These are the questions you should try to work through over the mid-term break. Attempt as much of it as you can after revising the related chapters.
If you get stuck, check here over the course of next week, as I may post some tips for each section.
In the classes after the mid-term break, we will cover as much ground as possible and hopefully resolve any issues that you got stuck on.
I will add some help in blue text like this for some of the questions over the next few days.
I have now finished adding my notes - you can add a comment here if you need more help.
Paper 1
Question 1 - Arithmetic
Page 135 q4, 5 (In q4b, put all money amounts into either euro or cent.In q4c, remember the way we laid out these questions - you need to navigate backwards from the net tax, adding pack tax credits etc.)
Question 2,3 - Algebra
Page 22 q2 (In part b and c you should tidy up the equations with fractions by multiplying across by the number under the line. In the evaluation questions, write on the page e.g. 2(-2)² - 3(-2)(1/3) before you go to your calculator.
Page 40 q6 (part a: find LCM of the denominators. Part b: remember that (x-2)² = (x-2)(x-2). Part c: remember that if say x=2 is a root of the equation, it means that if you sub x=2 in to the equation it will balance. There will be only 1 unknown which you can solve.)
Page 156 q3, 7 (Part b: Use the rules on page 140 to work out how to write 1/27 as 3 to the power of something. A clue is that 3³=27. Also note that 3²=9. Write both sides of the equation as 3 to the power of something and then equate the powers.)
Question 4 - Complex Numbers
Page 78 q6 (For all these questions, you need to revise argand diagrams, complex conjugates, modulus and division of complex numbers. For part b see the top of page 76 for a similar question.)
Question 6,7,8 - Functions and Calculus
Page 281 q 6 (For part bi) the coordinates of a and b are both going to be (something, 0) as they are both on the x-axis, or y = 0. So if you solve the equation x² +2x -3 = 0 you will get the two values of x. c is on the Y-axis, so this means that x=0. Sub this value of x in to x² +2x -3 and you will get the output value y. Part iii is like a repeat of part i with =0 replaced by = -3. Remember that you can only solve quadratic equations with = 0 on the right hand side so you need to rearrange it before solving.)
Page 306 q2, 4 and 5(a) (For q2, you need to use the quotient rule for ii and iii. In iii you can write as a single fraction, but you can't simplify the top of the fraction. First principles is coming up on the mock exam, worth 20 marks, so practice a few of these. Make sure you always check your answer f(x) =x² -4x so f'(x) = 2x -4. For part c, you need to get the 1st derivative and set up an equation with this = 2. Note you are asked for the point on the curve, so this means you have to give an (x,y) pair and you can find y by subbing x into the original function. Turning point occurs where slope = 0, check your notes on the shape of a function (wine-glass upsidedown or right way up) to work out if it is a max or a min.)
Paper 2
Question 1 - Perimeter area and volume
Page 175 q14 (Remember the relationship between the radius, the height and the slant height of a cone - a right angle triangle. For the last part of the question, you need to set up an equation: vol of cylinder = 5 × vol of cone. Leave both volumes in terms of π.)
Page 184 q5, page 185 q7 (Straightforward Simpson's Rule questions, the second one has an unknown as one of the heights.)
Page 189 q5 (Part b works out exactly if you use π = 22/7)
Question 2 and 3 - Coordinate Geometry of Line and Circle
Page 58 q5, 13 (For q5: remember that the corners of a parallelogram are always given either clockwise or anti-clockwise in order. Try drawing a rough sketch and solve using translations. For q 13 you check if a point is on a line by subbing. The word "verify" means that the point is on the line. To find the equation of the image of a line you have to find the two points on the line, find the slope (*) and plug values into y -y1 = m(x - x1).
(*) Note that the slope won't change under translation so that step isn't necessary.)
Page 61 q1 (Remember how to calculate the slopes of perpendicular lines - invert and change sign. For part c, find the distance between a and b. Then set up an equation with the distance formula subbed in for a(1,6) and c(2,y) on the left hand side and the distance between a and b on the right hand side. To solve this equation, your first step will be to square both sides before re-arranging it as ...x² ...x....a number =0 and solving.)
Page 208 q3 (Very straightforward circles question if you revise the chapter. You prove if a point is on a circle exactly the same way that you prove a point is on a line - by subbing in (x,y).) To prove the last part, you need to find the equation of the line that the 2 points (centres of the circles) are on .... slope, y-y1= m(x-x1) etc.)
Question 6 - Probability
Page 239 q2 (Revise the notation. Large brackets with e.g. 5 on top and 2 below means "5 choose 2" - in other words how many ways can you choose 2 people from a group of 5. The answer is (5×4)/(2×1) and you can work this out using the nCr button on your calculator.)
Page 256 q6 (Part b could be solved with a sample space B1 B2 G1 G2. If you write the sample space correctly (should have 24 rows) then you can just count the number of cases where the 2 girls are in the middle and express P(2 girls in middle)=this number/24. For part c, this is like the one we did in class. If the number has to be bigger than 9000 then there is only 1 option for the first position, 4 left for the second position etc. If the number must be even then there are 3 options for the last position. Remember to always start with the most restricted positions.)
Question 7 - Statistics
Page 105 q2 (Note the means are different, but the two sets of data are equally "spread out". Make a table with the values in the first column, the difference between it and the mean in the next column, this value squared next and so on. Follow the steps on the previous page.)
Page 110 q7 (For the median, you have to imagine the data written out long (i.e. not in a frequency distribution. It would be 2 2 2 3 3 3 3 5 6 6 and the median value would be the middle value. When drawing the ogive in part b, be sure to put the frequency (i.e. number of students) on the vertical axis. If there are 40 students, then the last entry in the cumulative frequency table should be 40. The median student will be the 20th student, lower quartile will be the 10th student and upper quartile will be the 30th student. Find these students on the vertical axis and then draw lines across to your curve and down to read the corresponding marks on the horizontal axis. Remember the interquartile range is a number (not an upper and lower number as in the range of a periodic function.))
Question 11 - Linear Programming
Page 396 q3, 6 (For both of these, use points to find slope, then use slope and one point to find equation of the line, then use a test point to decide the direction of the inequality).
Page 414 q6 (For part b, your first inequality has to do with "number of people the chalet types can accomodate. Don't forget that x is the number of type A that the builder is going to build - so if he builds 10 of them then they can accommodate 6x or 6%times;10 = 60 people. When graphing don't forget to shade in the appropriate region, taking the two implicit inequalities x ≥ 0 and y ≥0 into account. For part ii, you need to work out the intersection point of the two diagonal lines on your graph using simultaneous equations and then maximise the expression you create using the additional information you're given about rent.)
Friday, February 13, 2009
Thursday, February 12, 2009
Statistics question
P 111 q 8.
(a) This table is a frequency distribution and the second row in tables like this is always the frequency.
You can either be asked to find the mean in which case you multiply each number in the top row by its frequency and add these together and then divide by the sum of the frequencies.
In this case you are given the mean, so, kind of like the simpsons rule example we did today, you set up an equation and solve it.
Left hand side of equation:
You will have 0(1) + 1(x) + etc on top and 1 + x + 1 etc on the bottom.
Right hand side of equation:
This will be the mean which you have been given.
Then you will have an equation with a fraction in it, so you multiply across by whatever is under the line and solve.
(b) See page 103.
The standard deviation is a measure of how spread out data is.
You could have 2 sets of numbers with the same mean eg
1 2 3 4 5 6 7 8 9 10
and
5 5 5 5 5 6 6 6 6 6
both have mean = 5.5, but the standard deviation of the first set would be bigger because the numbers are more spread out.
To find standard deviation you need to first find the mean value and then make a little table:
First column is all the numbers in your list.
Second column is the difference between the number in the 1st column and the mean.
Third column is the number in the 2nd column squared.
Add the values of the third column.
Then divide this number by the number of values.
Then get the square root of the result.
When you have done this take a look at how to find the standard deviation of a frequency distribution.
(c) This is explained on page 96.
When you are making the cumulative frequency table, note that the last number should be 240 (as you are told there are 240 villagers).
When drawing the ogive, make sure that you put the frequency (which is no. of villagers) on the vertical axis. On the horizontal axis you put the values you have in the top row, without the ≤ or < signs.
You graph cumulative frequencies on an ogive.
(i) For the median, you need to find the "middle" villager. Start on the vertical axis.
(ii) The interquartile range is found by finding the difference between the ages of the lower quartile villager and the upper quartile villager.
For both these questions you start on the vertical axis, draw a horizontal line until it reaches the curve, then drop a vertical line to read the age from the horizontal axis.
(iii) For this question, you need to start on the age axis at 55. Then go up to the curve and then across to the vertical axis. Then re-read what the question is asking you.
(a) This table is a frequency distribution and the second row in tables like this is always the frequency.
You can either be asked to find the mean in which case you multiply each number in the top row by its frequency and add these together and then divide by the sum of the frequencies.
In this case you are given the mean, so, kind of like the simpsons rule example we did today, you set up an equation and solve it.
Left hand side of equation:
You will have 0(1) + 1(x) + etc on top and 1 + x + 1 etc on the bottom.
Right hand side of equation:
This will be the mean which you have been given.
Then you will have an equation with a fraction in it, so you multiply across by whatever is under the line and solve.
(b) See page 103.
The standard deviation is a measure of how spread out data is.
You could have 2 sets of numbers with the same mean eg
1 2 3 4 5 6 7 8 9 10
and
5 5 5 5 5 6 6 6 6 6
both have mean = 5.5, but the standard deviation of the first set would be bigger because the numbers are more spread out.
To find standard deviation you need to first find the mean value and then make a little table:
First column is all the numbers in your list.
Second column is the difference between the number in the 1st column and the mean.
Third column is the number in the 2nd column squared.
Add the values of the third column.
Then divide this number by the number of values.
Then get the square root of the result.
When you have done this take a look at how to find the standard deviation of a frequency distribution.
(c) This is explained on page 96.
When you are making the cumulative frequency table, note that the last number should be 240 (as you are told there are 240 villagers).
When drawing the ogive, make sure that you put the frequency (which is no. of villagers) on the vertical axis. On the horizontal axis you put the values you have in the top row, without the ≤ or < signs.
You graph cumulative frequencies on an ogive.
(i) For the median, you need to find the "middle" villager. Start on the vertical axis.
(ii) The interquartile range is found by finding the difference between the ages of the lower quartile villager and the upper quartile villager.
For both these questions you start on the vertical axis, draw a horizontal line until it reaches the curve, then drop a vertical line to read the age from the horizontal axis.
(iii) For this question, you need to start on the age axis at 55. Then go up to the curve and then across to the vertical axis. Then re-read what the question is asking you.
Tuesday, February 10, 2009
Probability question
This is some help for q1 on page 254.
(a) Use the same approach as we used today for "roll a die - what is the probability of getting a 4?" type of question. There are 6 sides, one of them is a 4 so the probability P(4) = 1/6.
If you were asked for the probability of getting a 4 or a 3, then it would be P(4or3)= 2/6 = 1/3
(b) Arrangements. There are 6 letters, so there are 6 options for the first position, 5 for the second and so on.
If the first position must contain B, then there is 1 option for that position, 5 for the second and so on.
For the third part of this question, there are further restrictions. Start with the most restricted positions and then work out how many options remain for the other positions.
(c) Similar to the arrangements questions, but with a bit of extra logic required.
For part (iii) remember that there are only 4 digit positions to be filled, so there are 6 options for the first position, 5 for the next, 4 for the next and 3 for the last.
For part (iv) if the number is to be >4000 then the first digit must be either 4 or 5 - in other words there are only 2 options.
We won't get much more of a chance to revise probability so look over the whole chapter 11 now.
(a) Use the same approach as we used today for "roll a die - what is the probability of getting a 4?" type of question. There are 6 sides, one of them is a 4 so the probability P(4) = 1/6.
If you were asked for the probability of getting a 4 or a 3, then it would be P(4or3)= 2/6 = 1/3
(b) Arrangements. There are 6 letters, so there are 6 options for the first position, 5 for the second and so on.
If the first position must contain B, then there is 1 option for that position, 5 for the second and so on.
For the third part of this question, there are further restrictions. Start with the most restricted positions and then work out how many options remain for the other positions.
(c) Similar to the arrangements questions, but with a bit of extra logic required.
For part (iii) remember that there are only 4 digit positions to be filled, so there are 6 options for the first position, 5 for the next, 4 for the next and 3 for the last.
For part (iv) if the number is to be >4000 then the first digit must be either 4 or 5 - in other words there are only 2 options.
We won't get much more of a chance to revise probability so look over the whole chapter 11 now.
Monday, February 9, 2009
Complex Numbers Question
To do question 4 on page 78 you need
(a) to be able to draw an Argand Diagram (see page 72) and to plot the complex numbers
z = 1-4i and iz = i(1-4i) on it.
Before you plot the second number you have to multiply out i(1-4i). This works the same way as multiplying say, x(1+4x) would in algebra, but there is a second step you need to do - resolving the i²=-1
(b) To do this you need to remember division of complex numbers (page 68).
10-2i
------
2-3i
You multiply above and below by the complex conjugate of the bottom.
[If z = 1 + 7i, then z¯, the complex conjugate of z = 1-7i]
So you multiply above and below by (2+3i)
(10-2i)(2+3i)
-------------
(2-3i)(2+3i)
If you don't get a real number below (after you have adjusted i²=-1) then you've made a mistake.
Whatever you get for your result, you have to factor out (1+i) in order to find the missing value k. If you don't get something like 2+2i which could be written as 2(1+i) or 3+3i which could be written as 3(1+i) then you have made a mistake.
Note also that they say k∈N which is a further clue - telling you that it is positive and not a fraction.
(c)
(i) for this you need to revise equality of complex numbers (p70).
If you are told 2a + 2i = 6 +bi then you can equate the real parts of the equation
2a = 6 so a = 3
and the imaginary parts
2 = b (Note - leave the i's out of the imaginary equation)
to solve.
Sub in w = 3-4i and solve.
(ii) ¦w¦ is the modulus of w (page 73).
Sub in all the values you are given, and solve the real and imaginary equations using the same rules as above (equate the reals, equate the imaginary parts).
In this case, the solution is much more complicated. You end up with s's and t's in both equations, so you have to solve using simultaneous equations.
The answers turn out to be fractions which look a bit unlikely - but again there is a clue in the question where you are told s,t ∈ R.
(a) to be able to draw an Argand Diagram (see page 72) and to plot the complex numbers
z = 1-4i and iz = i(1-4i) on it.
Before you plot the second number you have to multiply out i(1-4i). This works the same way as multiplying say, x(1+4x) would in algebra, but there is a second step you need to do - resolving the i²=-1
(b) To do this you need to remember division of complex numbers (page 68).
10-2i
------
2-3i
You multiply above and below by the complex conjugate of the bottom.
[If z = 1 + 7i, then z¯, the complex conjugate of z = 1-7i]
So you multiply above and below by (2+3i)
(10-2i)(2+3i)
-------------
(2-3i)(2+3i)
If you don't get a real number below (after you have adjusted i²=-1) then you've made a mistake.
Whatever you get for your result, you have to factor out (1+i) in order to find the missing value k. If you don't get something like 2+2i which could be written as 2(1+i) or 3+3i which could be written as 3(1+i) then you have made a mistake.
Note also that they say k∈N which is a further clue - telling you that it is positive and not a fraction.
(c)
(i) for this you need to revise equality of complex numbers (p70).
If you are told 2a + 2i = 6 +bi then you can equate the real parts of the equation
2a = 6 so a = 3
and the imaginary parts
2 = b (Note - leave the i's out of the imaginary equation)
to solve.
Sub in w = 3-4i and solve.
(ii) ¦w¦ is the modulus of w (page 73).
Sub in all the values you are given, and solve the real and imaginary equations using the same rules as above (equate the reals, equate the imaginary parts).
In this case, the solution is much more complicated. You end up with s's and t's in both equations, so you have to solve using simultaneous equations.
The answers turn out to be fractions which look a bit unlikely - but again there is a clue in the question where you are told s,t ∈ R.
Sunday, February 8, 2009
Lead up to Mock Exams
As preparation for the mock exams, we will revise the following topics in this order.
Monday 9:
Coordinate geometry of the line and circle
Tuesday 10:
Complex Numbers
Wednesday 11:
Probability
Thursday 12:
Perimeter area and volume
Friday 13:
Statistics
Over the mid-term break, I will set some work to do covering algebra and revising functions.
Monday 9:
Coordinate geometry of the line and circle
Tuesday 10:
Complex Numbers
Wednesday 11:
Probability
Thursday 12:
Perimeter area and volume
Friday 13:
Statistics
Over the mid-term break, I will set some work to do covering algebra and revising functions.
Monday, May 26, 2008
Paper 2 Question 11 - Linear Programming
Here are the notes on Linear Programming:
Introduction - here and here
Looking at part a) half-planes
Worked example of a part b).
NB: when answering part b) (i) which is usually "write down two inequalities in x and y and illustrate these on graph paper" you must
a) use graph paper and
b) shade in the region which represents all the inequalities (including x≥0 and y≥0) or at least indicate the half planes with arrows. If you just draw the lines you could lose 3 marks per inequality ... or 1%!
Introduction - here and here
Looking at part a) half-planes
Worked example of a part b).
NB: when answering part b) (i) which is usually "write down two inequalities in x and y and illustrate these on graph paper" you must
a) use graph paper and
b) shade in the region which represents all the inequalities (including x≥0 and y≥0) or at least indicate the half planes with arrows. If you just draw the lines you could lose 3 marks per inequality ... or 1%!
Paper 2 Question 7 - Statistics
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