Mathematics: Post your doubts here!

[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? :/

 

[daredevil]

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

y=2 is the equation of the straight line
the area under the shaded region = area under line - area under graph

that is why we had to subtract the two integrals

 

[Dug]

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 solved it but the range will be different for the second. It's 1 ≤ f(x) < 3 unless you made a typo with the inequality...

 

[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

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.

Thanks a lot.. I have one more question.
What if the functions were:
f(x) = 3 - 2sinx for 0 ≤ x < pi
f(x) = 3 - 2sinx for 0 ≤ x < 2pi
Would there be any difference in the ranges? (Notice the domain has changed)..

 

[Rutzaba]

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

wen we square a line and integrate fr given limits it gives us the volume under that p[articular line or curve
wen there is no line given and its a straight line parralel to x axis the eq of line wud be y=2

now see volume under y=2 gives us the complete vol under line ie the whole rectangle
no we dun need the non shaded region of the rectangle
non shaded region of rectangle = vol under the curve

now u kno
vol of shaded + vol of non shaded = vol of total rectangle
so vol of shaded = total (under y=2) - non shaded (under curve)
ask again if u dun get it

 

[PANDA-]

Thanks a lot.. I have one more question.
What if the functions were:
f(x) = 3 - 2sinx for 0 ≤ x < pi
f(x) = 3 - 2sinx for 0 ≤ x < 2pi
Would there be any difference in the ranges? (Notice the domain has changed)..

I see you haven't understood my answer through PM's.
Anyways, the range won't change.

 

[Rutzaba]

guys wch is the toughest p1 ever set? gimme the link...

 

[syed1995]

I see you haven't understood my answer through PM's.
Anyways, the range won't change.

It would change.

there is no sin 270 here.. so no sinx=-1..

so the limits would be 3-2(0) and 3-2(1)
so
3 to 1 would be the limits.. instead of the 5 to 1 in the case of 0<x<2pi.

 

[syed1995]

guys wch is the toughest p1 ever set? gimme the link...

All 2012 papers were the toughest ones I encountered.

 

[Rutzaba]

All 2012 papers were the toughest ones I encountered.

i kno that of experience
p3 vector question was soo confusing may june last q variant 2
un mai se bhi wch one is most?