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Miscellaneous Questions (1 Viewer)

pomsky

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Yooo :) So in prep for trials, i managed to stub my metaphorical brain into COUGHalmosteverythingCOUGH. Will be much appreciated if you guys could give me hints and tips to approaching these questions (and a walkthrough if you're feeling particularly generous). Or if you'd just like to have a mental workout..? :p

 

Crisium

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Thanks Integrand! :)
Factor out the two in the denominator

(x^2) / 2(1 - cos2x)

Since 1 - cos2x = 2sin^2(x)

(x^2) / 2(2sin^2(x))

(x^2) / 4sin^2(x)

Factor out the one-quarter

[1 / 4] x [ (x^2) / sin^2(x)]

Take the squared outside of the second bracket

[ 1 / 4] x [ x / sinx ]^2

You know that the limit of x / sinx approaches zero = 1 from the trigonometric function topic

NOTE: Keep that limit sign of x approaches zero should be placed in front of the working out until this point (I can't LaTex LOL)

Therefore

[ 1 / 4] x [ 1 ]^2

= [ 1 / 4] x 1

= 1 / 4

Is this right?
 
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Can you guys please help me with a question I'm currently stuck on? file:///Users/kim/Desktop/Outlook.com%202/IMG_5506.JPG
Specifically questions C) and G) and how do I tackle inequations and equations questions that has double or triple absolute values?
 
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Could you explain why x doesn't equal to one? Is it because the denominator can't be zero or else it would be undefined?
 
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What confuses me is that, there is two absolute values. Why isn't there two restrictions, but you only mentioned one "x cannot equal to 1", could you explain that to me?
 

rand_althor

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I'm assuming you mean the restriction where x cannot be -1. I didn't consider it as when you end up with your final answer, you realise that x is always positive and hence it won't ever be -1 anyway.
 

iamaloser17

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Part g (Sorry for no LaTeX):

|x-1| + |x+3| > 6

Check the critical points
x=1: |1-1| + |1+3| /> 6 --> x=1 is not a solution
x=-3: |-3-1| + |-3+3| /> 6 --> x=-3 is not a solution

For all x<-3, x-1<0 and x+3<0
-(x-1) - (x+3) > 6
-x+1 - x-3 >6
-2x > 8
x < -4
Therefore x<-4 is a solution.

For -3 < x < 1, x-1 < 0 and x+3 > 0
-(x-1) + x+3 >6
4 > 6
Therefore -3 < x < 1 is not a solution as 4 /> 6

For x>1, x-1>0 and x+3>0
x-1 + x+3 >6
2̶x̶ ̶>̶ ̶8̶ ̶
x̶ ̶>̶ ̶4̶ ̶
T̶h̶e̶r̶e̶f̶o̶r̶e̶ ̶x̶>̶4̶ ̶i̶s̶ ̶a̶ ̶s̶o̶l̶u̶t̶i̶o̶n̶.̶
Edit:
2x > 4
x > 2
Therefore x>2 is a solution.

Therefore the solution to |x-1| + |x+3| > 6 is x<-4, x̶>̶4̶ x>2. I hope I'm right.
 
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For g, these are the conditions for which the inequality holds

<a href="http://www.codecogs.com/eqnedit.php?latex=|x-1|&space;&plus;&space;|x&plus;3|&space;>&space;6&space;\\&space;(|x-1|&space;&plus;&space;|x&plus;3|)^{2}&space;>&space;36&space;\\&space;\\&space;x^{2}&space;-2x&space;&plus;&space;1&space;&plus;&space;2|x-1||x&plus;3|&space;&plus;&space;x^{2}&plus;6x&plus;9>36&space;\\&space;$Now,&space;middle&space;term&space;is&space;positive&space;for&space;any&space;value&space;of&space;x&space;because&space;of&space;the&space;absolute&space;value&space;signs&space;\\&space;\\&space;\therefore&space;x^{2}&space;-2x&space;&plus;&space;1&plus;&space;2(x^{2}&plus;2x-3)&space;&plus;&space;x^{2}&plus;6x&plus;9>36&space;\\&space;\\4x^{2}&plus;&space;8x-32>&space;0&space;\\&space;4(x^{2}&plus;2x&space;-8)>&space;0&space;\\&space;4(x&plus;4)(x-2)>0&space;\\&space;\therefore&space;x>2\\&space;x<-4" target="_blank"><img src="http://latex.codecogs.com/gif.latex?|x-1|&space;&plus;&space;|x&plus;3|&space;>&space;6&space;\\&space;(|x-1|&space;&plus;&space;|x&plus;3|)^{2}&space;>&space;36&space;\\&space;\\&space;x^{2}&space;-2x&space;&plus;&space;1&space;&plus;&space;2|x-1||x&plus;3|&space;&plus;&space;x^{2}&plus;6x&plus;9>36&space;\\&space;$Now,&space;middle&space;term&space;is&space;positive&space;for&space;any&space;value&space;of&space;x&space;because&space;of&space;the&space;absolute&space;value&space;signs&space;\\&space;\\&space;\therefore&space;x^{2}&space;-2x&space;&plus;&space;1&plus;&space;2(x^{2}&plus;2x-3)&space;&plus;&space;x^{2}&plus;6x&plus;9>36&space;\\&space;\\4x^{2}&plus;&space;8x-32>&space;0&space;\\&space;4(x^{2}&plus;2x&space;-8)>&space;0&space;\\&space;4(x&plus;4)(x-2)>0&space;\\&space;\therefore&space;x>2\\&space;x<-4" title="|x-1| + |x+3| > 6 \\ (|x-1| + |x+3|)^{2} > 36 \\ \\ x^{2} -2x + 1 + 2|x-1||x+3| + x^{2}+6x+9>36 \\ $Now, middle term is positive for any value of x because of the absolute value signs \\ \\ \therefore x^{2} -2x + 1+ 2(x^{2}+2x-3) + x^{2}+6x+9>36 \\ \\4x^{2}+ 8x-32> 0 \\ 4(x^{2}+2x -8)> 0 \\ 4(x+4)(x-2)>0 \\ \therefore x>2\\ x<-4" /></a>
 

DatAtarLyfe

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Part g (Sorry for no LaTeX):

|x-1| + |x+3| > 6

Check the critical points
x=1: |1-1| + |1+3| /> 6 --> x=1 is not a solution
x=-3: |-3-1| + |-3+3| /> 6 --> x=-3 is not a solution

For all x<-3, x-1<0 and x+3<0
-(x-1) - (x+3) > 6
-x+1 - x-3 >6
-2x > 8
x < -4
Therefore x<-4 is a solution.

For -3 < x < 1, x-1 < 0 and x+3 > 0
-(x-1) + x+3 >6
4 > 6
Therefore -3 < x < 1 is not a solution as 4 /> 6

For x>1, x-1>0 and x+3>0
x-1 + x+3 >6
2x >8
x > 4

Therefore x>4 is a solution.

Therefore the solution to |x-1| + |x+3| > 6 is x<-4, x>4. I hope I'm right.
Minor error in your algebra. The answer is x<-4, x>2
 
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Do you guys have any advice on tackling double/triple absolute equations/inequations??/
 

iamaloser17

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Do you guys have any advice on tackling double/triple absolute equations/inequations??/
Make sure you understand the concepts and the methods in tackling these questions. Look for a variety of questions, doing them and just practicing will help. And don't make silly mistakes like me lel. Nothing is better than simply practicing honestly.
 

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