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Since you only need the coefficient of x^8How did you expand (x+3x^2)^4?
You can skip the full expansion and only do (3x^2)^4 Since no other terms in the expansion will give x^8
So it gives 81x^8
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Since you only need the coefficient of x^8How did you expand (x+3x^2)^4?
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 ...From your above results you can see that x^8 is when (x+3x^2)^4 and (x+3x^2)^5
Similarly in (1+(x+3x^2))^5 x^8 would be at when the expression has power of 4 and 5
So 5C4 (1) x (x+3x)^4 + 5C5 (x+3x^2)^5
Therefore 450x^8 + 270x^8 = 675x^8
Hope you understand![]()
Welcome!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!![]()
http://onlineexamhelp.com/wp-content/uploads/2014/08/9709_s14_qp_32.pdf
Q5, Q6(i), Q7(b) , Q9(ii) and (iii), Q10 (iii)
Smaller deviation, higher peakSomeone 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
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
The range of the original function is -2.5 ≤ f(x) < 0http://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.
Well since the domain is x ≥ 1But why is the range less than 0? If x ≥ 1, then shouldn't the range just be ≥-2.5?
Thaaaaaaaaaaaaaaaaanks ^_^
Which year is this?
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