Mathematics: Post your doubts here!

[redapple20]

Thankooooooo!

 

[aaakhtar19]

can someone please solve question 8 : ii (b)
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_33.pdf

Hope u got it @sea_princess and @2pac

 

[PhoenixAsh12]

Assalamualaykom, I have some questions... I hope someone can answer them. And please tell me if there are rules that I should know about them.

and

Thank you!!

 

[aaakhtar19]

Hey,I also can't solve these kind of questions.Posted a question which had 2 theeta instead of 0.5 and sadly haven't gotten no response but its understandable as p1 examination is near.If you do at all get the answer for your query,please forward to me as well.
Thanks

I have posted answer for O/N 2010 P33 Q8
hope u'll be now able to solve the other question with 2 theeta if not
Reply me with the question and Paper no

 

[PhoenixAsh12]

I have posted answer for O/N 2010 P33 Q8
hope u'll be now able to solve the other question with 2 theeta if not
Reply me with the question and Paper no

Hello! Can you please try and solve my questions? They're above your reply

 

[PhoenixAsh12]

Oh that...Here is an example.
We have 2 lines with gradients m=2 and m=3 respectively.
To find the angle between them,
tan theta = 2
theta = 63.4

tan theta = 3
theta = 71.6

Angle between them = 71.6 - 63.4 = 8.2

Is that for any two lines or tangents?
Very helpful thank you

 

[Dug]

Is that for any two lines or tangents?
Very helpful thank you

Yup

 

[aaakhtar19]

Hello! Can you please try and solve my questions? They're above your reply

Are these from P1 ??

 

[PhoenixAsh12]

Yup

Thanks! Can you please try and solve my questions?

and

if you have the time please

 

[PhoenixAsh12]

Are these from P1 ??

Yes

 

[PhoenixAsh12]

Are these from P1 ??

They're o/n 05 question 9 and m/j 09 queation 8

 

[sea_princess]

Hope u got it @sea_princess and @2pac

thanx a lot

 

[ffaadyy]

can someone please solve question 8 : ii (b)
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_33.pdf

Hey,I also can't solve these kind of questions.Posted a question which had 2 theeta instead of 0.5 and sadly haven't gotten no response but its understandable as p1 examination is near.If you do at all get the answer for your query,please forward to me as well.
Thanks

I'll do the ii(a) part of this question first so that you can understand this method.



(√6) cos θ + (√10)sin θ


R=4
a=52.23

4 cos(θ − 52.23)

In the ii(a) part of this question which is (√6) cos θ + (√10)sin θ = −4, we'll be needing to modify the range first.

0 < θ < 360
0 - 52.23 < θ < 360 - 52.23
-52.23 < θ < 307.76

With the above range, our angle may fall either in the -1 quadrant (cos positive), 1st quadrant (all positive), 2nd quadrant (sin positive), 3rd quadrant ( tan positive) or the 4th quadrant (cos positive).

(√6) cos θ + (√10)sin θ = −4
4 cos(θ − 52.23) = −4
cos (θ − 52.23) = -1

Now, we need to see that in which quadrant is 'cos' negative from the range of quadrants we found above. We have the following quadrants: -1, 1, 2, 3 & 4 and the quadrants in which 'cos' is negative are the 2nd quadrant and the 3rd quadrant. In the 2nd quadrant, its going to be '180 - (cos^-1) (1)' and in the 3rd quadrant, its going to be '180 + cos (1)'. If we solve both these equation, we get the same answer.

cos (θ − 52.23) = -1
θ − 52.23 = 180
θ = 180 + 52.23
θ = 232.23

Coming to the ii(b) part of this question, we'll again be needing to modify the range.

0/2 < θ/2 < 360/2
0 < θ/2 < 180
0 - 52.23 < θ/2 - 52.23 < 180 - 52.23
-52.23 < θ/2 - 52.23 < 127.77

Now with the above range, our angle may fall either in the -1 quadrant (cos positive), 1st quadrant (all positive) or the 2nd quadrant (sin positive).

(√6) cos θ/2 + (√10)sin θ/2 = 3
4 cos(θ/2 − 52.23) = 3

Now, for this part,we need to see that in which quadrant is 'cos' positive from the range of quadrants we found above. We have the following quadrants: -1, 1 and 2 and the quadrants in which 'cos' is positive is the -1 quadrant and the 1st quadrant. In the -1 quadrant, its going to be '- (cos^-1)(3/4) ' and in the 1st quadrant, its going to be '(cos^-1)(3/4) '. In the -1 quadrant, we get θ = - 41.41 and in the 1st quadrant, we get θ = 41.41. θ = -41.41 being the smaller angle, we'll use it.


(√6) cos θ/2 + (√10)sin θ/2 = 3
4 cos(θ/2 − 52.23) = 3
θ/2 − 52.23 = (cos^-1)(3/4)
θ/2 − 52.23 = -41.41
θ/2 = -41.41 + 52.23
θ/2 = 10.82
θ = 21.7°

 

[ffaadyy]

Hope u got it @sea_princess and @2pac

No hard feelings but you've used the wrong range; albeit your answer is correct but the method isn't. The range which needs to be modified is '0 < θ < 360', not '0 < θ < 90'. The range '0 < θ < 90' is for 'a'.

 

[Becky250892]

Isn't 2x^2-8x + 10 = 2((x^2-4x)+10) =2(x^2-4x+4-4) +10 = 2((x-2)^2-4)+10 = 2(x-2)^2-8+10 =

2(x - 2)^2+2 ?

Then F^-1 will be y= 2(x-2)^2+2 = under root y-2/2 +2

It's easy convert it into the general form
2(x^2 - 8x +10 )
2( (x)^2 -2(x)(4) + (4)^2 - (4)^2 + 10)
2( (x -4)^2 -16 + 10)
2(x-4)^2 -12
y = 2(x-4)^2 -12
y + 12/2 = (x-4)^2
underoot +-(y + 12)/2 +4 = x
F^-1(x) = +-(x + 12)/2 +4

 

[aaakhtar19]

No hard feelings but you've used the wrong range; albeit your answer is correct but the method isn't. The range which needs to be modified is '0 < θ < 360', not '0 < θ < 90'. The range '0 < θ < 90' is for 'a'.

Oops Sorry i misread thanks for correcting me

 

[Dug]

They're o/n 05 question 9 and m/j 09 queation 8

I solved the second question thinking someone would solve the first and then we would have two identical solutions. (Happens )
For the second question,
i)
y=12/x --------1
y=-2x + 11 -------2
Set them equal and you get,
12/x = -2x + 11
2x^2 -11x + 12 = 0
Using the quadratic formula gives,
x = 4 , 3/2
y = 3 , 8
(4,3) (3/2,8)

ii) 12/x = -2x + k
Simplify to get 2x^2 - kx + 12 = 0
b^2 - 4ac < 0
k^2 - 4(2)(12) < 0
k^2 < 96
k < +√96
k > -√96

iii)
y=12x^-1
dy/dx= -12/x^2
At P, dy/dx = -12/(2)^2 = -3
tan theta = -3
theta = -71.6

For the line, tan theta = -2
theta = -63.4

Angle between them = -63.4-(-71.6) = 8.2

 

[2pac]

I have posted answer for O/N 2010 P33 Q8
hope u'll be now able to solve the other question with 2 theeta if not
Reply me with the question and Paper no

lol,was too confusing,ill have another look at it again and as for the 2 theeta question.
There is one in q6 part ii nov 2011 p3.
Appreciate your effort.
thanks

 

[PhoenixAsh12]

I solved the second question thinking someone would solve the first and then we would have two identical solutions. (Happens )
For the second question,
i)
y=12/x --------1
y=-2x + 11 -------2
Set them equal and you get,
12/x = -2x + 11
2x^2 -11x + 12 = 0
Using the quadratic formula gives,
x = 4 , 3/2
y = 3 , 8
(4,3) (3/2,8)

ii) 12/x = -2x + k
Simplify to get 2x^2 - kx + 12 = 0
b^2 - 4ac < 0
k^2 - 4(2)(12) < 0
k^2 < 96
k < +√96
k > -√96

iii)
y=12x^-1
dy/dx= -12/x^2
At P, dy/dx = -12/(2)^2 = -3
tan theta = -3
theta = -71.6

For the line, tan theta = -2
theta = -63.4

Angle between them = -63.4-(-71.6) = 8.2

Thank you very much! And unfortunately no one solved the other one =P

 

[PhoenixAsh12]

Can anybody solve question 6(ii)- finding the coordinates of point D please? Also question 8(i)...
http://www.xtremepapers.com/papers/...AS Level/Mathematics (9709)/9709_s07_qp_1.pdf