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[CIE] All Pure Mathematics (P1, P2 and P3) doubts here !

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cs:

--- Quote from: Shoshou..Mony on June 14, 2011, 08:41:36 am ---@cs :

5)iii) -5 < 4sinx - 3cox < 5

1/4sinx-3cosx+6 has greatest value = 1/(-5+6)

i.e equal 1

8)b)iii) (OA/OB) = OC

--- End quote ---

Thank you so much for your help :)

+REP

Shoshou..Mony:

--- Quote from: cs on June 15, 2011, 06:33:27 am ---Thank you so much for your help :)

+REP

--- End quote ---

If you have any more doubts then post before I forget my P3 syllabus. :P

Thanks for the +rep. =]

Arissa_04:
Hi
Could someone please explain how to do June 2010 Paper 3 variant 1-Question 4 ii)
and also Paper 3 June 2009 Question 10(the whole question).

Thanks in advance.:)

dlehddud:
4 ii) 2010

As we know that sin 3x sin x= 0.5 (cos 2x- cos 4x) from the previous part of the question, we can substitute this into our given integral:

  integral sign (0.5 cos 2x- 0.5 cos 4x)
 
if we integrate this, we get

(sin2x)/4 - (sin4x)/8

substituting limits, we get

((sin (2pi/3))/4 - (sin(4pi/3))/8)  -  ((sin (2pi/6))/4 - (sin(4pi/6))/8)

=> (sqrt 3)/8 - (-(sqrt3)/16) - (sqrt 3)/8 - (-(sqrt 3)/16)
=> 2(sqrt 3)/16
=>(sqrt 3)/8

question 10 2009

i) To find x at M, we must find the maximum of the curve, which is simply the derivative of the curve equated to zero.

y=x^2*(sqrt(1-x^2))

we find the derivative by using the product rule,

dy/dx= 2x*(sqrt(1-x^2))+0.5*(-2x)/(sqrt(1-x^2))*x^2
        = 2x(sqrt(1-x^2)) - x^3/(sqrt(1-x^2))

dy/dx=0

2x(1-x^2)/(sqrt(1-x^2)) - x^3/(sqrt(1-x^2))=0 ......... i've equated the denominators if u were unsure

(2x- 3x^3)/(sqrt(1-x^2))=0

Now we only require the numerator to be zero. Therefore,

2x-3x^3=0
x(2-3x^2)=0

Hence, x=0 and 2-3x^2=0
                   => 3x^2=2
                   => x^2=2/3
                   => x= sqrt(2/3)

therefore, the x coordinate of M is sqrt(2/3). It cannot be 0, as M is obviously not centered at x=0


ii)As we are integrating by substituting, we must also change the variables around a bit:

x=sin(theta)
dx/d(theta)=cos(theta)
dx=cos(theta) d(theta)

To do this, we merely substitute in sin (theta) and the new variables to replace dx into the equation given:

Area=integral sign   sin^2(theta)*sqrt(1-sin^2(theta))*cos(theta)  d(theta)

Area=integral sign   sin^2(theta)*sqrt(cos^2(theta))*cos(theta)  d(theta)

Area=integral sign   sin^2(theta)*cos(theta)*cos(theta) d(theta)

Area=integral sign   0.25*(sin2x)^2  d(theta)


the limits are given by substituting into the equation x=sin(theta)
we are finding theta, so we use   theta=sin^-1  x

iii)Do not think that because they give you an equation, that you can use it straight away. We cannot integrate most trigonometric integrals if they are squared, such as this. So we convert it to a form we CAN integrate, namely, a cos4x form:

0.25*(sin^2(2theta))

cos(4theta)= 1-2*(sin^2(2theta))


thus,

2*(sin^2(2theta))=1-cos(4theta)

and so

0.25*(sin^2(2theta))=(1-cos(4theta))/8

Hence we are integrating this, NOT 0.25*(sin^2(2theta))


(1-cos(4theta))/8= 1/8 - (cos(4theta))/8

So, we integrate this:


Area= integral sign  1/8 - cos(4theta))/8  d(theta)

Area= 1/8*(theta) - (sin(4theta))/32       with limits pi/2 and 0


Area= 1/8*(pi/2) - (sin(2pi))/32 - 1/8*(0) + (sin(0))/32

Area=pi/16



Hope this wasn't too confusing =P
It is however up to the standard of working you require at CIE A-level so it shouldn't be tooo hard :)
Ask any questions if you don't get certain parts of it!

TBT:
I dont knw how to go about question 5 for maths p1 June 2006... can anyone help?? its part of my homework... thanks!!

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