Mathematics: Post your doubts here!

[Rutzaba]

I think this will help:

• If (b^2-4ac) > 0 , the equation will have 2 roots.
• If (b^2-4ac) < 0 , the equation will have no roots.
• If (b^2-4ac) = 0 , the equation will have 1 root.

graph hunz ... they need graph

 

[Maz]

graph hunz ... they need graph

Graphs for what?

 

[Rutzaba]

Graphs for what?

never mind

 

[Maz]

never mind

huh??? now i'm confused.

 

[Maz]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_12.pdf

There's a problem in 6(i) right?

 

[Rutzaba]

Guys tell me wat u dun get? panda @syed1996
U kno 2 values of k but dunno wat in equality to put ryt?

 

[Magenta]

Somebody please do this!
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_11.pdf
Q7 b)

 

[Maz]

Somebody please do this!
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_11.pdf
Q7 b)

You should remember that |r| < 1 for a convergant sequence
In this question r = 1/3 tan^2 θ
therefore, 1/3 tan^2 θ < 1
tan^2 θ < 3
tan θ < [sqrt]3
θ < π/3

 

[doctorofelectronics]

kx^2+1=kx
kx^2-kx+1=0 for this b^2-4ac must be less than zero (<0)
(-k^2)-(4*k*1)<0
k^2-4k<0
k(k-4)<0
so k can be 0 or 4
since the coefficient of k^2 is positive the graph should be a parabola.
The answer is 0<k<4
Please check the graph
The graph is of y=k(k-4)
since we want k(k-4)<0
we want y<0 in our graph
Sorry for the bad drawing. Hope it helps

 

[ZainH]

If f(x)=x²-4x+7 for x>2 and g(x)= x-2 for x>2

The function h is such that f=hg and the domain of h is x>0
Q. Obtain an expression for h(x) .

 

[doctorofelectronics]

Here is the graph

 

[doctorofelectronics]

If f(x)=x²-4x+7 for x>2 and g(x)= x-2 for x>2

The function h is such that f=hg and the domain of h is x>0
Q. Obtain an expression for h(x) .

h(x)=x^2+3

 

[ZainH]

h(x)=x^2+3

How? Explain please =)

 

[Manobilly]

Guys tell me wat u dun get? panda @syed1996
U kno 2 values of k but dunno wat in equality to put ryt?

Can u explain the graph part? How we put the equality sign

 

[daredevil]

If f(x)=x²-4x+7 for x>2 and g(x)= x-2 for x>2

The function h is such that f=hg and the domain of h is x>0
Q. Obtain an expression for h(x) .

this question was solved a while ago in this thread i think ...

 

[PANDA-]

Guys tell me wat u dun get? panda @syed1996
U kno 2 values of k but dunno wat in equality to put ryt?

Exactly

 

[asd]

Somebody please do this!
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_11.pdf
Q7 b)

Well for part one you only have to assume that the common ratio r is greater that 0 and less than 1. That goes for a convergent progression.
so your r is= 1/3tan^2(theta)
0<1/3tan^2(theta)<1u'd get the answer.
for (ii) Sum to infinity= a/(1-r)
-> 1/[1-1/3tan^2(pi/6)]

 

[A star]

this question was solved a while ago in this thread i think ...

yeah i dont get why they donot search the thread

 

[haha101]

someone help mee

 

[doctorofelectronics]

Its easy. just imagine in which function of x would you substitute x-2 to get x^2-4x+7
here you need to square the x in x-2 because it is x squared in f(x)
So you square the bracket (x-2) ie: (x-2)(x-2)
You get x^2-4x+4
Now to get f(x) just add 3 to the equation!