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Mathematics: Post your doubts here!

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PhyZac

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sma786 said:
Hii, can anyone solve this question in detail for me please?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s06_qp_1.pdf Question 5, thanks, :love:
Click to expand...
You are lucky, i just finished this paper!


They said to find the distance between two points of intersection.

First we should find what are the points.

To do this we should do simultaneous equation.

3y= 4x+6 -----(1)
y^2 = 12x ----(2)

In equation (2)
x= y^2/12

Sub this to equation (1)

3y = 4(y^2/12) + 6
3y = y^2/3 +6
9y = y^2 + 18
y^2 - 9y +18 = 0
(y-3) or (y-6) = 0

y = 3 or y = 6

when y = 3 [sub to (1)]
3(3) = 4 x + 6
x = 3/4

so first point (3/4 , 3)

when y = 6
3(6) = 4x + 6
x = 3

so second point (3, 6)

Distance √(x1-x2)^2 + (y1-y2)^2
√(3-(3/4))^2 + (6-3)^2
= 3.75
 
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sma786

sma786

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th
PhyZac said:
You are lucky, i just finished this paper!


They said to find the distance between two points of intersection.

First we should find what are the points.

To do this we should do simultaneous equation.

3y= 4x+6 -----(1)
y^2 = 12x ----(2)

In equation (2)
x= y^2/12

Sub this to equation (1)

3y = 4(y^2/12) + 6
3y = y^2/3 +6
9y = y^2 + 18
y^2 - 9y +18 = 0
(y-3) or (y-6) = 0

y = 3 or y = 6

when y = 3 [sub to (1)]
3(3) = 4 x + 6
x = 3/4

so first point (3/4 , 3)

when y = 6
3(6) = 4x + 6
x = 3

so second point (3, 6)

Distance √(x1-x2)^2 + (y1-y2)^2
√(3-(3/4))^2 + (6-3)^2
= 3.75
Click to expand...
thnku, thnku so much :) what grade to expect? are you giving A2 aswell?!
 
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PhyZac

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Asad said:
I will be immensely grateful...
Click to expand...
Okay Alhamdulilah I finished. I really hope you gain help with this...took time :p

First they ask to solve.
(a)
iw^2 = (2-2i)^2 [using this=> a^2 - 2ab + b^2]
iw^2 = 4 - 8i +4(i^2) [using i^2 = -1]
iw^2 = 4 - 8i -4
iw^2 = -8i [ divide both sides with i]
w^2 = -8
w = √-8 or -(√-8)
w = √8 x √-1 = i√8 [√-1 = i]
or
w = -i√8

(b)
Now see,
|z − 4 − 4i| ≤ 2
so first it should be in form z - (x + iy)
therefore
|z - (4 + 4i)| ≤ 2

now take the point 4 + 4i on Argand diagram
and then take 2 as radius

and you will get something like this

Capture.PNG





(b)(ii)
So we have to find the value of z which is in R (the shaded place)

the |z| ( is the mod of z)

so the smallest |z| which is p is the green line below [since this is shortest line from origin to circle] to fine smallest value, we can use the triangle that form, which i made with purple color. usig pythagorus theorom the hypotenuse is √4^2 + 4^2 = 5.657 but tht is the hypotenuse,to make it th green line we should minus the radius which is 2 making p = 3.66. To find q which is largest |z| is the brown line in next picture, so to find it, easy, add to the hypotenuse a radius so q = 7.66
11.PNG 2e3.PNG sdf.PNG

Now to find the smallest and largest arg z
see the light blue shows smallest angle and dark blue shows largest angle to find tht, we should know tht the line from origin to centre (like the above brown line) cuts the 90 degree to half tht is 45 degree.
Okay first make a triangle with the sides as hypotenuse(brown down , as before is 5.657) and 2 (radius) and the light blue line, and to find brown angle we make this equation sin(brown angle) = opp/hyp = 2/5.657
angle = 0.3613 (in radians)
then to find light blue angle 1/4pi - 0.3613 = 0.424 [1/4pi is 45 degree]
and to find the dark blue angle 1/4pi + 0.3613 = 1.15

lat.PNG
 
ahmed abdulla

ahmed abdulla

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sma786 said:
Hii, can anyone solve this question in detail for me please?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s06_qp_1.pdf Question 5, thanks, :love:
Click to expand...
i got an easier way !
first ANY case you find point of intersection .. EQUATE the two equations by making Y as the subject in both cases .....
so y = root of 12x----1
y=4x+6/3--------2
by equationg them you get two x values from a quadratics eqation ... plug this two x values in one of the equation and you get the y values! and then find the distance
 
sma786

sma786

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ahmed abdulla said:
i got an easier way !
first ANY case you find point of intersection .. EQUATE the two equations by making Y as the subject in both cases .....
so y = root of 12x----1
y=4x+6/3--------2
by equationg them you get two x values from a quadratics eqation ... plug this two x values in one of the equation and you get the y values! and then find the distance
Click to expand...
thnkuuu :)
 
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PhyZac

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sma786 said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s06_qp_1.pdf
Question 7, (i) how to do this? there are these types of questions in everryy paper!
i dunno the tan formula, can anyone explain?
and part (iii) what formula did they take for area of AOBT ?
Click to expand...
I will give you idea, if you still have doubt i will try to solve.

7 (i) see make a triangle OAT use watever sin cos or tan then find angle then multiply by 2 to got AOB angle.
(iii)the same triangle as up can be used and find area 1/2bh formula then multiply by 2
 
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