Mathematics paper 3, modulus inequalities.

[OakMoon!]

How do you find the critical value? My concept is unclear.

An example question: Q2 of
http://www.xtremepapers.me/CIE/Internat ... 6_qp_3.pdf

 

[WellWIshER]

U DIDNT GIV ALEVELS DIS YEAR?>

 

[ks136]

take x-1 greater than 2x and then less than 2x....u'll have range of x...
after dat...put one value of x from each range and see whether or not they both fulfils the given requirements ..... the one which does...is d only range of x

 

[OakMoon!]

@Wellwisher=No, I'll be giving As and A2 together next year. I am preparing for the finals of my school exams.
ks136=I am looking for the number line method, this is quite obvious. Do you know that? And what about the critical value? Plus, why didn't you use -2x in one of the inequalities you made? :S

 

[ks136]

greater dan -2x and less than 2x is d same

 

[OakMoon!]

Are you sure? x-1<2x and x-1>-2x is same? I don't think so. x-1>-2x will become 1-x<2x, which is NOT the same as x-1<2x.

 

[ks136]

yup sorry my mistake...

 

[ks136]

yup so solve dem and u'll get
x>-1 or x<1/3
now draw on number line.... somewhat like this 1 below....hope u understand
========================............
..................================
.................-1...............1/3
____________________________________

now d range will be 1 dominated by both lines.... i.e -1<x<1/3

 

[OakMoon!]

You are doing it wrong man. It'll be x-1<2x and x-1>-2x. And so you get x>1/3 and x>-1. The final answer is x>1/3 and x<-1. My question is that why do you eliminate x<-1 and the final answer is only x>1/3. I found out the answers correctly but no idea why x<-1 was eliminated. You didn't find the correct answers man. Thanks, anyways.

 

[ruzache]

This method works:

Square both sides of the inequality. Don't care if both sides don't have the modulus sign. THen you'd get: (2x)^2 = (x-1)^2 which is when expanded: 4x^2 =x^2 -2x +1 => 3x^2+2x-1=0 (3 x-squared plus 2x minus 1 equals 0)

and then you solve the quadratic equation and you'll get two roots which are x=1/3 and x=-1

sketch a graph and youll find out that the region suiting the inequality would be x<-1 and x>1/3

Now we have to check this with the original inequality. Therefore substitute a value lesser than -1 first:

2(-2) > |-2-1| <----- This is not true because -4 is not greater than 3

Then substitute a value in between the crit values:

2(0) > | 0-1| <----- This is also not true because 0 is not greater than 1

Then substitute a value greater than 1/3

2(2) > | 2-1 | <---- This is true because 4 is greater than 1

and hence the answer is only x>1/3 because that's the only one which suits the original modulus inequality. hope that was helpful. this method always works guaranteed

 

[OakMoon!]

Yes, ruzache, that's what is taught in the school. Thankyou, for the long reply. I know this method and I have been solving modulus questions like this for the past 6 months. What I want to know is that how do you find the critical value and stuff. I need to learn the exact critical value method and not only "a" method.

 

[ruzache]

So you make the inequality into a quadratic equation with one side being zero and solve the quadratic equation using the calculator or the formula? What's another method to that? Are you talking about |x|=2 being evaluated as -2<x<2 ? That method is only good for linear ones. For AL standard the quadratic solving method is enough right?

 

[OakMoon!]

No, this method also works on other questions. And no, it's not enough. What if you square on both sides and you get a cubic or a quartic equation. So then you need the other method. The marking scheme mentions a 'critical' value. I know what a critical value is but I don't get how you can eliminate the other value by just knowing the critical value. You don't need to put in the values and check with that method.

Help from anyone else with the knowledge of the method of critical value will be appreciated.

 

[ruzache]

Nope never came across a sum that results in a cubic or quartic when you square. And I did P3 this year and there was a simple modulus question with modulus sign on both sides eliminating the need to substitute and check even.

 

[OakMoon!]

Okay! Thankyou for your help. But I already know this method. I don't want to do the questions on the paper (I'll deal with them), I want to learn; the method.