# OCR MEI C2 june 2017

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#1
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?
1
3 years ago
#2
(Original post by Unefleur)
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!
1
#3
(Original post by charlottesacha99)
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!
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
0
3 years ago
#4
(Original post by Unefleur)
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
0
#5
(Original post by gemma.louise)
w = 1/12 a³ x^4
That's what I got
Do you remember the exact question for the log question?
0
3 years ago
#6
i think it was logw = 3 + logx^5 - log2x + log6
i got
3a^3x^4
(Original post by Unefleur)
Do you remember the exact question for the log question?
2
3 years ago
#7
(Original post by Unefleur)
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
0
3 years ago
#8
Guys what did you get for the last question for n???? I got 26 years???
2
3 years ago
#9
(Original post by Levisual360)
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
0
3 years ago
#10
(Original post by Aminah53)
Would it be okay if you rounded 3.99 to 4.0 but left the minimum point as a surd
rounding is fine as long as youve shown somewhere in your working you got the value of r to be 3.99.... etc,
if you just straight worked out r to be 4, then im not sure if you would get the mark.

As for the surd form, im not too sure, it depends on the examiner. I think as long as youve shown ur substitution of r back into the equation for A, and simplified that should be fine.
0
3 years ago
#11
How do you find the minimum radius r?? I took the second differential and made it equal to 0 and got r as 5 and that was the wrong answer 0
3 years ago
#12
What did everyone get for the car park field percentage?
0
3 years ago
#13
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)
Logaw = 3 + logax^5 + loga6 – loga2x
Logaw = logaa^3 + logax^5 + loga6 – loga2x
Logaw = loga(6a^3x^5) – loga2x
Logaw = loga(6a^3x^5/2x)
Logaw = loga(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…
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.
10th year – Asif = £39000. Bettina = £38783. Asif had more.
11th 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 = 12x^3 - 7
Equation of curve: y = 3x^4 -7x + c
c = -24
y = 3x^4 -7x - 24

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) ?

12)
a)
Sum of (3r + 2) from 1 to 5 = 5 + 8 + 11 + 14 + 17 = 55
b)
AP First term = 4.2. Sixth term = 1.8.
4.2 + 5d = 1.8
5d = -2.4
d = -0.48
2
3 years ago
#14
[QUOTE=gemma.louise;71942372]w = 1/12 a³ x^4
That's what I got[/QUOTE
Nicee I got that as well
1
3 years ago
#15
(Original post by Eldronyx)
What did everyone get for the car park field percentage?
I think I got 92.7%
0
3 years ago
#16
(Original post by wmirza)
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)
Logaw = 3 + logax^5 + loga6 – loga2x
Logaw = logaa^3 + logax^5 + loga6 – loga2x
Logaw = loga(6a^3x^5) – loga2x
Logaw = loga(6a^3x^5/2x)
Logaw = loga(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…
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.
10th year – Asif = £39000. Bettina = £38783. Asif had more.
11th 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
0
3 years ago
#17
Cheers (Original post by Levisual360)
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
0
#18
(Original post by wmirza)
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)
Logaw = 3 + logax^5 + loga6 – loga2x
Logaw = logaa^3 + logax^5 + loga6 – loga2x
Logaw = loga(6a^3x^5) – loga2x
Logaw = loga(6a^3x^5/2x)
Logaw = loga(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…
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.
10th year – Asif = £39000. Bettina = £38783. Asif had more.
11th 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😱
0
3 years ago
#19
(Original post by Unefleur)
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 is the base
0
#20
(Original post by wmirza)
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
Thanks goodness. So we didn't have to show that the pond lies in the triangle? Thanks btw
0
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