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4U Maths Question Thread

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esteban:

--- Quote from: milie10 on March 05, 2020, 12:24:24 pm ---Hi!!

I have reached the dreaded topic of proving ~inequalities~ :')

I have a general question about working out layout:
e.g. RTP \( (a+b)( \frac{1}{a} + \frac{1}{b} ) \ge 4\)
I like to work backwards by assuming that the proof is correct, shuffling it around until it gives me a property that I know is correct (e.g. in this case I ended up with \( (a-b)^2 \ge 0 \)) and then writing the proof out backwards in my final answer- if I do this, where should I put my original assumption? I know we're not meant to write it as an inequalities since that means that you're assuming that it's true
I was thinking about doing LHS= and RHS= but then that doesn't let me move numerals from either side

orrr is there a better method to solving these that doesn't involve me working backwards?

Thanks so much  ;D

--- End quote ---

You can also just start with the LHS and manipulate it directly. This doesn't involve writing down any inequalities that you have not yet proven.

Eg/

LHS = (a+b)(1/a+1/b) = (a+b)^2/ab = ((a-b)^2+4ab)/ab >= 4ab/ab = 4 = RHS, where the inequality follows from non-negativity of squares.

milie10:
hii back again



really struggling with these two - I made a start on the first one but not sure where to go from there

thank you!

fun_jirachi:
Hey there!

Just notice for the first question when transitioning from the second line to the third, you've made a small computational error: the third line should read \(\frac{1}{n} < \ln (n) - \ln (n-1) < \frac{1}{n-1}\) instead of what you've got. From here, we can manipulate the expressions with log laws amongst other things to obtain the result in the question. Perhaps work from these lines here:



For the second question, consider a similar idea where instead of upper rectangles and lower rectangles, we have a lower rectangle, an integral and a trapezium.

Hope this helps :)

milie10:
hi!


I did this using X~bin(24, 1/3) - does this method work?


These are the answers and I'm not sure how they got the 16/81

Thank you!

fun_jirachi:
Hey there!

'I did this using X~bin(24, 1/3) - does this method work?' - yes, for X=0, since we want there to be no tagged sheep.

However, doing parts (a) and (b) are supposed to lead you on to the answer given! Multi-step questions are supposed to lead you into discovering a result - make sure you look to incorporate earlier results in the later parts of the question :)

Having said that, from (a) you would have found that the probability of having no tagged sheep on a day is \(\left(\frac{2}{3}\right)^4 = \frac{16}{81}\), which is where that comes from. They simply did it on a per day basis, which makes more intuitive sense given they asked for no sheep on six consective days, not 24 consecutive selections.

Hope this makes sense :)

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