Mathematics: Post your doubts here!

[Nibz]

Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!

Simple!

gf(x) = (2x^2 + 8x + 8 + a^2)/a^2
This can be written as: (2x^2)/a^2 + (8x)/a^2 + (8+a^2)/a^2 = 2x^2 -bx + 9
Compare the co-efficients of x^2 and x!
We have => 2/a^2 = 2
so a=1 or a= -1
and
8/a^2 = -b (put a=1 and a= -1; in both cases the answer will be -8)
so a = 1; b = -8 and a = -1; b = -8

 

[Nibz]

Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!

You should have edited your post before :|

Look, when you take square root the answer can be + or -
The square root of 1 in this question gives you +1 or -1... and putting these two values, you get b = -8

 

[justina]

Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!

oh, you mean we can just compare the left and right? no need further workings? thanks!

(sorry about the later edit ....)

 

[Nibz]

Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!

Yes, just compare the co-efficients! ( Co-efficient of x^2 with that of x^2; co-efficient of x with x)

You got that point of two values?

 

[justina]

Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!

Here's another one:

f-1 (inverse f - how do i write that?) = px + q
g-1 (inverse g) = 3-2x. Find p and q if gf9x) = 5/2 - x

So I worked that f(x) = (x-q)/p and that g(x) = (3-x)/2 ..... correct?

So that makes gf(x) = (3-((x-q)/p))/2 .... right?

Then how?

 

[Nibz]

Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!

What's the answer?
Is it p= 1/2 and q=1 ?

 

[justina]

Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!

yes

 

[Nibz]

Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!

Here's another one:

f-1 (inverse f - how do i write that?) = px + q
g-1 (inverse g) = 3-2x. Find p and q if gf9x) = 5/2 - x

So I worked that f(x) = (x-q)/p and that g(x) = (3-x)/2 ..... correct?

So that makes gf(x) = (3-((x-q)/p))/2 .... right?

Then how?

Then => (3p + q - x)/2p = (5 - 2x)/2
=> -2x + 6p + 2q = -4px + 10p
=> (again compare the co-efficents) = -2 = -4p => p=1/2
and 6p + 2q = 10p
here 2q = 2 => q=1
Clear?

 

[justina]

Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!

much thanks

 

[XPFMember]

Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!

Assalamoalaikum!!

Thnx Nibz 4 solving it...sorry i was away...so just saw the post =)

 

[basimnazir]

Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!

Hey I'm stuck on May/June 2008, I got some doubts in here, and I'm sure once cleared I will be able to do the other papers quite easily, here;
Q.1.b.iii.
Q.2.c.iii.
Q.3.b (both i and ii)
Q.4.c.
Q.7.a.ii.iii.iv.
Q.10.b.i.
If you answer me I would appreciate it very much. Anticipating a quick reply,
Basim

 

[abcde]

Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!

Paper 1 or paper 2? IGCSE or O-level?

 

[SalmanPakRocks]

Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!

which paper 1 or 2 and IGCSE or CIE ??

 

[basimnazir]

Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!

Paper 1 or paper 2? IGCSE or O-level?

which paper 1 or 2 and IGCSE or CIE ??

Whoops! It would be Paper 4;
0580/04/MJ/08

 

[basimnazir]

Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!

Hey! I solved a question;
2.c.iii.
I don't know if my procedure is correct but here's how I did:
y=m^2-4n^2 (given)
(iii) m=2x+3... n=x-1. Find y in terms of x, in its simplest forms.
I substituted the 2 equations in the given one here:
y= (2x+3)^2- 4(x-1)^2
I won't go deep, the answer I'm getting is 2x+5...is it correct? :$

 

[Nibz]

Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!

Wrong!
It's 5(4x + 1) or 20x + 5

y = (4x^2 + 12x + 9) - (4x^2 - 8x + 4)
=> (4x^2 and -4x^2 cancelled out)
=> 12x + 8x + 9 - 4
=> 20x + 5

 

[basimnazir]

Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!

I didn't understand a word :$ Won't we substitute the equations in y=m^2-4n^2? :S

 

[Nibz]

Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!

Substitute m = 2x + 3 and n = x-1 in the equation!
It would become like this => y = (4x^2 + 12x + 9) - (4x^2 - 8x + 4)

 

[basimnazir]

Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!

Mm I'm not sure if I understood it :$

 

[Nibz]

Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!

Okay!
Here's the full solution:
Substitute the values of m and n in the equation!
y = ( 2x + 3) ^2 - 4 (x-1)^2
(use this formula (a + b)^2 => a^2 + 2ab + b^2) and ( (a-b)^2 => a^2 -2ab + b^2)
=> (4x^2 + 2(2x)(3) +9) - 4(x^2 -2x + 1)
=> (4x^2 + 12x + 9) - (4x^2 -8x + 4)
=> 4x^2 + 12x + 9 - 4x^2 + 8x - 4
=> 20x + 5