OCR MEI C2 june 2017

What did every one get for the log questions where you have to find w?

Also for integration question on the first page. When the second part didn't have the limit, did you put +c at the end?

How did you find the paper?

I found the log w question a bit weird, I got something like w=a^3 + 3x^4? Can't really remember though!
And yes you have to put +c for one mark
Overall I thought it was ok, although time was a big issue with section B!

What did every one get for the log questions where you have to find w?

Also for integration question on the first page. When the second part didn't have the limit, did you put +c at the end?

How did you find the paper?

w = 1/12 a³ x^4
That's what I got

Do you remember the exact question for the log question?

I didn't get that for the log quesions. Think that's the correct answer as all the other people I talked to got that.

I though section B was good compared to section A. For the surface area question i got r as 3.99 and surface area as 101. For the car park meadow and pond question I got 93.7% or 94.7% can't remember which one though. And for the last question I got n as 25

if you put r as 3.99, back into the equation for A you should have ended up with 300.49... which should be rounded to 301 or 300.5

Would it be okay if you rounded 3.99 to 4.0 but left the minimum point as a surd

What did everyone get for the car park field percentage?

This is what I can remember, random order :/

1)
a) Integral 5 to 1: 4xdx = [2x^2] = 50 – 2 = 48.
b) Integral: 6x^1/2 = 4x^3/2 + c

2)
a) (log0.2)-(log0.1)/(0.2-0.1) = 3.01
b) Plot c in between a and b

3)
Log_aw = 3 + log_ax^5 + log_a6 – log_a2x
Log_aw = log_aa^3 + log_ax^5 + log_a6 – log_a2x
Log_aw = log_a(6a^3x^5) – log_a2x
Log_aw = log_a(6a^3x^5/2x)
Log_aw = log_a(3a^3x^4)
W = 3a^3x^4

4) Y=2x^3. Normal at x = 2. y = 16. Dy/dx = 6x^2. Dy/dx = 24. Normal = -1/24. Equation: x + 24y = 386.

5)
a) Cosine rule. a^2 = 32^2 + 15^2 - (2 x 32 x 15 x cos116). AE = 40.86….
b)
Perp. Distance from AE to D.
Area ADE = 1/2 absinC = 1/2 x 32 x 15 x sin116 = 215.7.
Area ADE = b x h x 1/2.
215.7... = 40.86... x h x 1/2
h = 10.55… > 10. Therefore pond lies in triangle.
c)
Area of pond = 116/360 x Pi x 10^2 = 101.22…
Area of ADE = 215.7…
Area of meadow = 114.48…
d)
Angle C = 360 - 90 - 90 - 116 = 64.
Use that to work out total length of base of trapezium = 70 something.
Area of trapezium = 1/2 x (32 + 70 something) x 80 = 4120.74…
Area of car park = trapezium – ADE = 3905.033…
90% of trapezium = 3708.66…
3905>3708 so carpark takes up more than 90% of field.

6)
6cosx^2 = 5 - sinx
6(1-sinx^2) = 5 - sinx
6 - 6sinx^2 = 5 - sinx
6sinx^2 - sinx - 1 = 0
(3sinx + 1)(2sinx - 1) = 0
sinx = -1/3 or 1/2
x = 3.48, 5.94, 1/6PI, 5/6Pi

7)
a)
Asif = 30000 + (n-1)1000. Bettina = 25000 x 1.05^n-1.
10^th year – Asif = £39000. Bettina = £38783. Asif had more.
11^th year – Asif = £40000. Bettina = £40722. Bettina had more
b)
Total after 17 years.
Asif = 646000. Bettina = 646000 (to the nearest hundred). The same
c)
Total > £M.
25000(1.05^n – 1)/1.05-1 > M
25000(1.05^n – 1)/0.05 > M
25000(1.05^n – 1) > 0.05M
500000(1.05^n – 1) > M
(500000 x 1.05^n) – 500000 > M
500000 x 1.05^n > M + 500000
1.05^n > (M + 500000)/500000
log1.05^n > log (M + 500000) – log500000
nlog1.05 > log (M + 500000) – log500000
n > (log (M + 500000) – log500000)/log1.05
n = 25 or 26 :/

8)
a)
Stretch parallel to y-axis SF 2.
b)
Translation (3 0)

9)
Curve goes through (2,10)
Dy/dx = ? - 7.
Equation of curve: y = ? -7x + c
c = ?

10)
a)
V = 400
400 = Pi r^2 h
h = 400/Pir^2
A = 2Pir^2 + 2Pirh
A = 2Pir^2 + 800/r
b)
dA/dr = 4Pir -800/r^2
d2A/dr2 = 4Pi +1600/r^3
c)
Min value of r = 3.992..
d2A/dr2 = 37.6... > 0 Therefore minimum.
A with this value = 300.53.. = 301

11)
a) Sketch graph y = 2^x. Goes through (0,1)
b) ?

For the integral question it was asking to integrate
dy/dx = 12x^3 - 7, given that the curve goes through (2,10)

so c should be =-24

This is what I can remember, random order :/

1)
a) Integral 5 to 1: 4xdx = [2x^2] = 50 – 2 = 48.
b) Integral: 6x^1/2 = 4x^3/2 + c

2)
a) (log0.2)-(log0.1)/(0.2-0.1) = 3.01
b) Plot c in between a and b

3)
Log_aw = 3 + log_ax^5 + log_a6 – log_a2x
Log_aw = log_aa^3 + log_ax^5 + log_a6 – log_a2x
Log_aw = log_a(6a^3x^5) – log_a2x
Log_aw = log_a(6a^3x^5/2x)
Log_aw = log_a(3a^3x^4)
W = 3a^3x^4

4) Y=2x^3. Normal at x = 2. y = 16. Dy/dx = 6x^2. Dy/dx = 24. Normal = -1/24. Equation: x + 24y = 386.

5)
a) Cosine rule. a^2 = 32^2 + 15^2 - (2 x 32 x 15 x cos116). AE = 40.86….
b)
Perp. Distance from AE to D.
Area ADE = 1/2 absinC = 1/2 x 32 x 15 x sin116 = 215.7.
Area ADE = b x h x 1/2.
215.7... = 40.86... x h x 1/2
h = 10.55… > 10. Therefore pond lies in triangle.
c)
Area of pond = 116/360 x Pi x 10^2 = 101.22…
Area of ADE = 215.7…
Area of meadow = 114.48…
d)
Angle C = 360 - 90 - 90 - 116 = 64.
Use that to work out total length of base of trapezium = 70 something.
Area of trapezium = 1/2 x (32 + 70 something) x 80 = 4120.74…
Area of car park = trapezium – ADE = 3905.033…
90% of trapezium = 3708.66…
3905>3708 so carpark takes up more than 90% of field.

6)
6cosx^2 = 5 - sinx
6(1-sinx^2) = 5 - sinx
6 - 6sinx^2 = 5 - sinx
6sinx^2 - sinx - 1 = 0
(3sinx + 1)(2sinx - 1) = 0
sinx = -1/3 or 1/2
x = 3.48, 5.94, 1/6PI, 5/6Pi

7)
a)
Asif = 30000 + (n-1)1000. Bettina = 25000 x 1.05^n-1.
10^th year – Asif = £39000. Bettina = £38783. Asif had more.
11^th year – Asif = £40000. Bettina = £40722. Bettina had more
b)
Total after 17 years.
Asif = 646000. Bettina = 646000 (to the nearest hundred). The same
c)
Total > £M.
25000(1.05^n – 1)/1.05-1 > M
25000(1.05^n – 1)/0.05 > M
25000(1.05^n – 1) > 0.05M
500000(1.05^n – 1) > M
(500000 x 1.05^n) – 500000 > M
500000 x 1.05^n > M + 500000
1.05^n > (M + 500000)/500000
log1.05^n > log (M + 500000) – log500000
nlog1.05 > log (M + 500000) – log500000
n > (log (M + 500000) – log500000)/log1.05
n = 25 or 26 :/

8)
a)
Stretch parallel to y-axis SF 2.
b)
Translation (3 0)

9)
Curve goes through (2,10)
Dy/dx = ? - 7.
Equation of curve: y = ? -7x + c
c = ?

10)
a)
V = 400
400 = Pi r^2 h
h = 400/Pir^2
A = 2Pir^2 + 2Pirh
A = 2Pir^2 + 800/r
b)
dA/dr = 4Pir -800/r^2
d2A/dr2 = 4Pi +1600/r^3
c)
Min value of r = 3.992..
d2A/dr2 = 37.6... > 0 Therefore minimum.
A with this value = 300.53.. = 301

11)
a) Sketch graph y = 2^x. Goes through (0,1)
b) ?

In 5)b) answer you've given did we have to compare the height to something? Did I not read the questions properly omg😱

So you needed to find the perpendicular distance from AE to D, which is the height of the triangle ADE, and AE being the base