Mathematics: Post your doubts here!

[HubbaBubba]

T

I expanded Σ(x-36)^2 it to pull out the term Σx^2. That is what we are supposed to find.

Consider this example:

(a+b)^2 = 4
a^2 + 2ab + b^2 = 4

If we are given 'ab' and 'b^2', we can easily find a^2 by putting the values into the equation. My solution is exactly the same. Did you get it now?

Yes I understand now! Thank you so much!

 

[Alice123]

AOA wr wb
can someone pls help me with question 5(i)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s04_qp_3.pdf

cos^2θ=(1+cos2θ)/2
sin^2θ=(1-cos2θ)/2
cos^2(2θ)={1+cos2(2θ)}/2=(1+c0s4θ)/2
LHS=sin^2θcos^2θ
={(1-cos2θ)/2}{(1+cos2θ)/2}
=(1-cos^2(2θ))/4
={1-(cos2θ)^2}/4
={1-(1+c0s4θ)/2}/4
=(1-4cosθ)/8=RHS
Hope this helps . Ask if u dnt understand

 

[littlecloud11]

Paper 3 Help needed
Paper: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_31.pdf

Question No. 10 - Part (iii)

Someone please help me with it. Please explain with basic concepts.. Paper 3 is killing me!! Please please help :'(

10 iii) You have the equation of two lines that the plane contains.
l= <1,1,1> + s<1,-1,2>
m= <4, 6, 1> +t <2, 2, 1>
The normal vector n, of the plane is perpendicular to the directional vectors of both of the lines contained in the plane. Therefore to find the normal vector you take the cross product of the direction vectors of the two lines.
n = <1,-1, 2>x<2, 2,1>
1 -1 2
2 2 1
=(-1 -4)i - (1-4)j + (2+2)k
= -5i + 3j +4k
So the normal of the plane is -5i + 3j +4k
Use r.n =a.n to find the equation of the plane. You can use the position vector of any point of the plane as 'a'. like when s= 0 the position vector of a point on line l = <1 1 1>. You can take and value of s or t to find a point.
So, <x y z>.<-5 3 4> = <1 1 1>.<-5 3 4>
-5x + 3y + 4z = 2
That's the plane equation.

 

[Jspake]

5iii) y= a- bCos x, The max and min value of Cos x are 1 and -1 respectively.
For this equation, y has a maximum value when Cos x =-1 (and b is negative) i.e. y= a -(b* -1) = a+b.
y has a min value when Cox x =1 (and b is positive) i.e. y = a- (b*1) = a-b
After you have found the max and min values you have to find the values of x at which these occurs. Cos x is equal to -1 only when x is equal to 180 in the given range so the max occurs at 180 degree. Cos x is +1 for two values of x 0 and 360, therefore the min occurs at these values.
Now sketch the curve.
View attachment 21794

Hope this helps. If you have any questions feel free to ask.

Sorry for late reply..

Thanks a lot for your help

 

[Light Yagami]

The detailed solution of:-

 

[Iffat]

can sum1 plz help with the foll qs
2007 may-jun:
q5
q8q11(i),(ii),(iii)

 

[badrobot14]

Assalamualikum
Can someone please help me with question 7(ii), 8(i) & 11(ii).
Thanks a bunch in advance.

11 (ii).... well b is the x value where max point of curve occurs.... for max point.. all you do is take derivative of curve... and set it equal to zero.... you solve for x.. and that gives u x value at max pt...

 

[Rutzaba]

the awkward mo wen u solve ur own question before sumone else does

 

[Silent Hunter]

Aoa anybody got some good notes according to the syllabus (M1) ? need them urgent?

JazakAllah

 

[PhyZac]

Assalamu Alikum Wa Rahmatullahi Wa Barakatoho.

Can anyone solve this, please?

 

[inquisitiveness]

Factorise 24x^2+34x-45 (2 marks)

 

[Dug]

Assalamu Alikum Wa Rahmatullahi Wa Barakatoho.

Can anyone solve this, please?

Walaikum AsSalam Wa Rahmatullahi Wa Barakatoho!!

-> Separate variables: 1/(x + 1) dx = cos2Θ/sin2Θ dΘ
-> Integrate: ln|x + 1| = (1/2)ln|sin2Θ| + c
-> Put values: ln|1| = (1/2)ln(1/2) + c
c = (-1/2)ln(1/2)
-> Simplify: ln|x + 1| = (1/2)ln|sin2Θ| - (1/2)ln(1/2)
ln|x + 1| = (1/2)ln|(sin2Θ)/(1/2)|
ln|x + 1| = ln|2sin2Θ|^1/2
x + 1 = √(2sin2Θ)
x = √(2sin2Θ) - 1

 

[PhyZac]

Walaikum AsSalam Wa Rahmatullahi Wa Barakatoho!!

-> Separate variables: 1/(x + 1) dx = cos2Θ/sin2Θ dΘ
-> Integrate: ln|x + 1| = (1/2)ln|sin2Θ| + c
-> Put values: ln|1| = (1/2)ln(1/2) + c
c = (-1/2)ln(1/2)
-> Simplify: ln|x + 1| = (1/2)ln|sin2Θ| - (1/2)ln(1/2)
ln|x + 1| = (1/2)ln|(sin2Θ)/(1/2)|
ln|x + 1| = ln|2sin2Θ|^1/2
x + 1 = √(2sin2Θ)
x = √(2sin2Θ) - 1

Thank you so much sir! Jazaka Allah khairan, May Allah give you success in this life and hereafter....Alhamdulilah, i got it now..!

 

[Dug]

Factorise 24x^2+34x-45 (2 marks)

24x^2+34x-45
= 24x^2 + 54x - 20x - 45
= 6x(4x + 9) - 5(4x + 9)
= (6x - 5)(4x + 9)

 

[Kumkum]

cos^2θ=(1+cos2θ)/2
sin^2θ=(1-cos2θ)/2
cos^2(2θ)={1+cos2(2θ)}/2=(1+c0s4θ)/2
LHS=sin^2θcos^2θ
={(1-cos2θ)/2}{(1+cos2θ)/2}
=(1-cos^2(2θ))/4
={1-(cos2θ)^2}/4
={1-(1+c0s4θ)/2}/4
=(1-4cosθ)/8=RHS
Hope this helps . Ask if u dnt understand

thank u so much....I understood

 

[inquisitiveness]

thanks alot!Really appreciate your help!!Is there anyway you found it the two numbers are 54 and 20?or do you just keep guessing?

 

[Dug]

thanks alot!Really appreciate your help!!Is there anyway you found it the two numbers are 54 and 20?or do you just keep guessing?

I did it with my calculator (fx-991es). You can solve for the roots and then work your way backwards. It's very helpful !!

 

[Iffat]

can sum1 plz help with the foll qs
2007 may-jun:
q5
q8q11(i),(ii),(iii)

can sum1 plz reply asap!

 

[Dug]

The detailed solution of:-
View attachment 22107

x̄ = Σx/n = 645/150 = 4.3

Expand:
Σ(x-x̄)^2 = Σx^2 - 2Σxx̄ + Σx̄^2

Put values, Σx^2 = 8287.5 ; Σx = 645 ; x̄ = 4.3
Σ(x-x̄)^2 = 8287.5 - 2(645)(4.3) + (150)(4.3)^2
Σ(x-x̄)^2 = 5514

 

[josephsai]

Assalamualikum
Can someone please help me with question 7(ii), 8(i) & 11(ii).
Thanks a bunch in advance.

Q7i) first of all u should know that p is the x coordinate
and q is the y coordinate so substitute y=11-x^2 into other eqn & arrive at
11-x^2=5-x
rearrange them
x^2 - x - 6 =0
the reason why we equate them is bcuz they intersect @ a pt
so solve this quadratic eqn &
x=3 & x= -2
we take x=3 cause it's on the de positive axis
so thats our p
to find q put the value of p which is x=3 into
y=5-x
y=5-3
=2
meaning q is 2 and p is 3
simple inform me if it helped