davidgoes4wce
Well-Known Member
Re: HSC 2015 3U Marathon
Another attempt at getting the 2003 Q7 (iii)
Another attempt at getting the 2003 Q7 (iii)
-
Finished your HSC? Check out our University Discussion, Non-School forums or help out next year's HSC students! -
YOU can help the next generation of students in the community! Share your trial papers and notes on our Notes & Resources page
HSC 2015 MX1 Marathon (archive) (2 Viewers)
- Thread starter Trebla
- Start date
- Status
davidgoes4wce
Well-Known Member
Re: HSC 2015 3U Marathon
continued:
continued:
davidgoes4wce
Well-Known Member
Re: HSC 2015 3U Marathon
Will do it again and again and again till it gets through my head.
Will do it again and again and again till it gets through my head.
davidgoes4wce
Well-Known Member
Re: HSC 2015 3U Marathon
Special mention to Jeff Geha. I understood his working out better than the text.
Special mention to Jeff Geha. I understood his working out better than the text.
davidgoes4wce
Well-Known Member
Re: HSC 2015 3U Marathon
Had a deep think about this question tonight from Sharon Kennedy's trials. Having a quick look at the solutions my way of thinking was to equate the x^2 term on the LHS and the sum of the x^2 terms on the RHS. Any other suggestions for a question such as this?
Had a deep think about this question tonight from Sharon Kennedy's trials. Having a quick look at the solutions my way of thinking was to equate the x^2 term on the LHS and the sum of the x^2 terms on the RHS. Any other suggestions for a question such as this?
davidgoes4wce
Well-Known Member
Re: HSC 2015 3U Marathon
I also say that I managed to equate the x^1 term on the LHS with the sum of x^1 on the RHS.
I also say that I managed to equate the x^1 term on the LHS with the sum of x^1 on the RHS.
davidgoes4wce
Well-Known Member
Re: HSC 2015 3U Marathon
To keep things simple I have managed to prove x^0, x^1 and the x^2 on both sides (LHS and RHS ) to be equal. No need to answer this question anymore.
To keep things simple I have managed to prove x^0, x^1 and the x^2 on both sides (LHS and RHS ) to be equal. No need to answer this question anymore.
dan964
what
Re: HSC 2015 3U Marathon
yes, you equate the coefficients of both sides. you are on the right track.
also note that nCn=nC0 and similarily for other values
yes, you equate the coefficients of both sides. you are on the right track.
also note that nCn=nC0 and similarily for other values
davidgoes4wce
Well-Known Member
Re: HSC 2015 3U Marathon
I am looking at 7 b(ii)
my question is why do we have a plus/minus square root sign for cos a and sin b? I was thinking we would use just the positive sign.
The rest of the working out I am fine with.
I am looking at 7 b(ii)
my question is why do we have a plus/minus square root sign for cos a and sin b? I was thinking we would use just the positive sign.
The rest of the working out I am fine with.
davidgoes4wce
Well-Known Member
Re: HSC 2015 3U Marathon
I tried to visualise cos a and sin b by drawing out the triangle method and working out the unknown length but can you have a negative length ?
I tried to visualise cos a and sin b by drawing out the triangle method and working out the unknown length but can you have a negative length ?
Re: HSC 2015 3U Marathon
I am looking at 7 b(ii)
my question is why do we have a plus/minus square root sign for cos a and sin b? I was thinking we would use just the positive sign.
The rest of the working out I am fine with.
davidgoes4wce
Well-Known Member
Re: HSC 2015 3U Marathon
I also think because of the trig identity stated that the values could be positive or negative. I guess I didn't look at the whole question when evaluating the answer.
I also think because of the trig identity stated that the values could be positive or negative. I guess I didn't look at the whole question when evaluating the answer.
dan964
what
Re: HSC 2015 3U Marathon
yep,
positive values only occur for angles between 0 and
(or 90 degrees)
although I would also note each of the inverse functions do have a limited domain and range (which comes into your answer to the second part)
yep,
positive values only occur for angles between 0 and
although I would also note each of the inverse functions do have a limited domain and range (which comes into your answer to the second part)
Last edited:
davidgoes4wce
Well-Known Member
Re: HSC 2015 3U Marathon
Doing the derivative of something along the lines of
I know the answer but I think the formula can be a bit misleading and can trick alot of students.
You should bring the derivative of d/dx(3x) =3. Is there any other good ways to remember this?
Doing the derivative of something along the lines of
I know the answer but I think the formula can be a bit misleading and can trick alot of students.
You should bring the derivative of d/dx(3x) =3. Is there any other good ways to remember this?
AztecWarrior
Member
Re: HSC 2015 3U Marathon
you do d/dx tan^1(x/a) =(1)/(1+x^2/a^2)*(d/dx(x))
or
you do d/dx tan^1(x/a) =(1)/(1+x^2/a^2)*(d/dx(x))
or
Last edited:
AztecWarrior
Member
Re: HSC 2015 3U Marathon
true
true
davidgoes4wce
Well-Known Member
Re: HSC 2015 3U Marathon
When doing a question such as this I think its important to define what your numerator term and denominator term are. (in this case 'x' and 'a')
I broke it up into two parts, the first part is the derivative (given by the formula) and the 2nd part is the derivative of the numerator term.
When doing a question such as this I think its important to define what your numerator term and denominator term are. (in this case 'x' and 'a')
I broke it up into two parts, the first part is the derivative (given by the formula) and the 2nd part is the derivative of the numerator term.
sharoooooo
Active Member
Re: HSC 2015 3U Marathon
pls help with probability :'(
Q. Four families each have four children.
What is the probability that exactly two of these families have two boys and two girls?
(Assume that each child is equally likely to be a boy or a girl).
I've done:
Family 1: (4C2) (0.5)^2 (0.5)^2
Family 2: (4C2) (0.5)^2 (0.5)^2
dunno what to do for family 3 and 4 :/
they could have like 3boys1girl, 1boy3girls, all boys, all girls...
idk what to do. do i even need to take those possibilities of family 3 and 4 into account?
help pls
pls help with probability :'(
Q. Four families each have four children.
What is the probability that exactly two of these families have two boys and two girls?
(Assume that each child is equally likely to be a boy or a girl).
I've done:
Family 1: (4C2) (0.5)^2 (0.5)^2
Family 2: (4C2) (0.5)^2 (0.5)^2
dunno what to do for family 3 and 4 :/
they could have like 3boys1girl, 1boy3girls, all boys, all girls...
idk what to do. do i even need to take those possibilities of family 3 and 4 into account?
help pls
Drsoccerball
Well-Known Member
- Joined
- May 28, 2014
- Messages
- 3,650
- Gender
- Undisclosed
- HSC
- 2015
Re: HSC 2015 3U Marathon
^2(\frac{1}{2})^2 =\frac{3}{8} )
Probability of exactly two familys having 2 boys:
^2(\frac{5}{8})^2 =\frac{675}{2048} )
Probability of one family having 2 boys:
Probability of exactly two familys having 2 boys:
- Status