Mathematics: Post your doubts here!

[Haya Ahmed]

http://onlineexamhelp.com/wp-content/uploads/2014/08/9709_s14_qp_32.pdf
Q5, Q6(i), Q7(b) , Q9(ii) and (iii), Q10 (iii)

 

[Ram97]

Ohhh! this makes so much sense now! the mark scheme just confused me as hell as to why they just added part i and ii ...
Thanks a lot!

Welcome!
Haha... They do confuse a lot

 

[sara kamal]

http://maxpapers.com/wp-content/uploads/2012/11/9709_s14_qp_31.pdf
please anyone?
explain q5 part 2

 

[Debonny1]

if x-y is a factor of a polynomial and if you use substitute y in the differential equation of the polynomial does the answer have to be zero?? Why?

 

[_Ahmad]

http://onlineexamhelp.com/wp-content/uploads/2014/08/9709_s14_qp_32.pdf
Q5, Q6(i), Q7(b) , Q9(ii) and (iii), Q10 (iii)

http://justpastpapers.com/9709_s14_ms_32/

 

[Mayisha M]

Someone help me with this S1 problem. It's from O/N/13 - 61. I solved it (attached the picture) - my distribution for X is wrong; can someone please explain why the distribution for X is lower than that for Y? I attached the markscheme screenshot too, haha. Please help!

 

[Ram97]

Someone help me with this S1 problem. It's from O/N/13 - 61. I solved it (attached the picture) - my distribution for X is wrong; can someone please explain why the distribution for X is lower than that for Y? I attached the markscheme screenshot too, haha. Please help!
View attachment 52744
View attachment 52746

Smaller deviation, higher peak
Larger deviation, low peak

 

[Debonny1]

http://justpastpapers.com/9709_s14_ms_32/

I don't get 5(2). If x-2 is a factor of the equation, why is it also a factor of the differential of the equation :S

 

[flesh-eater]

http://maxpapers.com/wp-content/uploads/2012/11/9709_s13_qp_12.pdf
Can someone help me with Q9(iii)? Why is the domain -2.5 ≤ x < 0 and not just x ≥ -2.5?

 

[nehaoscar]

http://maxpapers.com/wp-content/uploads/2012/11/9709_s13_qp_12.pdf
Can someone help me with Q9(iii)? Why is the domain -2.5 ≤ x < 0 and not just x ≥ -2.5?

The range of the original function is -2.5 ≤ f(x) < 0
Therefore the domain of the inverse is the range of the original.

 

[flesh-eater]

The range of the original function is -2.5 ≤ f(x) < 0
Therefore the domain of the inverse is the range of the original.

But why is the range less than 0? If x ≥ 1, then shouldn't the range just be ≥-2.5?

 

[nehaoscar]

But why is the range less than 0? If x ≥ 1, then shouldn't the range just be ≥-2.5?

Well since the domain is x ≥ 1
The denominator will always be negative
so f(x) will also always be negative, so it's also <0

 

[flesh-eater]

Oh okay, I get it now. Thanks!

 

[Haya Ahmed]

http://justpastpapers.com/9709_s14_ms_32/

Thaaaaaaaaaaaaaaaaanks ^_^

 

[Haya Ahmed]

http://onlineexamhelp.com/wp-content/uploads/2014/08/9709_s14_qp_31.pdf

Q5 (ii) and Q10(ii)

_Ahmad

 

[nehaoscar]

Can someone please explain and solve part ii with workings please!!!

 

[ali_jaliawala]

Best Revision Portal!
Please help us in spreading the page so that maximum students can avail this opportunity!
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[Ram97]

View attachment 52809

Can someone please explain and solve part ii with workings please!!!

Which year is this?

 

[nehaoscar]

Which year is this?

Variant 12 - Winter 2012

 

[_Ahmad]

http://maxpapers.com/wp-content/uploads/2012/11/9709_s14_qp_31.pdf
please anyone?
explain q5 part 2

z=9 e^i(1/3)

z=9(e^0(cos(1/3)+i sin(1/3))

square root of z=(9 (e^0(cos(1/3)+i sin(1/3)) )^1/2

=9^(1/2)( cos(0.5*(1/3 π))+i sin(0.5*(1/3 π)) )

=+-3(cos(1/6 π) + i sin(1/6 π))

one square root is =3(cos(1/6 π) + i sin(1/6 π))
= (3√3/2)+i (3/2) (open the bracket)
r=3
θ=1/6 π
so 3 e^i (1/6) π

other root = - 3(cos(1/6 π) + i sin(1/6 π))
= - (3√3/2)- i (3/2)
r=3
θ = (1/6 π)
θ= (1/6 π)-π = -(5/6)π
so 3 e^i (-5/6) π