these equations is done by finding the slopes at the points first using the derivative, and then using point-slope form). Then, solve these two equations simultaneously to show that the $x$-value of the point of intersection is $\frac{\alpha + \beta}{2}$.$

drsabz101 · Feb 25, 2016

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

Can someone show the working out to this question , thanks

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

InteGrand · Feb 25, 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

$\noindent Consider separately the cases where $x <-3$, $-3\leq x \leq 2$, $x >2$. In each case, eliminate the absolute value bars appropriately. Then we just need to solve a quadratic equation for each case.$

drsabz101 · Feb 25, 2016

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

I am still confused on what to do in each case

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

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