O-Level Additional Math

[SKT]

Hi iFruit, tks for the solution.

I think y = -10 is rejected.

Need help for another Q:
Obtain the range of values of x for which │2x² - 7│ > 1/2 (x² + 1), for -2 ≤ x ≤ 5.

TIA.

[iFruit]

Hi SKT,


1) y = -10 is valid but it just means α, β are complex numbers. For real numbers it can be ignored.

2)

│2x² - 7│ > 1/2 (x² + 1) ----> │4x² - 14│ > (x² + 1)


i.e. 4x² - 14 > (x² + 1) or -(4x² - 14) > (x² + 1)

For 4x² - 14 > (x² + 1),


4x² - x² > 15 —> x² > 5----> x < -√5 or x > √5


For -(4x² - 14) > (x² + 1),

-(4x² - 14) > (x² + 1) —> 4x² - 14 < - (x² + 1) —> 5x² < 13

so x² < 13/5 –> x < √(13/5) or x > -√(13/5)

so we have,

x < -√5 or x > √5; x < √(13/5) or x > -√(13/5)

and

-2 ≤ x ≤ 5.


Combining all we get,

-√(13/5) < x < √(13/5) and √5 < x ≤ 5

[SKT]

Can anyone help to clear my DD’s doubt for the following:


(x+1)/(x-1) = (-x-1)/(1-x), but the remainder of (x+1)/(x-1) is 2, and the remainder of (-x-1)/(1-x) is -2. Why are they not the same?

TIA.

[iFruit]

(x+1)/(x-1) is of the form


D = d.q + r where D = x+1, q =1, d = x-1, r =2

whereas (-x-1)/(1-x) is of the form

-D = -d.q -r ==> (-D) = (-d).q + (-r) where (-D) = (-x-1), q = 1, (-d) = (1-x) and -r = -2


Because the divisor is changed to -ve sign, the remainder also has to change to -ve sign so that dividend also changes sign

HTH.

SKT:

Can anyone help to clear my DD's doubt for the following:

(x+1)/(x-1) = (-x-1)/(1-x), but the remainder of (x+1)/(x-1) is 2, and the remainder of (-x-1)/(1-x) is -2. Why are they not the same?

TIA.

[benorito]

Hi,


Can you kindly help with this question :
Given that 280 and a number y have a LCM of 6160 and a HCF of 40, find the number y.

Thanks in advance!

[iFruit]

benorito:

Hi,


Can you kindly help with this question :
Given that 280 and a number y have a LCM of 6160 and a HCF of 40, find the number y.

Thanks in advance!

You need to know that product of two numbers = HCF x LCM of those two numbers


280 * y = 6160 * 40

y = (6160 * 40)/280 = 880

[benorito]

iFruit:

You need to know that product of two numbers = HCF x LCM of those two numbers


280 * y = 6160 * 40

y = (6160 * 40)/280 = 880

Thank you !!

[SKT]

Hi,

Given that x and y satisfy the simultaneous equations
mx + (m-1)y = 10,
(m-2)x + 3my = 20.

(a) If the equations have no unique solution, find the values of m.
(b) If the equations have no solutions, find the value of m.

TIA.

[iFruit]

SKT:

Hi,

Given that x and y satisfy the simultaneous equations
mx + (m-1)y = 10,
(m-2)x + 3my = 20.

(a) If the equations have no unique solution, find the values of m.
(b) If the equations have no solutions, find the value of m.

TIA.

Using Cramer’s rule

x = [30m - 20(m-1)]/ [3m² – (m² -3m +2)] = (10m+20)/ (2m² + 3m -2)

y = [20m -10(m-2)] / (2m² + 3m -2) = (10m+20)/ (2m² + 3m -2)


Recall that equation has
1) infinite solutions when (10m+20) = 0 and (2m²+ 3m -2)=0
2) No solutions when (10m+20) # 0 and (2m² + 3m -2)=0

(2m² + 3m -2)= (2m-1)(m+2) =0----> m=-2 or m=1/2


When m=-2, 10m+20 = 0,
when m=1/2, 10m+20 = 25


So infinite solutions when m=-2
No solutions when m=1/2

[atutor2001]

SKT:

Hi,

Given that x and y satisfy the simultaneous equations
mx + (m-1)y = 10,
(m-2)x + 3my = 20.

(a) If the equations have no unique solution, find the values of m.
(b) If the equations have no solutions, find the value of m.

TIA.

Below is an ELEMENTARY approach

For linear equation : y = mx +c

(a) 2 lines will have no unique solution if they are the same line i.e. both lines have the same gradient, m and the same y-intercept, c

mx + (m-1)y = 10 that is: y = -mx/(m-1) +10/(m-1)
(m-2)x + 3my = 20 that is: y = -(m-2)/(3m) + 20/(3m)

If the gradients are the same :

-mx/(m-1) = -(m-2)/(3m)
3m² = m²-2m-m+2
2m²+3m-2 = 0
(2m-1)(m+2) = 0
m = 1/2 or -2

If the y-intercepts are the same :

10/(m-1) = 20/(3m)
30m = 20m - 20
m = -2

Therefore, there is no unique solution if m = -2

(b) 2 lines will have no solution if they are parallel i.e. same gradient but different y-intercept.

Therefore, there is no solution if m = 1/2