Editorial - Matlan - TO TKA 1
Problem 1
(A)
Problem 2
Subproblem 2.1
x+y=-7
The statement is false.
Subproblem 2.2
x2-y2=-48
The statement is true.
Subproblem 2.3
[[2*x-1,4],[2,1]]*[[y-6,2],[4,1]]=[[13,-2],[6,5]]
The statement is true.
Problem 3
(B) key different?
Problem 4
Problem 4.1
Divide
ƒ(x) byx-2 and the remainder is3 True, as
ƒ(2)=3
Problem 4.2
Divide
ƒ(x) by2*x2-3*x-2 and the remainder is1-2*x
That means
ƒ(2)=1-4=-3 Which is not true, so the statement is false.
Problem 4.3
Divide
ƒ(x) by2*x2+17*x+8 and the remainder is2*x-1
That means
ƒ(-1/2)=-2⋅1/2-1=-2 Which is true, so the statement is true.
Problem 5
I hate Horner's method of synthetic division, so I won't use it.
(D)
Problem 6
Fast.
If this year it's
87 , then next year at worst it's87-8=79 Impossible, as
78<79 Impossible,
78<79 Yeah, the maximum is
87+8=95
|T-0.87|≤0.08 yeahThis problem is quite bad as it's quite ambiguous.
Problem 7
Limit problems are free if you know how to solve them.
Whatever's inside the log cannot be smaller or equal to
0
Test at
x=0
-2>0
The first statement is automatically invalidated.
The second statement is clearly false.
The graphic never touches the
y axis asx=0 is indeterminate. Therefore, the third statement is true.The fourth statement is false, as logs can have negative values if we just test out the inner function.
Therefore the REAL answer is that of just the third statement is true.
Problem 8
Personally, I used functional analysis (made-up term)
2x forx≤1 will approachy=0 as it approaches-∞
3*x-1 will approachy=∞ as it approaches∞ Therefore, the minimum value is most likely at the second part of the function, at
x2-1
The first statement is true.
The second statement is also true, as the maximum value is at
∞ The third statement can be proven false just by testing, which I won't do as it's tedious to write here.
Problem 9
Well, since there are
4 unknowns, we need to utilize all4 points.
There we go. The first statement is true,
g(x)=2*x-1
The second statement is clearly false.
The graph also never touches
y=1 , as the original asymptote ofy=0 has been moved up1 unit. So, the third statement is true!
Problem 10
Subproblems 10.1 to 10.3
The best way to judge transformations is via comparing the distance between
2 minima and/or maxima.
Therefore, the first three statements are true. If you move
sin(2*x) byπ/4 in any direction, you get-cos(2*x)
Subproblem 10.4
Try
n=1
Substitute that to
sin(2*x)
Wait. This subproblem's statement is actually true, as both graphs go through
(3/8*π,√(,2)/2)
Subproblem 10.5
ƒ(x) andg(x) both fall at3/8*π≤x≤π/2 Clearly false, and you can prove that just by knowing that
-cos(2*x) peaks atπ/2 , so in this interval it's actually rising.
Problem 11
Well, only the first one is actually needed. If you look at the image, you can find the amplitude of the first function, because we know the peak is at
π/4 From there, we can guess the shift of the function, therefore making the second statement obsolete.
Problem 12
I saw another problem like this once, using hexagons. If you encounter something like that but you can't solve it outright, try tiling the shape!
Problem 13
Just do it.
Problem 14
First statement is obvious.
Second statement is obvious.
Third statemnt is obvious.
Fourth statement is obvious, just extend one vector after another.
Fifth statement can be found via pythagoras.
Problem 15
Can't draw. But, if you visualize, it's easy.
Problem 16
First statement is clearly false.
Second statement is clearly true.
Third statement can be checked easily to be false
Fourth statement is obviously true.
(C)
Problem 17
The first statement's probably useless, let's try using the second.
We can find the rest. The answer is B.
Problem 18
Subproblem 18.1
|u|=3
True.
Subproblem 18.2
u⋅v=7
True.
Subproblem 18.3
|v|=7/3
False.
Subproblem 18.4
False.
Subproblem 18.5
True.