Mathematics: Post your doubts here!

[Dug]

hey can any1 please help me in http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_11.pdf question 9 part ii. Thank you

Set both eqs equal and use b^2-4ac=0.
After that u get a quadratic equation...Solve it and you have the answer

 

[Bama Boy]

Set both eqs equal and use b^2-4ac=0.
After that u get a quadratic equation...Solve it and you have the answer

Thank you so much ! Bless you Mate

 

[allysaleemally]

Hey could please explain me question 1 in P1 may/june 2010 V1?
i cant understand anything of it!
http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s10_qp_11.pdf

marksheet: http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s10_ms_11.pdf

 

[allysaleemally]

It is because the shaded region is above the curve and if you try to find the volume directly by integrating square of given equation about x-axis it gives the volume of substance formed when the area bounded by the curve and x-axis (which is below the curve) is rotated about x-axis.
So first you have to rotate a rectangle formed by line AB with x-axis so that a cylinder is formed then find its volume and finally subtract the volume of curve rotated about x-axis which will give the volume of required region...
GOT IT????

Yes got it, thanks alot!

 

[Dug]

Hey could please explain me question 1 in P1 may/june 2010 V1?
i cant understand anything of it!
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_11.pdf

marksheet: http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_ms_11.pdf

Pg 86

 

[allysaleemally]

Pg 86

Whaat?

 

[Dug]

Whaat?

Ur question is already answered on page 86.

 

[ousamah112]

may/june 10 p31 q9 part 2.

 

[reina81]

ON 2009 P3 Q10 i Don't get how to connect the equations
http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w09_qp_31.pdf

 

[Esme]

Yeah, like mentioned above you just need to suggest a suitable value. There's a range of answers (like the mean being 40-80 kg and deviation being like 10 kg) so you don't really need to worry about it.

Thanks a lot !!

 

[Esme]

no u dont have to go for calculation
question says to suggest so u are to estimate some parameters
but your values sud fit the expected weights too

Thank you !! =D

 

[aaakhtar19]

may/june 10 p31 q9 part 2.

 

[Hazkeel Rizvi]

Please solve this question , i tried it doing so many times but i am unable to get the same gradient ( as per marking scheme ) and equation required to be shown in part (i)

 

[aaakhtar19]

ON 2009 P3 Q10 i Don't get how to connect the equations
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_qp_31.pdf

 

[Dug]

Please solve this question , i tried it doing so many times but i am unable to get the same gradient and equation required to be shown in part (i)

AoA wr wb

First find the coordinates of A:
At A, y=0
2-(18/2x+3) = 0
u get x=3

Now find the derivative of the curve:
dy/dx = -18(2x+3)^-1-1 (-1) (2)
dy/dx = 36/(2x+3)^2

At A, x=3
dy/dx = 36/(2(3)+3)^2 = 4/9

For the line AC:
m= -9/4
Insert coordinates of A
0 = -9/4 (3) +c
c=27/4
y=(-9/4)x +27/4
4y+9x=27

No need of thanks but I need prayers.

 

[Lyfroker]

how to sketch the graph of 5-3sin2x ????????? (0<=x<=pi)
does it have an inverse?

 

[aaakhtar19]

Please solve this question , i tried it doing so many times but i am unable to get the same gradient ( as per marking scheme ) and equation required to be shown in part (i)

(i)
y=2-18/(2x+3)
find the value of x when y=0
0=2-18/(x+3)
x=3 So,Coordinates of A (3,0)

find y' it would be (16x-21)/(2x+3)^2
find y' at A = put 3 in place of x in above equation :
=4/9
as m1*m2=-1
so gradient of normal (AC) will be -9/4

Equation of AC:
(y-y1)=m(x-x1)
use A (3,0) and m=-9/4
y-0=(-9/4)(x-3)
4y=-9x+27
4y+9x=27 Done Hope u got it

 

[Hazkeel Rizvi]

AoA wr wb

First find the coordinates of A:
At A, y=0
2-(18/2x+3) = 0
u get x=3

Now find the derivative of the curve:
dy/dx = -18(2x+3)^-1-1 (-1) (2)
dy/dx = 36/(2x+3)^2

At A, x=3
dy/dx = 36/(2(3)+3)^2 = 4/9

For the line AC:
m= -9/4
Insert coordinates of A
0 = -9/4 (3) +c
c=27/4
y=(-9/4)x +27/4
4y+9x=27

No need of thanks but I need prayers.

thanks buddy btw even i was doing it the same way but you see when i was solving it the answer was coming to 0.444 as i have a MS calculator and when i was checked now with ab/c i could see it means the same as 4/9 i guess i need to get myself an ES calculator now or get used to using the ab/c button more often anyways thanks man (despite you mentioned it but i should say this as you helped me freakout less and helped me figure out the problem and well sure i pray for your success best of luck

 

[Dug]

how to sketch the graph of 5-3sin2x ????????? (0<=x<=pi)
does it have an inverse?

Y-intercept = 5
Amplitude = 3
The function is reflected in x-axis (inverted) because of the negative sign.
cycles = 2

If the domain was 0 <=x <= 2pi then you would have to fit 2 inverted sine cycles in that period. But its not...u just have to construct an inverted sine graph in the stated domain.

And prayers needed!!!

 

[Lyfroker]

Y-intercept = 5
Amplitude = 3
The function is reflected in x-axis (inverted) because of the negative sign.
cycles = 2

If the domain was 0 <=x <= 2pi then you would have to fit 2 inverted sine cycles in that period. But its not...u just have to construct an inverted sine graph in the stated domain.

And prayers needed!!!

tnx a lot
sure