HL Maths 2018 discussion

Thr000w

PatsyJ wrote: »

Anyone?

Multiply both sides by (the denominator (bottom part of the fraction) that contains x). In this case, you multiply both sides by (x + 2) squared. Because it's squared, it has to be positive, so the inequality signs don't flip like they would if you multiplied by a negative number. You can't cross multiply, because you don't know if the denominator containing x is negative or not, and if it is then you'd have to flip the inequality sign.

Once you solve the ensuing quadratic, the answer is -9 ≤ x ≤ -2 - but the question tells us that x can't equal 2, as the denominator would then be zero and make the fraction undefined, so it's -9 ≤ x < 2.

japester

PatsyJ wrote: »

Anyone?

I would have looked at it this way:


(2x-3)/(x+2) >= 3


so either



1. 2x-3 >= 3(x+2), but this is possible only provided both (2x-3) and (x+2) are simultaneously >=0


or


2. 2x-3 <= 3(x+2), but this is possible only provided both (2x-3) and (x+2) are simultaneously <=0


Examining possibility 1 further and solving the inequality will lead to x <= -9
However (2x-3) and (x+2) must also be simultaneously >= 0 in order for this possibility to work out. If (2x-3) >=0 then x >= 3/2 and if (x+2) >=0 then x >= -2. In order for x to be both >= 3/2 and x >= -2 simultaneously, the only possibility is that x >= 3/2. However we already found that the inequality solution is that x<= -9 and as there is no overlap between the sets x >= 3/2 and x<= -9 then there are no possible solutions for x in this case.


Examining possibility 2 further and solving the inequality will lead to x >= -9
However (2x-3) and (x+2) must also be simultaneously <= 0 in order for this possibility to work out. If (2x-3) <=0 then x <= 3/2 and if (x+2) <=0 then x <= -2. In order for x to be both <= 3/2 and x <= -2 simultaneously, the only possibility is that x <= -2. However we already found that the inequality solution is that x>= -9 and as there is overlap between these sets then there we have possible solutions for x in this case. So the possible solution set is -9 <= x <= -2 but the question tells us x cannot be -2 (as division by zero would ensue in this case) and finally we have the possible solution set as -9 <= x < -2


hope this helps

jeonahr

Thr000w wrote: »

Multiply both sides by (the denominator (bottom part of the fraction) that contains x). In this case, you multiply both sides by (x + 2) squared. Because it's squared, it has to be positive, so the inequality signs don't flip like they would if you multiplied by a negative number. You can't cross multiply, because you don't know if the denominator containing x is negative or not, and if it is then you'd have to flip the inequality sign.

Once you solve the ensuing quadratic, the answer is -9 ≤ x ≤ -2 - but the question tells us that x can't equal 2, as the denominator would then be zero and make the fraction undefined, so it's -9 ≤ x < 2.

I was just doing the papers there (it's been assigned to my 5th year class, now 6th year :rolleyes: ) and I totally missed the fact that x can't be equal to 2.

Boomer18

Can anybody please tell me what constitutes a pass in Foundation Maths with the new marking scheme?

ger664

Q5 Paper 2
(a)
sub in (-2,1) into 2x+3y+1=0 and verify its true
(b)
Line AB is perpendicular. Slope m = -2/3 slope AB = 3/2
Equ of AB 3x-2y+8 = 0
Point B got from sim equ with n 2x+3y-51 = 0 B = (6,13)
(c)
Radius ratio is 1:3 so is the Diameter thus the |AB| = Radius of small cicle * 8
Length AB = 4sqrt13 radius = sqrt13/2
Center lies on the point thats splits the line AB into a ratio of 1:7
center = (-1,5/2)
Equation of circle (x+1)^2 + (y-5/2)^2 = 13/4
Expanded out to x^2 + y^2 +2x -5y + 4 =0

Graph of Question

japester

ger664 wrote: »

Q5 Paper 2
(a)
sub in (-2,1) into 2x+3y+1=0 and verify its true
(b)
Line AB is perpendicular. Slope m = -2/3 slope AB = 3/2
Equ of AB 3x-2y+8 = 0
Point B got from sim equ with n 2x+3y-51 = 0 B = (6,13)
(c)
Radius ratio is 1:3 so is the Diameter thus the |AB| = Radius of small cicle * 8
Length AB = 4sqrt13 radius = sqrt13/2
Center lies on the point thats splits the line AB into a ratio of 1:7
center = (-1,5/2)
Equation of circle (x+1)^2 + (y-5/2)^2 = 13/4
Expanded out to x^2 + y^2 +2x -5y + 4 =0

Graph of Question

Many thanks for that Ger, I can confirm I got the same results as you for it