Mathematics: Post your doubts here!

[mohmed ahmed soliman]

help in 3 iii

 

[mohmed ahmed soliman]

in I why o.256^4 why not 0.256*4
i cant do anything in II

 

[Saad the Paki]

View attachment 59991 View attachment 59992
help in B

 

[mohmed ahmed soliman]

View attachment 59997

thanks

 

[farhan141]

Find AB.
AB = 2rcos@
Area of sector ABC = 4@r^2cos^2@
Area of triangle BOA = r^2cos@sin@
Area of sector OBA = 1/2r^2(pi - 2@)
Aera of segment = area of sector OBA - Area of triangle BOA = r^2(1/2(pi - 2@) - cos@sin@)
Area of 2 segements = 2 *r^2(1/2(pi - 2@) - cos@sin@)
Total shaded region = Aera of sector ABC + area of 2 segments = 2r^2(pi/2 - @ - cos@sin@) + 4@r^2cos^2@
Now manipulate it to get r^2(pi - sin2@ + 4@cos^2@ - 2@)
Now we know Total area of shaded region = 1/2 * pir^2
So r^2(pi - sin2@ + 4@cos^2@ - 2@) = 1/2 * pir^2
So manipulating this will give us cos2@ = 2sin2@ - pi/4@

P.S. I am in hurry.. If u have any doubts in this ask me. Just to make sure in triangle OBA there are 2@ drawn in diagram below the third angle will be pi - 2@ so maybe if this can arise as ur doubt, this is the solution to it. (area of segment OBA))

Thanks a lot. My mistake was confusing the identity of sin(pi-2@) with cos compound angle identity

 

[iSean97]

I dont know how to do this but seriously though how do you use that spoiler button?

[spoiler] Insert Image/Stuff here [/spoiler][code]

 

[farhan141]

How is v=pi*h^3

 

[qwertypoiu]

How is v=pi*h^3

Spoiler: A pic of the diagram

To find the volume of the cone, the formula is V = 1/3 * h * pi * r^2.
From above diagram you can see r = h*tan60 = sqrt(3) * h
So substituting this into above formula gives: V = 1/3 * h * pi *(sqrt(3) * h)^2 = pi*h^3

 

[The Sarcastic Retard]

Spoiler is interesting thing... xD LOLWA ka JALWA xD

Spoiler: A pic of the diagram

View attachment 60014

To find the volume of the cone, the formula is V = 1/3 * h * pi * r^2.
From above diagram you can see r = h*tan60 = sqrt(3) * h
So substituting this into above formula gives: V = 1/3 * h * pi *(sqrt(3) * h)^2 = pi*h^3

 

[farhan141]

Part B.

 

[farhan141]

I got the equation in terms of y and x but i dont know how to find the stationary points.

 

[qwertypoiu]

Part B.

Spoiler: Large Image

 

[farhan141]

Spoiler: Large Image

View attachment 60021

I didnt get the modulus of AC and AB part. Can u explain it a bit

 

[qwertypoiu]

I got the equation in terms of y and x but i dont know how to find the stationary points.

Maybe spoiler buttons are getting annoying for some.

 

[qwertypoiu]

I didnt get the modulus of AC and AB part. Can u explain it a bit

Okay forget modulus.
We wanna find the area of triangle, so we'll take AC as base. To find the distance AC, we can use the distance formula (from coordinate geometry) or simply think about it. The line is vertical, so we just need to find the difference in i components of A and C. You'll get 2*sqrt(2).

For the height part, we just measure the horizontal distance between B and A, which is simply the difference in x coordinates of them, or the difference in real part of the numbers. That gives you 5*sqrt(2). Hope that's clear.

Edit: since you know all the coordinates, you can also use the Shoelace Formula for finding its area.

 

[Layla..]

http://papers.gceguide.com/A Levels/Mathematics (9709)/9709_s14_qp_31.pdf Can anyone please help solve 7ii and 7iii
Thanks

 

[Layla..]

http://papers.gceguide.com/A Levels/Mathematics (9709)/9709_s14_qp_32.pdf Q 7b as well please

 

[Saad the Paki]

Q9 (iii). When you solve the equation the two values of x are +2.5 and -2.5. So -2.5 <x <+2.5 should be the answer but in the mark scheme it's -2.5 <x <0. Why is the upper limit 0 and not 2.5?

 

[Rizwan Javed]

Q9 (iii). When you solve the equation the two values of x are +2.5 and -2.5. So -2.5 <x <+2.5 should be the answer but in the mark scheme it's -2.5 <x <0. Why is the upper limit 0 and not 2.5?View attachment 60042

Because x = 2.5 is out of the domain of f(x).

 

[Saad the Paki]

Because x = 2.5 is out of the domain of f(x).

Ummm but it is in the domain of g (x) why not consider that?