2018 HSC Marathon 3U Q and A (1 Viewer)

HoldingOn

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I thought it would be nice to have a thread for answering past HSC questions and any other for that matter, in the lead up to the exam in a few weeks. So here it is.

I'll start off with one from the 2001 HSC which I have been struggling with. Any help would be appreciated.

http://prntscr.com/kumrq3
 

fan96

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Consider .





By the cosine rule,



Equating and dividing out by (the height cannot be zero),

 
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HoldingOn

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I thought it would be nice to have a thread for answering past HSC questions and any other for that matter, in the lead up to the exam in a few weeks. So here it is.

I'll start off with one from the 2001 HSC which I have been struggling with. Any help would be appreciated.

http://prntscr.com/kumrq3
Could someone try part iii as well
 

fan96

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Could someone try part iii as well


Applying the auxiliary angle transformation:



* Note the restriction in the diagram - we do not need to take a general solution here.



.





(Since )

Since this is the only turning point within the given domain and , this is clearly a minimum turning point.

If we put on the vertical axis, the graph of this function will be similar in shape to a concave up parabola.

Note when sketching that are actually excluded from the domain, so you should draw an open circle at the endpoints of the domain to indicate that they are not part of the function.
 
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HoldingOn

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i) you expand , differentiate both sides then sub in x = 1

ii) you expand , integrate both sides twice (make sure you can find the constants) then sub x = -1
Ok thanks I'll try that. Just out of curiosity is there anything you look for when approaching these more left-field binomials, or just intuition?
 

integral95

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Ok thanks I'll try that. Just out of curiosity is there anything you look for when approaching these more left-field binomials, or just intuition?
Mostly the latter for me really, there's a lot of pattern recognition going on when I approach them.
 

fan96

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The horizontal range of the ball is maximised when . This can be easily shown by examining the equation in ii).

Case 1: The ceiling allows for an angle of projection of .

If the ceiling allows for at least this angle of projection, then is the optimal angle to throw the ball.

It doesn't matter how tall the ceiling is after a certain point - because the best angle is and any higher angle will simply lower .

The horizontal range obtained by projecting at this optimal angle is then .

But when does the ceiling allow for this angle? It is when the maximum height of the projectile (with angle of projection ) is less than or equal to the height allowed by the ceiling (which is ).

Mathematically, this can be represented by the condition

.

Case 2: The ceiling does NOT allow for an angle of projection of .

As decreases past , so too does , from ii).

Therefore we want the largest value of we can get. This value occurs when the tip of the ball's trajectory coincides with the ceiling, i.e.



Rearrangement gives:





Finally, to get the answer, substitute (1) and (2) into:

 
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fan96

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i) Hint: the expression you need to prove can be rearranged to



ii) Starting from the inductive hypothesis:





From i) by rearranging we get .
 

HoldingOn

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This seems to be my thread haha.

If anyone could help with this, and maybe some tips on how to approach these sort of problems- I'm not great at them.

A die is loaded in such a way that in 8 throws of the die, the probability of getting 3 even numbers is four times the probability of getting 2 even numbers. Find the probability that a single throw of the die results in an even number.
 

Drongoski

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This is a Binomial Probability question.



Solving you get the P(even) in a single throw = p = 2/3.
 
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HeroWise

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May I ask where you found that proj motion question?
 

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