Preliminary mathematics marathon (1 Viewer)

[ohexploitable]

WTF. My school is delayed.

Same here.

 

[hscishard]

Yea.
Cambridge Ext1 books lays the entire 3unit couse out differently. Some topics in the HSC course are included in the Prelim one and some of the prelim are in the Year 12 one. How odd is that?

 

[super.muppy]

Yea.
Cambridge Ext1 books lays the entire 3unit couse out differently. Some topics in the HSC course are included in the Prelim one and some of the prelim are in the Year 12 one. How odd is that?

yea thats y the cambridge ext1 books get smaller. year 11 compared to year 12 ext1 book is huge

 

[hscishard]

Evaluate

 

[fullonoob]

this is integrating trig functions, more of yr 12. which topic do you want me to post a question on, i'll do it after i watch tv

 

[ohexploitable]

which topic do you want me to post a question on, i'll do it after i watch tv

Any Year 11 topic will do

 

[nikkifc]

By expressing in the form Rcos(2x+a) (where a is in radians), or otherwise: Find all solutions to:

for .

 

[hscishard]

this is integrating trig functions, more of yr 12. which topic do you want me to post a question on, i'll do it after i watch tv

Shit. Cos0 is 1

 

[hscishard]

By expressing in the form Rcos(2x+a) (where a is in radians), or otherwise: Find all solutions to:

for .

R = root 1+3
r=2
tanx=b/a
tanx=root 3 /1
x=pi/3

2cos(2x+pi/3)=1
cos(2x+pi/3)=1/2
x= 0and 2pi/3

Yep 2pi-pi/3 = 11pi/3..

 

[Trebla]

Explain carefully why | x + a | + | x + b | (for some a =/= b) is always strictly positive.

 

[hscishard]

Isn't that a bit obvious but... they both can't be negative. So they must positive.

(I'm pretty sure that's not supposed to be the answer. Lol)

 

[Trebla]

You're right in the fact that both expressions cannot be negative, but that does not necessarily mean that it must therefore be positive.

Can it be zero? Why or why not? You need to come up with a good mathematical argument to justify your conclusion.

 

[fullonoob]

R = root 1+3
r=2
tanx=b/a
tanx=root 3 /1
x=pi/3

2cos(2x+pi/3)=1
cos(2x+pi/3)=1/2
x= 0and 2pi/3

Yep 2pi-pi/3 = 11pi/3..

can show me the formulas of how you do this xD
or just private message me with the long steps
i kinda forgot prelims

 

[super.muppy]

Isn't that a bit obvious but... they both can't be negative. So they must positive.

(I'm pretty sure that's not supposed to be the answer. Lol)

they must be strictly positive in that the solution is positive and also not equal to zero (since a=/= b). LOL iono. :L

 

[hscishard]

It wont be zero if a is not equal to b.

ooo I guess =/= is not equal to?

 

[Trebla]

The conclusion you have is correct, but it hasn't been justified with some form of rigourous argument. (That's the not so obvious part)

 

[ohexploitable]

can show me the formulas of how you do this xD
or just private message me with the long steps
i kinda forgot prelims

asin(x)+bcos(x)=Rsin(x+theta)
asin(x)-bcos(x)=Rsin(x-theta)
acos(x)+bsin(x)=Rcos(x-theta)
acos(x)-bsin(x)=Rcos(x+theta)

 

[fullonoob]

Explain carefully why | x + a | + | x + b | (for some a =/= b) is always strictly positive.
assume | x + a | > 0 | x + b | > 0 for some a =/= b
a =/= b =/=0 , as well as all real values,
| a | =/= | b |
since | x + a | > 0 , | a | =/= 0
therefore | x + a | > 0 is true
similarly, | x + b | > 0 is true
therefore, | x + a | + | x + b | (for some a =/= b) is always strictly positive.

 

[hscishard]

can show me the formulas of how you do this xD
or just private message me with the long steps
i kinda forgot prelims

 

[fullonoob]

R = root 1+3
r=2
tanx=b/a
tanx=root 3 /1
x=pi/3

2cos(2x+pi/3)=1
cos(2x+pi/3)=1/2
x= 0and 2pi/3

Yep 2pi-pi/3 = 11pi/3..

ehhh how you get R = root 1+3
i thought a = root 6 b = root 2
R = root (6+2)
i'm lost
link me to some tutorial site so i can learn how to deal with the (2x) part maybe