Hey does anyone remember the questions for 5i), 5ii), 6i) 6ii) and 9i) please
Oh thats right thanks do you know the questions for 6, 7 and 9?
6i) was that arithmetic progression where she used 6g then 7.8g for an experiment
6ii) was the geometric series where she had 1800g and you had to first prove that the maximum number of experiments she could do could be expressed as 1.3^N<91 and then use logarithms to find the value of N
7i) you had to show that the area between the curve y = x^3/2 - 1 and the x axis was 9.4
ii) you had to find the area between the curve, it's tangent at the point (4,7) and the x axis
9i) you had to find the remainder when 4x^3 -7x - 3 was divided by (x-2)
ii) you had to prove that (2x + 1) was a factor and express the cubic as the product of three linear factors
iii) you had to find out the possible values of x when 4cos^3x - 7cosx - 3 = 0
(Original post by kiraeil) For the area under curve question would I get some marks for workings as i worked out equation of tangent and area of small triangle but got complete wrong answer? thanks
Since there's not one in the OP: Worked Solution to 8ii) ax=4bx[br]⇒xlog2a=2+xlog2b[br]⇒x(log2ba)=2[br]Asab=2⇒b=a2[br]⇒x(log22a2)=2[br]⇒x=log2a−12
And: For the sake of completion: Worked solution to 7)i) I=∫14x23−1dx=[52x25−x]14[br]⇒I=(52(425)−4)−(52(125)−1)[br]⇒I=564−4+53=567−20=547=952
c) Any value where a > 1 and any value where 0 < b < 1 (2 marks)
(ii) Show ... (5 marks)
9. (i) 15 (2 marks)
(ii) Show f(-0.5)=0 and f(x)=(2x+1)(2x−3)(x+1) (6 marks)
(iii) θ=32π and θ=34π and θ=π (4 marks)
I'll be around a bit this afternoon and then part of this evening to answer questions. Please post in this thread rather than sending me private messages.
Heeeey for (4b) I never got this (i) (2 marks) I got -6x^-4+k can u explain what I did rong
Heeeey for (4b) I never got this (i) (2 marks) I got -6x^-4+k can u explain what I did rong
When integrating, you're supposed to add 1 to the index and then divide the coefficient by that new number. You got the index and took 1. -3 + 1 = -2, so you had to divide 24 by -2 not -4.
c) Any value where a > 1 and any value where 0 < b < 1 (2 marks)
(ii) Show ... (5 marks)
9. (i) 15 (2 marks)
(ii) Show f(-0.5)=0 and f(x)=(2x+1)(2x−3)(x+1) (6 marks)
(iii) θ=32π and θ=34π and θ=π (4 marks)
I'll be around a bit this afternoon and then part of this evening to answer questions. Please post in this thread rather than sending me private messages.
I got 59-62 marks do you think this will be an A? Everyone is saying it was easy, could you hazard a guess at the grade boundaries?
and also for 9ii) I got wrote (2x-3)(x+1) without the previous answer but then I wrote so x= -1/2 +2/3 or -1 will I lose a mark for having not written (2x+1)(2X-3)(X+1) ???
For 4b ii) evaluated it as lim n -> infinity 12/n = 0 therefore it is (0) - (-12/a) = 3 therefore a = 4, as I forgot the square sign on the a and left it as a = 4 how many marks would I lose for this?