Must know new 2008 "O" level Add Maths questions

Ahm97sic SG

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.

eagle: edited to make it bigger and easier to read

Ahm97sic SG

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.

Ahm97sic SG

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.

Ahm97sic SG

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 1

Express in partial fraction

(17x^2 +23x + 12)/[(3x+4)(x^2+4)]

Ahm97sic SG

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.

Ahm97sic SG

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.

eagle SG

O levels got graphics calculator???
blur...

but thanks a lot of your update! :D

Ahm97sic SG

O levels got graphics calculator???
blur...

but thanks a lot of your update! :D

Hi Eagle,

You have made a mistake. The question has stated clearly that without using a graphics calculator, solve the inequality.

No graphics calculator is allowed in the "O" Add Maths. Graphics calculator is only allowed in the "A" level Maths.

Thank you for your kind attention.

Regards,

ahm97sic

eagle SG

Actually, my understanding is that if they state without using a graphics calculator, it means that graphics calculator is allowed in the paper.

Or did you take the question from past year Maths C paper? That would explain....

btw, are you a tuition teacher?

Ahm97sic SG

Hi Eagle,

My purpose of writing the "must know new 2008 "O" level Add Maths questions" in the forum is just to share where to look for exam questions to practise on the new topics introduced in the new 2008 syllabus.

Thank you for your kind attention.

Regards,

ahm97sic

eagle SG

Ok stickied :D

wishboy SG

Just to add on to ur examples, this question on circles came out in my sch prelim exams.

Find the equation of the circle which passes through A(8, 1) and B(7, 0) and has, for its tangent at B, the line 3x - 4y - 21 = 0

Ans:
(x - 4)² + (y - 4)² = 25
or
x² + y² - 8x - 8y + 7 = 0

Ahm97sic SG

Hi Wishboy,

Step 1 : Rearrange the line 3x - 4y - 21 = 0 into the form of y = mx + c

4y = 3x - 21

y = (3/4)x -21/4

Hence, m = 3/4

Step 2 Find the gradient of BC where B(7,0) and centre C(-g, -f)

m1 = [0-(-f)]/[7-(-g)]

= f/[7+g]

Step 3 Since the line 3x - 4y - 21 = 0 is tangent to the circle at point B, so the line

3x - 4y = 0 is perpendicular to BC ie m = - 1/m1

3/4 = - (7+g)/f

f = - (4/3)(7 + g) ------------------ (0)

Step 4 Since the points A(8,1) and B(7,0) are points on the circle, substitute them

into the equation of the circle to form 2 simultaneous equations

Equation of the circle is x^2 + y^2 +2gx +2fy + c = 0

Substitute point A (8,1), 8^2 + 1^2 + 2g(8) + 2f(1) + c = 0

65 + 16g + 2f + c = 0 ------------- (1)

Substitute point B (7, 0), 7^2 + 0^2 + 2g(7) + 2f(0) + c = 0

49 + 14g + c = 0 ------------------ (2)

Equation (2) - Equation (1)

16 + 2g + 2f = 0

8 + g + f = 0

f = - 8 - g ---------------------------- (3)

Step 5 Susbtitute equation (3) into equation (0) to solve for g

Substitute f = - 8 - g into equation (0)

- 8 - g = - (4/3)(7 + g)

- 24 - 3g = - 28 - 4g

g = - 4

Step 6 Substitute g = - 4 into equation (3) to solve for f

f = - 8 - (- 4)

f = - 4

Step 7 Substitute g = - 4 into equation (2) to solve for c

49 + 14 (- 4) + c = 0

c = 7

Step 8 Substitue g = - 4 , f = - 4 and c = 7 into the equation of the circle

x^2 + y^2 +2gx +2fy + c = 0

x^2 + y^2 + 2(- 4)x + 2(- 4)y + 7 = 0

x^2 + y^2 - 8x - 8y + 7 = 0

Hope that I did not make any calculation mistakes.

Thank you for your kind attention.

Regards,

ahm97sic

jc123 SG

Rearrange the equation to (x-h)^2+(y-k)^2 = r^2 form to find the coordinates of the circle centre. Since the circle is a reflection on x-axis, the x coordinate and radius remain the same, but y-coordinate changes signs, so just substitute the new y coordinate into the equation to get the equation of the new circle.

Ahm97sic SG

Hi jc123,

Yes, the short-cut to the question 1 is

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)

when the circle is reflected in the x-axis,

The old equation of the circle is x^2 + y^2 - 4x + 6y - 12 = 0

The new equation of the circle is x^2 + y^2 - 4x - 6y - 12 = 0

Thank you for your kind attention.

Regards,

ahm97sic

PS : Have you tried the second 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.

jc123 SG

Question 2

(a) Just solve the two equations to get x and y values.

(b) Solve the equation at x=3 to get the two y values, P is (3, y1) and Q is (3, y2).

A. You will have to find the coordinates of the center of the circle.

B. The gradients of the lines from P and Q to the center.

C. Basic geometry, tangent is perpendicular to the radius, so the gradient of line P to center * gradient of the tangent at P is -1. You have got the gradient of tangent at P and point P, you know how to find the equation of the tangent at P now.

jinjia SG

For question one, if the question is changed to be "reflection to the y-axis" instead? change -4x to 4x?? so the equation will be x^2 + y^2 + 4x + 6y - 12 = 0. Am i correct?

Ahm97sic SG

Hi Jinja,

Yes, Jinjia, you are correct.

Regards,

ahm97sic

Note :

It is to be noted that reflected in the y axis,

The old equation of the circle is x^2 + y^2 - 4x + 6y - 12 = 0

The new equation of the circle is x^2 + y^2 + 4x + 6y - 12 = 0

It is to be noted that reflected in the x axis,

The old equation of the circle is x^2 + y^2 - 4x + 6y - 12 = 0

The new equation of the circle is x^2 + y^2 - 4x - 6y - 12 = 0

If the circle is reflected in the line y = x,

The old equation of the circle is x^2 + y^2 - 4x + 6y - 12 = 0

The new equation of the circle is x^2 + y^2 + 6x - 4y - 12 = 0

If the circle is reflected in the line y = - x,

The old equation of the circle is x^2 + y^2 - 4x + 6y - 12 = 0

The new equation of the circle is x^2 + y^2 - 6x + 4y - 12 = 0

eagle SG

Question 2

(a) Just solve the two equations to get x and y values.

(b) Solve the equation at x=3 to get the two y values, P is (3, y1) and Q is (3, y2).

A. You will have to find the coordinates of the center of the circle.

B. The gradients of the lines from P and Q to the center.

Taking a look back now, there's actually a short cut for this question, without needing to go through the troublesome steps A, B and C.

You can see the solution here in ExamWorld
http://examworld.blogspot.com/2008/11/o-lvl-maths-equation-of-circle.html

Ahm97sic SG

Taking a look back now, there's actually a short cut for this question, without needing to go through the troublesome steps A, B and C.

You can see the solution here in ExamWorld
http://examworld.blogspot.com/2008/11/o-lvl-maths-equation-of-circle.html

Hi Eagle,

You have copied the question for part (b) incorrectly.

The correct equation of the circle is x² + y² - 8x - 10y - 9 = 0.

You have incorrectly copied and solved the question using

the incorrect equation of the circle as x² + y² - 4x - 8y + 3 = 0.

Hence, you use the equation 4y = x - 3 as the equation of the tangent at the point P(3, 0) and using this gradient of the tangent to help to find the equation of tangent at the point Q(3, 8).

Since the correct equation of the circle is x² + y² - 8x - 10y - 9 = 0, then you cannot use 4y = x - 3 as the equation of the tangent at the point P(3, 0) and using this gradient of this equation to help to find the equation of tangent at the point Q(3, 8).

Thank you for your kind attention.

Regards,

ahm97sic