NAME: TOPIC: O level Conceptual Questions (A levels)

IG: @musclemath_ 1

NAME:

TOPIC: O level Conceptual Questions (A levels)

Website: www.musclemathtuition.com

Instagram: musclemath_

Facebook: www.facebook.com/musclemathtuition

Email: [email protected]

Concepts Overview:

• Discriminant

• Remainder and Factor Theorem

• Sum and Product of Roots

• Laws of Logarithm

• Coordinate Geometry

• Trigonometry

• Plane Geometry

• Motion of Particle (Kinematics)

mailto:[email protected]

www.musclemathtuition.com 2

Discriminant

Example 1 [NYJC Prelim/2018/P1/Q3] A curve 𝐶 has equation 𝑦2 + 14𝑦 + 4𝑥2 + 16𝑥 + 16𝑥𝑦 + 13 = 0. (i) If a real value of 𝑥 is substituted into the equation, it becomes a quadratic equation in 𝑦. Given

that there are two distinct values of 𝑦 for this equation, show that 5𝑥2 + 8𝑥 + 3 > 0, and hence find the set of possible values of 𝑥. [5]

(ii) Find the coordinates of the points where 𝐶 cuts the 𝑦-axis. State with a reason whether 𝐶 is a graph of a function. [2]

IG: @musclemath_ 3

Factor Theorem

(𝑎𝑥 − 𝑏), where 𝑎 ≠ 0, is a factor of the polynomial 𝑓(𝑥) if and only if 𝑓 (𝑏

𝑎) = 0.

Remainder Theorem When a polynomial 𝑓(𝑥) is divided by a linear polynomial (𝑎𝑥 − 𝑏) where 𝑎 ≠ 0, the remainder is

𝑓 (𝑏

𝑎).

Note: • The remainder theorem helps you find remainders. • Remember, it only works if 𝑓(𝑥) is divided by a linear polynomial. • Just imagine: What value of 𝑥 will make the divisor equal 0?

Example 2

(a) The expression 𝑓(𝑥) = 𝑥3 + 𝑎𝑥2 + 𝑏𝑥 −2

3 is exactly divisible by 𝑥 + 2 but leaves a remainder of 4

when divided by 𝑥 − 1. Find the value of 𝑎 and of 𝑏. (b) Prove that 𝑥 + 2 is a factor of 4𝑥3 − 13𝑥 + 6.


Example 3 [N2017/P1/Q5] When the polynomial 𝑥3 + 𝑎𝑥2 + 𝑏𝑥 + 𝑐 is divided by (𝑥 − 1), (𝑥 − 2) and (𝑥 − 3), the remainders are 8, 12 and 25 respectively. (i) Find the values of 𝑎, 𝑏 and 𝑐. [4] A curve has equation 𝑦 = 𝑓(𝑥), where 𝑓(𝑥) = 𝑥3 + 𝑎𝑥2 + 𝑏𝑥 + 𝑐, with the values of 𝑎, 𝑏 and 𝑐 found in part (i). (ii) Show that the gradient of the curve is always positive. Hence explain why the equation

𝑓(𝑥) = 0 has only one real root and find this root. [3]

(iii) Find 𝑥-coordinates of the points where the tangent to the curve is parallel to the line 𝑦 = 2𝑥 − 3. [3]

IG: @musclemath_ 5

Example 4 [RJC Promo/2018/Q1] It is given that 𝑓(𝑥) = 𝑥3 + 𝑎𝑥2 + 𝑏𝑥 + 𝑐. When 𝑓(𝑥) is divided by (𝑥 + 1) and (𝑥 + 2), the remainders are 24 and 36 respectively. Given also that (𝑥 − 1) is a factor of 𝑓(𝑥), find the values of 𝑎, 𝑏 and 𝑐. [3]


Example 5 [DHS Prelim/2018/P1/Q7(b)] When the polynomial 𝑎𝑥4 + 𝑏𝑥3 + 𝑐𝑥2 + 24𝑥 − 44, where 𝑎, 𝑏 and 𝑐 ∈ ℝ, is divided by (𝑥 − 1), (𝑥 + 1) and (𝑥 − 2), the remainders are −18, −54 and 0 respectively. (i) Find the values of 𝑎, 𝑏 and 𝑐. [3] (ii) The equation 𝑎𝑥4 + 𝑏𝑥3 + 𝑐𝑥2 + 24𝑥 − 44 = 0, with the values 𝑎, 𝑏 and 𝑐 found in part (i), has

a root 3 − (√2)𝑖. Find the other roots of the equation, showing your working clearly. [3]

IG: @musclemath_ 7

Roots and Coefficients of Quadratic Equations If 𝛼 and 𝛽 are the roots of a quadratic equation 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0, this means that,

Since (𝑥 − 𝛼)(𝑥 − 𝛽) = 0, Since 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0,

(𝑥 − 𝛼)(𝑥 − 𝛽) = 0 ⇒ 𝑥2 − 𝛽𝑥 − 𝛼𝑥 + 𝛼𝛽 = 0 ⇒ 𝑥2 − (𝛼 + 𝛽)𝑥 + 𝛼𝛽 = 0

𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0

⇒ 𝑥2 + (𝑏

𝑎) 𝑥 + (

𝑐

𝑎) = 0

We can see that 𝑥2 − (𝛼 + 𝛽)𝑥 + 𝛼𝛽 = 𝑥2 + (𝑏

𝑎) 𝑥 + (

𝑐

𝑎), which means:

Firstly,

𝑆𝑢𝑚 𝑜𝑓 𝑅𝑜𝑜𝑡𝑠 = 𝛼 + 𝛽 = −𝑏

𝑎

𝑃𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑅𝑜𝑜𝑡𝑠 = 𝛼𝛽 =𝑐

𝑎

Secondly,

𝑥2 − (𝑆𝑢𝑚 𝑜𝑓 𝑅𝑜𝑜𝑡𝑠)𝑥 + (𝑃𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑟𝑜𝑜𝑡𝑠) = 0

Other Useful Identities Since our sum of roots and product of roots are in the form 𝛼 + 𝛽 and 𝛼𝛽 respectively, questions will often require students to deal with expressions that are not as simple as 𝛼 + 𝛽 and 𝛼𝛽. Hence, here are some useful identities that can be tested in a question. • 𝛼3 + 𝛽3 = (𝛼 + 𝛽)(𝛼2 − 𝛼𝛽 + 𝛽2) • 𝛼3 − 𝛽3 = (𝛼 − 𝛽)(𝛼2 + 𝛼𝛽 + 𝛽2)


Example 6 The roots of the equation 5𝑥2 + 7𝑥 − 3 = 0 are 𝛼 and 𝛽. (i) State the value of 𝛼 + 𝛽 and of 𝛼𝛽. [2] (ii) Find, in the form 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0, where 𝑎, 𝑏 and 𝑐 are integers, the equation whose roots are

𝛼2𝛽 and 𝛼𝛽2. [5]

IG: @musclemath_ 9

Laws of Logarithm

Example 7 Evaluate ∫ log3(3𝑥 − 1)d𝑥.

From example 1, we learn that: • log𝑎𝑎 = 1 • log𝑎1 = 0

• log𝑎 (1

𝑥) = log𝑎(𝑥−1) = −log𝑎𝑥

There are also laws of logarithm that we can apply to make our lives simpler: • log𝑎𝑥𝑟 = 𝑟log𝑎𝑥 • log𝑎(𝑥𝑦) = log𝑎𝑥 + log𝑎𝑦

• log𝑎 (𝑥

𝑦) = log𝑎𝑥 − log𝑎𝑦

Important Note: • You cannot take ‘log’ of a negative number or 0 (i.e. lg(−3) or lg 0 does not make sense) • The symbol ‘lg’ represents ‘ log10′.

The symbol ‘ln’ represents ‘ log𝑒′. Logarithms to base 10 and base 𝑒 are known as common logarithms and natural logarithms respectively.

• log𝑎(𝑥 ± 𝑦) ≠ log𝑎𝑥 ± log𝑎𝑦 • log𝑎(𝑥 ± 𝑦) ≠ (log𝑎𝑥)(log𝑎𝑦)

For any positive real numbers 𝑎, 𝑏 and 𝑥 such that 𝑎, 𝑏 ≠ 1,

• log𝑎𝑥 =log𝑏𝑥

log𝑏𝑎

• log𝑎𝑏 =log𝑏𝑏

log𝑏𝑎=

1

log𝑏𝑎


Example 8 [ACJC Prelim/2014/P2/Q4] (i) Solve the equation

ln(𝑥 + 5) = 1 + ln 𝑥 giving your answer in terms of 𝑒. [2]

(ii) Sketch the curve 𝐶 with equation 𝑦 = ln(𝑥 + 5) − 1 − ln 𝑥, stating the exact value of the 𝑥 −coordinate of its intersection with the 𝑥 −axis and showing clearly the equation of any asymptotes. [3]

(iii) Use your graph to solve the inequality ln(𝑥 + 5) > 1 + ln 𝑥. [1]

IG: @musclemath_ 11

Coordinate Geometry Let 𝐴(𝑥1, 𝑦1) and 𝐵(𝑥2, 𝑦2) be two random points on the 𝑥𝑦 plane. We should think of gradient as the value of representing the ‘slope’. The gradient of a straight line can be found using two different methods. Firstly, we can use the equation

𝑚 =𝑦2 − 𝑦1

𝑥2 − 𝑥1=

𝑦1 − 𝑦2

𝑥1 − 𝑥2 (𝑥1 ≠ 𝑥2)

Secondly, we can also use the equation

𝑚 = tan 𝜃

where 𝜃 is the anticlockwise positive angle the straight line with gradient 𝑚 makes with the positive 𝑥 −axis.

• Midpoint = (𝑥1+𝑥2

2,

𝑦1+𝑦2

2)

• The perpendicular bisector of the line segment 𝐴𝐵 is the line which passes through the mid-

point of the line segment 𝐴𝐵, is perpendicular to the line segment 𝐴𝐵 (i.e. 𝑚1 = −1

𝑚2)

• The distance between any two points 𝐴(𝑥1, 𝑦1) and 𝐵(𝑥2, 𝑦2) is known as the length or

magnitude of line segment 𝑨𝑩. The distance/magnitude of 𝐴𝐵 is given by:

√(𝑥1 − 𝑥2)2 + (𝑦1 − 𝑦2)2

𝐴(𝑥1, 𝑦1)

𝐵(𝑥2, 𝑦2)

𝑦

𝑂 𝑥

𝜃


• Area of Triangle Given 3 points 𝐴(𝑥1, 𝑦1), 𝐵(𝑥2, 𝑦2) and 𝐶(𝑥3, 𝑦3), the area of ∆𝐴𝐵𝐶 is given by:

1

2|𝑥1

𝑦1 𝑥2

𝑦2 𝑥3

𝑦3 𝑥1

𝑦1|

which also equals to:

1

2|(𝑥1𝑦2 + 𝑥2𝑦3 + 𝑥3𝑦1) − (𝑥2𝑦1 + 𝑥3𝑦2 + 𝑥1𝑦3)|

Unofficially, this method is known at the “shoelace” method and it works with any 𝑛 −sided polygon. Vertices must be taken in an anti-clockwise order.

Example 9 A quadrilateral 𝐴𝐵𝐶𝐷 is such that point 𝐴 is (0, 4), 𝐵 is (10, 10), 𝐶 is (8, 1) and 𝐷 is (4, −3). Find the (i) equation of the perpendicular bisector of 𝐵𝐶, [4] (ii) area of quadrilateral 𝐴𝐵𝐶𝐷. [2]

𝐴(𝑥1, 𝑦1)

𝐵(𝑥2, 𝑦2)

𝐶(𝑥3, 𝑦3)

𝑦

𝑥 𝑂

IG: @musclemath_ 13

Example 10 [N2019/P1/Q7] A curve 𝐶 has equation 𝑦 = 𝑥𝑒−𝑥. (i) Find the equation of the tangents to 𝐶 at the points where 𝑥 = 1 and 𝑥 = −1. (ii) Find the acute angle between these tangents.


Example 11 [HCI Prelim/2010/P1/Q4] A curve is defined by the parametric equations

𝑥 =𝑡

1 + 𝑡2, 𝑦 =

𝑡

1 − 𝑡2, where 𝑡 ≠ −1, 1.

(i) Show that the tangent to the curve at any point with parameter 𝑡 has equation (1 − 𝑡2)3𝑦 = (1 + 𝑡2)3𝑥 − 4𝑡3.

[3]

(ii) Find the gradient of the tangent to the curve at 𝑡 =1

√2. Hence determine the acute angle

between this tangent and the line 𝑦 = 𝑥 + 3. [3]

IG: @musclemath_ 15

Example 12 [VJC Prelim/2018/P1/Q10(a)]

A line passes through the point (4, 5) and cuts the 𝑥-axis and 𝑦-axis at points 𝑃 and 𝑄 respectively. It is

given that angle 𝑂𝑃𝑄 = 𝜃, where 0 < 𝜃 <𝜋

2 and 𝑂 is the origin. You may assume that the same scale is

used on both axes. (i) Show that the equation of line 𝑃𝑄 is given by 𝑦 = (4 − 𝑥) tan 𝜃 + 5. [3] (ii) Hence or otherwise, show that 𝑂𝑃 + 𝑂𝑄 = 9 + 4 tan 𝜃 + 5 cot 𝜃. By differentiation, find the

stationary value of 𝑂𝑃 + 𝑂𝑄 as 𝜃 varies. Determine the nature of this stationary value. [5]


Trigonometry

• sin(90° − 𝜃) = cos 𝜃 • cos(90° − 𝜃) = sin 𝜃

• tan(90° − 𝜃) =1

tan 𝜃

Example 13

Given that sin 𝐴 =3

5, where 𝐴 is obtuse, express each of the following in fractions.

(i) sec 𝐴 [2] (ii) cos 2𝐴 [2]

IG: @musclemath_ 17

Example 14 Solve, for angles between 0 and 2𝜋, cos 2𝑥 + sin 𝑥 − 1 = 0. [5] Example 15 [TJC Promo/2005]

Given cos 𝑥 + sin 𝑦 = 3

4, find the exact value of

d𝑦

d𝑥 when 𝑥 =

𝜋

3 and

𝜋

2< 𝑦 < 𝜋 .


Example 16 [DHS Prelim/2018/P1/Q4] Show that the following inequality

tan 𝑥 + cot 𝑥 > 4 for 0 < 𝑥 <1

2𝜋

can be simplified to

0 < sin 2𝑥 <1

2.

[4] Hence solve the inequality leaving your answer in terms of 𝜋. [2]

IG: @musclemath_ 19

R-Formulae (Trignometry) The R-Formulae is a very common concept that is almost always tested in the O levels. It is, however, not given in the help sheet and is important for students to memorize this formula. • 𝑎 sin 𝜃 ± 𝑏 cos 𝜃 = 𝑅 sin(𝜃 ± 𝛼) • 𝑎 cos 𝜃 ± 𝑏 sin 𝜃 = 𝑅 cos(𝜃 ∓ 𝛼)

where 𝑅 = √𝑎2 + 𝑏2 and tan 𝛼 =𝑏

𝑎

Example 17 (i) Without the use of the graphic calculator and by using the R-formulae, solve the equation

5 sin 𝑥 + 12 cos 𝑥 = −6.5, for 0° < 𝑥 < 360°. [2]

(ii) State the greatest value of 5 sin 𝑥 + 12 cos 𝑥. [1]

𝑅

𝑏

𝑎


Example 18 [DHS J2 MYE/2017/P1/Q7]

(a) Show that tan𝛼

2=

sin 𝛼

1+cos 𝛼 . [2]

Hence find the exact value of arg(𝑤) where 𝑤 = (1 +√3

2) −

1

2𝑖. [4]

(b) Given that 𝑧 = √2𝑒𝑖𝜃 where 0 ≤ 𝜃 ≤ 𝜋. By considering 𝑅 sin(𝜃 + 𝛼), where 𝑅 is a positive constant and 𝛼 is a non-zero constant, find the exact value of 𝜃 such that Re(𝑧) + Im(𝑧) = 1. [4]

IG: @musclemath_ 21

Example 19 [SAJC Prelim/2018/P2/Q3] (i) Using the 𝑅-formula, express sin 𝜃 + 𝑚 cos 𝜃, for 𝑚 > 0, in the form 𝑅 sin(𝜃 + 𝛼) where 𝑅 > 0

and 0 < 𝛼 <𝜋

2 are constants to be determined in terms of 𝑚.

Hence show that 2(sin 𝜃 + 𝑚 cos 𝜃) sin(𝜃 − 𝛼) = 𝑅(cos 2𝛼 − cos 2𝜃). [3]

(ii) Given that 𝛼 =𝜋

2, evaluate ∫

cos 2𝜃

(sin 𝜃+𝑚 cos 𝜃) sin(𝜃−𝛼) d𝜃 in terms of 𝑚 and 𝜃 . [4]


Plane Geometry

• Congruency Tests Two triangles ∆𝐴𝐵𝐶 and ∆𝑋𝑌𝑍 are congruent if they have the same shape and size. i.e. their corresponding sides and angles must be congruent. Congruent triangles must satisfy one of the following tests.

SAS

If two sides and the included angle on one triangle are congruent to the corresponding parts of another, then the triangles are congruent.

ASA

If two angles and the included side of one triangle are congruent to the corresponding parts of another, then the triangles are congruent.

AAS

If two adjacent angles and a side of one triangle are congruent to the corresponding parts of another, then the triangles are congruent.

SSS If the three sides of one triangle are congruent to the three sides of another, then the triangles are congruent.

RHS

If the hypotenuse and one side of a right-angled triangle are congruent to the corresponding parts of the second right-angled triangle, the two triangles are congruent.

Note: Do not use ‘tests’ like ‘AAA’, ‘ASS’ or ‘SSA’.

2

3. Congruent Triangles Two triangles ABC and PQR are said to be congruent if they have the same shape and size.

i.e. their corresponding sides and angles must be congruent.

Triangles Corresponding congruent angles Corresponding congruent sides

PQRABC

PA

QB

RC

PQAB

QRBC

PRAC

4. Tests for Congruent Triangles

Test

1 SAS

If two sides and the included angle on one

triangle are congruent to the corresponding

parts of another, then the triangles are

congruent.

2 ASA

If two angles and the included side of one



congruent.

3 AAS

If two adjacent angles and a side of one



congruent.

4 SSS

If the three sides of one triangle are

congruent to the three sides of another, then

the triangles are congruent.

5 RHS

If the hypotenuse and one side of a right-angled triangle are congruent to the

corresponding parts of the second right-

angled triangle, the two triangles are

congruent.

A C

B

P Q

R

A

B

P

Q

A

B

P

Q

R

A C

B

P R

Q

A C

B

P R

Q

C R

C

2




PQRABC

PA

QB

RC

PQAB

QRBC

PRAC


Test

1 SAS




congruent.

2 ASA




congruent.

3 AAS




congruent.

4 SSS




5 RHS




congruent.

A C

B

P Q

R

A

B

P

Q

A

B

P

Q

R

A C

B

P R

Q

A C

B

P R

Q

C R

C

2




PQRABC

PA

QB

RC

PQAB

QRBC

PRAC


Test

1 SAS




congruent.

2 ASA




congruent.

3 AAS




congruent.

4 SSS




5 RHS




congruent.

A C

B

P Q

R

A

B

P

Q

A

B

P

Q

R

A C

B

P R

Q

A C

B

P R

Q

C R

C

2




PQRABC

PA

QB

RC

PQAB

QRBC

PRAC


Test

1 SAS




congruent.

2 ASA




congruent.

3 AAS




congruent.

4 SSS




5 RHS




congruent.

A C

B

P Q

R

A

B

P

Q

A

B

P

Q

R

A C

B

P R

Q

A C

B

P R

Q

C R

C

2




PQRABC

PA

QB

RC

PQAB

QRBC

PRAC


Test

1 SAS




congruent.

2 ASA




congruent.

3 AAS




congruent.

4 SSS




5 RHS




congruent.

A C

B

P Q

R

A

B

P

Q

A

B

P

Q

R

A C

B

P R

Q

A C

B

P R

Q

C R

C

𝐴

𝐵

𝐶 𝑃

𝑅

𝑄

IG: @musclemath_ 23

• Similarity Tests Two triangles ∆𝐴𝐵𝐶 and ∆𝑋𝑌𝑍 are similar if they satisfy one of the following tests.

Side-Side-Side (SSS Similarity)

𝐴𝐵

𝑋𝑌=

𝐵𝐶

𝑌𝑍=

𝐴𝐶

𝑋𝑍

Side-Angle-Side (SAS Similarity)

𝐴𝐵

𝑋𝑌=

𝐵𝐶

𝑌𝑍

∠𝐵 = ∠𝑌

Angle-Angle-Angle (AA/AAA Similarity)

∠𝐴 = ∠𝑋 ∠𝐵 = ∠𝑌 ∠𝐶 = ∠𝑍

• Other Useful Theorems

1. Mid-Point Theorem

𝑆𝑇//𝑄𝑅

𝑆𝑇 =1

2𝑄𝑅

2. Alternate Segment Theorem The angle between the tangent and a chord through the point of contact is equal to the angle in the alternate theorem.

𝑁�̂�𝐵 = 𝑀�̂�𝑁

2

3

1 3

4.5

1.5

2

3

3

4.5

𝐶 𝑌

𝐴

𝐵

𝑋

𝑍


• Circle Theorems

6

Circle Theorems – Link to what you have learned in E Math

AVOID abbreviations! Spell in full for words such as “segment”, “tangent”, “radius”, “opposite”, “parallelogram”, etc. You must

ensure that Cambridge markers know what you are writing. They may not know our abbreviations.

Look at the arc AB that subtends the angles!

Look at the arc AB that subtends the angles!

Also known as opposite angles of cyclic quad

IG: @musclemath_ 25

Angles Properties of Quadrilaterals • Parallelogram

Quadrilateral with 2 pairs of parallel sides

Quadrilateral with 2 pairs of equal sides

Quadrilateral with 2 pairs of equal opposite angles

Quadrilateral with 1 pair of equal and parallel opposite sides

• Rectangle

Quadrilateral with 4 right angles

Parallelogram with 1 right angle

• Rhombus

Quadrilateral with 4 equal sides

Parallelogram with equal adjacent sides

• Square

Quadrilateral with 4 equal sides and 4 right angles

Rectangle with equal adjacent sides

Rhombus with 1 right angle

18

Singapore Chinese Girls’ School Secondary 4

Additional Mathematics Plane Geometry

Name: ( ) Class: Sec 4_____

Date: ________________

Worksheet 5: Angles Properties of Quadrilaterals

The table below shows some properties commonly used for proving special quadrilaterals.

(TBPg251) 1. Parallelogram

quadrilateral with 2 pairs of parallel sides

quadrilateral with 2 pairs of equal sides

quadrilateral with 2 pairs of equal opposite angles

quadrilateral with 1 pair of equal and parallel opposite sides

2. Rectangle

quadrilateral with 4 right angles

parallelogram with 1 right angle

3. Rhombus

quadrilateral with 4 equal sides

parallelogram with equal adjacent sides

4. Square

quadrilateral with 4 equal sides and 4 right angles

rectangle with equal adjacent sides

rhombus with 1 right angle

18




Date: ________________








2. Rectangle



3. Rhombus



4. Square




18




Date: ________________








2. Rectangle



3. Rhombus



4. Square




18




Date: ________________








2. Rectangle



3. Rhombus



4. Square




18




Date: ________________








2. Rectangle



3. Rhombus



4. Square




18




Date: ________________








2. Rectangle



3. Rhombus



4. Square




18




Date: ________________








2. Rectangle



3. Rhombus



4. Square




18




Date: ________________








2. Rectangle



3. Rhombus



4. Square




18




Date: ________________








2. Rectangle



3. Rhombus



4. Square




18




Date: ________________








2. Rectangle



3. Rhombus



4. Square




18




Date: ________________








2. Rectangle



3. Rhombus



4. Square





Example 20 [BPGHS Prelim/2016/P2]

In the diagram, 𝑆𝑇 is a tangent at 𝐴 to the circle and 𝐵𝑃 is parallel to 𝑆𝑇. 𝐶 is the point where the straight line 𝐴𝑃 intersects the circle. (i) Show that ∠𝐴𝐵𝑃 = ∠𝐴𝐶𝐵. [2] (ii) Prove that ∆𝐴𝐵𝑃 and ∆𝐴𝐶𝐵 are similar. [2] (iii) Hence, or otherwise, show that 𝐴𝐵2 = 𝐴𝐶 × 𝐴𝑃. [2]

IG: @musclemath_ 27

Example 21 [SAJC Prelim/2018/P2/Q2]

In the figure above, points 𝐴 and 𝐶 are fixed points on the circle which form the diameter passing through the centre 𝑂. The circle has a fixed radius 𝑟 units. The variable point 𝐵 moves along the circumference of the circle between 𝐴 and 𝐶 in the upper half of the circle. The chord 𝐴𝐵 makes an angle of 𝜃 radians with the diameter 𝐴𝐶. Use differentiation to find, the exact angle 𝜃, such that 𝑆, the area of triangle 𝐴𝐵𝐶, is a maximum. [6]


Example 22 [ACJC Prelim/2018/P1/Q3] (a) The quadrilateral 𝐴𝐵𝐶𝐷 is such that 𝑃, 𝑄, 𝑅 and 𝑆 are the midpoints of 𝐴𝐵, 𝐵𝐶, 𝐶𝐷 and 𝐷𝐴

respectively. Prove that 𝑃𝑄𝑅𝑆 is a parallelogram. [3] (b) Referred to the origin 𝑂, points 𝐴, 𝐵 and 𝐶 have position vectors 𝐚, 𝐛 and 𝐜 respectively, where 𝐚 is

a unit vector, |𝐛| = 3, |𝐜| = √3 and angle 𝐴𝑂𝐶 is 𝜋

6 radians. Given that 3𝐚 + 𝐜 = 𝑘𝐛 where 𝑘 ≠ 0, by

considering (3𝐚 + 𝐜) ∙ (3𝐚 + 𝐜), find the exact values of 𝑘. [4]

IG: @musclemath_ 29

Example 23 [DHS J2 MYE/2017/P1/Q2] As shown in the diagram below, a ball is dropped from a point 2 metres horizontally away from a lightsource that is 10 metres above the horizontal ground. Light from the lightsource casts a shadow of the falling ball on the ground. The point where the ball eventually hits the ground is denoted by 𝑂 is

denoted by 𝑦 m and the distance from the moving shadow to 𝑂 is denoted by 𝑥 m. Show that 𝑥 =2𝑦

10−𝑦.

[1]

When the ball has dropped 5 metres from its initial position, its speed (downward) is given to be 10 metres per second. Find the rate at which its shadow is moving at this instant. [4]


Motion of a Particle in a Straight Line/Kinematics

There are 3 different concepts we will learn in this section: • Displacement • Velocity • Acceleration Displacement Displacement (𝑠) is defined as the straight-line distance of a moving particle measured from a fixed reference point. - The displacement at any point in time is represented by the equation 𝑠 = 𝑓(𝑡). - When a particle starts from or reaches back its original point, the displacement is zero. - Negative value of displacements indicates that the particle is at the left side of the starting point. - A formula for the displacement 𝑠 is given by

𝑠 = ∫ 𝑣 d𝑡

This formula also refers to the area enclosed under the velocity-time graph.

Velocity Velocity (𝑣) is defined as the rate of change of displacement with respect to time. - Denoting velocity as 𝑣, we have to differentiate displacement to find the velocity function.

i.e.

𝑣 =d𝑠

d𝑡

- ‘Initial’ means when time (𝑡) equals 0. - When 𝑣 > 0, the particle is moving in the positive direction, vice versa. - ‘Instantaneously at rest’ simply means velocity (𝑣) equals 0. - A formula for the velocity 𝑣 is also given by

𝑣 = ∫ 𝑎 d𝑡

This formula also refers to the area enclosed under the acceleration-time graph.

IG: @musclemath_ 31

Acceleration Acceleration is defined as the rate of change of velocity with respect to time. - Denoting acceleration as 𝑎, we have to differentiate the velocity function with respect to time.

i.e.

𝑎 =d𝑣

d𝑡

- ‘Retardation’ or ‘Deceleration’ simply means negative acceleration value.

Note: - Instantaneous speed = |𝑣|

- Average speed = Total Distance travelled

Total Time taken

- Instantaneous rest does not mean 𝑎 = 0 - There is a difference between displacement and distance

Example 24 A particle travelling in a straight line passes a fixed point 𝑂 with a velocity of 0 m/s. Its acceleration 𝑎 m/s2 is given by 𝑎 = 2(𝑡 − 2). Find (a) its velocity after 3 seconds, [2] (b) the time when it is next instantaneously at rest, [2] (c) its distance from 𝑂 after 7s, [2] (d) its total distance moved in the first 7s. [2]


Sine Rule/Cosine Rule

Example 25 [RVHS Prelim/2017/P1/Q8] (a) Using differentiation, find the exact dimensions of the rectangle of largest area that can be

inscribed in the ellipse, 𝑥2

9+

𝑦2

36= 1. Hence, find the area of this largest rectangle. [8]

(b) In the triangle 𝐷𝐸𝐹, angle 𝐸𝐷𝐹 =𝜋

3 and angle 𝐷𝐹𝐸 =

𝜋

3+ 𝛼 and 𝐸𝐹 = 6. Given that 𝛼 is sufficiently

small. Show that 𝐷𝐹 − 𝐷𝐸 ≈ 𝑑𝛼,

where 𝑑 is an exact constant to be determined. [5]

IG: @musclemath_ 33

Example 26 [TJC J2 MYE/2018/P1/Q5] A property developer decides to divide a triangular plot of land 𝑃𝑄𝑅 into two plots of equal area. One section, 𝑀𝑁𝑃, is to be used for commercial development and the other, 𝑀𝑁𝑄𝑅, is to be used for high-rise residential development. It is given that 𝑃𝑁 = 𝑥 km, 𝑃𝑀 = 𝑦 km, 𝑀𝑁 = 𝑧 km, 𝑃𝑅 = 1.2 km, 𝑃𝑄 = 1.4 km, and ∠𝑄𝑃𝑅 = 𝛼, where 𝛼 is a constant.

Show that 𝑧2 = 𝑥2 +

0.7056

𝑥2 − 1.68 cos 𝛼. [3]

To minimize construction costs, the architects have decided that the boundary 𝑀𝑁 should preferably be of the minimum possible length. Using differentiation, find the value of 𝑥 which will minimize the construction cost. [5]


Example 27 [EJC Prelim/2019/P2/Q2]

The diagram shows a mechanism for converting rotational motion into linear motion. The point 𝑃 is on the circumference of a disc of fixed radius 𝑟 which can rotate about a fixed point 𝑂. The point 𝑄 can only move on the line 𝑂𝑋, and 𝑃 and 𝑄 are connected by a rod of length 2𝑟. As the disc rotates, the point 𝑄 is made to slide along 𝑂𝑋. At time 𝑡, angle 𝑃𝑂𝑄 is 𝜃 and the distance 𝑂𝑄 is 𝑥. (i) State the maximum and minimum values of 𝑥. [1]

(ii) Show that 𝑥 = 𝑟(cos 𝜃 + √4 − sin2 𝜃). [2]

(iii) At a particular instant, 𝜃 =𝜋

6 and

d𝜃

d𝑡= 0.3. Find the numerical rate at which point 𝑄 is moving

towards point 𝑂 at that instant, leaving your answer in terms of 𝑟. [3]

END

2019 JC2 H2 Mathematics Preliminary Examination

Section A: Pure Mathematics [40 marks]

1 (a) Find the complex numbers z and w that satisfy the equations

[3]

(b) It is given that is real, where c is also real. By first expressing in Cartesian form, find

all possible values of c. [3]

2

The diagram shows a mechanism for converting rotational motion into linear motion. The point P is on the

circumference of a disc of fixed radius r which can rotate about a fixed point O. The point Q can only move

on the line OX, and P and Q are connected by a rod of length 2r. As the disc rotates, the point Q is made to

slide along OX. At time t, angle POQ is and the distance OQ is x.

(i) State the maximum and minimum values of x. [1]

(ii) Show that 2cos 4 sinx r . [2]

(iii) At a particular instant, 6

and d

0.3.dt

Find the numerical rate at which point Q is moving towards

point O at that instant, leaving your answer in terms of r. [3]

3 The position vectors of points P and Q, with respect to the origin O, are p and q respectively. Point R, with

position vector r, is on PQ produced, such that .

(i) Given that and 11p.r , find the length of projection of OQ onto . [4]

(ii) S is another point such that PS r . Given that and 2 3r i j k , find the area of the

quadrilateral OPSR. [3]

O

P

Q

x X

r

2r

NAME: TOPIC: O level Conceptual Questions (A levels)

Documents

Transcript of NAME: TOPIC: O level Conceptual Questions (A levels)