Hi everyone. This is the official thread for the following exams
Maths Pure Core 1 JUN13/MPC1
Maths Pure Core 2 JUN13/MPC2
Guys, if you want to make corrections or anything, please copy and paste the whole thing, and get the answers in the right order. I won't be online for now, most of you should be able to get most of this mark scheme done. Please make corrections |etc, and at the end, I will update my own post to reflect this.
THANKS
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Core 1
question 2)a)ii) the question asked x√12 = 7√3 - √48
x= 3/2
For Q1, p was -3, k = 10, D = (-7, -4).
For the surds question, x = 1.5 and for part b the answer they were looking for was 8 - (15)^0.5.
The complete the square question was 2(x+1.5)^2 + 0.5 and the length of AB was 0.5(10)^0.5.
Dividing the polynomial by x+3 gave x^2 - 3x + 5.
The integral was 31.5 or 63/2 and the area of the shaded region was 27.
The point furthest away from the x-axis was (5, -14) and the translation was by vector (6, -7).
The last question was 20/7 ( < or =) k (< or =) 4.
Equation of normal
X=4 Y=20 normal -4/9 So Y-20=-4/9(x-4)
3. (Circle question)
(a) (x-5)^2 + (y+7)^2 = 49
Radius was 7 so to draw the circle, you plot the middle point (5, -7) and go out 7 from the centre.
This implies a shift to the right of 6 units in x and 7 down in y. This is represented by the vector [6, -7].
Also, for question 5)b)ii), it asked you to use the coordinates that we got from part a)ii) (-3,-13) or something I think,
and find a minimum value putting it into 1/2√n
any done that one? And did you get 1/2√100? - DID ANYONE ELSE GET THIS BECAUSE I DID?
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Core 2
Trapezium Rule Question - 7.12
1.
The first term of a geometric series in 80 and the common ratio is 0.5
a) work out the third term in the series - 20
b) work out the sum to infinity - 160
c) work out the sum of the first 12 terms in the series to 2 decimal places - 159.96
2. Work out Arc --> 20 x 0.8 = 16
ii) work out area of sector ----> 0.5 x 20^2 x 0.8 = 160
b) work out obtuse angle ABD
Ad was 15 I think, and rad was 0.8
Other length was 20
Sin0.8/15 = SinD/20
20sin0.8 = 15sinD
SinD= 20sin0.8/15
D= sin^-1 ( 20sin0.8/15) in rads
D= 1.27 rads
Since its obtuse, it's Pi -1.27
D= 1.87
3.
a) (2+Y)^3 = 8 + 12y + 6y^2 + Y^3
B) 16 +12x^-4
(c) Integral = 39/2
6.
a. Write log(a)b = c in index form. - a^c = b
b. Show that only one value of x satisfies 2log(2)(x+7) - log(2)(x-5) = 3
log2(x+7)2 - log2(x+5) - 3 = 0
→log2(x2+14x+49) = 3 + log2(x+5)
→log2(x2+14x+49) = log223 + log2(x+5)
→log2(x2+14x+49) = log2(8x+40)
→x2+14x+49 = 8x+40
→x2+6x+9 = 0
→(x+3)2
7.
a) Show p=3/2
72=p96+q
24=p24+q
48=p72
48/72= 2/3
b) Find u3
72= 2/3*96 +q
72=64 + q
Q=8
u3= 2/3*72 +q
U3= 48 +q
u3= 48 + 8
= 56
The transformation of [2 -0.7] was √((x-2)3 +1) -0.7
2k question k = 6/5
x = 16, 104 and 136 for the last one
also.. tan x = -1
solves to
x = 135, 315