Mathematics: Post your doubts here!

[~`Heba`~ :)]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_qp_41.pdf #6 (ii) I don't understand why we use sin5/sin1 here, please explain. Thanks a lot

 

[iKhaled]

please need help with may june 2011 31 question 8(ii) and 8(iii)

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s11_qp_31.pdf

 

[Iffat]

Plz help!
2012 may june p11 q5(ii) and q8(i)
p12 q4 and q10

 

[Iffat]

Plz help!
2012 may june p11 q5(ii) and q8(i)
p12 q4 and q10

Can sum1 plz relpy asap an theres 1 more qustion p13 q7

 

[Dug]

Plz help!
2012 may june p11 q5(ii) and q8(i)
p12 q4 and q10

Q5(ii)

6x+k = 7√x
Let y = √x
6x^2 - 7y + k = 0

b^2 - 4ac = 0 (Since its a tangent)
(-7)^2 - 4(6)(k) = 0
k = 49/24

Q8(i)
x^2 - 4x + k
= x^2 - 4x + (-2)^2 - (-2)^2 + k
= (x - 2)^2 - 4 + k

 

[Iffat]

Q5(ii)

6x+k = 7√x
Let y = √x
6x^2 - 7y + k = 0

b^2 - 4ac = 0 (Since its a tangent)
(-7)^2 - 4(6)(k) = 0
k = 49/24

Q8(i)
x^2 - 4x + k
= x^2 - 4x + (-2)^2 - (-2)^2 + k
= (x - 2)^2 - 4 + k

Thanx but for q5 how did it becum 6x^2?
n plz plz if u can ans d other 2 questions (2012 may-jun p12 q4 n q10) also

 

[Dug]

Thanx but for q5 how did it becum 6x^2?
n plz plz if u can ans d other 2 questions (2012 may-jun p12 q4 n q10) also

I copy-pasted that from a previous post and it seems the typo tagged along. It's 6y^2.

 

[Dug]

Thanx but for q5 how did it becum 6x^2?
n plz plz if u can ans d other 2 questions (2012 may-jun p12 q4 n q10) also

Q4

AB:
m = [1 - (-5)] / [7 - (-1)] = 3/4

Coordinates of Mid-point = -1+7/2 , -5+1/2 => 3 , 2

CD:
m = -4/3

Eq of CD: 2 = 3(-4/3) + c => c = -2
y = (-4/3)x - 2

At C, y = 0 and x = -3/2

At D, x = 0 and y = -2

Length of CD = √[(-3/2)^2 + (-2)^2] = 5/2 = 2.5 units

 

[Iffat]

I copy-pasted that from a previous post and it seems the typo tagged along. It's 6y^2.

k but wat iwas actually askin is how does it become 6y^2

 

[Dug]

k but wat iwas actually askin is how does it become 6y^2

I supposed that y = √x. So x = y^2 and 6x = 6y^2. Hope you got it this time.

 

[Kumkum]

can someone plz help me with question 2
http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s08_qp_3.pdf

 

[Iffat]

I supposed that y = √x. So x = y^2 and 6x = 6y^2. Hope you got it this time.

oooh thanx so much!

 

[Dug]

can someone plz help me with question 2
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s08_qp_3.pdf

e^x + e^2x - e^3x = 0
e^x (1 + e^x - e^2x) = 0
Reject e^x = 0

1 + e^x - e^2x = 0
let y = e^x
y^2 - y - 1 = 0
I am sure you can solve the rest yourself. You have to use the quadratic formula and in the end, take log of the roots to find x.

 

[Kumkum]

e^x + e^2x - e^3x = 0
e^x (1 + e^x - e^2x) = 0
Reject e^x = 0

1 + e^x - e^2x = 0
let y = e^x
y^2 - y - 1 = 0
I am sure you can solve the rest yourself. You have to use the quadratic formula and in the end, take log of the roots to find x.

Jazakallah! This part confused me e^x + e^2x - e^3x, I used substitution straight away saying let u=e^x....and got stuck. but now I understand.

 

[snowbrood]

3 The polynomial p(x) is defined by
p(x) = x3 − 3ax + 4a,
where a is a constant.
(i) Given that (x − 2) is a factor of p(x), find the value of a. [2]
(ii) When a has this value,
(a) factorise p(x) completely, [3]
(b) find all the roots of the equation p(x2) = 0. [2]
can ne1 pls help with 3b?
thanks in advance

is this as level and from which past paper?

 

[Alice123]

is this as level and from which past paper?

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_31.pdf

 

[Dug]

3 The polynomial p(x) is defined by
p(x) = x3 − 3ax + 4a,
where a is a constant.
(i) Given that (x − 2) is a factor of p(x), find the value of a. [2]
(ii) When a has this value,
(a) factorise p(x) completely, [3]
(b) find all the roots of the equation p(x2) = 0. [2]
can ne1 pls help with 3b?
thanks in advance

p(x) = (x - 2)^2 (x + 4)
p(x^2) = (x^2 - 2)^2 (x^2 + 4)
Roots:
x^2 = 2
x = ±√2

x^2 = -4
x = ±√4i
x = ±2i

 

[snowbrood]

can anyone tell me how to derive quadratic equation for this curve
i know y intercept is 0 so c =0
0.36a+0.6b=0
b=-0.6a
(0.3^2)a +0.3(-0.6a)=8
a=-88.9
b=53.33
-88.9x^2+53.33x i get x intercepts vertex and area under the curve right but when i plug in a value of x other than intercept or vertex i get a wrong answer need help guys

 

[Dug]

View attachment 21464
can anyone tell me how to derive quadratic equation for this curve
i know y intercept is 0 so c =0
0.36a+0.6b=0
b=-0.6a
(0.3^2)a +0.3(-0.6a)=8
a=-88.9
b=53.33
-88.9x^2+53.33x i get x intercepts vertex and area under the curve right but when i plug in a value of x other than intercept or vertex i get a wrong answer need help guys

Use completing square.

 

[snowbrood]

Use completing square.

done but no help why dont u try to solve it and see if u can get an equation?>