yr 11 revision - questions (1 Viewer)

[jyu]

i tried that but with the second step, when you square both sides, simplify and transpose....
i managed to get the RHS, but wasn't sure how to get the LHS?

....what's "transpose"?

Transpose, change or rearrange the terms and factors around in the equation.

:wave:

 

[-pari-]

thanks [again] jyu =)

 

[-pari-]

i should be able to do this one now:

(1) |3y - 1| + |2y- 3| > 5

but i'm getting stuck...?
answer: y< -1, y >3/5

i used |a| = a^(2/2)

squared both sides etc, got to

(3y - 1)^2 - (2y - 3)^2 > 25 - 10 (2y - 3)^ (2/2)

i expanded and simplified RHS & brought the 25 over to the RHS getting

5y^2 - 18y - 33 > - 10 |2y - 3|

then i tried finding the 3 solutions for (2y - 3) but they all turn out wrong ?

(2) how do u find the range of y = 1/(x^2-1) algebraically?
answer: y>0 and y</= -1

i tried making x the subject, but i'm getting the wrong answer...

(3)

y = |x|/x^2 can be written as
y = |x|/|x|^2.

Cancel common factor |x| to obtain
y = 1/|x| = |1/x|

Hence both branches are above the x-axis.

y = 1/|x| = |1/x| <-- how does that work?

& wouldn't you still say
when x > 0
y = 1/x

when x < 0
y = -1/x

?

(4) Find:

lim (2Öx)/(x-1)
x -> ¥

(5) how do you graph an hyperbola/asymptote? :-(

eg. y = (x + 1) / (x^2 - 1)

- what steps do i need to take?
i found domain x=/= 1, -1
and y =/= 0

how do i know its a hyperbola in all three sections of the plane? (after dividing up into x>1, x<1 and -1<x<1)
i mean, how do i know its not a parabola in the middle section and a hyperbola in the other two or something??

Asymptotes are quite problematic

thats an understatment. uf.

 

[bos1234]

i should be able to do this one now:

(1) |3y - 1| + |2y- 3| > 5

but i'm getting stuck...?
answer: x = -2

i used |a| = a^(2/2)

squared both sides etc, got to

(3y - 1)^2 - (2y - 3)^2 > 25 - 10 (2y - 3)^ (2/2)

i expanded and simplified RHS & brought the 25 over to the RHS getting

5y^2 - 18y - 33 > - 10 |2y - 3|

then i tried finding the 3 solutions for (2y - 3) but they all turn out wrong ?

(2) how do u find the range of y = 1/(x^2-1) algebraically?
answer: y>0 and y</= i -1<>

i tried making x the subject, but i'm getting the wrong answer...

(3)


y = 1/|x| = |1/x| <-- how does that work?

& wouldn't you still say
when x > 0
y = 1/x

when x < 0
y = -1/x

?

(4) Find:

lim (2Öx)/(x-1)
x -> ¥

(5) how do you graph an hyperbola/asymptote? :-(

eg. y = (x + 1) / (x^2 - 1)

- what steps do i need to take?
i found domain x=/= 1, -1
and y =/= 0

how do i know its a hyperbola in all three sections of the plane? (after dividing up into x>1, x<1 and -1<X<1)< font>
i mean, how do i know its not a parabola in the middle section and a hyperbola in the other two or something??

thats an understatment. uf.

r u sure x = -2 is the answer??
there are no x's in it?

 

[-pari-]

r u sure x = -2 is the answer??
there are no x's in it?

ahh...gud point..

- will edit that

 

[fishy89sg]

this thread is leaning towards mathematics extension 1 stuff..

 

[bos1234]

i should be able to do this one now:

(1) |3y - 1| + |2y- 3| > 5

but i'm getting stuck...?
answer: y< -1, y >3/5

i used |a| = a^(2/2)

squared both sides etc, got to

(3y - 1)^2 - (2y - 3)^2 > 25 - 10 (2y - 3)^ (2/2)

i expanded and simplified RHS & brought the 25 over to the RHS getting

5y^2 - 18y - 33 > - 10 |2y - 3|

then i tried finding the 3 solutions for (2y - 3) but they all turn out wrong ?

(2) how do u find the range of y = 1/(x^2-1) algebraically?
answer: y>0 and y</= i -1<>

i tried making x the subject, but i'm getting the wrong answer...

(3)


y = 1/|x| = |1/x| <-- how does that work?

& wouldn't you still say
when x > 0
y = 1/x

when x < 0
y = -1/x

?

(4) Find:

lim (2Öx)/(x-1)
x -> ¥

(5) how do you graph an hyperbola/asymptote? :-(

eg. y = (x + 1) / (x^2 - 1)

- what steps do i need to take?
i found domain x=/= 1, -1
and y =/= 0

how do i know its a hyperbola in all three sections of the plane? (after dividing up into x>1, x<1 and -1<X<1)< font>
i mean, how do i know its not a parabola in the middle section and a hyperbola in the other two or something??

thats an understatment. uf.

1) |3y - 1| + |2y- 3| > 5

-------1/3---------3/2---------

If x<1/3 then
-3y+1-2y+3 >5
-5y+4>5
-5y>1
y<-1/5

If x>3/2 then
3y-1+2y-3>5
5y-4>5
5y>9
y>9/4

You got different answers. I dont know what I did wrong. Some1 please correct it

---------------------------------------------------------------------------------------

2)y = 1/(x^2-1)
y(x^2-1)=1
x^2-1=1/y
x^2=1/y+1
x^2 = (1+y)/y
x= rt[(1+y)/y]

You know that a square root cannot be negative.
rt[(1+y)/y] > 0
(1+y)/y> 0

Now draw a number line and see what happens when y<-1, -1<Y<0 y and>0

------------------------

(3) They are both the same thing
Its like say rt(16/4) = rt16/rt4

And yes you can say that

4)Divide by the highest power of x in the denominator which is a half and then simplify from there

5)
1. Find where the curve cuts the x and y-axis. Let x = 0 and y = 0 and solve

2. Find the domain and range which you already have. Dot these lines in

3. Find the horizontal asymptote ( Examine what happens to the curve as x approaches positive and negative infinity )

4. Find the Vertical asymptotes. Examine what happens to the curve as it approaches the domains from the left hand sides and the right hand sides



----------------------------------

THIS IS JUST A BASIC GUIDE. COULD SOME1 PLEASE MODIFY/CHANGE WHERE INCORRECT!!


ASK AGAIN IF YOU DIDNT UNDERSTAND ANYTHING!!!!



WTF IS THIS? you guys c what i c?

 

[SoulSearcher]

WTF IS THIS? you guys c what i c?

Yes.

 

[-pari-]

i think u need to change ur font?

4)Divide by the highest power of x in the denominator which is a half and then simplify from there

wouldn't the highest power be 1?

 

[bos1234]

sry.. so u divide by x.. i forgot the statment

 

[bos1234]

Do u understand everything else??

 

[-pari-]

um...yes and no :$

(2) Now draw a number line and see what happens when y<-1, -1<Y0

everythin else is the same as i did, i ended up with (1 +y)/y as well, but didn't get what to do from hree. how'd u get y< -1 ?
-----

was just doin trig n came up w/a few things....any help muchly appreciated

(1)Simplify: 1 - 2sin<SUP>2</SUP>10q
answer: cos 20q <-- how do u get that?

(2) 2sin105cos105
got the right answer: -0.5

but shudn't u be able to simplify it into 2(sin 60 + sin45)(cos 60 + cos45)
to get exact ratios
2(1/rt2 + rt3/2)(1/rt2 + 1/2) ? how come it doesn't work?

(3) Simply and answer in exact ratios: tan 120
wudnt that be tan 60 + tan 60 = rt(3) + rt(3) = 2(rt)3 ?
answer is : - rt(3)

(4) solve for -180(smaller.than.or.equal.to) x (smaller.than.or.equal.to)180
--> tan2x =1/(rt)3

i got 2 out of4 of the right answers, but how do u knw when to test for both positive & negative ? coz in this case x =15 so wud'nt u only test for when 'tan' is positive? ie in quadrants 1&3?

(5) find the general equation of 6sinq - 8cosq = 5
what? ....general solution....?
-----

 

[bos1234]

can you turn your private messages on!!!!!!

i put -1 becduase it makes the function 0

 

[bos1234]

(2) 2sin105cos105
got the right answer: -0.5

but shudn't u be able to simplify it into 2(sin 60 + sin45)(cos 60 + cos45)
to get exact ratios
2(1/rt2 + rt3/2)(1/rt2 + 1/2) ? how come it doesn't work?

(3) Simply and answer in exact ratios: tan 120
wudnt that be tan 60 + tan 60 = rt(3) + rt(3) = 2(rt)3 ?
answer is : - rt(3)

(4) solve for -180(smaller.than.or.equal.to) x (smaller.than.or.equal.to)180
--> tan2x =1/(rt)3

i got 2 out of4 of the right answers, but how do u knw when to test for both positive & negative ? coz in this case x =15 so wud'nt u only test for when 'tan' is positive? ie in quadrants 1&3?

(5) find the general equation of 6sinq - 8cosq = 5
what? ....general solution....?
-----

You cant change the angles. They are fixed!
sin105 is sin(60+45) = What is that identity? Whats the expansion of sin(a+b)? its the same thing here. Its like SinaSinb + CosaCosb


3)Do you know the all stations to central thing?
tan120 is in the 2nd quadrant
tan120 is the same as tan(180-60)
tan(180-60) = tan-60 = -rt3

 

[Riviet]

(1)Simplify: 1 - 2sin<SUP>2</SUP>10q
answer: cos 20q <-- how do u get that?

1-2sin²A = cos(2A)

Also, 2cos²A-1=sin(2A)

(2) 2sin105cos105
got the right answer: -0.5

but shudn't u be able to simplify it into 2(sin 60 + sin45)(cos 60 + cos45)
to get exact ratios
2(1/rt2 + rt3/2)(1/rt2 + 1/2) ? how come it doesn't work?

sin(A+B) =/= sinA + sinB
cos(A+B) =/= cosA + cosB

(3) Simply and answer in exact ratios: tan 120
wudnt that be tan 60 + tan 60 = rt(3) + rt(3) = 2(rt)3 ?
answer is : - rt(3)

tan(A+B) =/= tanA + tanB

tan(120) = tan(180-60)
=-tan60 [since tan is negative in the second quadrant]
=-rt(3)

(4) solve for -180(smaller.than.or.equal.to) x (smaller.than.or.equal.to)180
--> tan2x =1/(rt)3

i got 2 out of4 of the right answers, but how do u knw when to test for both positive & negative ? coz in this case x =15 so wud'nt u only test for when 'tan' is positive? ie in quadrants 1&3?

Given -180<x<180,
then -360<2x<360

Now tan2x =1/rt3
so 2x=30, 210, -150, -330
x=15, 105, -75, -165

 

[ih8skool]

1-2sin²A = cos(2A)

Also, 2cos²A-1=sin(2A)

sin(A+B) =/= sinA + sinB = Sin(A+B) use double angle formula
cos(A+B) =/= cosA + cosB

Sin(A+B)

 

[-pari-]

hmkay..... i thnk i'm doin something wrong..
what steps do u take when u do gte something like....

tan 120?

you first look at the angle.....120. lies in 2nd quad.
then u look at the tan...and say well in the 2nd quad...tan is negative.....
so..then.....? :S

 

[-pari-]

ooohh okay so for things like tan 120 u use the identities

tan (180-A)
tan (180 + A)
tan (360 - A)

and take it from there.

amirite?

 

[bos1234]

yes. now u got it how about if the agnle is bigger than 360 degrees....

practice these alot because soon the school will start doing integration of trig and differentiation of trig and it requires alot of these ideas

goodluck cya man

 

[-pari-]

^ if the angle is >360, do you say for eg sin (360 + A) ?


~ and if it's something like sin^2(225)

.....do you take the square root of the whole thing including the angle?