Editorial - Matlan - Soal TKA 1

Problem 1

True False Fasle

Subproblem 1.1

True because degree 2 + degree 3 is 5 degree.

Subproblem 1.2

False because it's -64

Subproblem 1.3

False

Problem 2

ƒ(x)=p*x2025-q*x2023+r*x-47

ƒ(3)=2024=p⋅32025-q⋅32023+3*r-47

3*(p⋅32024-q⋅32022+r)=2024+47

We have that to substitute.

ƒ(-3)=p⋅(-3)2025-q⋅(-3)2023-3*r-47

-3*(p⋅(-3)2024-q⋅(-3)2022+r)-47

-1⋅3*(p⋅32024-q⋅32022+r)-47

-(2024+47)-47

-2118

Problem 3

Luas kebun jagung: 4*t2+20*t+2

Luas kebun kedelai: 2*t2+3*t+1

1. Total: 6*t2+23*t+26 ⊥

2. Difference: 2*t2+17*t+24 ⊤

3. Total on 2nd year: 24+46+26 ⊥

4. Difference on 3rd year: 18+51+24=93 ⊤

Problem 4

ƒ(-2)=-7

ƒ(4)=11

ƒ(x)=(x2-2*x-8)*Q(x)+a*x+b

ƒ(-2)=-2*a+b=-7

ƒ(4)=4*a+b=11

a=3,b=-1

∴R(x)=3*x-1

Problem 5

A(-1)=1/(6-5)*[[3,1],[5,2]]=[[3,1],[5,2]]

Problem 6

(a_ij)=i+2*j+k

(b_ij)=2*i+j+k

k2+k+3∈ℙ?

A=[[k+3,k+5],[k+4,k+6]]

B=[[k+3,k+4],[k+5,k+6]]

First hint

A-B=[[0,1],[-1,0]]

It's not enough because all the k cancels out if you subtract between the 2 matrices.

Second hint

AB=[[52,62],[62,74]]=[[(3+k)2+(5+k)2,(3+k)*(4+k)+(5+k)*(6+k)],[(4+k)*(3+k)+(6+k)*(5+k),(4+k)2+(6+k)2]]

(3+k)2+(5+k)2=52→=2*k2+16*k-18=0

∴k=-9,1

(3+k)*(4+k)+(5+k)*(6+k)=62→2*k2+18*k-20=0

∴k=10,1

2is enough because there can only be 2 possibilities.

∴k=1

So, just the 2nd hint is enough. B.

Finding k isn't necessary to answer the question.

Problem 7

[[1,13],[1,14]]-[[0,1],[7,13]]=[[1,12],[-6,1]]

Problem 8

3*x+2*y=12500

5*x+4*y=21000

x=|[[12500,2],[21000,4]]|/|[[3,2],[5,4]]|

y=|[[3,12500],[5,21000]]|/|[[3,2],[5,4]]|

Problem 9

y=x2

y=(x-3)2-2

y=x2-6*x+7

Problem 10

y=a*x+b

Mirror sumbu x

[[(x^′)],[(y^′)]]=[[1,0],[0,-1]]*[[x],[y]]=[[x],[-y]]

-y=a*x+b

y=-a*x-b=4*x-7

a=-4

b=7

∴2*a+b=-8+7=-1

Problem 11

[[(x^′)],[(y^′)]]=[[-1,0],[-1,2]]*[[-1,0],[0,1]]*[[x],[y]]=[[1,0],[1,2]]*[[x],[y]]=[[x],[x+2*y]]

(y^′)=x+2*y

2*y=(y^′)-(x^′)

Substitute.

(y-x)/2=2*x+3

y-x=4*x+6

y=5*x+6=m*x+c

∴m=5,c=6

• m+c=6+5=11≠10

• 2*m-c=10-6=4

• m+2*c=5+12=17

• m2-c2=25-36=-11≠11

• m2+c2=25+36=61

Problem 12

(x+2)2+(y-3)2=r2

Jarak terhadap (1,-1) adalah r

r2=9+16=25

r=5

False True False

Subproblem 12.1

Rotate 90 clockwise.

(3,2)

Then scale 2 from center.

(6,4)

False statetement. The center is not (-6,4)

Subproblem 12.2

Radius is doubled, so it's 10. True!

Subproblem 12.3

(x-6)2+(y-4)2=100=x2+y2-12*x-8*y-48=0

False. Not equal to what is written in the question.

Subproblem 12.4

A=102*π=100*π

True.

Problem 13

ƒ(x)=√(,x2+x-2)=√(,(x+2)*(x-1))

(x+2)*(x-1)≧0

∴{x≤-2∪x≥1}

Problem 14

ƒ(3)=3*p-q=3

ƒ(6)=(p-1)⋅6-(q+1)=-7-q+6*p=2

6*p-q=9

3*p-q=3

∴3*p=6,p=2

∴-q=9-12=-3,∴q=3

ƒ(1)=2+1=3

ƒ(ƒ(1))=ƒ(3)=3

• p=2 True

• q=1 False

• ƒ(ƒ(1))=3 True

Problem 15

Use the y intercept

b/d=-1/2

d=-2*b

Use the x intercept

0=(a/2+b)/(c/2+d)=(a+2*b)/(c+2*d)

d=-2*b

Use the inverse function and the function itself for the asymptotes

d+2*b=0

a+2*b=0

2*c+d=0

-2*c-a=0

a=-2*b

d=-2*b

c=b

(a,b,c,d)=(-2*b,b,b,-2*b)

Yang paling kecil harus 1, b=1 agar FPB bisa 1.

Aside: Karena ada 2 yang sama, maka jika angka tersebut bukan 1, FPBnya juga bukan 1.

Or something like that, right?

(-2,1,1,-2)

a+b-2*c-3*d=-2+1-2+6=3

Can also be -3 if the signs are reversed.

Problem 16

(y_1)=3*sin(x)

Period: 2*π

Amplitude: 3

(y_2)=sin(3*x)

Period: (2*π)/3

Amplitude: 1

• The period of (y_1) is triple the period of (y_2), True

• False

• The amplitude of (y_1) is triple the amplitude of (y_2), True

• False

Problem 17

ƒ(x)=2*sin(3*x+90)+4

g(x)=2*cos(3*x+60)

What transformations do we need to get g(x) from ƒ(x)?

ƒ(x)=2*cos(3*x)+4

We can see clearly that we need to move ƒ(x) by 20 to the left and 4 units down.

Problem 18

ƒ(x)=20*cos(30*(x-7))+8

We know that 12≤x<13

In this interval we know that the function is decreasing. Which means that the peak is probably at the left.

ƒ(12)=8-10√(,3)≈-9.32

Problem 19

ABCD.EFGH

BD+AH-FH

BD=FH

BD+AH-FH=AH=BG

Problem 20

The plane traverses:

• +9 in the x axis

• -3 in the y axis

• -6 in the z axis

In 3 minutes. We can just divide by 3 to find the so called slope.

That means, the coordinates of C are (2+3,3-1,-1-2)=(5,2,-3)

Problem 21

u=[[-2],[3],[1]],v=[[3],[1],[-4]]

p=u-3*v=[[-2-9],[3-3],[1+12]]=[[-11],[0],[13]]

q=2*u+v=[[-4+3],[6+1],[2-4]]=[[-1],[7],[-2]]

Subproblem 21.1

p⋅q=11+0-26=-15

True!

Subproblem 21.2

What is the cosine of the angle between u and v

First we need to find the magnitude of each vector.

|u|=√(,4+9+1)=√(,14)

|v|=√(,9+1+16)=√(,26)

Then find the dot product.

u⋅v=-6+3-4=-7

cos(θ)=(-7)/√(,14⋅26)=(-7)/(2√(,91))

So it's false because it's supposed to be negative.

Subproblem 21.3

(-7)/√(,14)=(-√(,14))/2

True because we take the absolute value.

Subproblem 21.4

-7/√(,26)=-7/26√(,26)

False even if we take absolute value.

Subproblem 21.5

(-15)/54⋅[[-1],[7],[-2]]=-5/18⋅[…]=[[5/18],[-35/18],[5/9]]

So, the statement is correct.

Problem 22

4*i+2*j-3*k

Problem 23

Where is the point between (0,4) and (12,0)?

It's very clearly at (6,2)

Then, what is the distance between the 2?

Via pythagoras it's clearly 4√(,10)

So the radius of the circle is 2√(,10)

(x-6)2+(y-2)2=40

x2+y2-12*x-4*y=0

It's A.

Problem 24

x2+y2=10

We can just observe from the gradient of 3 that the point must be in the first or third quadrant.

That narrows it down to (-3,1),(-1,3),(3,-1)

(-1,3) is impossible because that requires a smaller gradient.

Just by observation, the answer is (-3,1) and (3,-1)

Problem 25

(-5,10) because it has a distance of 7.

Problem 26

1. Radius is 2, False!

2. Circumference is 4*π, False!

3. Radius 4*π, False!

4. Center is (-2,3), True!

Problem 27

(r_1):(r_2):(r_3)=1:8:10

False False True

Subproblem 27.1

2*π*(r_2)⋅1/2=16*π

r2=16

Is this probable? No, because then (r_1)=20

That doesn't leave a gap.

Subproblem 27.2

3/4*π*(r_1)2=12*π

(r_1)=4

Is this probable? No, because then the other circles will be too big.

Subproblem 27.3

True because the radii are squared.

Problem 28

ƒ(x)=(x2-25)/(x-5)

True True False

Subproblem 28.1

(lim_x→5)((x2-25)/(x-5))=10

True!

Subproblem 28.2

True, it isn't defined

Subproblem 28.3

False, we just found it earlier.

Problem 29

1+p+1+p=-p+14

p=4

Problem 30

Very clearly 204/4=51

Yippee!