Anglo-Chinese Junior College 2012 JC 2 H2 Math Mid-Year Examination Solution & Comments

 Anglo-Chinese Junior College 2012 JC 2 H2 Mathematics Mid-Year Exam Solution  Remarks Essay

Anglo-Chinese Junior College

2012 JC 2 H2 Math Mid-Year Exam Solution & Comments

Qn

1

(i)

Solution

The //gram is

A

M

Note that STRYGE: EC = 2: one particular

E

To

C

пЈ« 2пЈ¶

From your diagram, OCCITAN = STOMACH = пЈ¬ 3 пЈ·

пЈ¬ пЈ·

пЈ¬1пЈ·

пЈ­ пЈё

пЈ® пЈ« a couple of пЈ¶ пЈ« 1пЈ¶ пЈ№

пЈ«5пЈ¶

2OC + OA 1 пЈЇ пЈ¬ пЈ· пЈ¬ пЈ· пЈє one particular пЈ¬ пЈ·

= пЈЇ 2 пЈ¬ 3 пЈ· + пЈ¬ 1пЈ· пЈє = пЈ¬ 7 пЈ·

Using MF15, OE =

3

three or more пЈ¬ пЈ· пЈ¬ пЈ·

3пЈ¬ пЈ·

пЈЇ пЈ­ 1 пЈё пЈ­ 1пЈё пЈє

пЈ­ 3пЈё

пЈ°

пЈ»

пЈ«5пЈ¶ пЈ«3пЈ¶

пЈ«4пЈ¶

1пЈ¬ пЈ· пЈ¬ пЈ·

1пЈ¬ пЈ·

Therefore , BECOME = STOCK в€’ HINSICHTLICH = пЈ¬ 7 пЈ· в€’ пЈ¬ 4 пЈ· = в€’ пЈ¬ a few пЈ·. 3пЈ¬ пЈ· пЈ¬ пЈ·

3пЈ¬ пЈ·

пЈ­ 3пЈё пЈ­ 2пЈё

пЈ­3пЈё

(ii)

PURSE Г— BC represents the area of parallelogram OABC.

Note: BA Г— BC offers a normal to the plane HURUF (no modulus sign).

Comments

For vectors, argument

of complex figures,

P& C, and likelihood,

sketch a diagram to

help visualize the

scenario.

The plan helps you

find immediately that

OC sama dengan AB.

Rather than MF15, you

may use AE = a couple of AC therefore

3

that BE = BA & AE.

The modulus signal tells

you that the consequence is a

amount, not a vector, so

the response cannot be

" normal towards the plane”.

(iii) Method 1:

пЈ« в€’2 пЈ¶ пЈ« 1 пЈ¶

пЈ¬ пЈ· пЈ¬ пЈ·

Since пЈ¬ 1 пЈ·. пЈ¬ 2 пЈ· sama dengan 0, the queue L is usually parallel to plane OABC. пЈ¬ you пЈ· пЈ¬0пЈ·

пЈ­ пЈё пЈ­ пЈё

Note the statements you

need to compose to explain

/ describe what you are

undertaking!

пЈ« you пЈ¶ пЈ« в€’2 пЈ¶

пЈ¬ пЈ· пЈ¬ пЈ·

Since пЈ¬ 1 пЈ·. пЈ¬ 1 пЈ· в‰ 0, the idea (1, 1, 0 ) does not lie on plane OABC. пЈ¬ 0пЈ· пЈ¬ 1 пЈ·

пЈ­ пЈё пЈ­ пЈё

Hence, aircraft OABC does not contain the range L.

Approach 2:

пЈ« 1 + О» пЈ¶ пЈ« в€’2 пЈ¶

пЈ¬

пЈ·пЈ¬ пЈ·

пЈ¬ you + 2О» пЈ·iпЈ¬ 1 пЈ· = в€’2 в€’ 2О» & 1 & 2О» = в€’1 в‰ 0 пЈ¬ 0 пЈ·пЈ¬1пЈ·

пЈ­

пЈёпЈ­ пЈё

Since the position vector of every stage on the line will not fit into the equation of the plane, issues the plane OABC does not contain the series L. Approach 3:

пЈ« 1 пЈ¶ пЈ« в€’2 пЈ¶

пЈ¬ пЈ· пЈ¬ пЈ·

Since пЈ¬ one particular пЈ·. пЈ¬ 1 пЈ· в‰ zero, the point (1, 1, zero ) would not lie upon plane OABC. пЈ¬ 0пЈ· пЈ¬ one particular пЈ·

пЈ­ пЈё пЈ­ пЈё

Consequently, plane OABC does not develop the line D.

1

one particular

(iv)

Either

♦ use the formula

" perpendicular range

from stage V to plane sama dengan

пЈ«1пЈ¶

Z = пЈ¬ 1 пЈ·

пЈ¬ пЈ·

пЈ¬0пЈ·

пЈ­ пЈё

пЈ« в€’2 пЈ¶

пЈ¬ пЈ·

OD iпЈ¬ 1 пЈ·

пЈ¬1пЈ·

пЈ­ пЈё

Shortest Distance =

пЈ« в€’2 пЈ¶

пЈ¬1пЈ·

пЈ¬ пЈ·

пЈ¬1пЈ·

пЈ­ пЈё

Л†

VA⋅ and ” where A is virtually any

пЈ« one particular пЈ¶ пЈ« в€’2 пЈ¶

пЈ¬ пЈ·пЈ¬ пЈ·

пЈ¬ 1 пЈ·iпЈ¬ 1 пЈ·

пЈ¬0пЈ· пЈ¬ 1 пЈ·

в€’1

1

6

sama dengan пЈ­ пЈёпЈ­ пЈё =

=

=

6

six

6

пЈ« в€’2 пЈ¶

пЈ¬1пЈ·

пЈ¬ пЈ·

пЈ¬1пЈ·

пЈ­ пЈё

point on the plane,

♦ or find the position

vector of the foot of the

perpendicular from D to

the plane first (see page

11 method one particular of notes).

(May create answer since 0. 408 (3 sf) as query did not request exact benefit. )

a couple of

( 2w + 1)

5

Solve for z5 = 4в€љ2 first,

after that solve for

(2w + 1)5 sama dengan 4в€љ2.

a few

= 4 2 = 2 two e0i

a few

= a couple of 2 electronic 2 kПЂ i

a couple of w & 1 = 2e

пЈ« 2 kПЂ

5

a couple of пЈ¬

w=

eпЈ­

two

пЈ« two kПЂ пЈ¶

пЈ¬

пЈ·i

пЈ­ a few пЈё

пЈ¶

пЈ·i

пЈё

в€’

O

4в€љ2

afeafef (4в€љ2) sama dengan 0 radians

, k = в€’2, в€’1, 0, 1, 2

List out all of the 5

answers in the actual Alevel examination. Don't leave

your answer like this.

one particular

k = в€’2, в€’1, 0, 1, 2

a couple of

пЈ« a couple of kПЂ пЈ¶

2 пЈ¬ 5 пЈ·i

1

1

2

=

eпЈ­ пЈё

в‡’

w+ =

2

2

two

2

The points symbolizing these a few complex roots lie on a circle with centre a couple of

пЈ« one particular пЈ¶

in пЈ¬ в€’, 0 пЈ· and radius

.

2

пЈ­ 2 пЈё

From above, watts +

3

(i)

в‡’

1+ i actually

−2 ≥2

z

(ii)

1 + i в€’ 2z

≥2

z

In shape formulas in to any

one of the three standard

loci initially (circle, halfline and verticle with respect

bisector). Be aware that

coefficient of z must

always be " 1”.

Switch to Cartesian

form if they don't fit.

2z −1+ we ≤ one particular + my spouse and i

2

пЈ«1 1 пЈ¶

⇒ z . − − i ≤

2

пЈ­2 2 пЈё

⇒ 1+ i − 2z ≥ 2 unces

пЈ«1 one particular пЈ¶

∴ z − + i ≥ z .

пЈ­2 a couple of пЈё

Im(z)

( 12, 12 )

Re(z)

пЈ« в€’1 в€’1 пЈ¶

пЈ¬, пЈ·

пЈ­ 2 a couple of пЈё

a couple of

2

( 1 two, в€’1 2 )

The question already

alerted you this is a circle,

so you just search for

centre and radius. Direct

to solution for positionnement:

|z в€’ z0| sama dengan r

so centre can be z0 and

radius is r.

Label the mid-point for

the perpendicular

bisector as (1/4, 1/4).

Indicate the 90В° angle.

This circle goes

through the source.

Tell marker where the

Positionnement answer is!

2

several

(iii)

пЈ« пЈ« 1 1 пЈ¶пЈ¶

пЈ« пЈ« 1 1 пЈ¶пЈ¶

afeafef ( two z & 1 + i ) = arg 2 + arg пЈ¬ z в€’ пЈ¬ в€’ в€’ i actually пЈ· пЈ· = arg пЈ¬ unces в€’ пЈ¬ в€’ в€’ i пЈ· пЈ· пЈ­ пЈ­ two 2 пЈёпЈё

пЈ­ пЈ­ 2 a couple of пЈёпЈё

Therefore , в€’

some

ПЂ

four

≤ afeafef...


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