HSC 2013 Maths Marathon (archive) (2 Viewers)

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omgiloverice

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2 unit people deserve a marathon thread too...
I'll start off with a typical hsc troll question (but abit harder)



Find the equation at which P moves
 
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Re: HSC 2013 2U Marathon



I haven't done 2U so I don't know how difficult the questions are supposed to be. This is an easy one.
 

Reloman

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Re: HSC 2013 2U Marathon

Consider the two functions: y=mx-4 and y=x^3
Find the value of 'm' such that it is a tangent to the function y=x^3
 
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Re: HSC 2013 2U Marathon

Look up the Align function in latex...it will save you all those tildes.
 
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Re: HSC 2013 2U Marathon

I agree with nightweaver...it'd be too hard without guidance in 2u.

Here is a solution:



Clearly this question is too involved for a 2u student, and would challenge 3u and 4u students as well.
 
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HeroicPandas

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Re: HSC 2013 2U Marathon

I agree with nightweaver...it'd be too hard without guidance in 2u.

Here is a solution:



Clearly this question is too involved for a 2u student, and would challenge 3u and 4u students as well.
Nice!

EDIT: misread ur post xD
 
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Re: HSC 2013 2U Marathon

Nice!

Why would it be hard without guidance? this question is similar to differentiating
You will see in all HSCs that 2u students are guided through questions with 'hints' to get to the final answer. You'd probably need to set the question up:

ai) Show that y=x^x=e^xlnx by log laws (1)
ii) Find dy/dx (2 or 3, depends on marking scheme. things to note here are: 1 mark for realising you have a chain rule in the exponential, and then differentiating xlnx with a product rule successfully. so this is 3 marks, really)

b) let z=x^x^x. By considering z=x^y, and using a), find dz/dx (4? ok so you'll give 1 mark for using part a at least. then another mark for doing the chain rule and realising the derivative of y is dy/dx (you'd be surprised), product rule again, and back substitution)
 

HeroicPandas

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Re: HSC 2013 2U Marathon

You will see in all HSCs that 2u students are guided through questions with 'hints' to get to the final answer. You'd probably need to set the question up:

ai) Show that y=x^x=e^xlnx by log laws (1)
ii) Find dy/dx (2 or 3, depends on marking scheme. things to note here are: 1 mark for realising you have a chain rule in the exponential, and then differentiating xlnx with a product rule successfully. so this is 3 marks, really)

b) let z=x^x^x. By considering z=x^y, and using a), find dz/dx (4? ok so you'll give 1 mark for using part a at least. then another mark for doing the chain rule and realising the derivative of y is dy/dx (you'd be surprised), product rule again, and back substitution)
yeh yeh i understand, i misread ur post haha

Finding the derivative of would be hard for 2U students
 

Sy123

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