Mathematics: Post your doubts here!

[Echelon94]

mmmm why cant 3^y=-3/4
Q4 ii) MAT/JUNE 2011/31

Because to calculate y you'll need to calculate ln(-3/4)/ln(3) & ln(-ve number) is never possible!

 

[Echelon94]

hey can anyone show & explain the complete working for q5(ii) that will earn all 3 marks in http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w08_qp_3.pdf ??
thanks...

(http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w08_ms_3.pdf
&
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w08_er.pdf)

 

[aliya_zad]

http://www.xtremepapers.com/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_s08_qp_3.pdf
Please help me with the 8th one!!

 

[reina81]

already did it here man...
http://www.xtremepapers.com/communi...st-your-doubts-here.9599/page-127#post-228526

thanks

 

[miss irfan]

how do we solve this z^6=-64

 

[Chooi1993]

Can anyone explain to me why q10 (ii) answer is 1800 ?
Please help me ! 2moro is my exam !
http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_s11_qp_31.pdf​

How to get the initial value ?​

http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_s11_qp_32.pdf​

 

[smzimran]

Time to announce my entry here!

 

[smzimran]

help needed in q6 ii

w = -1 + i
w^2 = 0 - 2i
To find centre of circle, use midpoint formula:
= [ {(-1 + 0) / 2} , {(i - 2i) / 2} ]
= [-0.5 , -0.5i ]

To find radius find half the length of w to w^2
= 0.5 * | {-1 - (0)} , {i - (-2i)} |
= 0.5 * | -1 , 3i |
Find length by taking modulus
radius = 0.5 * root [ (-1)^2 + (3)^2 ]
= 0.5 * root [1 + 9]
= 0.5 * root (10)

Now write the equation in the form |z − (a + bi)| = k
where (a + bi) is the coordinate of the centre of the circle
and k is the radius

 

[smzimran]

mmmm why cant 3^y=-3/4
Q4 ii) MAT/JUNE 2011/31
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_31.pdf
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_ms_31.pdf

12 × 27^y + 25 × 9^y − 4 × 3^y − 12 = 0
Let 3^y = a
12 × a^3 + 25 × a^2 - 4 × a - 12 = 0
12a^3 + 25a^2 - 4a - 12 = 0

This is comparable to the equation in (i) so the factorisation will be:
(a + 2)(4a + 3)(3a – 2) = 0

Your question was why cant 3^y=-3/4 ?
The answer is because to solve this equation we have to take ln or lg of both sides!
Check the value of lg (-3/4) or ln(-3/4)
It is undefined because there is no lg or ln of a negative number!
That is why it can't be 3^y=-3/4 !
Hope you got it!

 

[smzimran]

Here is a little note on logarithms by me, hope it helps!

ln is natural log, it stands for log base 'e' and you should know that:
ln e = 1
ln 1 = 0
if equation such as this is given:
ln x = 5
then
x = e^5
Similarly if an equation such as this is given:
e^x = 5
then taking ln on both sides
x lne = ln5 (lne = 1)
x = ln5

lg is general log, it stands for log base '10' and you should know that:
lg 10 = 1
lg 1 = 0
if equation such as this is given:
lg x = 8
then
x = 10^8
Similarly if an equation such as this is given:
10^x = 8
then taking lg on both sides
x lg10 = lg8 (lg10 = 1)
x = lg 8

Another type of question can be this:
3^x = 5
Now, its your choice either you can take ln of both sides or lg of both sides
Im taking ln of both sides
ln(3^x) = ln5
x ln3 = ln5
x = ln5/ln3

 

[smzimran]

hey can anyone show & explain the complete working for q5(ii) that will earn all 3 marks in http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w08_qp_3.pdf ??
thanks...

(http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w08_ms_3.pdf
&
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w08_er.pdf)

One of the factors is 2x^2 − 3x + 3
So
p(x) = 4x^3 − 4x^2 + 3x + 3 = (2x^2 − 3x + 3) (bx+c)

Comparing coefficients of x^3:
4 = 2(b)
b = 2

Comparing coefficients of x^0 (the constant term):
3 = 3(c)
c = 1

p(x) = 4x^3 − 4x^2 + 3x + 3 = (2x^2 − 3x + 3) (2x+1)
p(x) = (2x^2 − 3x + 3) (2x+1)

p(x) < 0
(2x^2 − 3x + 3) (2x+1) < 0

(2x^2 − 3x + 3) is never zero because:
b^2 - 4ac = (-3)^2 - 4(2)(3) = 9 - 24 = -15

So the only real root is
(2x+1) < 0
x< -1/2

 

[smzimran]

how do we solve this z^6=-64

Post the whole question with link to the p.p and m.s !
I think this is a question from complex numbers.

 

[miss irfan]

Post the whole question with link to the p.p and m.s !
I think this is a question from complex numbers.

https://docs.google.com/viewer?a=v&...ufarCv&sig=AHIEtbR25jP8EKBUeSdXW_ilg8B8yTawGw
Q7iii

 

[miss irfan]

https://docs.google.com/viewer?a=v&q=cache:IaRQJhRmAkwJ:www.xtremepapers.com/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_s11_qp_33.pdf june 2011 paper 33 9709&hl=en&gl=pk&pid=bl&srcid=ADGEESjYz73WA17qH7QAbaflbMEE_JFcN4JyTaUj6TO5ViQDUWn2LZD-aqAovg-bma5R5mQNrn2hclR6h4yfIz6jRIdKI7ObAhT6V7KAzkD0R3HccWZm71mN3G8dAA-yQspXn2ufarCv&sig=AHIEtbR25jP8EKBUeSdXW_ilg8B8yTawGw
Q7iii

https://docs.google.com/viewer?a=v&...91m51f&sig=AHIEtbT8Gd8fvDJMkYEoLWJzPLYRNl-4Wg

 

[smzimran]

Can anyone explain to me why q10 (ii) answer is 1800 ?​

Please help me ! 2moro is my exam !​

http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_s11_qp_31.pdf​

How to get the initial value ?​

http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_s11_qp_32.pdf​

Next time post the link to the m.s as well!

After a long time, as t becomes very large, lets suppose t = 200
e^(t/2) = e^(200/2) = e^100 = very large value so large that adding 5 to it won't make any difference,
So (5 + e^(t/2)) will become equal to e^(t/2)
The numerator and denominator will cancel out and the answer left will be 1800.
You may verify this using a calculator!

Initial value in which question ?
If its the iteration question, take initial value as pie

 

[aliya_zad]

http://www.xtremepapers.com/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_s04_qp_3.pdf
Q6..
ive reached till this point..dunno how to continue..
ln y + (y^3/3) = x +(1/3)

Please help!!

 

[smzimran]

https://docs.google.com/viewer?a=v&q=cache:IaRQJhRmAkwJ:www.xtremepapers.com/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_s11_qp_33.pdf june 2011 paper 33 9709&hl=en&gl=pk&pid=bl&srcid=ADGEESjYz73WA17qH7QAbaflbMEE_JFcN4JyTaUj6TO5ViQDUWn2LZD-aqAovg-bma5R5mQNrn2hclR6h4yfIz6jRIdKI7ObAhT6V7KAzkD0R3HccWZm71mN3G8dAA-yQspXn2ufarCv&sig=AHIEtbR25jP8EKBUeSdXW_ilg8B8yTawGw
Q7iii

https://docs.google.com/viewer?a=v&q=cache:TmYSFWZpho0J:www.xtremepapers.com/CIE/index.php?dir=International%20A%20And%20AS%20Level/9709%20-%20Mathematics/&file=9709_s11_ms_33.pdf june 2011 paper 33 9709&hl=en&gl=pk&pid=bl&srcid=ADGEESikYPmcahCHYpqpmGx_zgPTgXf_SJXKtNhcu9Q0nGP0gNBAP50IcXUQ3kEUv7RmzrkRabDkIfZcW2qUTqgxflAkqBDQlYGgsWerPrhESNfCyBnS930jvtREmYId4Li7GH91m51f&sig=AHIEtbT8Gd8fvDJMkYEoLWJzPLYRNl-4Wg

Its simple:
Break the power 6 into 3 times square and solve for both roots one by one
Like this:
(− root(3) + i)^2 * (− root(3) + i)^2 * (− root(3) + i)^2

Btw, are you a teacher ?

 

[miss irfan]

Its simple:
Break the power 6 into 3 times square and solve for both roots one by one
Like this:
(− root(3) + i)^2 * (− root(3) + i)^2 * (− root(3) + i)^2

Btw, are you a teacher ?

thanks a bunch
and no i am NOT!

 

[miss irfan]

View attachment 9581
w = -1 + i
w^2 = 0 - 2i
To find centre of circle, use midpoint formula:
= [ {(-1 + 0) / 2} , {(i - 2i) / 2} ]
= [-0.5 , -0.5i ]

To find radius find half the length of w to w^2
= 0.5 * | {-1 - (0)} , {i - (-2i)} |
= 0.5 * | -1 , 3i |
Find length by taking modulus
radius = 0.5 * root [ (-1)^2 + (3)^2 ]
= 0.5 * root [1 + 9]
= 0.5 * root (10)

Now write the equation in the form |z − (a + bi)| = k
where (a + bi) is the coordinate of the centre of the circle
and k is the radius

how did you find the radius?

 

[smzimran]

http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_s04_qp_3.pdf
Q6..
ive reached till this point..dunno how to continue..
ln y + (y^3/3) = x +(1/3)

Please help!!