Database - Must know new 2008 "O" level Add Maths questions
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The topic of equation of a circle is a new topic introduced in the new syllabus 2008 of Additional Mathematics. Hence, there are no past year exam questions on this topic in the ten years series.
The equation of the circle was once set in the late 60s and early 70s.
If you need more practices, you can go to Bras Basah Complex second hand bookshop to look for the old, old ten years series.
Question 1
The equation of a circle is x^2 + y^2 - 4x + 6y - 12 =0
Find the equation of the new circle, which is a reflection of the circle
in the x-axis. (Hint : There is a short-cut to this question)
Question 2
(a) Show that the line 4y = x - 3 touches the circle
x^2 + y^2 - 4x - 8y + 3 = 0
(b) The straight line x = 3 intersects the circle
x^2 + y^2 - 8x - 10y - 9 = 0 at the points P and Q.
Find the equations of tangents at P and Q.Another new topic is the solving of an inequality with modulus introduced in the new syllabus 2008 of Additional Mathematics. Hence, there are no past year exam questions on this topic in the ten years series.
In the previous syllabus, students are required to sketch the inequality and state the number of solutions but in the new syllabus, students are required to solve an inequality with modulus.
Question 1
Without using a graphics calculator, solve (not sketch) the inequality
l x^2 - 3x - 7 l < 3.Another new topic is partial fraction introduced in the new syllabus 2008 of Additional Mathematics.
The topic of partial fraction was previously in the old syllabus of "A" level Maths C. Hence, students will have to look for the old ten years series of A" level Maths C to practise on the topic of partial fraction.
Question 1Express in partial fraction
(17x^2 +23x + 12)/[(3x+4)(x^2+4)]Another new topic is the R-formula, addition and subtraction of angles, double angles and factor formulae introduced in the new syllabus 2008 of Additional Mathematics.
The topic of R-formula, addition and subtraction of angles, double angles and factor formulae was previously set in the 80s and 90s. If you need more practices, you can go to Bras Basah Complex second hand bookshop to look for the old, old ten years series.
Question 1
Solve the equation 3cosx - 2sinx = 2 for all angles between 0 degree and 360 degrees inclusive.Another new topic is the midpoint theorem, intercept theorem, alternate segment or tangent chord theorem, intersecting chords theorem and tangent secant theorem introduced in the new syllabus 2008 of Additional Mathematics.
The midpoint theorem, intercept theorem, alternate segment or tangent chord theorem were previously in the old syllabus of E. Maths. If you need more practices, you can go to Bras Basah Complex second hand bookshop to look for the old, old ten years series.
However, the intersecting chords theorem and tangent secant theorem have never appeared before in the "O" and "A" level Maths exam questions. (I have all the "A" level Maths C exam papers from 1968 to 2007, "O" level Add Maths (includes "AO" level Maths) and E. Maths exam questions from 1958 to 2007, "N" level Maths exam questions from first year 1984 to 2007) Hence, there will be no past exam questions on this topic in the ten years series. If you need more practices, you can only try practising on the question in the specimen papers provided by MOE for the new syllabus 2008 of additional mathematics.Another new topic is the use of a (alpha) and b (beta) in the quadratic equation introduced in the new syllabus 2008 of Additional Mathematics.
The use of a (alpha) and b (beta) in the quadratic equation was once set in the late 70s and early 80s.If you need more practices, you can go to Bras Basah Complex second hand bookshop to look for the old, old ten years series.
Question 1
Given that one root of the equation
x^2 + px + q = 0
is twice the other, express q in terms of p.
Question 2
Form the quadratic equation for which the sum of the roots is 2 and the sum of the squares of the roots is 18. -
Question 1
Without using a graphics calculator, solve (not sketch) the inequality
l x^2 - 3x - 7 l < 3.Is it the same as doing a normal modulus question?
Alright, why not you check for my mistakes and thus answers my doubt at the same time.
x^2 - 3x - 7 < 3 x^2 - 3x - 7 < -3
x^2 - 3x -10 < 0 x^2 - 3x - 4 < 0
(x-5)(x+2) < 0 (x-4)(x+1) < 0
-2 < x < 5 -1 < x < 4
Therefore, -1< x < 4 (falls between both ranges)
Is this correct? Somehow I don't feel right...
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Originally posted by Garrick_3658:
Is it the same as doing a normal modulus question?
Alright, why not you check for my mistakes and thus answers my doubt at the same time.
x^2 - 3x - 7 < 3 x^2 - 3x - 7 < -3
x^2 - 3x -10 < 0 x^2 - 3x - 4 < 0
(x-5)(x+2) < 0 (x-4)(x+1) < 0
-2 < x < 5 -1 < x < 4
Therefore, -1< x < 4 (falls between both ranges)
Is this correct? Somehow I don't feel right...
Dear Darrick_3658,
There is a trick in this type of question. The usual approach to solve it wil give the wrong answer as you have realised it in your answer ie if you choose a x value in your answer range -1< x < 4 eg x = 1 and substitute to the original question
l x^2 - 3x - 7 l < 3.
l 1^2 - 3(1) - 7 l ie 9 not less than 3 . (Incorrect)
Question
Without using a graphics calculator, solve (not sketch) the inequality
l x^2 - 3x - 7 l < 3.Answer
l x^2 - 3x - 7l < 3
Square both sides
(x^2 - 3x -7)^2 < 3^2
(x^2 - 3x -7)^2 - 3^2 < 0
[x^2 - 3x - 7 - 3][x^2 - 3x - 7 + 3] < 0
[x^2 - 3x - 10][x^2 - 3x - 4] < 0
[(x-5)(x+2)][(x-4)(x+1)] < 0
(x-5)(x+2)(x-4)(x+1) < 0
Use the number line method,
the answer is -2 < x < -1 and 4 < x < 5.
Regards,
ahm9s7ic
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(x-5)(x+2)(x-4)(x+1) < 0
Once again i am not very good in math so I dont understand how you can solve and get an inequality through this the above. Iis a quadriple equation nia XD
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Dear Davidche,
(x-5)(x+2)(x-4)(x+1) < 0
Step 1 : Find the critical points first.
This means x = 5, x = -2, x = 4 and x = -1
Step 2 : Draw a line marking the critical points on the line ie
-----------l-----------l-----------------------l-----------l--------
-2 -1 4 5
Step 3 : Substitute x < -2 eg x=-3
-2 < x < -1 eg x = -1.5
-1 < x < 4 eg x= 1
4 < x < 5 eg x = 4.5
x > 5 eg x = 6
into the (x-5)(x+2)(x-4)(x+1) to see whether the answers will be + or -
Step 4 : Put in the + and - signs on top of the number line
+ - + - +
-----------l-----------l-----------------------l-----------l--------
-2 -1 4 5
Step 5 : Choose the range of x values that will give the - sign since we are asked
to solve the inequality (x-5)(x+2)(x-4)(x+1) < 0
Hence, the solution is
-2 < x < -1 , 4 < x< 5
Regards,
ahm97sic
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Oh cool. This ''checking'' method is the best way to doing inequalities! even for quadratic.
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Originally posted by davidche:
Once again i am not very good in math so I dont understand how you can solve and get an inequality through this the above. Iis a quadriple equation nia XD
can still solve even if you reach this...
This is a little bit of E maths involved... Such a x^4 curve will give you a W shape graphAnd the points... you should know...
if it is a -x^4 curve, it will be M shaped
Ahm97sic's way is very good, but if you know how the graph looks like, i.e. W shaped, you can do it even faster.
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Hi Eagle,
The "O" level E.Maths and Add Maths Students have not learnt the curve sketching techniques yet. They still do not know the shapes of many curves yet.
For example, these students will not know the shape of the curve
[(x-5)(x+2)]/[(-4)(x+1) and so on.
The number line method will be easier for these students as this method can be used even if they do not know the shape of the curve.
Nevertheless, I do agree with you that if the students know shape of the curve, they can get the inequality faster as the curve y = (x-5)(x+2)(x-4)(x+1) is a W-shaped curve since the coefficient of x^4 is positive.
Hence, the solution for the inequality (x-5)(x+2)(x-4)(x+1) < 0 is
-2 < x < -1 , 4 < x< 5
as for -2 < x < -1, the W-shaped curve is below the x-axis and for 4 < x < 5, the W-shaped curve is also below the x-axis.
Regards,
ahm97sic
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Originally posted by Ahm97sic:
Hi Eagle,
The "O" level E.Maths and Add Students have not learnt the curve sketching techniques yet. They still do not know the shapes of many curves yet.
For example, these students will not know the shape of the curve
[(x-5)(x+2)]/[(-4)(x+1) and so on.
The number line method will be easier for these students as this method can be used even if they do not know the shape of the curve.
Nevertheless, I do agree with you that if the students know shape of the curve, they can get the inequality faster as the curve y = (x-5)(x+2)(x-4)(x+1) is a W-shaped curve since the coefficient of x^4 is positive.
Hence, the solution for the inequality (x-5)(x+2)(x-4)(x+1) < 0 is
-2 < x < -1 , 4 < x< 5
as for -2 < x < -1, the W-shaped curve is below the x-axis and for 4 < x < 5, the W-shaped curve is also below the x-axis.
Regards,
ahm97sic
actually, it does not take more than a few minutes to learn the basic shapes of x^4, x^3 curves, etc...
that's why I will always teach it to my students ;)
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Hi Eagle,
Please come into my forums (addmaths.sgforums.com and emaths.sgforums.com)
more often. OK.
Regards,
ahm97sic
PS : I am in the process of setting up the "A" level (H2 and H1) Economics forum
too. Please come to visit the economics forum too. OK.
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