yr 11 revision - questions (1 Viewer)

[-pari-]

absolute values:

(1)
textbook:

if a > 0 then |a| = a
if a< 0 then |a| = -a <-- i never get that :mad1: . help?

so by the 2nd definition when solving equations with absolute values we always need to test the positive and negative...

but i dont get the 2nd definitiion so i dont really see why we test for: -|a|

eg

|x-3| = x
3 - x

i got the answer right but i dnt get why i test for

a) x-3

& b) - x + 3 ? why not just (a)?

(2)
| t + 2|+ |3t -1| < 5

answer: -1 < t <1

(3)
|x-3| + |x-4| = |x-2|

(answer: no solutions.... why?)

4) solve:
x^(3/2) = 1/8

 

[ronaldinho]

what are the answers to these?

hey how did u get for question 3. as no solution??

how about if x = 5?

 

[-pari-]

those are textbook answers so thats what i'm asking.....how'd those answers come about

answer to (1) was x = 1

answer to (4) ...x = 1/4

 

[SoulSearcher]

For (4), take natural logarithms of each side, and then simplify using log laws to get x = 1/4

 

[bos1234]

i dont udnerstand y they say like this

if a > 0 then |a| = a
if a< 0 then |a| = -a

but for question 1...

denom. cannot = 3

so draw a number line

12345

so take a number less than 3... 2

if u take 2 then denominator is positive.. so just multiply out like usual and solve...(x-3)(x+1).. and we said x cannot = 3 so -1 is a possible answer

take a number greater than 3.. 4

denominator is negative.. so at the numerator its -x+3 = x
.....................................................................................3 - x

and solve..

so u should have x = 1 and x = -1

if u sub in -1 it doesnt work so 1 is the answer...

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

correct any mistakes if found plz

 

[jyu]

absolute values:

(1)
textbook:

if a > 0 then |a| = a
if a< 0 then |a| = -a <-- i never get that :mad1: . help?

Consider e.g. a = -108, do you agree that -108 < 0.

Now what is |-108|?

Of course |-108| = 108 = -(-108),

i.e. |a| = -a if a < 0.

:wave:

 

[jyu]

absolute values:

(1)
|x-3| = x
3 - x


(2)
| t + 2|+ |3t -1| < 5



(3)
|x-3| + |x-4| = |x-2|



(4) solve:
x^(3/2) = 1/8

Let's do the last one first.

x^(3/2) = 1/8 , .: x = (1/8)^(2/3)
= (1/2^3)^(2/3)
=(2^(-3))^(2/3)
=2^(-2)
=(1/2)^2
=1/4

:wave:

 

[jyu]

(1)

|x-3|/(3-x) = x,
.: |x-3| = x(3-x) and x =/= 3.

Two possibilies: x-3 > 0 or x-3 < 0.

If x-3 > 0, then x-3 = x(3-x) and x > 3.
x-3 = 3x - x^2
x^2 -2x - 3 = 0
(x-3)(x+1)=0
.: x = 3 or x = -1.
Both are not the right solutions because x > 3.

If x-3 < 0, then -(x-3) = x(3-x) and x < 3.
x^2 - 4x + 3 =0
(x-3)(x-1) = 0
.: x = 3 or x = 1
x = 1 is the only correct solution because x < 3.

Take a break :wave:

The other two require a fair bit of typing. I shall post them tomorrow/night unless some one does them before.

 

[jyu]

:wave:

 

[chousta]

lol jyu you are a machine, you should get paid for answerin questions. I think you answer nearly all questions posted these days.


nice work

 

[jyu]

:wave:

 

[jyu]

lol jyu you are a machine, you should get paid for answerin questions. I think you answer nearly all questions posted these days.


nice work

Thanks for your kind words

 

[-pari-]

legend jyu


thanks heaps!

 

[-pari-]

*sigh* mkay i'm back lol

functions and graphs:


draw the graph of y = |x| + 3x - 4

with this, since there's |x| ...an absolute value, dont we have to draw the curve of both
y = x + 3x - 4 as well as
y = -x + 3x - 4 ?

shouldn't there be two curves on the graph?
so why does the answer only show the curve for y = x + 3x - 4?

find the domain and range of
y = 1
x^2 - 1

i know that
x =/= 1 or -1
y =/= 0

but it ALSO says ( in the answers) y (smaller than or equal to) -1 ... <-- why this?

how do u find the range of something?

eg:

y = 2x + 7
x + 3

x =/= -3

but for range it says y =/= -2....
how do u get that?



sketch y = |x|/x^2

shouldn't there be a hyperbola in every quadrant?
coz y = 1/x gives a hyperbola in quadrants 1 & 3
y = -1/x gives a hyperbola in quads 2& 4?

answer shows hyperbolas only in quads 1& 2.

 

[jyu]

*sigh* mkay i'm back lol

functions and graphs:


(1) draw the graph of y = |x| + 3x - 4

with this, since there's |x| ...an absolute value, dont we have to draw the curve of both
y = x + 3x - 4 as well as
y = -x + 3x - 4 ?

shouldn't there be two curves on the graph?
so why does the answer only show the curve for y = x + 3x - 4?

(2) find the domain and range of
y = 1
x^2 - 1

i know that
x =/= 1 or -1
y =/= 0

but it ALSO says ( in the answers) y (smaller than or equal to) -1 ... <-- why this?

(3) how do u find the range of something?

eg:

y = 2x + 7
x + 3

x =/= -3

but for range it says y =/= -2....
how do u get that?



(4) sketch y = |x|/x^2

shouldn't there be a hyperbola in every quadrant?
coz y = 1/x gives a hyperbola in quadrants 1 & 3
y = -1/x gives a hyperbola in quads 2& 4?

answer shows hyperbolas only in quads 1& 2.

(1) The graph consists of two sections.

For x < 0, y = -x + 3x - 4
For x >/= 0, y = x + 3x - 4

(2) Sketch the graph of y = 1/(x^2 - 1), see that y > 0 or y < -1.

.: range is {x: x > 0} U {x: x < -1}.

(3) Similar to (2), graph the function to find out.

(4) 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.

:wave:

 

[bos1234]

*sigh* mkay i'm back lol

how do u find the range of something?

eg:

y = 2x + 7
x + 3

x =/= -3

but for range it says y =/= -2....
how do u get that?



quote]

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

I thought range would b y not equal to 0??

 

[jyu]

*sigh* mkay i'm back lol

how do u find the range of something?

eg:

y = 2x + 7
x + 3

x =/= -3

but for range it says y =/= -2....
how do u get that?



quote]


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

I thought range would b y not equal to 0??

y = (2x + 7)/(x + 3) can be written as y = 2 + 1/(x + 3).

The graph of y = (2x + 7)/(x + 3) is the translation of the graph of 1/x left 3 and up 2. The asymptotes are x = -3 and y = 2.

Hence the range is R\{2}.

Note: not -2.

:wave:

 

[-pari-]

(2) Sketch the graph of y = 1/(x^2 - 1), see that y > 0 or y < -1.
.: range is {x: x > 0} U {x: x < -1}.
(3) Similar to (2), graph the function to find out.

The graph of y = (2x + 7)/(x + 3) is the translation of the graph of 1/x left 3 and up 2. The asymptotes are x = -3 and y = 2.

is there a way other than graphing the function to find out the range?
like is there an algebraic method?


also...

:wave:

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"?

 

[Rekkusu]

Wow nice work with the detailed answers Jyu!!

Asymptotes are quite problematic, especially during HSC periods, it's a good pointer to look out for these in your exam. [Uni mathematic lecturers loved placing asymptote-related questions in exams]

 

[jyu]

is there a way other than graphing the function to find out the range?
like is there an algebraic method?

:wave: