Mathematics: Post your doubts here!

[sma786]

th

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

thnku, thnku so much what grade to expect? are you giving A2 aswell?!

 

[sma786]

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 ?

 

[Esme]

What is limiting equilibrium ?

 

[PhyZac]

I will be immensely grateful...

Okay Alhamdulilah I finished. I really hope you gain help with this...took time

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







(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


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

 

[Asad]

Okay Alhamdulilah I finished. I really hope you gain help with this...took time

YOU MY FRIEND ARE A LIFE SAVER. MAY ALLAH BLESS YOU WITH LOTS AND LOTS OF MORE KNOWLEDGE.
THANK YOUUU!

 

[PhyZac]

YOU MY FRIEND ARE A LIFE SAVER. MAY ALLAH BLESS YOU WITH LOTS AND LOTS OF MORE KNOWLEDGE.
THANK YOUUU!

Aameeeeen, and you are welcome. More doubts do post but tomorrow i will solve. In Sha Allah.

 

[ahmed abdulla]

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,

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]

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

thnkuuu

 

[sma786]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s06_qp_1.pdf
Question 9, part i, i did uptil finding the equation, but what about after that?! how did they find the points?

 

[PhyZac]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s06_qp_1.pdf
Question 9, part i, i did uptil finding the equation, but what about after that?! how did they find the points?

Those points are when x = 0 and when y = 0
then take mid point.

 

[PhyZac]

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 ?

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

 

[ahmed abdulla]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s06_qp_1.pdf
Question 9, part i, i did uptil finding the equation, but what about after that?! how did they find the points?

since both sides of tangent equal and you studied in ur IGCSE"S that angle between tangent and radi is 90 degree!
so < OAT and <OBT are 90 degree,,,, so you got a triangle OAT or OBT and use trignometry to find the half of < AOT and then multiply with 2 !

same thing for area ... find area of one of the tiangle and multiply with two A=0.5 *bas*height

 

[ahmed abdulla]

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

u gotta now the reason! its a congruent triangle .. all sides are equal ...so use trignometry

 

[Asad]

hey PhyZac, Is there any rule about when to use the dot product and when to use cross product in vectors?

Because I get confused in these two sometimes.

 

[surinr]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_qp_31.pdf
Really need some help with Q9
Thank you

 

[PhyZac]

hey PhyZac, Is there any rule about when to use the dot product and when to use cross product in vectors?

Because I get confused in these two sometimes.

Ofcourse, they are so different.

Dot product, gives you the cos of the angle between lines.

Cross product, gives you a vector, this vector is perpendicular to both line.

I will see if i can gather when is each used, but see when a vector perpendicular to two line is to be found cross product is used.
Dot is used for angles.

 

[PhyZac]

Asad

You see I will give a tip for vectors.

When ever you solve make a sketch of all the stuff they told you, and then try to figure out how will you get an answer.

You see in the question of complex you gave I couldn't figure the answer until I made a sketch and tried to use all values I have. So in such questions do try making sketchs, mainly needed to know which two line needed to use cross product to find the perpendicular line to both.

 

[Asad]

PhyZac:

True. But I find it a bt hard to sketch a 3d vector with x, y and z components.
how do u suggest i should go about it?

 

[PhyZac]

Asad

Well, see when you draw sketch don't try to assume where or how the numbers will make the plane or a line look like. Just make and x and y axis and draw a rectangle dont see what is the number or anything, just make a rectangle, and if they say a line is parallel make it as if it is parallel or perpendicular etc.

Check this as an example. https://www.xtremepapers.com/commun...st-your-doubts-here.9599/page-352#post-504194

 

[PANDA-]

Asad
Draw an upside down Y, something like this

and use it like a template to guide you as to where the axis are