Mathematics: Post your doubts here!

[Rutzaba]

The answer I get if i do it this way is 4pi/9 or 1.89 whereas the marking scheme gives the answer 1.5Pi or 4.71. :/

one eq is y=2
square = 4
integrate 4x for the limits 0 and 1
you get 4
4(1) - 4(0) =4
then sec eq (3x+1) ^1/2
square = 3x+1
integrate (3/2) x^2 +x apply limits 0 and 1
(3/2) (1)^2 +(1) - (3/2) (0)^2 +(0)

5/2
then (4- 5/2) pi
1.5 pi ^_^

 

[A star]

Can someone please solve (Nov 2012, P13) Q2, and Q7 part (ii), with full working and explanations? Thank you. There seems to be some problem with the mark scheme.

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_13.pdf

Q2 find f '(x) and see that you will get two terms in x with neg sign hence for all positive values of x it will be a decreasing function
7i
y=5-x
y=11-x^2
sub y you will get 5-x=11-x^2 solve it and take the positive value of x as p and its y coordinate as q

 

[tom ed]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_qp_63.pdf
q2 and q6 plz

 

[qwerty123123]

I usually get confused when we are asked to find the range of trigonometric functions.. its just too complicated for me! What do I have to consider when solving such questions?

Look at these two functions:
f(x) = 3 - 2sinx for 0 ≤ x < pi
f(x) = 3 - 2sin(1/2x) for 0 ≤ x < pi.

How different would the range of the two functions be?
regards salvatore

The range for those two functions is going to be the exact same, the only thing that is changing is the period.
-2sin(x) means the graph will be flipped and the amplitude will be twice.
3+ means the graph will be translated +3 in the y direction so the range for both is 1=<f(x)<=5

 

[Maz]

I usually get confused when we are asked to find the range of trigonometric functions.. its just too complicated for me! What do I have to consider when solving such questions?

Look at these two functions:
f(x) = 3 - 2sinx for 0 ≤ x < pi
f(x) = 3 - 2sin(1/2x) for 0 ≤ x < pi.

How different would the range of the two functions be?
regards salvatore

qwerty123123 gave you the correct answer, but if you are still confused as to how to go about it then here goes.
What i do is substitue the whole sinx, sin(1/2x), or for that matter cosx or any other cos or sin function by 1 for the maximun value and -1 for the minimum value. I guess you can check it.
And then it's child's play.
3 - 2(1) = 1 & 3-2(-1) = 5 for both of them.
It's only applies for sin/cos not for tan.

 

[Rutzaba]

The range for those two functions is going to be the exact same, the only thing that is changing is the period.
-2sin(x) means the graph will be flipped and the amplitude will be twice.
3+ means the graph will be translated +3 in the y direction so the range for both is 1=<f(x)<=5

qwerty123123 gave you the correct answer, but if you are still confused as to how to go about it then here goes.
What i do is substitue the whole sinx, sin(1/2x), or for that matter cosx or any other cos or sin function by 1 for the maximun value and -1 for the minimum value. I guess you can check it.
And then it's child's play.
3 - 2(1) = 1 & 3-2(-1) = 5 for both of them.
It's only applies for sin/cos not for tan.

salvatore

 

[qwerty123123]

qwerty123123 gave you the correct answer, but if you are still confused as to how to go about it then here goes.
What i do is substitue the whole sinx, sin(1/2x), or for that matter cosx or any other cos or sin function by 1 for the maximun value and -1 for the minimum value. I guess you can check it.
And then it's child's play.
3 - 2(1) = 1 & 3-2(-1) = 5 for both of them.
It's only applies for sin/cos not for tan.

Good point, that would be a quicker method.

 

[Yousif Mukkhtar]

Can someone explain Q7) b) i)
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_11.pdf

 

[igcsedude_Jor]

hello
can someone please explain to me everything i need to know about completing the sqaure and coefficient?
please can someone explain to me this: w12 qp 11 question 10 parts 1 and 2 (btw whats domain?) and question 4
and this: w2008 q1, q5, and q10 part 3
thank u soo much

 

[A star]

Can someone explain Q7) b) i)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_11.pdf

please take the bother of reading past thread as i have already solved this for another person i should think it is between 8 -5 pages back.

 

[qwerty123123]

Can someone explain Q7) b) i)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_11.pdf

u1 = a = 1
u2 = ar = (1/3)tan^2(theta)
a = 1 therefore 1/3tan^2(theta) = r
For a convergent series abs(r) < 1 or -1< r<1 but the limits define are 0<theta<pi/2
So for r must be greater than 0 and less than 1 because our limits say we can't have a negative value for theta.
so 1/3(tan(theta))^2 < 1
So (tan(theta))^2 < 3
so theta < tan^-1(sqrt(3))
So theta < pi/3
And 1/3(tan(theta))^2 > 0
So theta > tan^-1(sqrt(3x0)) = tan^-1(0) = 0
So 0<theta<pi/3

 

[daredevil]

u1 = a = 1
u2 = ar = (1/3)tan^2(theta)
a = 1 therefore 1/3tan^2(theta) = r
For a convergent series abs(r) < 1 or -1< r<1 but the limits define are 0<theta<pi/2
So for r must be greater than 0 and less than 1 because our limits say we can't have a negative value for theta.
so 1/3(tan(theta))^2 < 1
So (tan(theta))^2 < 3
so theta < tan^-1(sqrt(3))
So theta < pi/3
And 1/3(tan(theta))^2 > 0
So theta > tan^-1(sqrt(3x0)) = tan^-1(0) = 0
So 0<theta<pi/3

ohh wow nice explaination but are we suppposed to know the convergent thing?? I never did it!! :O A star

 

[Khatmalization]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_qp_13.pdf

Q2 please anyone with full steps
will be very grateful!

 

[PANDA-]

ohh wow nice explaination but are we suppposed to know the convergent thing?? I never did it!! :O A star

I solved that question as syed1995 asked for it a few pages back.
Yes you're supposed to know that a convergent geometric sequence is a sequence that has a sum to infinity, this is available if the common ratio is less than 1 and greater than -1.

 

[qwerty123123]

ohh wow nice explaination but are we suppposed to know the convergent thing?? I never did it!! :O A star

Yes, It says on the formula sheet That |r|<1 for a sum to infinity. And a convergent series is a series which converges to 1 value and thus is the sum to infinity series.
Are we supposed to know r for an oscillating divergent and divergent series too? I guess for oscillating divergent r < -1 or r <= -1 for non-divergent and non-convergent and for divergent r > 1 or r < -1 right?

 

[qwerty123123]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_13.pdf

Q2 please anyone with full steps
will be very grateful!

OK so (x-(2x^-1))^6
It asks you for decending powers of x so you want x^6 to be your first term
So then its x^6 + 6C1(((x)^5)((-2x)^1)) +6C2(((x)^4)((-2x^-1)^2))
that simplifies to x^6 -12x^4 + 60x^2

ii)
so (1+x^2)(x-2x^-1)^6 = (1+x^2)(x^6-12x^4+60x^2......)
You only want the powers of 4 so you need x squared multiplied by x squared and x multiplied by x to the power of four.
So = -12x^4 + 60x^4
= 48x^4
So co-efficient = 48

 

[Khatmalization]

OK so (x-(2x^-1))^6
It asks you for decending powers of x so you want x^6 to be your first term
So then its x^6 + 6C1(((x)^5)((-2x)^1)) +6C2(((x)^4)((-2x^-1)^2))
that simplifies to x^6 -12x^4 + 60x^2

ii)
so (1+x^2)(x-2x^-1)^6 = (1+x^2)(x^6-12x^4+60x^2......)
You only want the powers of 4 so you need x squared multiplied by x squared and x multiplied by x to the power of four.
So = -12x^4 + 60x^4
= 48x^4
So co-efficient = 48

Thanks alot dear, may Allah bless you.
My concepts were quite foggy in this topic and now I get it.
Thanku

 

[qwerty123123]

Thanks alot dear, may Allah bless you.
My concepts were quite foggy in this topic and now I get it.
Thanku

your welcome

 

[syed1995]

Yes, It says on the formula sheet That |r|<1 for a sum to infinity. And a convergent series is a series which converges to 1 value and thus is the sum to infinity series.
Are we supposed to know r for an oscillating divergent and divergent series too? I guess for oscillating divergent r < -1 or r <= -1 for non-divergent and non-convergent and for divergent r > 1 or r < -1 right?

When I solved that question .. the only thing I knew that

Convergent graph means .. it is |r| < 1 it can be negative fraction as well the only thing which I didn't knew in that question was that tan^2(x) = (tan(x))^2 ,, which was the only thing which created a problem for me.. but a little research on the net solved that for me And then you confirmed my working ^^

 

[mominzahid]

one eq is y=2
square = 4
integrate 4x for the limits 0 and 1
you get 4
4(1) - 4(0) =4
then sec eq (3x+1) ^1/2
square = 3x+1
integrate (3/2) x^2 +x apply limits 0 and 1
(3/2) (1)^2 +(1) - (3/2) (0)^2 +(0)

5/2
then (4- 5/2) pi
1.5 pi ^_^

I dont get it.. Where did that 1st equation come from? and why are u subtracting the second answer from the first? :/