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Maths P12 !!!!!!!

How Much Minimum You Think Would Be The Curve For grade A For Maths P12

  • 65+

    Votes: 4 6.8%

  • 60+

    Votes: 29 49.2%

  • 55+

    Votes: 20 33.9%

  • or below (which isnt possible as per my thoughts)

    Votes: 6 10.2%
  • Total voters
    59
Y

youngsinoi

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For the question where it says prove a^2 <3b (if i remember). Why do you use b^2 - 4ac??? This is how i proceeded: disregarding part ii) of the question, The curve can either be increasing or decreasing , that is dy/dx is either greater or less than zero. Then i find dy/dx and then d^2 y/dx^2 , I equate the second derivative to zero, then i solve for x (because we don't know if it is increasing or decreasing).
I take this value of x i substitute in dy/dx, then i assume dy/dx>0 or dydx<0. For dy/dx>0 i do get the answer as the negative cancels and it becomes less than. Do you think i will get any mark for my method?
 
A

Abdel Moniem

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249
Points
53
Maths Paper 12 1) y=x^2+1, x=(y-1)^1/2 use ∫ → π x^2dy Set the limits from 5 to 1. 1/2 and ^2 gets cancelled so we are left with (y-1)^1. (y-1)^2/2. (5-1)^2/2-(1-1)^2/2*pi=8pi 2) Use Pythagorean theorem to find AB=((12)^2+(5)^2)^1/2=13 Use Sin Rule 13/sin90=5/Sinx Sin^-1=5/13=22.619*pi/180=0.394 We are find the angle in radian form. ii) AOB=AQO+OQB, AQO=(1/2*base*height)-(1/2r^2Ѳ(B)=1/2*12*5-1/2*(5)^2*1.179=15.30125 APO is a sector with radius 12 and with Angle A=1/2(12)^2(0.3948)=28.4256 QPO = 28.4256 - 15.3013 = 13.1243 3) (1+x)^5=1+5(x)+10x^2, ii) (1+(px+x^2))^5 = 1 + 5(px+x^2) + 10(px+x^2)^2=1+5px+5x^2+10p^2x^2+20px^3+10x^4 collect the terms x^2 5+10p^2=95, 10p^2=90 p=3. 4) y=12/3-2x dy/dx=12*-2(3-2x)^-1=-12*-2/(3-2x)^2 ii) dy/dt=dy/dx*dx/dt=0.15/0.4=dy/dx=1/2 dy/dt divided by dx/dt=8/3 8/3=24/(3-2x)^2 x=0 or x=3 6) 1+sinxtanx=5 cosx 1+sinx(sinx/cosx)=5cosx 1+sinx^2/cosx=5cosx cosx+(1-cos^2x)=5cos^2x 6cos^2x-cosx-1=(2cosx-1)(3cosx+1) cosx=-1/3 or 1/2 cosine is negative in the 2nd and 3rd quadrant. 180-70.5 and cos^-1(1/2)=60 6) x^3+ax^2+bx dy/dx<0 3x^2+2ax+b<0 b^2-4ac<0 (2a)^2-4(3)(b)^2, 4a^2-12b<0 4(a^2-3b)<0 a^2<3b. 3x^2-12x+9<0 3(x^2-4x+3)<0 3(x-3)(x-1)<0 1<x<3
 

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A

Abdel Moniem

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n/2(2a+(n-1)d)=32n-n^2
2an/2+dn^2/2-dn/2=32n-n^2
an+dn^2/2-dn/2=32n-n^2
Calculate the common difference by collecting only the terms which are n^2.
dn^2/2=-n^2 n^2 are cancelled out, d/2=-1 d=-2
Now, collect the "n" terms and "n"s are cancelled out.
-dn/2+an=32n
-(-2)/2+a=32
a=32-1=31

ar^n-1
a2=a1*r
a1+a2=12.8=a1+a1*r
12.8=a1(1+r)
a1=12.8/(1+r)
Sn=a/1-r =12.8/1+r/1-r
20=12.8/(1-r^2)
12.8/20=(1-r^2)
-r^2=12.8/20-1
r=3/5
a1=12.8/(1+3/5)=8
 
Last edited:
A

Abdel Moniem

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162
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y2-y1/x2-x1=-3-6/5-2=-9/3=-3 Parallel lines share the same gradient y-y1=m(x-x1)=y-6=-3(x-2)=y-6=-3x+6=y=-3x+12 AD is perpendicular to CD the gradient is the reciporical to the gradient of equation AB. Equation of CD and AD
CD=y-3=-3(x-8) y-3=-3x+24, y=-3x+27
AD=y-6=1/3(x-2)=3y-18=x-2=3y=x+16
3(-3x+27)=x+16
-9x+81=x+16
-10x=-65
x=6.5
y=7.5
For a parallelogram diagonals bisect each other, M is the midpoint of AC and BE
A(2,6) C(8,3)
(8+2/2,3+6/2)=(5,4.5)
B(5,-3) E(x2,y2)
E(5+x2/2=5, -3+y2/2=4.5)
E(5+x2=10,-3+y2=9)
E(5,12)
Length=((x2-x1)^2+(y2-y1)^2))^1/2=(5-5)^2+(12--3)^2)^1/2=15
 
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