Mathematics: Post your doubts here!

[shazmina]

The types of questions that came in the 1990s don't come in papers now but they were harder than the current papers so if you want to solve them for practice then go ahead, it'll only make your Maths better.

JazakAllah khairan saad bhai thank you sooo much

 

[biscuitbiscuit]

Someone please solve 'Q9' and do explain why (i) is 6 and skip (ii) if you want

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w09_qp_12.pdf

 

[ahmed abdulla]

show differtial of 1 /cosx = 1/ cosecx -sinx .... HELP

 

[Saad Mughal]

Someone please solve 'Q9' and do explain why (i) is 6 and skip (ii) if you want

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_qp_12.pdf

(i) Explanation: AC is the diagonal of the rectangle, the midpoint of which lies on the line y = k (where k is a constant).
Now, Midpoint of AC = (6,6), therefore the midpoint of AC, B and D lie on the line y = 6.
Hence, y-coordinate D = 6.

Solution:
(ii) Gradient AD = 0-h/-2-6 = -h/-8 = h/8
Gradient CD = h-12/6-14 = (h-12)/-8

(iii) Using law of perpendicular lines,
(h/8)*(h-12)/-8 = -1
h^2 - 12h = 64
h^2 - 12h - 64 = 0
(h-16)(h+4) = 0
Therefore, h = 16 (since D lies in positive axis).

Midpoint BD = Midpoint AC
(x+16)/2 , (6+6)/2 = (6,6)
By comparing,
(x+16)/2 = 6
x + 16 = 12
x = -4 (x-coordinate B).

Therefore, B is point (-4,6) & D is point (16,6).

(iv) Length AD = root [(16-0)^2 + (6+2)^2] = 17.9 units
Length CD = root [(12-16)^2 + (14-6)^2] = 8.94 units
Area of ABCD = AD * CD = 17.9*8.94 = 160 units^2 (correct to 3 s.f.)

 

[Saad Mughal]

show differtial of 1 /cosx = 1/ cosecx -sinx .... HELP

y = (cos x)^-1
dy/dx = sinx (cos x)^-2
dy/dx = sinx (1/cos^2 x)
dy/dx = sinx (1/(1-sin^2 x))
dy/dx = sinx/(1-sin^2 x)
This can also be written as,
dy/dx = 1/[(1-sin^2 x)/sin x]
dy/dx = 1/[(1/sin x) - (sin^2 x/sin x)]
dy/dx = 1/[(cosec x) - sin x]
dy/dx = 1/(cosec x - sin x)

 

[biscuitbiscuit]

(i) Explanation: AC is the diagonal of the rectangle, the midpoint of which lies on the line y = k (where k is a constant).
Now, Midpoint of AC = (6,6), therefore the midpoint of AC, B and D lie on the line y = 6.
Hence, y-coordinate D = 6.

Solution:
(ii) Gradient AD = 0-h/-2-6 = -h/-8 = h/8
Gradient CD = h-12/14-6 = (h-12)/8

(iii) Using law of perpendicular lines,
(h/8)*(h-12)/8 = -1
h^2 - 12h = -64
h^2 - 12h + 64 = 0
(h-16)(h+4) = 0
Therefore, h = 16 (since D lies in positive axis).

Midpoint BD = Midpoint AC
(x+16)/2 , (6+6)/2 = (6,6)
By comparing,
(x+16)/2 = 6
x + 16 = 12
x = -4 (x-coordinate B).

Therefore, B is point (-4,6) & D is point (16,6).

(iv) Length AD = root [(16-0)^2 + (6+2)^2] = 17.9 units
Length CD = root [(12-16)^2 + (14-6)^2] = 8.94 units
Area of ABCD = AD * CD = 17.9*8.94 = 160 units^2 (correct to 3 s.f.)

I think you took the wrong factors for the equation 'h^2 - 12h + 64 = 0' because when you multiply -16*4 you get -64 instead of getting +64

 

[Saad Mughal]

I think you took the wrong factors for the equation 'h^2 - 12h + 64 = 0' because when you multiply -16*4 you get -64 instead of getting +64

I've edited it. Sorry for the mistake.
Edit: Gradient CD was incorrectly calculated (it should have been -ve).

 

[biscuitbiscuit]

(i) Explanation: AC is the diagonal of the rectangle, the midpoint of which lies on the line y = k (where k is a constant).
Now, Midpoint of AC = (6,6), therefore the midpoint of AC, B and D lie on the line y = 6.
Hence, y-coordinate D = 6.

Solution:
(ii) Gradient AD = 0-h/-2-6 = -h/-8 = h/8
Gradient CD = h-12/6-14 = (h-12)/-8

(iii) Using law of perpendicular lines,
(h/8)*(h-12)/-8 = -1
h^2 - 12h = 64
h^2 - 12h - 64 = 0
(h-16)(h+4) = 0
Therefore, h = 16 (since D lies in positive axis).

Midpoint BD = Midpoint AC
(x+16)/2 , (6+6)/2 = (6,6)
By comparing,
(x+16)/2 = 6
x + 16 = 12
x = -4 (x-coordinate B).

Therefore, B is point (-4,6) & D is point (16,6).

(iv) Length AD = root [(16-0)^2 + (6+2)^2] = 17.9 units
Length CD = root [(12-16)^2 + (14-6)^2] = 8.94 units
Area of ABCD = AD * CD = 17.9*8.94 = 160 units^2 (correct to 3 s.f.)

Thanks

 

[biscuitbiscuit]

Someone solve 'Q5" http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s05_qp_1.pdf

 

[Rutzaba]

Someone solve 'Q5" http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s05_qp_1.pdf

want the procedure or the entire thing?

 

[biscuitbiscuit]

want the procedure or the entire thing?

Well tell me the procedure and if i am not able to solve than the whole thing

 

[Rutzaba]

Well tell me the procedure and if i am not able to solve than the whole thing

see first of all u know this that if we know the gradient ofa straight line we can know the gradient of the perpendicular gradient? so if we find out the grad of BD which is -2
the m1*m2=-1 if m1 is perpendicular to m2
-2*m2=-1
m2=.5
midpoint is (x1+x2)/2 must give u the x coordinate
(y1+y2)/2 will give u y
it will be (4,6)
now make the eq of line u have a grad u have a coordinate
2y= x+ 8----i
thisis the eq of AC
now A lies on the x axis on the x axis y is 0
putting y =0 into the eq u can get
x=-8
A(-8,0)

 

[Rutzaba]

is that much clear? if yes then il proceed

 

[Rutzaba]

see. if we have 2 thins and we need to find the third thing we can ryt?
like if x+y= z aand we know two of them then we can fynd the third ryt??
similarly if the midpoint of A and C is M
and we know A and M we can fynd C
see the x coordinate of all three variables and put them into the formula of the midpoint
(-8+ x) /2= 4
x=16
and
(0 +y)/2 = 6
y=12
then the coordinates of C are (16, 12)

 

[biscuitbiscuit]

see. if we have 2 thins and we need to find the third thing we can ryt?
like if x+y= z aand we know two of them then we can fynd the third ryt??
similarly if the midpoint of A and C is M
and we know A and M we can fynd C
see the x coordinate of all three variables and put them into the formula of the midpoint
(-8+ x) /2= 4
x=16
and
(0 +y)/2 = 6
y=12
then the coordinates of C are (16, 12)

Thank you so much. For this question tell me if I am doing correct, first I'll find the gradient of AB and use it to find the gradient of DA and its equation. Than the equation of DC and for the co-ordinates i'll solve them.

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_w07_qp_1.pdf

 

[Rutzaba]

Thank you so much. For this question tell me if I am doing correct, first I'll find the gradient of AB and use it to find the gradient of DA and its equation. Than the equation of DC and for the co-ordinates i'll solve them.

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

yes so now u know the equation of the line AD
now u also know that parallel lines have the same gradient
so u have a gradient and the coordinates C so find the equation of the line CD
then solve the eq to line CD and AD to get the coordinates of D.

 

[biscuitbiscuit]

yes so now u know the equation of the line AD
now u also know that parallel lines have the same gradient
so u have a gradient and the coordinates C so find the equation of the line CD
then solve the eq to line CD and AD to get the coordinates of D.

Okay. For 'Q7' I'll find the mid-point of AB than the gradient of AB and use it to find the equation of the perpendicular bisector. Can you tell how will i find the co-ordinate of D"

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_w05_qp_1.pdf

 

[Rutzaba]

Okay. For 'Q7' I'll find the mid-point of AB than the gradient of AB and use it to find the equation of the perpendicular bisector. Can you tell how will i find the co-ordinate of D"

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w05_qp_1.pdf

First u will find the midpoint of AB and the gradient of AB then the eq of line AB
then find the equation of line BC then solve simultaneously

 

[Rutzaba]

Okay. For 'Q7' I'll find the mid-point of AB than the gradient of AB and use it to find the equation of the perpendicular bisector. Can you tell how will i find the co-ordinate of D"

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w05_qp_1.pdf

First u will find the midpoint of AB and the gradient of AB then the eq of line AB
then find the equation of line BC then solve simultaneously

 

[biscuitbiscuit]

First u will find the midpoint of AB and the gradient of AB then the eq of line AB
then find the equation of line BC then solve simultaneously

Thanks. For 'Q6' how to do the second sub-part, http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s07_qp_1.pdf