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Mathematics: Post your doubts here!

Talha Khatri

Talha Khatri

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MiniSacBall said:
xD, I just typed it randomly. :D .
Thank you so much <3

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s13_qp_11.pdf
Q1, Am i supposed to find the maximum and minimum point to prove that it is an increasing function.
how do i do this, help on this one thanks :)
God Bless you!
Click to expand...
Haha! Cool :D

No you don't need to find the maximum point, just differentiate the function once and you'll get f '(x) = 6(2x+3)^2 + 1 and if you look at it, no matter what the value of x you input, the square will always turn it into a positive value so the gradient always stays positive hence we can be assured that f(x) is an increasing function.
 
sudeep1097

sudeep1097

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sitooon said:
needs Second part only . when meet x-axis what happens , just tell me this
I got first one as 12x+6y-6z=48 , and second one is ( -2 , 7 , -5 )
Tell me what happens when parallel to x-axis only .
For complex no. , i got step except how they used y=x for lz -3l=lz-3il
midha.ch

View attachment 40534

View attachment 40535
Click to expand...
Sure!
Midpoint of PQ is (p+q)/2 and get the position vector plane passes through this point
now (q-p) gives direction of pq vector this is perpendicular to the plane so this direction can be considered as normal of the plane.
Now you can substitue a,b,c as the normal direction you obtained. and since equation satisfies the position vector of midpoint you can substitute for x y and z to get k hence the equation
Straight line through P and parallel to x axis has equation R = (7i+7i-5i) + t(1i+0+0) (this is direction of vector parallel to x axis)
now obtain position vector of R that is (7+t, 7, -5)and substitute in the equation of plane to get B.
Use distance formula from A to B to get distance AB.

|z-3|=|z-3i| Square both sides to get (x+yi-3)^2=(x+yi-3i)^2 (I supposed z as x+yi)
solve to get x=y ( i guess cause i have already done this check mark scheme)
now arg(z-2i)= pi/6 (this is the angle )
arg (x+yi-2i) = pi/6
arg (x+(y-2)i) = pi /6
y-2/x = tan pi/6 (tan theta = y component/x component)
solve and substitute x = y from before to get values of x and y
now r = square root of (x^2 + y^2)
and tan theta = y/x = 1 = pi/ 4 ( since y=x and tan 45 = 1)


Hope this helps! :D
 
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sitooon

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soo ep1097 said:
Sure!
Midpoint of PQ is (p+q)/2 and get the position vector plane passes through this point
now (q-p) gives direction of pq vector this is perpendicular to the plane so this direction can be considered as normal of the plane.
Now you can substitue a,b,c as the normal direction you obtained. and since equation satisfies the position vector of midpoint you can substitute for x y and z to get k hence the equation
Straight line through P and parallel to x axis has equation R = (7i+7i-5i) + t(1i+0+0) (this is direction of vector parallel to x axis)
now obtain position vector of R that is (7+t, 7, -5)and substitute in the equation of plane to get B.
Use distance formula from A to B to get distance AB.

|z-3|=|z-3i| Square both sides to get (x+yi-3)^2=(x+yi-3i)^2 (I supposed z as x+yi)
solve to get x=y ( i guess cause i have already done this check mark scheme)
now arg(z-2i)= pi/6 (this is the angle )
arg (x+yi-2i) = pi/6
arg (x+(y-2)i) = pi /6
y-2/x = tan pi/6 (tan theta = y component/x component)
solve and substitute x = y from before to get values of x and y
now r = square root of (x^2 + y^2)
and tan theta = y/x = 1 = pi/ 4 ( since y=x and tan 45 = 1)


Hope this helps! :D
Click to expand...
Thanks soo much bro ,
Got a doubt, why direction vector is
( 1'0'0) why the 1 is it alwaya 1 when
Parallel to x-axis?
When i squared both sides, didnt get
X=y
Thanks again
 
Last edited:
Talha Khatri

Talha Khatri

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sitooon

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mmmmmm said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_32.pdf

Question:10, part (iii). I am getting one position vector of "P" but not the other. Help me guys to solve this question.
Click to expand...
We have to find the point p where the perpendicular distance to the two planes is same.


first we will for an equation the way the paper gave.
therefore
for plane m
|x+2y-2z-1|/sqrt((1^2) + (2^2) + (-2^2) = |x+2y-2z-1|/3 [P.S, sqrt is square root]
for plane n
|2x-2y+z-7|/sqrt((2^2)+(-2^2)+(1^2) = |2x-2y+z-7|/3


since you are finding a point where both distance is same , therefore.
|x+2y-2z-1|/3 = |2x-2y+z-7|/3 (3 can be cancelled both sides)
=
|x+2y-2z-1| = |2x-2y+z-7|

now we will sub the line values in the equation.
the x component (1+2t) [P.S, t is lamda ]
the y comp. (1+t)
the z comp. (-1+2t)

therefore.
| (1+2t) +2 ( 1+t) -2(-1+2t) -1 | = |2(1+2t) - 2(1+t) + (-1+2t) - 7 |
simplifying it you get
|4| = |-8+4t|
now to solve modulus, we do squaring method
16 = 64-64t+16t^2
simplify
2 = 8 - 8t +2t^2
2t^2 - 8t + 6 = 0
solve it and get
t = 3 or t= 1
when t=3 the position of point is [P.S, u do this by sub t value in line equation)
(7, 4, 5) => OA
when t = 1 the position of point is
(3, 2, 1) => OB

BA (or AB, same thing) = OA - OB
= (4 , 2, 4)

now find the mod
sqrt(4^2 + 2^2 + 4^2)
= 6

I dint do every step in detail, (very long) if u dont get any step, just ask.
 
mmmmmm

mmmmmm

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thanks, brother.

sitooon said:
We have to find the point p where the perpendicular distance to the two planes is same.


first we will for an equation the way the paper gave.
therefore
for plane m
|x+2y-2z-1|/sqrt((1^2) + (2^2) + (-2^2) = |x+2y-2z-1|/3 [P.S, sqrt is square root]
for plane n
|2x-2y+z-7|/sqrt((2^2)+(-2^2)+(1^2) = |2x-2y+z-7|/3


since you are finding a point where both distance is same , therefore.
|x+2y-2z-1|/3 = |2x-2y+z-7|/3 (3 can be cancelled both sides)
=
|x+2y-2z-1| = |2x-2y+z-7|

now we will sub the line values in the equation.
the x component (1+2t) [P.S, t is lamda ]
the y comp. (1+t)
the z comp. (-1+2t)

therefore.
| (1+2t) +2 ( 1+t) -2(-1+2t) -1 | = |2(1+2t) - 2(1+t) + (-1+2t) - 7 |
simplifying it you get
|4| = |-8+4t|
now to solve modulus, we do squaring method
16 = 64-64t+16t^2
simplify
2 = 8 - 8t +2t^2
2t^2 - 8t + 6 = 0
solve it and get
t = 3 or t= 1
when t=3 the position of point is [P.S, u do this by sub t value in line equation)
(7, 4, 5) => OA
when t = 1 the position of point is
(3, 2, 1) => OB

BA (or AB, same thing) = OA - OB
= (4 , 2, 4)

now find the mod
sqrt(4^2 + 2^2 + 4^2)
= 6

I dint do every step in detail, (very long) if u dont get any step, just ask.[/quote
Click to expand...
 
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Snowysangel

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In integration, when were determining the area or the volume and the range is following over both positive and negative values, do we have to divide the values into positive and negative sections? For example if the value are from -1 to 3, do we do f the area or the volume from -1 to 0 and 0 to 3 separately, or can we collectively find the area/volume from -1 to 3?
 
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