Editorial - Matlan - Soal Detik-Detik

Problem 1

A-B=(C^T)

A=[[5,1],[7,-2]]

B=[[-1,1],[3,5]]

C=[[a,b],[c,d]]

(C^T)=[[5+1,1-1],[7-3,-2-5]]=[[6,0],[4,-7]]

C=[[6,4],[0,-7]]

∴a-b=6-4=2

Problem 2

A=[[8,2],[19,5]]

B=[[6,x],[-4,3]]

Subproblem 2.1

If x is positive, det(B)>0

det(B)=18-4*x>0

4*x<18

x<18/4

This is true because it doesn't contradict

Subproblem 2.2

If det(B)>det(A),x>-4

det(A)=40-38=2

18-4*x>2

4*x<16

x<4

False it's not the same.

Subproblem 2.3

If det(A*B)<0,x>9

det(A*B)=(18-4*x)⋅2=36-8*x

36-8*x<0

8*x>36

Clearly false.

Problem 3

A=[[4,1,-1],[2,2,x],[3,0,-2]]

det(A)=-16+3*x+0-0+4+6=-6+3*x

First statement: No inverse if x≠2

No inverse is when det(A)=0

-6+3*x=0

3*x=6

x=2

First statement is false.

Second statement: Has inverse if x≠2

Third statement: Has inverse if x>2

Both true since we just proved that.

All of the other statements are false.

Problem 4

A-B=C

A=[[4,1,5],[-2,3,1]]

(C^T)=[[9,4],[6,8],[7,2]]

B=[[4,1,5],[-2,3,1]]-[[9,6,7],[4,8,2]]

B=[[-5,-5,-2],[-6,-5,-1]]

Problem 5

The first statement is true.

100*x+40*y=500000

200*x+80*y=1000000

150*x+80*y=850000

50*x=150000

x=3000

40*y=500000-300000

y=5000

The second statement is true.

100*x+50*y=100⋅3000+50⋅5000>600000

The third statement is false.

Problem 6

KL-I=M

K=[[2,1],[-5,3]]

M=[[1,9],[6,-18]]

First statement is true. K as an inverse.

Second statement is false. The determinant is 11

K*L-I=M

K*L=I+M

K*L=[[2,9],[6,-17]]

L=1/11⋅[[3,-1],[5,2]]*[[2,9],[6,-17]]

L=1/11⋅[[0,44],[22,11]]=[[0,4],[2,1]]

Third statement is true.

L(-1)=1/(-6)*[[1,-4],[-2,0]]

Fourth statement is clearly false.

L*M=[[0,4],[2,1]]*[[1,9],[6,-18]]=[[24,-72],[8,0]]

Fifth statement is clearly false.

Problem 7

g(x)=3*x4-x3+6*x2-29

Subproblem 7.1

g(3)=3⋅81-27+54-29=21

True!

Subproblem 7.2

False

Subproblem 7.3

True

Subproblem 7.4

True.

Subproblem 7.5

False.

Problem 8

A(x)=4*x3+3*x2-7*x-6

B(x)=-6*x3+2*x2+9*x-8

Subproblem 8.1

A(x)+B(x)=-2*x3+5*x2+2*x-14

True.

Subproblem 8.2

A(x)-B(x)=10*x3+x2-16*x+2

True.

Subproblem 8.3

4*A(x)+3*B(x)

16*x3+12*x2-28*x-24-6*x3+15*x2+6*x-42

False.

Problem 9

Divide 2*x4-3*x3-10*x2+9*x-14 by 2*x+5

x3-4*x2+5*x-8

,leaving 26

Problem 10

P(x)=2*x3+3*x2+8*x-m

P(x)/(2*x-1)=?

After long division, we have this:

-m+5=-4

m=9

The first statement is false. m≠6

The second statement is true. The quotient is:

x2+2*x+5

The third statement is true.

P(x)/(x+3)=2*x2-3*x+17+C

The fourth statement is false.

The fifth statement is true.

Problem 11

ƒ(x)=x3-6*x2+9*x-4

Subproblem 11.1

Grafik ƒ(x) menyinggung sumbu x

Salah, bisa dicari dengan mencari semua faktor dari angka 4, sehingga hanya ada 2 akar real, yaitu 1 dan 4.

Subproblem 11.2

Grafik ƒ(x) mempunyai tiga akar real berbeda.

Tidak, dengan cara Subproblem 11.1 didapatkan bahwa ƒ(x) menyinggu sumbu x, maka hanya ada 2akar real.

Subproblem 11.3

Impas pada x=4

Benar, dari grafiknya. Juga bisa dilihat bahwa ƒ(x) memiliki derajat 3 dan koefisien utama yang positif, maka pada akar terakhirnya, grafik akan mengarah ke atas.

Problem 12

mod(ƒ(x),x2-4*x-12)=9*x+p=mod(ƒ(x),(x-6)*(x+2))

mod(ƒ(x),x+2)=8

mod(ƒ(x),x-8)=?

ƒ(x)=(x-6)*(x+2)*Q(x)+9*x+p

Evaluate at x=-2, has to be 8

9⋅-2+p=8

∴p=26

Evaluate at x=6,

ƒ(6)=9⋅6+26=80

Easy

Problem 13

ƒ(x)=3*cos(x)+2

Find the range.

We know that the normal range for cosine is -1 ≤y≤1

Multiply that by 3, then translate 2 units upwards.

-1≤y≤5

Problem 14

ƒ(x)=(log_2)(x-1)

ƒ(2)-ƒ(5)=?

ƒ(2)-ƒ(5)=(log_2)(1)-(log_2)(4)=0-2=-2

Problem 15

ƒ(x)=(x+1)/(x2-4)

Subproblem 15.1

ƒ(x) has asymptotes at x=2 and x=-2

This is true, as we can see just by observing that the denominator has roots 2 and -2

Subproblem 15.2

ƒ(x) has a diagonal asymptote of y=x+2

We can find this easily by dividing the numerator by the denominator. Clearly, the degree of the polynomial of the resulting quotient isn't of degree 2, as the degree of the denominator is higher than the degree of the numerator. So, this statement is false.

Subproblem 15.3

ƒ(x) has a horizontal asymptote at y=2

Clearly false because we can just find that ƒ(2) has a real value.

Problem 16

Subproblem 16.1

The graphic goes through the point (0,3)

Clearly false.

Subproblem 16.2

The graphic goes through (-1,2)

Clearly true.

Subproblem 16.3

For ƒ(x)

x≥0

Clearly false from the graph.

Subproblem 16.4

For ƒ(x)=y,y≥0

Clearly true from the graph.

Subproblem 16.5

ƒ(x)=2√(,x+2)

Let's use points from the graph, just in case.

ƒ(-2)=0

ƒ(2)=2⋅2=4

Probably true.

Subproblem 17

ƒ(x)=-2(2*x-1)

Subproblem 17.1

ƒ(x) goes through (1,-2)

Let's just test that.

ƒ(1)=-2(2-1)=-2=-2

This statement is true.

Subproblem 17.2

ƒ(x) is monotonically decreasing.

So, the exponent gets larger and larger while the value keeps decreasing. This statement is true.

WELL, technically ƒ(x) is not just monotonically decreasing, but strictly decreasing, but I think strictly decreasing functions are a subset of monotonically decreasing functions.

Subproblem 17.3

ƒ(x) is monotonically increasing.

False.

Problem 18

L(T)=0.01⋅|T-25|

L(T) must not exceed 0.15

Subproblem 18.1

L(T) =0.15 at T=10 and T=40

Let's test that.

0.15=0.01⋅|T-25|

15/100⋅100/1=15=|T-25|

Double cases.

T-25=15

T=40

Take the negative case.

T-25=-15

T=10

The statement is true.

Subproblem 18.2

Between T=10 and T=40, will L(T) ever exceeed 0.15?

We can say this is false, knowing that the shape of this linear absolute function is triangle shaped, opening upwards. That means that between 2 T that have the same value, the values must be less than L(T) (very hard to phrase this)

Subproblem 18.3

L(15)=0.1

L(15)=0.01⋅|15-25|=0.01⋅10=0.1

This statement is true.

Subproblem 18.4

L(20)=L(35)

0.01⋅5=0.01⋅10

False.

Subproblem 18.5

L(45) doesn't exceed bounds

Clearly false, as we found the bounds already at the first subproblem.

Problem 19

AB=[[-5],[-3],[5]]-[[-2],[7],[9]]=[[-3],[-10],[-4]]=-3*i-10*j-4*k

Problem 20

△K*L*M

K=(3,-2,1)

L=(2,4,-3)

M=(-6,0,5)

Subproblem 20.1

KL≤7

KL=√(,(3-2)2+(-2-4)2+(1+3)2)=√(,1+36+16)=√(,53)

√(,53)>√(,49)

This statement is false.

Subproblem 20.2

Among △K*L*M, the longest side is 12 units long.

KM=√(,81+4+16)=√(,101)

LM=√(,64+16+64)=12

√(,101)<√(,144)

Yes, this statement is true.

Subproblem 20.3

△K*L*M is an acute triangle (dan segitiga sembarang, tapi itu obvious)

To do this, we only need to find the largest angle.

53+101>144

154>144

So, this is an acute triangle, and this statement is true.

Problem 21

P=[[-2],[1],[5]]

Q=[[4],[-3],[8]]

P→Q for the drone.

Subproblem 21.1

PQ=6*i-4*j+3*k

PQ=[[4],[-3],[8]]-[[-2],[1],[5]]=[[6],[-4],[3]]=6*i-4*j+3*k

Therefore, this statement is true.

Subproblem 21.2

QP=3*i-4*j+6*k

Uh no, clearly false as the first component is wrong.

Subproblem 21.3

y component of PQ is 4

False, as it's -4

Subproblem 21.4

|PQ|=|QP|

Obviously.

Subproblem 21.5

z component of QP is -6

Clearly wrong.

Problem 22

Find (u+2*v)-(2*w-3*u)=4*u+2*v-2*w

4*(-4*i+2*j)+2*(3*i+2*j)-2*(-1*i-3*j)

-16*i+8*j+6*i+4*j+2*i+6*j=-8*i+18*j

Problem 23

u=2*i+3*j-2*k

v=a*i+-2*j+4*k

u⋅v=-4

Subproblem 23.1

a=5

Let's just prove this.

[[2],[3],[-2]]⋅[[a,-2,4]]=2*a-6-8=-4

2*a=-4+14

∴a=5

This statement is true.

Subproblem 23.2

u+v=7*i+j-2*k

u+v=2*i+3*j-2*k+5*i-2*j+4*k=7*i+j+2*k

False statement.

Subproblem 23.3

Switch j and k on v, u⋅v=26

[[2],[3],[-2]]⋅[[5,4,-2]]=10+12+4=26

True.

Problem 24

u=[[6],[4]]

v=[[-2],[3]]

Subproblem 24.1

u⋅v>0

u⋅v=-12+12=0

False.

Subproblem 24.2 and Subproblem 24.3

u,v are perpendicular

u,v are parallel

θ=arccos(0/…)=90

THIS MEANS THAT THEY'RE PERPENDICULAR???

Yeah, so the second statement is false, and the third statement is false.

Subproblem 24.4

|u|=√(,52)

|u|=√(,36+16)=√(,52)

This statement is true.

Subproblem 24.5

|v|>|u|

|v|=√(,4+9)=√(,13)<√(,52)

This statement is false.

Problem 25

○K→x2+y2+2*x-8*y+8=0

○L has double the diameter of ○K

sin(x)

Subproblem 25.1

○K is centered at (1,-4)

(x+1)2-1+(y-4)2-16+8=0

(x+1)2+(y-4)2=9

False, the center is actually at (-1,4)

Subproblem 25.2

○K diameter is 6

True, as the radius is √(,9)=3

Subproblem 25.3

○L→x2+y2+2*x-8*y-19=0

(x+1)2+(y-4)2=36

x2+2*x+1+y2-8*y+16-36=0

x2+y2+2*x-8*y-19=0

True.

Problem 26

(x+1)2+(y-2)2=8

Subproblem 26.1

x+2*y=7 is a tangent

Just substitute.

x=7-2*y

(8-2*y)2+(y-2)2=8

4*y2-32*y+64+y2-4*y+4-8=0

5*y2-36*y+60=0

Δ=36⋅36-4⋅5⋅60

Δ≠0

False statement.

Subproblem 26.2

x-y=-7 is a tangent.

y=7+x

(x+1)2+(x+5)2=8

x2+2*x+1+x2+10*x+25=8

2*x2+12*x+18=0

x2+6*x+9=0=(x+3)2

One solution. This statement is true.

Subproblem 26.3

x-2*y=2 intersects with the circle.

x=2*y+2

(2*y+3)2+(y-2)2=8

4*y2+12*y+9+y2-4*y+4-8=0

5*y2+8*y+5=0

Δ=8⋅8-4⋅5⋅5=-36

Negative discriminant. This statement is false, there is no intersection.

Subproblem 26.4

x+y=1 intersects with the circle.

x=1-y

(2-y)2+(y-2)2=8

2*(y2-4*y+4)=8

y2-4*y=0

y*(y-4)=0

There are solutions. The statement is true.

Subproblem 26.5

2*x+y=1 doesn't intersect with the circle.

y=1-2*x

(x+1)2+(-2*x-1)2=8

x2+2*x+1+4*x2+4*x+1-8=0

5*x2+6*x-6=0

Δ=36-4⋅5⋅(-6)=156

There are solutions. The statement is false.

Problem 27

○L is centered at (4,-2) and has a diameter of 7

Subproblem 27.1

○L→4*x2+4*y2-32*x+16*y+31=0

(x-4)2+(y+2)2=49/4

4*x2-32*x+64+4*y2+16*y+16-49=0

4*x2+4*y2-32*x+16*y+31=0

The statement is true.

Subproblem 27.2

The circumference is 44

C=7⋅22/7=22

False

Subproblem 27.3

The area is 154

A=49/4⋅22/7=77/2

False

Problem 28

Subproblem 28.1

If the arc AB=18 then BC=45

BC=18⋅100/40=45

The statement is true.

Subproblem 28.2

If the arc AC=56 then AB=14

AB=56⋅40/140=16

The statement is false.

Subproblem 28.3

If the the area of the sector OBC=75 then the area of the sector OAB=45

OAB=75⋅40/100=30

The statement is false.

Subproblem 28.4

If the area of the sector OAC=42 then the area of the sector OBC=30

OBC=42⋅100/140=30

The statement is true.

Subproblem 28.5

If the area of the sector OBC=15 then the area of the circle is 60

A=15⋅100/360=25/6

The statement is false.

Problem 29

A=[[4],[-2]]

B=[[5],[1]]

C=[[3],[3]]

△A*B*C

Reflect everything by the y axis.

A=[[-4],[-2]]

B=[[-5],[1]]

C=[[-3],[3]]

(B)

Problem 30

y=x+2

Is reflected by the x axis then

translated by [[-1],[3]]

Subproblem 30.1

After reflection, the equation is y=-x-2

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

-y=x+2

y=-x-2

The statement is true.

Subproblem 30.2

After reflection and translation, the equation is y=-x

[[(x^′)],[(y^′)]]=[[x-1],[y+3]]

y-3=-(x+1)-2

y=-x

The statement is true.

Subproblem 30.3

After reflection, the line intersects with (0,2)

False.

Subproblem 30.4

After fully transforming, the line intersects with (0,1)

False.

Subproblem 30.5

After fully transforming, the line intersects with (2,0)

False.

Problem 31

y=2*x+3

Reflected turns into

y=-2*x+3

What transformation has been done?

Since the x has been inverted, I think it's safe to say that the line has been reflected by the y axis, or x=0

(B)

Problem 32

A circle with the equation (x-3)2+(y+2)2=0

Is reflected along the x axis.

Subproblem 32.1

The resulting circle is (x-3)2+(y-2)2=9

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

Substitute that.

(x-3)2+(-y+2)2=9

Yes, this is equal to the what the statement said, as (a-b)2=(-a+b)2

So, the statement is true.

Subproblem 32.2

The center of the resulting circle is at (3,-2)

No, it's at (3,2)

The statement is false.

Subproblem 32.3

The radius of the circle is 6 units after reflection.

No, reflection doesn't scale the image. The statement is false.

Problem 33

The line 2*x-1=y

Is rotated α degrees pivoting at (0,0)

Subproblem 33.1

Rotate by 90 counter clockwise, turns into y=-x/2+1/2

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

Substitute that.

-x=2*y-1

2*y=-x+1

y=-x/2+1/2

Yes, the statement is true.

The rotation matrix is counter clockwise by default, by the way.

Subproblem 33.2

Rotate by 90 clockwise, turns into y=-x/2+1/2

No, clearly false, as proven by the previous subproblem.

Subproblem 33.3

Rotate by 180counter clockwise, turns into y=2*x+1

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

-2*x-1=-y

y=2*x+1

Indeed, the statement is true.

Subproblem 33.4

Rotate by 180 clockwise, turns into y=2*x+1

Clearly the same as the previous subproblem, as cos(180)=cos(-180)

Subproblem 33.5

Rotate by 0 turns into y=-2*x-1

No. False.

Problem 34

2*x+y=1

Reflected by y=-2

Then scaled by a factor of -2 centered around (0,0)

Subproblem 34.1

After reflection, the line turns into 2*x-y=5

[[(x^′)],[(y^′)]]=[[1,0],[0,-1]]⋅[[x],[y]]+[[0],[-4]]=[[x],[-y-4]]

Substitute that.

2*x-y-4=1

2*x-y=5

The statement is true.

Subproblem 34.2

After applying both transformations, the resulting line sis 2*x-y=-10

Just divide x and y by -2

(2*x)/(-2)-y/(-2)=5

2*x-y=-10

The statement is true.

Subproblem 34.3

After both transformations, the resulting line is parallel to the starting line.

No. They are not parallel, just by looking at the signs.

Problem 35

Let's answer rapidly.

1. (lim_x→-1-)(ƒ(x))=3 True

2. (lim_x→3-)(ƒ(x))=5 False

3. (lim_x→3+)(ƒ(x))=5 True

4. ƒ(x) has a limit at x=-1 True

5. ƒ(x) has a limit x=3 ⊥

Problem 36

ƒ(x)=(2*x2-14*x-60)/(x-10)=(2*(x+3)*(x-10))/(x-10)=2*x+6

Subproblem 36.1

(lim_x→12)(ƒ(x))=30

2⋅12+6=30

True.

Subproblem 36.2

(lim_x→15)(ƒ(x))=40

2⋅15+6=36

False.

Subproblem 36.3

(lim_x→20)(ƒ(x))=46

2⋅20+6=46

True.

Problem 37

Subproblem 37.1

(lim_x→5)(ƒ(x))=5

Clearly false!

Subproblem 37.2

(lim_x→∞)(ƒ(x))=5

True, just by observation.

Subproblem 37.3

(lim_x→-∞)(ƒ(x))=5

True, just by observation.

Problem 38

(lim_x→5)((2*x2-50)/(x2+5*x-50))=(lim_x→5)((2*(x-5)*(x+5))/((x+10)*(x-5)))=(2⋅10)/15=4/3

Problem 39

(lim_x→0)(((6*x)2*sin(6*x))/((tan^2)(3*x)*sin(12*x)))=6/3⋅6/3⋅6/12=2

Problem 40

E(x)=(20*x2-500)/(x-5)=(20*(x+5)*(x-5))/(x-5)=20*x+100

Subproblem 40.1

(lim_x→∞)(E(x))=20

It's divergent, so it's infinity. False statement.

Subproblem 40.2

(lim_x→∞)(E(x))=100

Same as before. False statement.

Subproblem 40.3

(lim_x→∞)(E(x))=∞

True statement.

Subproblem 40.4

Basically, is the function divergent? Yes, it is.

Subproblem 40.5

Basically, does the function converge to 500? No, it doesn't.