Mathematics: Post your doubts here!

[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?>

 

[Dug]

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

f(x) = a(x - p)^2 + k
Now plug in values from the graph:
4 = a(0.1 - 0.3)^2 + 8
a = -100

f(x) = -100(x - 0.3)^2 + 8

 

[salvatore]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_qp_11.pdf

Please help me solve question no. 3, I do not understand how to do it.
An explanation will be appreciated.
Thanks a lot!

 

[Dug]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_11.pdf

Please help me solve question no. 3, I do not understand how to do it.
An explanation will be appreciated.
Thanks a lot!

i) Fit 2 cosine cycles in the stated domain. y = 1/2 is a straight horizontal line passing through (0, 1/2).
ii) Number of roots = Number of Points of Intersection of the line and the cosine function.
iii) Simply use ratio and proportion. 2pi = 4 roots, 10pi = 20 roots

 

[Jspake]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_11.pdf

In no. 3, why is the value of AQ the radius of the sector APR? As stated in the question, isn't the radius 2 cm?
Please help!

 

[Dug]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_11.pdf

In no. 3, why is the value of AQ the radius of the sector APR? As stated in the question, isn't the radius 2 cm?
Please help!

What's stated in the question is the length of the side of triangle ABC. Length AB is not equal to length AQ. You have to use Pythagoras Theorem to find r or AQ.