TUMSO11th

7/27/2019 TUMSO11th

http://slidepdf.com/reader/full/tumso11th 1/34

Solution to TUMSO11th

Part 1: Choice

Question 1

Translation: Let x ∈ R such that

2 + log 9 = log (32x) + 2log(3x − 1)

Determine 31−x

Solution By K

2 + log 9 = log(32x) + 2log(3x

−1)

log 100 + log 9 = log (32x) + log (3x − 1)2

log 900 = log 32x(3x − 1)2

900 = 32x(3x − 1)2

(3x(3x − 1))2 − 302 = 0

(3x(3x − 1) + 30)(3x(3x − 1)− 30) = 0

First term: 3x(3x − 1) + 30 = 0 → 32x − 3x + 30 = 0 → No real number satisfies this equation.

Second term: 3x(3x − 1)− 30 = 0 → 32x − 3x − 30 = 0 → (3x − 6)(3x + 5) = 0 → 3x = 6 → 3−x = 1

6

31−x = 3

×3−x =

1

2

= 0.5

Question 2

Translation: Let z1, zx ∈ C such that

|z1| = |z1 + z2| =|z1 − z2|√

3

Determine(|z1|+ |z2|)2|z1 + z2|2

Solution By K

Let z1 = a + bi,z2 = c + di

From |z1| = |z1 + z2|

√ a2 + b2 =

√ (a + c)2 + (b + d)2

a2 + b2 = a2 + 2ac + c2 + b2 + 2bd + d2

c2 + d2 = −2ac− 2bd

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And from |z1| =|z1 − z2|√

3

√ a2 + b2 =

(a − c)2 + (b− d)2

3

a2 + b2 =a2 + b2 − 2ac − 2bd + (c2 + d2)

3

a2 + b2 =a2 + b2 − 2ac − 2bd− 2ac− 2bd

3

a2 + b2 =a2 + b2 − 4ac − 4bd

3

3a2 + 3b2 = a2 + b2 − 4ac − 4bd

a2 + b2 = −2ac− 2bd

a2 + b2 = c2 + d2

|z1| = |z2|Finally,

(|z1|+ |z2|)2|z1 + z2|2 = (|z1| + |z2|)2

|z1 + z2|2

=4|z1|2|z1|2

= 4

Question 3

Translation: Let a,b,c,d be one-digit positive integer such that a ≤ b ≤ c ≤ d and

343a + 49b + 7c + d = 415

Determine 343d + 49c + 7b + a + 514

Solution By K

343a + 49b + 7c + d = 73a + 72b + 71c + 70d = 415

Observe that the given equation, it looks like the conversion from 7-base number into 10-base number. The

solution to this problem is, firstly try to convert 41510 into 7-base number. By long division, we will have

a = b = 1, and we know that c <= d. Hence,

343 + 49 + 7c + d = 415

7c + d = 23

c = 1 → d = 16

c = 2 → d = 9

c = 3 → d = 2

d = 16 because d is one-digit integer.

d = 2 because that would make c > d.

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Therefore, c = 2, d = 9

343d + 49c + 7b + a = 343(9) + 49(2) + 7(1) + 1 + 514 = 3707

Question 4

Translation: Determine k

∈R such that the system of equations will have answers other than

w = x = y = z = 0

w + x− y + kz = 0

w − x− 3y + 2z = 0

w + 2x− 2y − 4z = 0

w − 3x + 4y + 6z = 0

Solution By K

For this question, all you need to do is finding the value of k that make determinant equals to zero.

det

1 1 −1 k

1 −1 −3 2

1 2 −2 −4

1 −3 4 6

= det

1 1 −1 k

0 −2 −2 2 − k

0 1 1 −4− k

0 −4 5 6 − k

= det

−2 −2 2 − k

1 1 −4 − k

−4 5 6 − k

= 0

−54 − 27k = 0

k = −2

Question 5

Translation: How many statements below are correct?1. If A ⊆ A× B, then P (A) ⊆ P (A ∩B)

2. If A− B = A −C then B = C

3. There exists a set that does not have proper subset.

4. Subset of ϕ is subset of every set.

Solution By K

1. (Still cannot prove why it is correct.)

2. If A = B = ϕ, then the statement is wrong.

3. That set is ϕ, because subset of ϕ is ϕ.

4. Subset of ϕ is ϕ, and it is subset of every set.There are 3 correct statements.

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Question 6

Translation: Let ABC be a triangle such that

(1− tan A4

)(1 − tan B4

)(1 − tan C 4

)(tan A4

+ tan B4

+ tan C 4

) = 518

Evaluate tanA

4+ tan

B

4+ tan

C

4

Solution By K

Let x = (1 − tanA

4)(1 − tan

B

4)(1 − tan

C

4), from the question we will have

tanA

4+ tan

B

4+ tan

C

4=

5

18x

In a triangle ABC , we know that

A + B + C = π

A

4+

B

4+

C

4=

π

4

tan(A

4+

B

4+

C

4

)= 1

tan A4

+ tan B4

+ tan C 4− tan A

4tan B

4tan C

4

1 − tan A4

tan B4− tan B

4tan C

4− tan A

4tan C

4

= 1

tanA

4+ tan

B

4+ tan

C

4− tan

A

4tan

B

4tan

C

4= 1 − tan

A

4tan

B

4− tan

B

4tan

C

4− tan

A

4tan

C

4

tanA

4

+ tanB

4

+ tanC

4

= 1

−tan

A

4

tanB

4 −tan

B

4

tanC

4 −tan

A

4

tanC

4

+ tanA

4

tanB

4

tanC

4

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After this, we will try to rearrange the term in RHS into x, we will have

tanA

4+ tan

B

4+ tan

C

4= 2− x− (tan

A

4+ tan

B

4+ tan

C

4)

tanA

4+ tan

B

4+ tan

C

4=

2− x

25

18x=

2− x

2

9x2 − 18x + 5 = 0

x =5

3,

1

3

We automatically know that x < 1, because A4

< π4→ tan A

4< 1.

tanA

4+ tan

B

4+ tan

C

4=

5

18(13

)=

5

6

Question 7

Translation: Let a,b,c,d be different real numbers, and they are the roots of this equation,

x4 − 12x3 + 21x2 + 76x + 24 = 0

Let P (x) be a 4th-polynomial such that

P (a) = b + c + d + bc + cd + db + bcd

P (b) = a + c + d + ac + cd + ad + acd

P (c) = a + b + d + ab + bd + ad + abd

P (d) = a + b + c + ab + bc + ac + abc

P (P (a) + P (b) + P (c) + P (d)) = abc + bcd + abd + acd

Determine P (a + b + c + d)

Solution By V

We automatically know that,

a + b + c + d = 12

ab + ac + ad + bc + bd + cd = 21

abc + bcd + abd + acd = −76

abcd = 24

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Consider first equation,

P (a) = b + c + d + bc + cd + db + bcd

= (b + c + d) + (bc + cd + db) + bcd

= (12− a) + (21 − ab− ac− ad) + (−76 − abc− abd− acd)

= 12

−a + 21

−a(b + c + d)

−76

−a(bc + cd + db)

= 12 − a + 21 − a(12 − a)− 76− a(21 − ab− ac− ad)

= −43− a − 12a + a2 − 21a + a2(b + c + d)

= −43− 34a + a2 + a2(12 − a)

P (a) = −43− 34a + 13a2 − a3

Because of the symmetry of first four equations, we also have

P (b) = −43− 34b + 13b2 − b3

P (c) = −43− 34c + 13c2 − c3

P (d) =

−43

−34d + 13d2

−d3

Thus,

P (x) = k(x− a)(x− b)(x− c)(x− d)− x3 + 13x2 − 34x− 43

P (x) = k(x4 − 12x3 + 21x2 + 76x + 24)− x3 + 13x2 − 34x− 43

We know that

P (a) + P (b) + P (c) + P (d) = 3(a + b + c + d) + 2(ab + ac + ad + bc + bd + cd) + (abc + bcd + acd + abd)

P (a) + P (b) + P (c) + P (d) = 3(12) + 2(21) + (−76)

P (a) + P (b) + P (c) + P (d) = 2

P (P (a) + P (b) + P (c) + P (d)) = abc + bcd + acd + abdP (2) = −76

Substitute x = 2 into P (x) that we has just derived,

−76 = k(24 − 12(2)3 + 21(2)2 + 76(2) + 24) − 67

k = − 1

20

P (x) = (− 1

20)(x4 − 12x3 + 21x2 + 76x + 24)− x3 + 13x2 − 34x− 43

P (a + b + c + d) = P (12) = −505

Question 8

Translation: Let ABC be an acute angle such that

cotA

2+ cot

B

2+ cot

C

2= 6.75

sin2A + sin 2B + sin 2C = 1.3824

Determine cos A + cos B + cos C

Solution By K

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Proof: tanA

2tan

B

2+ tan

B

2tan

C

2+ tan

A

2tan

C

2= 1

A + B + C = π

A

2+

B

2+

C

2=

π

2

tan( A2

+ B2

+ C 2

) = tan π2→∞

tan A2

+ tan B2

+ tan C 2− tan A

2tan B

2tan C

2

1 − tan A2

tan B2− tan B

2tan C

2− tan A

2tan C

2

→∞

1− tanA

2tan

B

2− tan

B

2tan

C

2− tan

A

2tan

C

2= 0

tanA

2tan

B

2+ tan

B

2tan

C

2+ tan

A

2tan

C

2= 1

Proof: sin2A + sin 2B + sin 2C = 4 sin A sin B sin C

sin2A + sin 2B + sin 2C = (sin2A + sin 2B) + sin 2C

= 2sin(A + B)cos(A −B) + sin 2C

= 2sin(π − C )cos(A− B) + 2 sin C cos C

= 2 sin C cos(A −B) + 2 sin C cos C

= 2 sin C (cos(A− B) + cos C )

= 2 sin C (2cos(A −B + C

2)cos(

A− B − C

2))

= 4 sin C cos(π − 2B

2)cos(

π − 2A

2)

sin2A + sin 2B + sin 2C = 4 sin A sin B sin C

Proof: cos A + cos B + cos C = 1 + 4 sinA

2 sinB

2 sinC

2

cos A + cos B = 2cos (A + B

2)cos(

A− B

2) = 2 sin

C

2cos(

A −B

2)

1− cos C = 2 sin2C

2= 2 sin

C

2sin

C

2= 2 sin

C

2cos(

A + B

2)

(cos A + cos B) − (1 − cos C ) = 2 sinC

2(cos(

A− B

2)− cos(

A + B

2))

(cos A + cos B) − (1 − cos C ) = 4 sinA

2sin

B

2sin

C

2

cos A + cos B + cos C = 1 + 4 sinA

2sin

B

2sin

C

2

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From the first equation,

cotA

2+ cot

B

2+ cot

C

2= 6.75 =

27

41

tan A2

+1

tan B2

+1

tan C 2

=27

4

tan A2

tan B2

+ tan B2

tan C 2

+ tan A2

tan C 2

tan A2

tan B2

tan C 2

=

27

4

1

tan A2

tan B2

tan C 2

=27

4

tanA

2tan

B

2tan

C

2=

4

27

From the second equation,

sin2A + sin 2B + sin 2C = 1.3824 =864

625

4sin A sin B sin C =864

625

sin A sin B sin C = 216625( 2tan A

2

1 + tan2 A2

)( 2tan B2

1 + tan2 B2

)( 2tan C 2

1 + tan2 C 2

)=

216

625

tan A2

tan B2

tan C 2

(1 + tan2 A2

)(1 + tan2 B2

)(1 + tan2 C 2

)=

27

625

4

27

(1 + tan2 A2

)(1 + tan2 B2

)(1 + tan2 C 2

)=

27

625

1

sec2 A2

sec2 B2

sec2 C 2

=272

502

cos2 A2

cos2 B2

cos2 C 2

= 272

502

Because it is an acute angle, all angle are less than 90◦. All of the cosine value is positive.

cosA

2cos

B

2cos

C

2=

27

50

tanA

2tan

B

2tan

C

2=

4

27

sin A2

sin B2

sin C 2

cos A2

cos B2

cos C 2

=4

27

sin A2 sin B2 sin C 2 = 225

cos A + cos B + cos C = 1 + 4 sinA

2sin

B

2sin

C

2

= 1 +8

25

= 1.32

Question 9

Translation: Let Z be a set such that

Z ={

b2c2

a2(a− b)(a− c)+

a2c2

b2(b− a)(b− c)+

b2a2

c2(c− a)(c− b) |a,b,c

∈R+, a

= b, b

= c, c

= a

}Determine the least value in Z

Solution By K

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After being trouble in rearranging the term, we will have

b2c2

a2(a− b)(a − c)+

a2c2

b2(b− a)(b− c)+

b2a2

c2(c − a)(c− b)= 1 +

(ab + ac + bc)(a2b2 + a2c2 + b2c2)

a2b2c2

= 1 + (a

b+

b

a+

c

b+

b

c+

a

c+

c

a) + (

ab

c2+

bc

a2+

ac

b2)

Second term is always greater than or equal to 6. Minimum of third term can be determined by A.M.-G.M.

inequality. Finally,b2c2

a2(a − b)(a − c)+

a2c2

b2(b− a)(b− c)+

b2a2

c2(c− a)(c − b)≥ 10

Question 10

Translation: Let u,v,w be vectors such that u · (v ×w) = v · (u× w) and

(u · v)2|w|2 + (v · w)2|u|2 + (w · u)2|v|2 = |u|2|v|2|w|2 + (u · v)(v · w)(w · u)

Let |u| = 1, |v| = 2, |w| = 3, w · u = 2, w · v < 0

Determine (4v − w) · ((u× v)× (v + 3w))

Solution By K

In this question, we need to know that a × (b × c) = (a · c)b − (a · b)c

Firstly, let’s consider what is asked by the question.

(4v −w) · ((u× v)× (v + 3w)) = (4v − w) · ((u× v)× v + 3(u× v)× w)

= 4v · ((u× v)× v)− w · ((u× v)× v) + 12v · ((u× v)× w)− 3w · (u× v)×w)

= 0 −w · ((u× v)× v) + 12v · ((u× v)×w)− 0

= 12v · ((u× v)× w)−w · ((u× v)× v)

= 12v · ((u · w)v − (w · v)u)−w · ((u · v)v − (v · v)u)

= 13(8 − (w · v)(u · v))

From basic vector knowledge, we know that u · (v ×w) = u · (v × w) = v · (w × u)

But from the question, there is something different. Observe these two equations (First is from basic vector

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knowledge, second is from the question)

u · (v × w) = v · (w × u)

u · (v × w) = v · (u×w)

Thus,

v · (w × u) = v · (u×w) = 0

From the given equation, it can be simplified into this,

9(u · v)2 − 2(u · v)(w · v) + (w · v)2 − 20 = 0

It is concluded that these three vectors are all in the same plane. Suppose the plane is on Cartesian coordinate.

This is where the complication comes from, we do not know how these three vectors lie on the plane. All we

know is that, the angle between each other must satisfy the above equation. Thus, we can rotate the orientation

of these three vectors as much as we want. Let w = 3i

With our assumption, we can determine u as follow,

u = 23

i ± √ 53

j

Let’s substitute all the value into above equation,

v = v1i + v2 j → v21 + v22 = 4

v · w < 0 → v1 < 0

9(u · v)2 − 2(u · v)(w · v) + (w · v)2 − 20 = 0

9(2

3v1 ±

√ 5

3v2)2 − 2(

2

3v1 ±

√ 5

3v2)(3v1) + 9v21 − 20 = 0

8v21 − 4v1 ∓ 2√

5v2 ± 4√

5v1v2 = 0

(4v1 ± 2√ 5v2)(2v1 − 1) = 0

2v1 ±√

5v2 = 0

13(8− (w · v)(u · v)) = 13(8− (2v21 ±√

5v1v2))

= 13(8− v1(2v1 ±√

5v2))

= 13× 8

= 104

Question 11Translation: Determine if the following statements are correct or not. Let Q′ = R −Q

1. ∀q ∈ Q∃x ∈ Q′∃y ∈ Q′[q = xy]

2. For a, b ∈ R+, if 3a + 13b = 17a and 5a + 7b = 11b, then a < b

3. For x,y,z ∈ R+, if xyz(x + y + z) = 1 then (x2 +

1

y2)(y2 +

1

z2)(z2 +

1

x2) = (x + y)(y + z)(z + x)

Solution By K,C

1. False, when q = 0

2. Roughly sketch a graph discretely, it can be seen easily that it is correct.

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3.

(x

2

+

1

y2 )(y2

+

1

z2 )(z2

+

1

x2 ) = (x2y2 + 1)(y2z2 + 1)(z2x2 + 1)

x2y2z2

x2y2 + 1 = x2y2 + xyz(x + y + z)

= xy(xy + z(x + y + z))

x2y2 + 1 = xy(x + z)(y + z)

y2z2 + 1 = yz(y + x)(z + x)

x2z2 + 1 = xz(x + y)(z + y) (x2 +

1

y2)(y2 +

1

z2)(z2 +

1

x2) =

(x2y2 + 1)(y2z2 + 1)(z2x2 + 1)

x2y2z2

=

x2

y2

z2

(x + z)2

(x + y)2

(y + z)2

x2y2z2

= (x + y)(y + z)(x + z)

Question 12

Translation: Determine the greatest integer that is less than

3

2× 6

5× 9

8× 12

11× ...× 2010

2009× 2013

2012

Solution By K

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(6

5× 9

8× .. × 2013

2012)3 =

6

5× 6

5× 6

5× 9

8× ...

>6

5× 7

6× 8

7× ... × 2014

2013× 2015

2014=

2015

3

> (22

3)3

6

5× 9

8× ..× 2013

2012>

22

33

2× 6

5× 9

8× ..× 2013

2012> 11

Let A1 = 6

5× 9

8× 12

11× ...× 2010

2009× 2013

2012

A2 = 9

8× 12

11× ... × 2010

2009× 2013

2012

and so on.

Let B1 = 5

4× 8

7× 11

10× ... × 2009

2008× 2012

2011

B2 = 8

7× 11

10× ...× 2009

2008× 2012

2011

and so on.

Let C 1 =4

3 ×7

6 ×10

9 × ... ×2008

2007 ×2011

2010

C 2 = 7

6× 10

9× ... × 2008

2007× 2011

2010

and so on.

It can be seen that An < Bn < C n

A31 < A1B1C 1 = 671

3

2A1 < 13.....

3

2× 6

5A2 < 12...

As n increases, LHS will be narrowed down to 12. (All of the cube root is approximated according to the closest

integer to get LHS value.) Hence,

11 ≤ 3

2× 6

5× 9

8× .. × 2013

2012≤ 12

Question 13

Translation: I will not translate this. However, this question can be concluded into “Find the ratio between

the tautology to the number of the statement while the statements are”

( p ∧ q ) → ( p → q )

( p ∧ ( p → q )) → q

(q ∧ ( p → q )) → p

Solution By K

Tautology can be checked by assume such statement is false. if the last result is conflicted, therefore it is

tautology. First statement: ( p ∧ q ) → ( p → q )

( p ∧ q ) → ( p → q ) ≡ F

∴ p → q ≡ F

p ≡ T

q ≡ F p ∧ q ≡ F

F → F ≡ F

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∴The first statement is tautology.

Second statement: ( p ∧ ( p → q )) → q

( p ∧ ( p → q )) → q ≡ F

∴ q ≡ F

( p ∧ ( p → q )) ≡ T

p ≡ T ( p → q ) ≡ T

∴The second statement is tautology.

Third statement: (q ∧ ( p → q )) → p

(q ∧ ( p → q )) → p ≡ F

∴ p ≡ F

(q ∧ ( p → q )) ≡ T

q ≡ T

( p

→q )

≡T

∴The third statement is not tautology.

The ratio is2

3

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Question 14-15

Translation: Arrow Sudoku has one another rule more than simple Sudoku. That is the number in the circle

has to be equal to the sum of the number that arrow passes through. Determine the value of X

Solution By K

Please don’t be panic. Arrow Sudoku will narrow down the possibility of numbers in the square. Observe the

longest arrow at the bottom, it can be seen that the circle in the bottom has to be 9 only, and the number is

1, 2, 1, 2, 3. Value of X can be found by trying to substitute 1 and 2 into the box. With Sudoku skill, we willget X = 4. And with Sudoku skill, we will get the result like this. The summation is 81

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7 9 5 1 6 3 2 8 4

4 6 2 5 9 8 3 1 7

1 3 8 7 2 4 5 6 9

3 2 4 9 7 6 8 5 1

5 8 1 3 4 2 7 9 6

9 7 6 8 5 1 4 3 2

2 5 9 6 3 7 1 4 8

6 1 7 4 8 5 9 2 3

8 4 3 2 1 9 6 7 5

Part 2: Short Answer(Less Marks)

Question 1

Translation: The relation between variable x and y can be expressed in the table. Linear regressions for x

and y can be made from this data. If we denote linear regression for y as L1 and for x as L2, where do these

two lines intersect?

Solution By K

Linear regression for y and x always intersect at (mx, my), where mx is mean of the data x and my is mean of

the data y. Therefore, they intersect at

(1 + 2 + 3 + 4 + 6 + 7 + 8 + 9 + 11 + 12

10,

4 + 6 + 9 + 13 + 15 + 16 + 18 + 22 + 24 + 27

10) = (6.3, 15.4)

Question 2

Translation: Let A, B ∈ [0, 2π), A = B are the roots of

cos x + 4 sin x = 3

Determine cos2 (A− B

2)

Solution By K

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cos x + 4 sin x = 3

1√ 17

cos x +4√ 17

sin x =3√ 17

Let θ = arccos 1√ 17

,

cos(x− θ) = 3√ 17

x = arccos 1√ 17

+ arccos 3√ 17

, arccos 1√ 17− arccos 3√

17

A −B = 2 arccos (3√ 17

)

cos2 (A− B

2) =

9

17

Question 3

Translation: Find the number of integers (x,y,z), that will have at least 3 possible sets of (a,b,c,d,e,f )

satisfy these conditions.

{a,b,c,d,e,f } = {2, 4, 6, 8, 10, 12}x y z

a b c

d e f

= 26

Solution By K

Let {A,B,C,D,E,F } = {1, 2, 3, 4, 5, 6}, we will havex y z

a b c

d e f

= 26

2

x y z

A B C

D E F

= 13

,which is not possible. The number of (x,y,z) is 0.

Question 4

Translation: There are 4 kinds of flowers. There are 2 pink flowers, 3 gold flowers, 4 orange flowers, and 5

orange-gold flowers. A girl wants to pick at least 4 flowers to decorate a crown. if the probability that a girl

will get all kinds of flowers ism

n, where m and n are positive integer, and great common divisor of m and n is

not equal to 1. Determine the least possible value of |m− n|Solution is needed

Question 5

Translation: Find the number of complex number pairs (w, z) that satisfy these equations.

w2 + z2 = w4 + z4 = w8 + z8

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Solution is needed

Question 6

Translation: To draw that picture, how many times do we need to stop writing and start writing at anotherpoints? (We can draw a line pass through one point several times, but we cannot draw a line pass through a

same line.)

Solution is needed

Question 7

Translation: Determine a positive real number c that satisfies these condition.

1. For any positive real number k less than c, kx5 − x + 1 = 0 has three distinct real roots.

2. For any positive real number k greater than c, kx5 − x + 1 = 0 has one real root.

Solution by K

Let f (x) = kx5 − x + 1, then we will determine two critical points.

f ′(x) = 5kx4

−1

xc1 = − 14√

5k→ f (xc1) = −k(

14√

5k)5 +

14√

5k+ 1 =

4k + (5k)54

4√

5k> 0

, xc2 =1

4√

5k→ f (xc2) = k(

14√

5k)5 − 1

4√

5k+ 1 =

−4k + (5k)54

4√

5k=

k4√

5k(−4 + (5k)

54 )

Then, we try to substitute random number from these three intervals (−∞, xc1), (xc1, xc2), (xc2,∞) to check

the trend of slope in each interval.

x < xc1 → f ′(x) > 0

xc1 < x < xc2 → f ′(x) < 0

x > xc2 → f ′(x) > 0

By these informations, if we sketch the graph, we will know that the number of roots depend on f (xc2). To

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have only one real root, f (xc2) > 0. Otherwise, f (x) will have 3 distinct real roots.

f (xc2) < 0

k4√

5k(−4 + (5k)

54 ) < 0

−4 + (5k)54 < 0

k <256

3125

c = 2563125

Question 8

Translation: Let a,b,c,d,e be complex numbers such that |a| + |b| + |c| + |d| + |e| = 1 + |a + b + c + d + e|.Determine the maximum value of

|a + b| + |a + c| + |a + d|+ |a + e| + |b + c|+ |b + d|+ |b + e|+ |c + d| + |c + e| + |d + e| − 4|a + b + c + d + e|

Solution is needed

Question 9

Translation: Let A = {1, 2, 3, 4, 5}. Define “Good Function” is a function that f : A → A and

1, f (1), f (f (1)), f (f (f (1))), f (f (f (f (1)))) have all different values. Find a number of (f, g) such that f,g,and f og are all Good Function

Solution is needed

Question 10

Translation: Three people decide to meet each other in the afternoon. Each one can show up in any time

between 12 : 00 and 13 : 00 and will wait for the others only 15 minutes. Determine the probability that 3

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people will meet each other.

Solution by K

Firstly, let consider this question in discrete case. We divide 1 hour into 60 minutes, therefore, a man can show

up any time he wants from 12 : 00 to 13 : 00 (61 choices).

- First person arrives at 12 : 00. If second person arrives at 12 : 00, that will give last person 16 choices.

- First person arrives at 12 : 00. If second person arrives at 12 : 01, that will give last person 15 choices....



...

From 12 : 00 to 12 : 45, the permutation gives (3

4× 60 + 1)(

1

4× 60 + 1

2)(

1

4× 60 + 1 + 1)






From 12 : 46 to 13 : 00, the permutation gives

i= 14

×60

i=1

i2

(i + 1)

For continuous case, let’s divide 1 hour into x intervals, the permutation gives a total result of

(3x

4+ 1)(

x4

+ 1

2)(

x

4+ 1 + 1) +

i=x

4i=1

i

2(i + 1) =

5

192x3 +

66

192x2 +

132

192x + 1

The probability that three man will meet is

P = limx→∞

3!(5

192x3

+66

192x2

+132

192x + 1)(x + 1)3

= 532

Part 3: Short Answer(More Marks)

Question 1

Translation: From the result of one room of students’ examination, 73 and 4 are mean and standard devi-

ation of male students’ scores, respectively. Female students have a mean at 69 and standard deviation at 2.

Determine the maximum possible value of total students’ variance.

Solution By K

Let m1 be a mean of male students’ score. m1 = 73

m2 be a mean of female students’ score. m2 = 69m be a mean of all students’ score.

M i be a score of male student.

F i be a score of female student.

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xi be a score of a student.

σ1 be a standard deviation of male students’ score. σ1 = 4

σ2 be a standard deviation of female students’ score. σ2 = 2

From the standard deviation formula, we know that

σ =

∑N i=1(xi −m)2

N

N σ2 =

N i=1

(xi −m)2

N σ2 =N i=1

(x2i − 2mxi + m2)

N σ2 =N i=1

x2i − 2mN i=1

xi + N m2

N σ2 =

N i=1

x2i − 2m(N m) + Nm2

N σ2 =

N i=1

x2i −N m2

16N 1 =

N 1i=1

M 2i − N 1(73

2

)

4N 2 =

N 2i=1

F 2i −N 2(692)

N i=1

x2i = 16N 1 + 4N 2 + 732N 1 + 692N 2

m =N 1m1 + N 2m2

N 11 + N 2

Now we will try to find the maximum variance of students’ score.

σ2 =

N i=1

x2i − (N 1 + N 2)(73N 1 + 69N 2

N 1 + N 2)2

= 16N 1 + 4N 2 + 732N 1 + 692N 2 − (N 1m1 + N 2m2)2

N 1 + N 2

σ2 =16N 21 + 36N 1N 2 + 4N 22

(N 1 + N 2)2

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Let N 2 = kN 1, k ∈ R+

σ2 =16k2 + 36k + 4

(k + 1)2

σ2 = 16 +4k − 12

(k + 1)2

dσ2

dk =

28

−4k

(k + 1)3 = 0

k = 7

σ2max = 16 +

4× 7− 12

82

= 16.25

Question 2

Translation: A fly is on the top floor of KingThong building. It looks up to the top of BaiYok building. It

finds out that the ground floor of KingThong building is 180m far away from the top of KingThong building.

If it looks to the top of the BaiYok building while it is on the ground floor of KingThong building, it will have

to look up at double angle from the angle it uses at the top of KingThong building. A spider is on the groundfar away from KingThong building and BaiYok building, 135m and 60m , respectively. The angle it needs to

look up to the top of BaiYok building is twice to KingThong. What is the distance between these two building?

Solution By O

According to the picture,

(a) A fly’s perspective (b) A spider’s perspective

Let H be a height of BaiYok building.

h be a height of KingThong building.

S be the distance between these two building.

θ be an angle that a fly look up.

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α be an angle that a spider look up.

H − h = S tan θ (1)

H = S tan2θ (2)

h = 135 tan α (3)

H = 60tan 2α = 120tan α1− tan2 α

(4)

H = 180 sin 2θ =360 tan θ

1 + tan2 θ(5)

(1) / (2);

H − h

H =

tan θ

tan2θ=

1− tan2 θ

2H − 135 tan α

H =

1− tan2 θ

2

H =270 tan α

1 + tan2 θ360 tan θ

1 + tan2 θ=

270 tan α

1 + tan2 θ

tan θ =3

4tan α

(4) = (5);

120 tan α

1 − tan2 α=

360 tan θ

1 + tan2 θ

120 tan α

1 − tan2 α=

360(34

tan α)

1 + ( 34

tan α)2

tan α =

2

3

tan θ =1

2

H = 144

h = 90

S =H − h

tan θ= 108

Question 3

Translation: A triangle ABC has AB = 33, BC = 15, CA = 20. Let A1, A2,...A32, B be a point on AB such

that

AA1 : A1A2 : A2A3 : · · · : A32B = 1 : 2 : 3 : · · · : 33

For i = 0, 1, ..., 33, let A0 = A, A33 = B, and ri be a radius of inscribed circle of △Ai−1AiC , Ri be a radius of

extended circle of △Ai−1AiC opposite to C . Determine

R1R2R3...R33

r1r2r3...r33

Solution By C

Consider △Ai−1AiC

tan θ = riXAi−1

tan α =Ri

W Ai

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(c) Triangle ABC (d) Triangle Ai−1AiC

From XAi−1 = W Ai → tan θ

tan α=

riRi

,

tan(C Âi−1Ai

2)

cot(C AiAi−1

2)

=riRi

riRi

= tan (C Âi−1Ai

2)tan(

C AiAi−12

)

Because tan (

C AiAi

−1

2 ) =

1

tan(C AiAi+1

2)

riRi

=tan(C

Âi−1Ai

2)

tan(C AiAi+1

2)

r1R1

× r2R2

× ... × r32R32

=tan(C A0A1

2)

tan(C A1A2

2)× tan(C A1A2

2)

tan(C A2A3

2)× ... × tan(C A31A32

2)

tan(C A32A33

2)

=tan(C AB

2)

tan(C A32A33

2)

r33

R33

= tan (C A32A33

2)tan(

C A33A32

2)

r1R1

× r2R2

× ... × r33R33

= tan (C AB

2)tan(

C BA

2)

s =a + b + c

2= 34

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tanA

2tan

B

2=

r2

(s− a)(s− b)

=s− c

sr1R1

× r2R2

× ...× r33R33

=1

34R1

r1 ×R2

r2 × ...×R33

r33 = 34

Question 4

Translation: A bijective function W : [−1

e,∞) → [1,∞) is defined as follow,

W = {(x, y)|x = yey}

Determine the negative real root of 2x = x2 in form of W

Solution By C

2x = x2

x2 × 2−x = 1

Let x = −k, k ∈ R

k2 × 2k = 1

k × 2k

2 = 1

k × ek

2ln 2 = 1

k

2ln 2 × e

k

2ln 2 =

ln 2

2

W (ln 2

2) =

k

2ln 2

k =2W ( ln 2

2)

ln 2

x = −2W ( ln 22

)

ln 2

Question 5

Translation: Let z be a complex number that satisfies these inequalities,

|z − 2i| + |z + 2 − 2i| ≥ 4

|z + 1 − 2i| ≤ 3

||z + 1

| − |z + 1

−4i|| ≤

2

Find a product of maximum and minimum of this value

(ℜ(z) + 1)2 + 2(ℑ(z)− 2)2

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Solution By O

Let z = x + yi, and rearrange the term in all inequalities. From the first inequality, it is an ellipse.

√ x2 + (y − 2)2 +

√ (x + 2)2 + (y − 2)2 ≥ 4 → (x + 1)2

22+

(y − 2)2

(√

3)2≥ 1

Second inequality, we will have a circle.

√ (x + 1)2 + (y

−2)2

≤3

→(x + 1)2 + (y

−2)2

≤9

From the last inequality, we will have a hyperbola.

|√

(x + 1)2 + y2 −√

(x + 1)2 + (y − 4)2| ≤ 2 → (y − 2)2

12− (x + 1)2

(√

3)2≤ 1

Let k = (ℜ(z) + 1)2 + 2(ℑ(z) − 2)2

(ℜ(z) + 1)2 + 2(ℑ(z)− 2)2 = k

(x + 1)2 + 2(y − 2)2 = k

(x + 1)2

(√

k)2+

(y − 2)2

( k2 )2= 1

Thus, we need to find a biggest ellipse and smallest ellipse that fit through given area. The smallest k is 4. The

biggest k can be found by substitute a point that a circle and a hyperbola intersects. It is 12. The product will

be 48.

Question 6

Translation: A stage in Angry Bird is to shoot a bird in projectile motion to destroy Green Pig. The leftmost

side is a bird. There is a 2m height obstacle to the right of a bird. There is an another obstacle behind the

first one, 5m far. This one swags from the ceiling. The tip of it is at 3m height from the ground. A pig is to

the right of second obstacle. A bird is shot from the ground at 45◦ with an uninitialized velocity. Evaluate thefarthest distance from first obstacle that a pig can be defeated by a bird. Let g = 10m/s2

Solution By K

Let S x1 be a horizontal position at point 1.

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Let S x2 be a horizontal position at point 2.

Let S y1 be a vertical position at point 1.

Let S y2 be a vertical position at point 2.

Let S xf be a horizontal position at point f .

Let S yf be a vertical position at point f .

Let t1 be a time that a bird reaches point 1.

Let t2 be a time that a bird reaches point 2.

Let tf be a time that a bird reaches a pig.u cos θ = u sin θ = u√

2

S yf = 0

u√ 2

tf − 5t2f = 0

tf =u

5√

2

S x1 = d1 =u√

2t1

S x2 = d1 + 5 =u√ 2 t2

u =5√

2

t2 − t1

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S xf = d1 + d2 + 5

u√ 2

tf = d1 + d2 + 5

u√ 2

(u

5√

2) = d1 + d2 + 5

d2 + 5 =u2

10− d1

d2 + 5 =5

(t2 − t1)2− u√

2t1

d2 + 5 =5

(t2 − t1)2− 5t1

t2 − t1

=5− 5t1t2 + 5t21

(t2 − t1)2

Thus, we need to maximize this equation.5− 5t1t2 + 5t21

(t2 − t1)2

From the point 1 and 2 in the figure, we can say that

S y1 =ut1√

2− 5t21 ≥ 2

ut1√ 2− 5t21 ≥ 2

3 ≥ S y2 =ut2√

2− 5t22

3 ≥ ut2√ 2− 5t22

ut1√ 2− 5t21 + 3 ≥ 2 +

ut2√ 2− 5t22

After we substitute the value of u, we will get this inequality,

t21 +4

5≤ t22

We are going to stop here, and derive another relationship, starting from,

S y1 =ut1√

2−21 ≥ 2

ut1√ 2− 5t21 ≥ 2

5t1t2 − t1

− 5t21 ≥ 2

t2 ≤7t1 + 5t3

12 + 5t21

t22 ≤ (7t1 + 5t31

2 + 5t21)2

t21 +4

5≤ t22 ≤ (

7t1 + 5t312 + 5t21

)2

t21 +4

5≤ (

7t1 + 5t312 + 5t21

)2

(5t21 + 4)(2 + 5t21)2 ≤ 5(7t1 + 5t31)2

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To solve this inequality, let A = 5t21 + 2 → t21 =A − 2

5

(5t21 + 4)(2 + 5t21)2 ≤ 5(7t1 + 5t31)2

(A + 2)(A)2 ≤ 5t21(A + 5)2

A3 + 2A2 ≤ (A− 2)(A + 5)2

6A2

+ 5A− 50 ≥ 0(2A − 5)(3A + 10) ≥ 0

(10t21 − 1)(15t21 + 16) ≥ 0

t1 ≥ 1√ 10

Because t22 ≥ t21 +4

5→ t2 ≥ 3√

10.

t2 − t1 ≥ 2√ 10

(t2 − t1)2

≥4

101

(t2 − t1)2≤ 5

2

t1t2 ≥ 3

10

−5t1t2 ≤ −3

2

−5t1t2 + 5 + 5t21 ≤ 4

5− 5t1t2 + 5t21(t2

−t1)2

≤ 10

Question 7

Translation: Let A,B,C,D,E,...,Y,Z,ψ are points on the coordinate. ABC is a triangle (counter-clockwise

direction) such that cos A : cos B : cos C = 25 : 33 : 39 and ADEB,BFGC,CHIA,EKLF,GMNH,IOJD,JPQK,LRSM

, N T U O , Q W X R , S Y Z T , U ψV P are squares (counter-clockwise sorting). Determine

[P V W Q] + [RXY S ] + TZψU

[ABC ]

Solution By O

When we try to draw such figure from the question, we will get something similar to below figure, Let −→k be aunit vector perpendicular to △ABC plane, and point out from the paper, and

−−→BC = −→a ,

−→CA =

−→b ,−−→AB = −→c

∴−→a +

−→b +−→c =

−→0

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∴−−→BF =

−−→CG = −→a ×−→k

−−→CH =

−→AI =

−→b ×−→k

−−→AD =

−−→BE = −→c ×−→k

∴−−→EF =

−−→KL =

−−→EB +

−−→BF = (−→a −−→c )×−→k

−−→GH =

−−→M N =

−−→GC +

−−→CH = (

−→b −−→a )×−→k

−→ID =

−→OJ =

−→IA +

−−→AD = (−→c −−→b )×−→k

∴

−−→EK =

−→F L =

−−→EF ×−→k = −→c −−→a

−−→GM =

−−→HN =

−−→GH ×−→k = −→a −−→b

−→IO =

−→DJ =

−→ID ×−→k =

−→b −−→c

∴−−→LM =

−→RS =

−→LF +

−−→F G +

−−→GM = 3−→a −−→b −−→c = 4−→a

−−→N O =

−→T U =

−−→N H +

−→HI +

−→IO = 3

−→b −−→c −−→a = 4

−→b

−−→JK =

−−→P Q =

−→JD +

−−→DE +

−−→EK = 3−→c −−→a −−→b = 4−→c

∴−→LR =

−−→M S =

−−→LM ×−→k = 4−→a ×−→k

−−→N T =

−−→OU =

−−→N O ×−→k = 4

−→b ×−→k

−→J P =

−−→KQ =

−−→JK ×−→k = 4−→c ×−→k

∴ −−→QR = −−→QK +−−→KL +−→LR = 5(−→a −−→c ) ×−→k−→ST =

−−→SM +

−−→MN +

−−→N T = 5(

−→b −−→a ) ×−→k

−−→U P =

−−→UO +

−→OJ +

−→JP = 5(−→c −−→b ) ×−→k

∴

−−→QW =

−−→RX =

−−→QR ×−→k = 5(−→c −−→a )

−→SY =

−→T Z =

−→ST ×−→k = 5(−→a −−→b )

−→Uψ =

−−→P V =

−−→U P ×−→k = 5(

−→b −−→c )

∴ −−→XY = −−→XR +−→RS + −→SY = 19−→a−→Zψ =

−→ZT +

−→T U +

−→U ψ = 19

−→b

−−→V W =

−−→V P +

−−→P Q +

−−→QW = 19−→c

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[RXYS ] =1

2|−−→RX ×−→RS |+

1

2|−−→Y X ×−→Y S |

=20

2|−→a ×−→c | +

95

2|−→a ×−→b |

= 20[ABC ] + 95[ABC ]

= 115[ABC ]

In the same manner, [P V W Q] = 115[ABC ], [TZψU ] = 115[ABC ][P V W Q] + [RXY S ] + TZψU

[ABC ]= 345

Question 8

Translation: Let F be a focal point on the left and V be a right vertices a hyperbola 3x2 − y2 = 3. If F

and E be points on the left side of hyperbola that makes F I V E be a circumscribable square and [F I V E ] = 99.

Determine F I + F E

Solution By O

From the hyperbolic equation we will have a = 1, b =√

3, c = 2. Let’s consider the relation between each

variable from △F IV and △F ′IV ,

cos α =−x2 + m2 + 32

2m(3)

cos(π − α) = − cos α =−(x + 2)2 + m2 + 12

2m(1)

−x2 + m2 + 32

2m(3)= −(

−(x + 2)2 + m2 + 12

2m(1))

m =√

x2 + 3x

cos α =2√

x + 3

2√

x

sin α =

√ 3√

x− 1

2√

x

By doing the same method, from △F EV and △F ′EV ,

n =

√ y2 + 3y

cos β = 2√ y + 32√

y

sin β =

√ 3√

y − 1

2√

y

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Apply sine’s rule to △F IV and △F EV ,

sin α

x=

sin β

y√ 3√

x− 1

2x√

x=

√ 3√

y − 1

2y√

y

(x−

y)(x2 + xy + y2) = xy(x−

y)(x + y)

We will get two equations from the above equation.

(y − 1)x2 + (y2 − y)x− y2 = 0

(x− 1)y2 + (x2 − x)y − x2 = 0

From the first equation, try to solve x in term of y,

x =−(y2 − y) +

√ (y2 − y)2 + 4y2(y − 1)

2(y − 1)

2x + y =y√

y + 3

√ y − 1(2x + y)(y − 1)

y=√

y + 3√

y − 1

Apply the same method to second equation also,

(2y + x)(x− 1)

x=√

x + 3√

x− 1

[F I V E ] = [F V I ] + [F V E ]

99 =1

2× 3 ×m sin α +

1

2× 3 × n sin β

44√ 3 = √ x + 3√ x− 1 +√

y + 3√

y − 1

44√

3 =(2y + x)(x− 1)

x+

(2x + y)(y − 1)

y

x + y = 44√

3

Question 9

Translation: Define f : R → R be a differentiable function such that f (1) = 1 and

f ′(x)

f (x)=

x

x2 + 1+

1

x3 + x− 8(f (x))3

Determine f (−1)

Solution By O

From the question, it can be concluded that f (x) is a continuous function, and f (x) > 0

∀x

∈R

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Let u = x2 + 1, dudx

= 2x

f ′(x)

f (x)=

x

u+

1

xu− 8(f (x))3

u(xu− 8(f (x))3)f ′(x) = xf (x)(xu − 8(f (x))3) + uf (x)

u2xf ′(x)− 8uf ′(x)(f (x))3 = ux2f (x)− 8x(f (x))4 + uf (x)

uf (x)(x2 + 1) − 8x(f (x))4 − u2xf ′(x) + 8uf ′(x)(f (x))3 = 0

u2f (x)− 8x(f (x))4 − u2xf ′(x) + 8uf ′(x)(f (x))3 = 0

u2(f (x) − xf ′(x)) + 4(f (x))2(2uf (x)f ′(x)− 2x(f (x))2) = 0

f (x)− xf ′(x)

(f (x))2+ 4(

(2uf (x)f ′(x)− 2x(f (x))2)

u2) = 0

d

dx(

x

f (x)) + 4

d

dx(

f (x)

x2 + 1) = 0

x

f (x)+

4(f (x))2

x2 + 1= C

From f (1) = 1→

C = 3, then substitute x =−

1 to determine f (−

1)

−1

f (−1)+

4(f (−1))2

1 + 1= 3

2(f (−1))3 − 3f (−1)− 1 = 0

f (−1) = −1,−1−√ 3

2,−1 +

√ 3

2

f (x) > 0 → f (−1) =−1 +

√ 3

2

Question 10

Translation: Evaluate ∫ π4

π

12

ln(8cos3 x + 2cos(2x)sec x− 2√

3sin(3x)) dx

Solution By O

PROOF : sin 3v = 4 sin v sin(v + π3

)sin(π3

−v)

sin3v = sin v(3 − 4sin2 v)

= 4 sin v((

√ 3

2)2 − sin2 v)

= 4 sin v(sin2π

3− sin2 v)

= 4 sin v(sinπ

3− sin v)(sin

π

3+ sin v)

= 4 sin v sin(v +π

3)sin(

π

3− v)

32

7/27/2019 TUMSO11th


8cos3 x + 2cos(2x)sec x− 2√

3sin(3x) = 2cos 3x + 6 cos x +2cos2x sin x

sin2x− 2

√ 3sin3x

=2cos3x sin2x + 6 cos x sin2x + 2cos 2x sin x− 2

√ 3sin3x sin2x

sin2x

=sin5x− sin x + 3sin 3x + 3 sin x + sin 3x− sin x +

√ 3cos5x−√ 3cos x

sin2x

=

2sin(5x + π3

) + 2 sin (x

−π3

) + 2 sin 3x + 2sin 3x

sin2x

=4sin(4x + π

6)cos(x + π

6) + 4sin(2x− π

6)cos(x + π

6)

sin2x

=4cos(x + π

6)(2sin3x cos(x + π

6))

sin2x

=4sin3x cos2 (x + π

6)

sin x cos x

=4sin3x sin2 (π

3− x)

sin x sin(π2− x)

=2(2sin3x)(2sin(π

3− x))2

2sin x(2sin(π2

−x))

∫ π4

π

12

ln(8cos3 x + 2cos(2x)sec x− 2√

3sin(3x)) dx

=

∫ π4

π

12

ln 2dx +

∫ π4

π

12

ln(2sin3x)dx +

∫ π4

π

12

2ln(2sin(π

3− x))dx

−∫ π

4

π

12

ln(2sin x)dx−∫ π

4

π

12

ln(2sin(π

2− x))dx

=π

6ln2 +

1

3

∫ 3π4

π

4

ln(2sin u)du− 2

∫ π

12

π

4

ln(2sin u)du

−∫ π

4

π

12

ln(2sin u)du +∫ π

4

5π12

ln(2sin u)du

=π

6ln2 +

2

3

∫ π2

π

4

ln(2sin u)du +

∫ π4

π

12

ln(2sin u)du

−∫ 5π

12

π

4

ln(2sin u)du

=π

6ln2 +

∫ π2

5π12

ln(2sin u)du +

∫ π6

π

12

ln(2sin u)du

+ ∫ π

4

π

6

ln(2sin u)du − 1

3 ∫ π

2

π

4

ln(2sin u)du

First integral let u = v + π3

, third let u = π3− v, last let u = 3v

=π

6ln2 +

∫ π6

π

12

ln(2sin(v +π

3))du +

∫ π6

π

12

ln(2sin v)dv

+

∫ π6

π

12

ln(2sin(π

3− v))du−

∫ π6

π

12

ln(2sin3v)du

=π

6ln2 +

∫ π6

π

12

ln ((2sin(v + π

3))(2 sin v)(2sin(v − π

3))

2sin3v)

=π

6

ln2 + ∫ π

6

π12

ln 1dv

=π

6ln 2

33

7/27/2019 TUMSO11th


Credit (ใชช อตามเฟสบ คนะครบ)

Solution by K:

Krittin Pachtrachai

Solution by V:

ว พรา Solution by C:

ชยานนท ชว ชว Solution by O:Nattajak Siribanluewuti

TUMSO11th

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