I added these questions in to give you a refresher course in using this function on your calculator.
There are differences between calculators so make sure that you know how to input numbers in scientific notation and how to convert to/from scientific notation on yours!
It is really important that you always practice with the calculator you will have with you during your exams.
Monday, February 2, 2009
Sets
Read over the chapter on sets in Book 1. It gives a good overview and covers all of the notation well. There is a little bit of higher level work on maximising and minimising sets which is very simple, but we'll do it after the mocks.
For now, make sure that you know all the notation and that you know how to deal with unknowns using a Venn diagram.
Don't forget that you fill a Venn diagram from the centre outwards.
For question 4:
Draw two circles marked A and B with a decent overlap.
The overlap area represents A ∩ B and the number of elements in the area is 6.
That means that the shape representing A not B or A\B is going to contain the number 13-6=7 and the 3rd region represented by B\A will contain the number 14-6= 8.
So reading left to right (assuming you've laid your work out that way with A on the left and B of the right) you will have the numbers 7 then 6 then 8.
That means that the total number of elements in A∪B is 7+6+8=21.
The total number of elements in A∪B is written as #(A∪B).
General point:
If you have numbers with dots beside them in a region of a Venn diagram then those numbers are the elements of a set.
If you have a single number and no dot beside it, then that number is the number of elements in that region.
For now, make sure that you know all the notation and that you know how to deal with unknowns using a Venn diagram.
Don't forget that you fill a Venn diagram from the centre outwards.
For question 4:
Draw two circles marked A and B with a decent overlap.
The overlap area represents A ∩ B and the number of elements in the area is 6.
That means that the shape representing A not B or A\B is going to contain the number 13-6=7 and the 3rd region represented by B\A will contain the number 14-6= 8.
So reading left to right (assuming you've laid your work out that way with A on the left and B of the right) you will have the numbers 7 then 6 then 8.
That means that the total number of elements in A∪B is 7+6+8=21.
The total number of elements in A∪B is written as #(A∪B).
General point:
If you have numbers with dots beside them in a region of a Venn diagram then those numbers are the elements of a set.
If you have a single number and no dot beside it, then that number is the number of elements in that region.
Perimeter Area and Volume
The 1st question in the chapter test is a perfect example of an exam type question, so you should take about 25 minutes to answer it.
For part a)
Use the area of a circle formula (area = πr²) twice for the larger circle and for the smaller one, making sure to do as you're told with π = 22/7.
Do not fall into the trap of thinking you can subtract the radii and find the area of a circle with radius = 7cm. If in doubt, sketch out the circles.
For part b)
You need to use the formulae for volume of a sphere and volume of a cylinder.
Part iii of this question is a little confusing. They are looking for the volume of air that is left in the cylinder when it contains the 3 balls.
Fort part c)
This is a tricky question. You are told the radius of the hemisphere so first step is to work out its volume. Note that the question doesn't mention whether π should be 3.14 or 22/7. That is because you can actually work out the answer conveniently without setting π equal to anything, just leave your work in terms of π.
When you find the volume of the hemisphere, you know that the volume of a cone is half of this.
So work out half the volume of the hemisphere (in terms of π) and then set up an equation with the formula for volume of a cone on the left (subbing r = 6) and the volume you just calculated on the right.
An example (not accurate for this question). Lets say the volume of the hemisphere turns out to be 20π. Then write down
1/3π r²h = 10π and sub in the value of the radius on the left.
To simplify your equation, first divide across by π (gets rid of both πs) and multiply across by 3 (gets rid of the fraction). Then solve for h, the only unknown.
Don't lose sight of what you were originally asked. The height of the entire shape is equal to this height plus the radius of the hemisphere. Make sure you can understand why.
General point:
If you are working out e.g. πr² make sure that you understand fully that it is only r which gets squared and that the result of r² gets multiplied by π.
You never square or cube π.
For part a)
Use the area of a circle formula (area = πr²) twice for the larger circle and for the smaller one, making sure to do as you're told with π = 22/7.
Do not fall into the trap of thinking you can subtract the radii and find the area of a circle with radius = 7cm. If in doubt, sketch out the circles.
For part b)
You need to use the formulae for volume of a sphere and volume of a cylinder.
Part iii of this question is a little confusing. They are looking for the volume of air that is left in the cylinder when it contains the 3 balls.
Fort part c)
This is a tricky question. You are told the radius of the hemisphere so first step is to work out its volume. Note that the question doesn't mention whether π should be 3.14 or 22/7. That is because you can actually work out the answer conveniently without setting π equal to anything, just leave your work in terms of π.
When you find the volume of the hemisphere, you know that the volume of a cone is half of this.
So work out half the volume of the hemisphere (in terms of π) and then set up an equation with the formula for volume of a cone on the left (subbing r = 6) and the volume you just calculated on the right.
An example (not accurate for this question). Lets say the volume of the hemisphere turns out to be 20π. Then write down
1/3π r²h = 10π and sub in the value of the radius on the left.
To simplify your equation, first divide across by π (gets rid of both πs) and multiply across by 3 (gets rid of the fraction). Then solve for h, the only unknown.
Don't lose sight of what you were originally asked. The height of the entire shape is equal to this height plus the radius of the hemisphere. Make sure you can understand why.
General point:
If you are working out e.g. πr² make sure that you understand fully that it is only r which gets squared and that the result of r² gets multiplied by π.
You never square or cube π.
Income Tax Questions
You should not have any problems with these questions.
Find a layout you like and stick to it.
E.g.
Amount above cut-off @ higher rate % = part 1 of tax
Total income <
Amount up to cut-off @ lower rate % = part 2 of tax
Add to give Gross Tax
Subtract Tax credits (minus)
Results in Net Tax
Find a layout you like and stick to it.
E.g.
Amount above cut-off @ higher rate % = part 1 of tax
Total income <
Amount up to cut-off @ lower rate % = part 2 of tax
Add to give Gross Tax
Subtract Tax credits (minus)
Results in Net Tax
Linear Inequalities
You are used to solving equations e.g. finding the value of x for which this is true
3x - 5 = 7
Treat it like a weighing scales, and do the same to both sides to solve the equation.
So
3x = 12 (+5 to both sides)
x = 4 (÷ 3 on both sides)
The same weighing scales analogy works for inequalities. If you have
3x - 5 ≤ 7
3x ≤ 12
x ≤ 4
In other words, the inequality is true for any number x as long as it is less than or equal to 4.
Try 4:
3(4) - 5 = 7 and 7 is ≤ 7
Try 2:
3(2) - 5 = 1 and 1 is ≤ 7
Try a number bigger than 4 e.g. 5 :
3(5) - 5 = 10 and 10 is not ≤ 7
When you solve an inequality you have to check which set of numbers the solution is in.
For example if you were told x ∈ N that means it is a natural number (whole number ≥ 0)
Then the solution set would be x = {0,1,2,3,4}
However, if it was an integer i.e. x ∈ Z, then the negative whole numbers could also be in the solution set.
Finally if you were told x ∈R, that means x can be any number (fraction, decimal or whole number).
In the last two cases you can't write out the solution set, so you will normally be asked to graph it on the number line.
To do this draw a numberline and indicate with dots and arrows which set of numbers are included.
NB Note that there is one important difference between solving equations and finding solution sets to inequalities.
If you have
-x = 3
then multiplying across by -1 gives you
x = -3
However, if you have
-x < 3
when you multiply accross by -1, you have to reverse the inequality
x > -3
The reason for this is simple. -1 < 2 but multiplying across by -1 gives 1 on the left and -2 on the right so the sign must turn around 1 > -2
3x - 5 = 7
Treat it like a weighing scales, and do the same to both sides to solve the equation.
So
3x = 12 (+5 to both sides)
x = 4 (÷ 3 on both sides)
The same weighing scales analogy works for inequalities. If you have
3x - 5 ≤ 7
3x ≤ 12
x ≤ 4
In other words, the inequality is true for any number x as long as it is less than or equal to 4.
Try 4:
3(4) - 5 = 7 and 7 is ≤ 7
Try 2:
3(2) - 5 = 1 and 1 is ≤ 7
Try a number bigger than 4 e.g. 5 :
3(5) - 5 = 10 and 10 is not ≤ 7
When you solve an inequality you have to check which set of numbers the solution is in.
For example if you were told x ∈ N that means it is a natural number (whole number ≥ 0)
Then the solution set would be x = {0,1,2,3,4}
However, if it was an integer i.e. x ∈ Z, then the negative whole numbers could also be in the solution set.
Finally if you were told x ∈R, that means x can be any number (fraction, decimal or whole number).
In the last two cases you can't write out the solution set, so you will normally be asked to graph it on the number line.
To do this draw a numberline and indicate with dots and arrows which set of numbers are included.
NB Note that there is one important difference between solving equations and finding solution sets to inequalities.
If you have
-x = 3
then multiplying across by -1 gives you
x = -3
However, if you have
-x < 3
when you multiply accross by -1, you have to reverse the inequality
x > -3
The reason for this is simple. -1 < 2 but multiplying across by -1 gives 1 on the left and -2 on the right so the sign must turn around 1 > -2
Solving Quadratics
There are 2 ways to Solve quadratics
1 - By factoring
2 - Using the quadratic formula.
Solving Quadratics by Factoring:
The main difference between solving quadratics on the Higher Level course is that you will often have a number in front of the x² term, and the quadratic won't always be presented as a nice tidy x² + 4x + 4 = 0 format.
Dealing with the number in front of the x² term:
Looking at q 13
7x² + 29x + 4 = 0
As 7 is prime, the two brackets will be of the form
(7x .... )(x .....)=0
The numbers that multiply to make 4 are 4 × 1 and 2 × 2.
Trying e.g.
??? (7x + 2)(x + 2)= 0
If we multiply the inside pair and outside pair to check the x term we get
inside pair = 2x
outside pair = 14x
Total = 16x ≠ 29x
Trying again
(7x + 1)(x +4) = 0
Gives x + 28x = 29x so these factors are correct.
Factors are (7x + 1)(x +4) = 0
Roots are
7x + 1 = 0
x = -1/7
x+4=0
x = -4
Ans: Roots of equation are x = -1/7 or -4
With question 16
x(x-4) = 21
you have to tidy it up first
x² - 4x = 21
x² - 4x -21 = 0 (Note the =0 part is critical, because we will be exploiting the zero law to solve the factored equation).
Then factor as usual.
With questions 41 and 44 there is a lot more tidying up to do. Get rid of brackets and then put each equation in the form ....x² ....x ... number = 0
While you are looking at this topic, note the questions in the next section. There is no sign of a quadratic in e.g. question 5 but when you multiply accross by x to get rid of the fraction, you end up with x(x) = x²
Solving Quadratics Using the Quadratic Formula
If the question asks for the solution correct to 1 or 2 etc places of decimals, you need to use the quadratic formula.
Note that all the tidying up stuff mentioned above still applies, so you need to make sure the equation is in the form ax² + bx + c = 0
Looking at p35 q 13 you have
4x² + x -1 = 0
Which means that
a = 4, b = 1 (because x is the same as 1x) and c = -1 (note the sign).
Plugging numbers into the formula you get x =
-1 ± √( 1² - 4(4)(-1))
----------------------
2(4)
= -1 ± √17
--------
8
The results are (-1 + 4.123)/8 = 0.3903 ≈ 0.39
or (-1 - 4.123)/8 = -0.6403 ≈-0.64
Checking your answers in quadratics
Checking your answer is always a good idea, especially if you have time in an exam. Sometimes you may be asked to verify your answer. This means taking the result you got and plugging it back in to the original equation for x.
That means that in the examples above:
1 - By factoring
2 - Using the quadratic formula.
Solving Quadratics by Factoring:
The main difference between solving quadratics on the Higher Level course is that you will often have a number in front of the x² term, and the quadratic won't always be presented as a nice tidy x² + 4x + 4 = 0 format.
Dealing with the number in front of the x² term:
Looking at q 13
7x² + 29x + 4 = 0
As 7 is prime, the two brackets will be of the form
(7x .... )(x .....)=0
The numbers that multiply to make 4 are 4 × 1 and 2 × 2.
Trying e.g.
??? (7x + 2)(x + 2)= 0
If we multiply the inside pair and outside pair to check the x term we get
inside pair = 2x
outside pair = 14x
Total = 16x ≠ 29x
Trying again
(7x + 1)(x +4) = 0
Gives x + 28x = 29x so these factors are correct.
Factors are (7x + 1)(x +4) = 0
Roots are
7x + 1 = 0
x = -1/7
x+4=0
x = -4
Ans: Roots of equation are x = -1/7 or -4
With question 16
x(x-4) = 21
you have to tidy it up first
x² - 4x = 21
x² - 4x -21 = 0 (Note the =0 part is critical, because we will be exploiting the zero law to solve the factored equation).
Then factor as usual.
With questions 41 and 44 there is a lot more tidying up to do. Get rid of brackets and then put each equation in the form ....x² ....x ... number = 0
While you are looking at this topic, note the questions in the next section. There is no sign of a quadratic in e.g. question 5 but when you multiply accross by x to get rid of the fraction, you end up with x(x) = x²
Solving Quadratics Using the Quadratic Formula
If the question asks for the solution correct to 1 or 2 etc places of decimals, you need to use the quadratic formula.
Note that all the tidying up stuff mentioned above still applies, so you need to make sure the equation is in the form ax² + bx + c = 0
Looking at p35 q 13 you have
4x² + x -1 = 0
Which means that
a = 4, b = 1 (because x is the same as 1x) and c = -1 (note the sign).
Plugging numbers into the formula you get x =
-1 ± √( 1² - 4(4)(-1))
----------------------
2(4)
= -1 ± √17
--------
8
The results are (-1 + 4.123)/8 = 0.3903 ≈ 0.39
or (-1 - 4.123)/8 = -0.6403 ≈-0.64
Checking your answers in quadratics
Checking your answer is always a good idea, especially if you have time in an exam. Sometimes you may be asked to verify your answer. This means taking the result you got and plugging it back in to the original equation for x.
That means that in the examples above:
- if you substitute x = -4 into 7x² + 29x + 4 you should get zero.
- if you substitute x = 0.39 into 4x² + x -1 you should get very close to zero.
Writing as a single fraction
Example 1:
If you are asked to add
2/3 + 1/4
you have to first convert them into the same type of fraction i.e. find a new denominator which is the LCM of 3 and 4 => 12
2/3 = 8/12 (you have to multiply the bottom by 4 to go from 3 to 12, so you must do the same with the top)
1/4 = 3/12
So the sum becomes
8/12 + 3/12 = 11/12
Example 2:
The exact same principle applies when you have to add (or write as a single fraction)
5x-1 - 2x + 3
----- -------
4 5
The new denominator will be 20.
You multiply 4 by 5 to get 20, so you must multiply 5x-1 by 5 also giving 5(5x-1) = 25x - 5
Repeat the corresponding operation on the other fraction and then add/subtract like terms.
Example 3:
In question 17 you have to find the LCM of 2x+1 and 2x-1.
This will be (2x+1)(2x-1) and exactly the same principle as in examples 1 and 2 applies.
Remember you don't have to multiply out the denominator (expression on the bottom).
If you are asked to add
2/3 + 1/4
you have to first convert them into the same type of fraction i.e. find a new denominator which is the LCM of 3 and 4 => 12
2/3 = 8/12 (you have to multiply the bottom by 4 to go from 3 to 12, so you must do the same with the top)
1/4 = 3/12
So the sum becomes
8/12 + 3/12 = 11/12
Example 2:
The exact same principle applies when you have to add (or write as a single fraction)
5x-1 - 2x + 3
----- -------
4 5
The new denominator will be 20.
You multiply 4 by 5 to get 20, so you must multiply 5x-1 by 5 also giving 5(5x-1) = 25x - 5
Repeat the corresponding operation on the other fraction and then add/subtract like terms.
Example 3:
In question 17 you have to find the LCM of 2x+1 and 2x-1.
This will be (2x+1)(2x-1) and exactly the same principle as in examples 1 and 2 applies.
Remember you don't have to multiply out the denominator (expression on the bottom).
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