Mathematics Question Thread

[sophiegmaher]

I strongly believe that this question was unfair and I'm actually appalled that they put that in a 2U paper. Please be advised the limit they made you evaluate was actually a 3U limit.

Thank you!

[sophiegmaher]

I'm also stuck on the question 8 multiple choice question and question 15a(ii) attached from the CSSA trial :/

[RuiAce]

I'm also stuck on the question 8 multiple choice question and question 15a(ii) attached from the CSSA trial :/

The other one just looks like an area between curves problem.

For 0≤x≤π/8, the upper curve is y=1 and the lower curve is y=cos(2x).
For π/8≤x≤π/4, the upper curve is y=1 and the lower curve is y=sin(2x).

[sophiegmaher]

The other one just looks like an area between curves problem.

For 0≤x≤π/8, the upper curve is y=1 and the lower curve is y=cos(x).
For π/8≤x≤π/4, the upper curve is y=1 and the lower curve is y=sin(x).

Yeah, but the answers say the integral has 1-cos2x dx to begin with and I don't understand how :/

[clarence.harre]

Question 7.b.(iii) from the 2009 HSC Paper.
Answers claim 6am to 11am. Everytime I do it I get 6am to 10am. Figured since it's BOSTES they aren't going to mess up the marking guidelines for a paper, but I still fail to see where I'm going wrong.
Question is attached as a screenshot.

[sidzeman]

Yes

Remember the log rule, log (M^k) = klog (M)

So if you raise (1/2) to the power (-1), you will get k = (ln2)/1600

Hmmmm I see. Could you explain how to do part ii then please

[RuiAce]

Yeah, but the answers say the integral has 1-cos2x dx to begin with and I don't understand how :/

Fixed up a typo in what I wrote earlier - forgot about the 2.

Symmetry can be exploited wherever possible, but it is not mandatory that it is used.

[RuiAce]

Question 7.b.(iii) from the 2009 HSC Paper.
Answers claim 6am to 11am. Everytime I do it I get 6am to 10am. Figured since it's BOSTES they aren't going to mess up the marking guidelines for a paper, but I still fail to see where I'm going wrong.
Question is attached as a screenshot.

\begin{align*}\sin \frac{\pi t}{6} &= \frac12\\ \frac{\pi t}{6} &= \frac\pi6, \frac{5\pi}6\\ t &= 1, 5\end{align*}

There weren't any traps here.

[sophiegmaher]

Fixed up a typo in what I wrote earlier - forgot about the 2.

Symmetry can be exploited wherever possible, but it is not mandatory that it is used.

Sorry I'm still not sure how that makes it equal that :/

[roygbivmagic]

Hi, can you please explain how to solve this?
'Find all values of m for which the equation |2x-3|=mx+1 has exactly one solution.'

[RuiAce]

Sorry I'm still not sure how that makes it equal that :/

The symmetry means the area between 0 and π/4, and the area between 0 and π/8, are the exact same. That is to say, the integrals actually equal each other: \( \int_0^{\frac\pi 4}(1-\cos 2x)\,dx = \int_{\frac\pi4}^{\frac\pi 8}(1-\sin 2x)\,dx \)

It's like with \( y=x^2 \). The curve is symmetric about the line \( y=0\). So if you compute the area between -1 and 0, and also the area between 0 and 1, by symmetry they will be the same. That is to say, \( \int_{-1}^0 x^2\,dx = \int_0^1 x^2\,dx \)

What happens when you add two of the exact same things? You just get double that thing.

Hi, can you please explain how to solve this?
'Find all values of m for which the equation |2x-3|=mx+1 has exactly one solution.'

Please find the solution in the compilation.

[RuiAce]

Hmmmm I see. Could you explain how to do part ii then please

[roygbivmagic]

\begin{align*}\sin \frac{\pi t}{6} &= \frac12\\ \frac{\pi t}{6} &= \frac\pi6, \frac{5\pi}6\\ t &= 1, 5\end{align*}

There weren't any traps here.

Wait but isn't 5 hours past 5AM = 10AM not 11AM? (I keep getting the answer as 10AM as well)

[kemi]

Wait but isn't 5 hours past 5AM = 10AM not 11AM? (I keep getting the answer as 10AM as well)

Hey!

Did this question and got the same answer as you. Lo and behold, Success One answers are incorrect. Here is the true BOSTES HSC sample solution.

[Justinhales]

Hey Guys!!

Sorry i'm quite stuck on a certain question! A textbook question mind u!!! 

Q13

The half-life of radium is 1600 years.
(a) Find the percentage of radium that will be decayed after 500 years.
(b) Find the number of years that it will take for 75% of the radium to decay.

Thanks in advance!!