Challenging +1.2 This is a solid of revolution problem requiring integration about the y-axis with a composite region. Students must express x in terms of y for both curves (x = √(y/2) and x = 3-y), set up the volume integral V = π∫[0 to 2](outer² - inner²)dy, and recognize the cone subtraction from y=2 to y=3. While it involves multiple steps and careful setup, the techniques are standard C4 material with straightforward integration, making it moderately above average difficulty.
Fig. 6 shows the region enclosed by part of the curve \(y = 2x^2\), the straight line \(x + y = 3\), and the \(y\)-axis. The curve and the straight line meet at P (1, 2).
\includegraphics{figure_6}
The shaded region is rotated through \(360°\) about the \(y\)-axis. Find, in terms of \(\pi\), the volume of the solid of revolution formed. [7]
[You may use the formula \(V = \frac{1}{3}\pi r^2 h\) for the volume of a cone.]
Answer: Vol = vol of rev of curve + vol of rev of line
\(\text{vol of rev of curve} = \int_0^{\pi} \pi x^2 \, dy\)
\(= \int_0^{\pi} \pi \frac{y^2}{2} \, dy\)
\(= \left[\pi\frac{y^2}{4}\right]_0^{\pi}\)
\(= \pi\)
height of cone = 3 - 2 = 1
so vol of cone = 1/3 π 1² x1 = π/3
so total vol = 4π/3
Answer
Marks
M1
(soi) at any stage
M1
substituting \(x^2 = y/2\)
B1
\(\left[\frac{x^2}{4}\right]\) even if π missing or limits incorrect or missing
A1
cao
B1
h=l soi
B1
www cao
A1
www cao
[7]
for M1 need π, substitution for x², (dy soi), intention to integrate and correct limits OR \(\pi\int_3^{(3-y)^2}dy\) M1(even if expanded incorrectly) \(= \pi/3\) A1 www
0.293 or cos ABC=correct numerical expression as RHS above, or better
A1
or rounds to 73.0° (accept 73° www)
[6]
condone rowssubstituted, ft their vectors AB, AC for method only need to see method for modulae as far as √…. use of vectors BA and CA could obtain B0 B0 M1 M1 A1 A1
Total: [6]
Question 7(ii):
Answer: A: \(x + y - 2z + d = 2 - 6 + d = 0\)
\(\Rightarrow d = 4\)
B: \(-2 + 0 - 2x1 + 4 = 0\) ✓
C: \(0 + 4 - 2x4 + 4 = 0\) ✓
Normal \(\vec{n} = \begin{pmatrix}1\\1\\-2\end{pmatrix}\)
substituting one point evaluating for other two points
DM1
A1
\(d = 4\) www
B1
stated or used as normal anywhere in part (ii)
M1
finding angle between normal vector and k allow ±2/√6 or 144.7° for A1
A1
A1
or rounds to 35.3° [7]
alternatively, finding the equation of the plane using any valid method (eg from vector equation, M1 A1 for using valid equation and eliminating both parameters, A1 for required form, or using vector cross product to get x+y-2z = oe M1 A1,finding c and required form, A1, or showing that two vectors in the plane are perpendicular to normal vector M1 A1 and finding d, A1) oedo not need to find 144.7° explicitly (or 0.615 radians)
Answer: The smallest possible PIN that does not begin with zero is 1000 and the largest is 9999, giving 9000.
However the 9 numbers 1111, 2222, … 9999 are disallowed.
The other disallowed numbers are 1234, 2345, … 6789 (6 numbers)
And 8876, 8765, … 3210 (7 numbers).
So, in all, there are 9000 – (9 + 6 + 7) = 8978 possible PINs
Answer
Marks
M1
from a correct starting point (eg 10,000 or 9000), clear attempt to eliminate (or not include) numbers starting with 0 clear attempt to eliminate all three of these categories (with approx correct values in each category)
M1
A1
SC 8978 www B3
All(1) for M1 (no 0 start). nos starting with 1,2,7,8,9 give 1000-2, nos starting with 3,4,5,6 give 1000-3 =5(900-2, nos starting with 1,2,7,8,9 give 1000-2, nos starting with 3,4,5,6 give 1000-3 =5(900-2+4(1000-3)=8978 M1,A1 or2) eg starting with1, 1,not2,any,any+1,2,not3,any +1,2,3,not4 =900+90+9=999-(1111term)=998 can lead to 5(900+90+9-1)=4990 oe
[3]
if unclear, M0 M marks not dependent SC 8978 www B3
Part (ii):
Answer: \(\frac{6,700,000,000}{8978} = 746,269\)
The average is about 750,000.
Answer
Marks
M1
ft from (i)
A1
ft
[2]
accept 2sf (or 1sf) only for A1
Total: [5]
Question 3 (4754B):
Answer: [Venn diagram showing]:
- Left circle (no breaches): 8
- Intersection: 1
- Right circle (breaches): 2
- Total: 11, All correct
Answer
Marks
M1
numbers total 11
A1
all correct
[2]
Total: [2]
Question 4 (4754B):
Answer: 100,000 transactions from 80 people over 3½ years with 365 days per year
allow approximate number of days in a year eg 360 for M1 A1
Total: [2]
Question 5 (4754B):
Answer: Allow any one of the following for 1 mark
- An attack can happen without a breach of the card's security.
- The probabilities that a successful attack followed or did not follow a breach of card security are so close that a court would look for other evidence before reaching a decision.
- In many cases of unauthorised withdrawals the banks refund the money.
- The banks' software does not detect all the attacks that occur.
Answer
Marks
B1
only accept versions of these statements
[1]
Total: [1]
Question 6 (4754B):
Part (i):
Answer
Marks
Guidance
Transactions
Authorised
Unauthorised
Queried
480
20
Not queried
499,460
40
Total
499,940
60
B1
for top row 480, 20, 500
B2
all five other entries correct
[3]
(500,000 is given) allow B1 for three or four correct from 499460,40,499500,499940,60
Total: [3]
Question 6 (4754B):
Part (ii):
Answer: \(\frac{480}{40} = 12\) or 12 to 1
Answer
Marks
B1
ft from (i)
[1]
their 480: their 40 isw accept unsimplified answers
Total: [1]
Question 6 (4754B):
Part (iii):
Answer
Marks
Guidance
Transactions
Authorised
Unauthorised
Queried
2,445
55
Not queried
497,495
5
Total
499,940
60
\(\frac{2445}{5} = 489\) or 489 to 1
Answer
Marks
M1
ft from (i)
DM1
their 2445 ft from (i) :5
A1
cao
[3]
NB they are not required to complete the table. {2500or 5xtheir 500}-(their 60-5) [=their 2445]
Total: [3]
**Answer:** Vol = vol of rev of curve + vol of rev of line
$\text{vol of rev of curve} = \int_0^{\pi} \pi x^2 \, dy$
$= \int_0^{\pi} \pi \frac{y^2}{2} \, dy$
$= \left[\pi\frac{y^2}{4}\right]_0^{\pi}$
$= \pi$
height of cone = 3 - 2 = 1
so vol of cone = 1/3 π 1² x1 = π/3
so total vol = 4π/3
| M1 | (soi) at any stage |
| M1 | substituting $x^2 = y/2$ |
| B1 | $\left[\frac{x^2}{4}\right]$ even if π missing or limits incorrect or missing |
| A1 | cao |
| B1 | h=l soi |
| B1 | www cao |
| A1 | www cao |
| [7] | for M1 need π, substitution for x², (dy soi), intention to integrate and correct limits **OR** $\pi\int_3^{(3-y)^2}dy$ M1(even if expanded incorrectly) $= \pi/3$ A1 www |
**Total: [7]**
---
## Question 7(i):
**Answer:** $\vec{AB} = \begin{pmatrix}-4\\0\\-2\end{pmatrix}, \vec{AC} = \begin{pmatrix}-2\\4\\1\end{pmatrix}$
$\cos BAC = \frac{\begin{pmatrix}-4\\0\\-2\end{pmatrix} \cdot \begin{pmatrix}-2\\4\\-1\end{pmatrix}}{|\vec{AB}| \cdot |\vec{AC}|} = \frac{(-4)(-2) + 0 \cdot 4 + (-2) \cdot 1}{\sqrt{20\sqrt{21}}}$
$= 0.293$
$\Rightarrow BAC = 73.0°$
| B1B1 | |
| M1 M1 | dot product evaluated **cos BAC= dot product / \|AB\|, \|AC\|** |
| A1 | 0.293 or cos ABC=correct numerical expression as RHS above, or better |
| A1 | or rounds to 73.0° (accept 73° www) |
| [6] | |
| | **condone rows** **substituted, ft their vectors AB, AC for method only need to see method for modulae as far as √…. use of vectors BA and CA could obtain B0 B0 M1 M1 A1 A1** |
**Total: [6]**
---
## Question 7(ii):
**Answer:** A: $x + y - 2z + d = 2 - 6 + d = 0$
$\Rightarrow d = 4$
B: $-2 + 0 - 2x1 + 4 = 0$ ✓
C: $0 + 4 - 2x4 + 4 = 0$ ✓
Normal $\vec{n} = \begin{pmatrix}1\\1\\-2\end{pmatrix}$
$\vec{n} \begin{pmatrix}0\\0\\1\end{pmatrix} = \frac{-2}{\sqrt{6}} = \cos\theta$
$\Rightarrow \theta = 144.7°$
$\Rightarrow$ acute angle = 35.3°
| M1 | substituting one point evaluating for other two points |
| DM1 | |
| A1 | $d = 4$ www |
| B1 | stated or used as normal anywhere in part (ii) |
| M1 | finding angle between normal vector and **k** allow ±2/√6 or 144.7° for A1 |
| A1 | |
| A1 | or rounds to 35.3° [7] | |
| | **alternatively, finding the equation of the plane using any valid method (eg from vector equation, M1 A1 for using valid equation and eliminating both parameters, A1 for required form, or using vector cross product to get x+y-2z = oe M1 A1,finding c and required form, A1, or showing that two vectors in the plane are perpendicular to normal vector M1 A1 and finding d, A1) oe** **do not need to find 144.7° explicitly (or 0.615 radians)** |
**Total: [7]**
---
## Question 7(iii):
**Answer:** At D, $-2 + 0 - 2k + 4 = 0$
$\Rightarrow 2k = 6, k = 3*$
$\vec{CD} = \begin{pmatrix}-2\\0\\-1\end{pmatrix} = \frac{1}{2}\vec{AB}$
$\Rightarrow$ CD is parallel to AB
$CD : AB = 1 : 2$
allow $CD:AB=1/2, \sqrt{5}:\sqrt{20}$ oe, AB is twice CD oe
| M1 | substituting into plane equation |
| A1 | **AG** |
| M1 | finding vector CD (or vector DC) |
| A1 | or DC parallel to AB or BA oe (or hence two parallel sides, if clear) but A0 if their vector CD is vector DC for B1 allow vector CD used as vector DC |
| B1 | mark final answer **www** allow CD:AB=1/2, √5:√20 oe, AB is twice CD oe |
| [5] | |
**Total: [5]**
---
## Question 8(i):
**Answer:** $\frac{dV}{dt} = -kx$
$V = \frac{1}{3}x^3 \Rightarrow \frac{dV}{dx} = x^2$
$\frac{dV}{dt} = \frac{dV}{dx} \cdot \frac{dx}{dt} = x^2 \cdot \frac{dx}{dt}$
$\Rightarrow x^2 \frac{dx}{dt} = -kx$
$\Rightarrow x\frac{dx}{dt} = -k$ *
| B1 | |
| M1 | oe eg dv/dt=dx /dV . dV/dt; dV/dt = 1/x² . -kx = -k/x |
| A1 | **AG** [3] |
**Total: [3]**
---
## Question 8(ii):
**Answer:** $x\frac{dx}{dt} = -k \Rightarrow \int x \, dx = \int -k \, dt$
$\Rightarrow \frac{1}{2}x^2 = -kt + c$
When $t = 0, x = 10 \Rightarrow 50 = c$
$\Rightarrow \frac{1}{2}x^2 = 50 - kt$
$\Rightarrow x = \sqrt{(100 - 2kt)}$ *
| M1 | separating variables and intention to integrate |
| A1 | condone absence of c |
| B1 | finding c correctly ft their integral of form ax²= bt+c where a,b non zero constants |
| A1 | **AG** [4] |
**Total: [4]**
---
## Question 8(iii):
**Answer:** When $t = 50, x = 0$
$\Rightarrow 0 = 100 - 100 k \Rightarrow k = 1$
| M1 | |
| A1 | [2] |
**Total: [2]**
---
## Question 8(iv):
**Answer:** $\frac{dV}{dt} = 1 - kx = 1 - x$
$\Rightarrow x^2 \frac{dx}{dt} = 1 - x$
$\Rightarrow \frac{dx}{dt} = \frac{1-x}{x^2}$ *
| M1 | for dV/dt= 1-kx or better |
| A1 | **AG** [2] |
**Total: [2]**
---
## Question 8(v):
**Answer:** $\frac{1}{1-x} - x - 1 = \frac{1-(1-x)x-(1-x)}{1-x}$
$= \frac{1-x+x^2-1+x}{x^2} = \frac{x^2}{1-x}$ *
$\int \frac{x^2}{1-x} \, dx = \int dt \Rightarrow \int\left(\frac{1}{1-x} - x - 1\right) dx = t + c$
$\Rightarrow -\ln(1-x) - \frac{1}{2}x^2 - x = t + c$
When $t = 0, x = 0 \Rightarrow c = -\ln 1 - 0 - 0 = 0$
$\Rightarrow t = \ln\left(\frac{1}{1-x}\right) + \frac{1}{2}x^2 - x$ *
| M1 | combining to single fraction |
| A1 | **AG** |
| M1 | separating variables & subst for x²/(1-x) and intending separating variables & subst for x²/(1-x) and intending |
| A1 | condone absence of c finding c for equation of correct form |
| B1 | accept ln(1/(1-x)) as -ln(1-x) www ie aln(1-x)+bx²+dx=et + c,a,b,d,e non zero constants |
| A1 | [6] cao **AG** | do not allow if c=0 without evaluation |
**Total: [6]**
---
## Question 8(vi):
**Answer:** understanding that ln(1/0) or 1/0 is undefined oe
| B1 | www |
| [1] | **In ln (1/0) = ln 0, 1/0 = ∞ and ln (1/0) = ∞ are all B0** |
**Total: [1]**
---
## Question 1 (4754B):
**Answer:** $\frac{16}{250} = 6.4\%$ or $\frac{16}{250} \times 100 = 6.4$ *
| B1 | or 250-(64+170) =6.4% oe |
| [1] | need evaluation |
**Total: [1]**
---
## Question 2 (4754B):
**Part (i):**
**Answer:** The smallest possible PIN that does not begin with zero is 1000 and the largest is 9999, giving 9000.
However the 9 numbers 1111, 2222, … 9999 are disallowed.
The other disallowed numbers are 1234, 2345, … 6789 (6 numbers)
And 8876, 8765, … 3210 (7 numbers).
So, in all, there are 9000 – (9 + 6 + 7) = 8978 possible PINs
| M1 | from a correct starting point (eg 10,000 or 9000), clear attempt to eliminate (or not include) numbers starting with 0 clear attempt to eliminate **all** three of these categories (with approx correct values in each category) |
| M1 | |
| A1 | SC 8978 www B3 |
| | **All(1) for M1 (no 0 start). nos starting with 1,2,7,8,9 give 1000-2, nos starting with 3,4,5,6 give 1000-3 =5(900-2, nos starting with 1,2,7,8,9 give 1000-2, nos starting with 3,4,5,6 give 1000-3 =5(900-2+4(1000-3)=8978 M1,A1 or2) eg starting with1, 1,not2,any,any+1,2,not3,any +1,2,3,not4 =900+90+9=999-(1111term)=998 can lead to 5(900+90+9-1)=4990 oe** |
| [3] | **if unclear, M0** M marks not dependent SC 8978 www B3 |
**Part (ii):**
**Answer:** $\frac{6,700,000,000}{8978} = 746,269$
The average is about 750,000.
| M1 | ft from (i) |
| A1 | ft |
| [2] | accept 2sf (or 1sf) only for A1 |
**Total: [5]**
---
## Question 3 (4754B):
**Answer:** [Venn diagram showing]:
- Left circle (no breaches): 8
- Intersection: 1
- Right circle (breaches): 2
- Total: 11, All correct
| M1 | numbers total 11 |
| A1 | all correct |
| [2] | |
**Total: [2]**
---
## Question 4 (4754B):
**Answer:** 100,000 transactions from 80 people over 3½ years with 365 days per year
$\frac{100,000}{(80 \times 3.5 \times 365)} (= 0.978...)$
Approximately 1 transaction per person per day
| M1 | |
| A1 | cao |
| [2] | allow approximate number of days in a year eg 360 for M1 A1 |
**Total: [2]**
---
## Question 5 (4754B):
**Answer:** Allow any one of the following for 1 mark
- An attack can happen without a breach of the card's security.
- The probabilities that a successful attack followed or did not follow a breach of card security are so close that a court would look for other evidence before reaching a decision.
- In many cases of unauthorised withdrawals the banks refund the money.
- The banks' software does not detect all the attacks that occur.
| B1 | only accept versions of these statements |
| [1] | |
**Total: [1]**
---
## Question 6 (4754B):
**Part (i):**
| Transactions | Authorised | Unauthorised | Total |
|---|---|---|---|
| Queried | 480 | 20 | 500 |
| Not queried | 499,460 | 40 | 499,500 |
| Total | 499,940 | 60 | 500,000 |
| B1 | for top row 480, 20, 500 |
| B2 | all five other entries correct |
| [3] | (500,000 is given) allow B1 for three or four correct from 499460,40,499500,499940,60 |
**Total: [3]**
---
## Question 6 (4754B):
**Part (ii):**
**Answer:** $\frac{480}{40} = 12$ or 12 to 1
| B1 | ft from (i) |
| [1] | their 480: their 40 isw accept unsimplified answers |
**Total: [1]**
---
## Question 6 (4754B):
**Part (iii):**
| Transactions | Authorised | Unauthorised | Total |
|---|---|---|---|
| Queried | 2,445 | 55 | 2,500 |
| Not queried | 497,495 | 5 | 497,500 |
| Total | 499,940 | 60 | 500,000 |
$\frac{2445}{5} = 489$ or 489 to 1
| M1 | ft from (i) |
| DM1 | their 2445 ft from (i) :5 |
| A1 | cao |
| [3] | **NB they are not required to complete the table.** {2500or 5xtheir 500}-(their 60-5) [=their 2445] |
**Total: [3]**
---
Fig. 6 shows the region enclosed by part of the curve $y = 2x^2$, the straight line $x + y = 3$, and the $y$-axis. The curve and the straight line meet at P (1, 2).
\includegraphics{figure_6}
The shaded region is rotated through $360°$ about the $y$-axis. Find, in terms of $\pi$, the volume of the solid of revolution formed. [7]
[You may use the formula $V = \frac{1}{3}\pi r^2 h$ for the volume of a cone.]
\hfill \mbox{\textit{OCR MEI C4 2011 Q6 [7]}}