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Mathematics: Post your doubts here!

Rutzaba

Rutzaba

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Faaiz Haque said:
Question 5 whole question
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_11.pdf

Full working please
Click to expand...
simple
a)
you integrate the dy/dx you will get
integral of 3x^0.5 -6
y=2x^1.5 -6x +c
put value of x and y that is given 9 and 2
2=2(9)^.5 -6(9) +c
2=54 -54 +c
c=2
putting values back ul get
y=2x^1.5 -6x +2

b) stationary point means at dy/dx =0
you differentiate the eq of curve that u have just made

2x^1.5 -6x +2=0
u will get
3x^0.5 -6+0=0
x^0.5=6/3
x^0.5= 2
x=4
to check the nature
d²y/dx²= differentiate this
3x^0.5-6
you have 1.5x^-0.5
put value of x-4
1.5(4^-0.5)
you 0.75 >0
thus its minimum....
 
F

Faaiz Haque

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Rutzaba said:
simple
a)
you integrate the dy/dx you will get
integral of 3x^0.5 -6
y=2x^1.5 -6x +c
put value of x and y that is given 9 and 2
2=2(9)^.5 -6(9) +c
2=54 -54 +c
c=2
putting values back ul get
y=2x^1.5 -6x +2

b) stationary point means at dy/dx =0
you differentiate the eq of curve that u have just made

2x^1.5 -6x +2=0
u will get
3x^0.5 -6+0=0
x^0.5=6/3
x^0.5= 2
x=4
to check the nature
d²y/dx²= differentiate this
3x^0.5-6
you have 1.5x^-0.5
put value of x-4
1.5(4^-0.5)
you 0.75 >0
thus its minimum....
Click to expand...
I said question 5 .. o_O
 
GCE As and a level

GCE As and a level

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hey guys
how are u all ?
solve as much as u can, even if only one question !! :)
i need a help in
1)
http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_w12_qp_12.pdf

Q4 all
Q6 ii)
Q7 ii)
Q8 all
Q10 ii)
Q11 all


2)
http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_s12_qp_12.pdf

Q7 a)

3)
http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_s09_qp_1.pdf

Q2 all
Q10 iii)


4)
http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_S08_qp_01.pdf

Q5 all
Q6 ii) Domain __ explain it plz


5)
http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_w08_qp_1.pdf

Q5 all
Q6 all
Q9 ii) and iii)
Q10 ii) and iii) and iv)


6)
http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_s07_qp_1.pdf

Q4 all
Q5 all
Q8 ii)
Q10 ii)
Q11 ii)

7)
http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_w07_qp_1.pdf

Q11 v)

8)
http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_w06_qp_1.pdf

Q10 v)


9)
http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_s12_qp_13.pdf

Q1 all
Q4 all
Q11 iii)

I really need your help :)
waiting your help
thanks in advance
wish u all good luck
:)
 
syed1995

syed1995

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Haya Ahmed said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w13_qp_13.pdf

Q7b please somone explain briefly .. thanks :)
Click to expand...

let me solve it in terms of degrees first and you'll understand better. Sin is Positive is 1st and 2nd Quadrant while negative in 3rd and 4th quadrant then again positive in the 5th and 6th Quadrant then negative in the 7th and 8th Quadrants.

1st Quad = Alpha 2nd = 180-Alpha 3rd = 180+Alpha 4th = 360-Alpha 5th = (1stQuad+360) 6th = (2ndQuad+360)

Sin(2x+60)=1/2
sin^-1(1/2) = 30 Degrees //1st quadrant and (180-30) //2nd quadrant and 360+30 //(5th Quadrant) 360+150 //(6th Quadrant)
(2x+60)= 30 or 2x+60= 150 or 2x+60=390 or 2x+60=510 Degrees
2x=-30 or 2x=90 Degrees or 2x=330 Degrees or 2x=450 Degrees
x=-15 Degrees or x= 45Degrees or x=165 Degrees or x=225 Degrees.
since first and last is outside the range.. their are only two valid answers x=45 Degrees or 0.25Pi and x= 165 Degrees or 11Pi/12

PS: the reason we took 5th and 6th Quadrant was because it said 2x in the question.. meaning the angles we get after sine inverse will be divided by 2.
so if the range was 0<x<180 for x
it will be 0*2<2x<2(180) for 2x which is 0<2x<360

Our question was 2x+60 .. so add 60 to the above range.
for 2x+60 it becomes 0+60<2x+60<360+60
for 2x+60 the range will be 60<2x+60<420 so all the angles in between that range will result in a value of x.
 
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