Lesson Trap problems

Trap problems

Problem 1

Let x  be the cost for varnish in the bottle (in dollars);

Then the cost of the bottle is (x-1) dollars.


Total cost equation

    x + (x-1) = 1.10  dollars

    x + x = 1.10 + 1

     2x   =   2.10

      x   =   2.10/2 = 1.05,


giving the above mentioned correct answer.

Problem 2

The standard formula  = 20 + (n-1)*35  does    work now - it should be modified.


    +----------------------------------------------+
    |   The correct formula is P[n] = 20 + n*35.   |
    +----------------------------------------------+


Again, your correct explicit formula is   = 20 + n*35.


You can    it:   = 20 + 1*35 = 20 + 35 = 55.


In particular,   = 20 + 100*35 = 3520.

Problem 3

Problem 4

The right way to simplify is THIS


     =  =  = .    ANSWER



According to commonly accepted agreements about square root, 
the expression  '  '   always goes from the square root as   |a|,  the absolute value of   'a'.


In this form,  this transformation works universally for positive and negative values of  ' a '.


The given problem is a standard  TRAP  to catch    of those,  who are unfamiliar with this trap for advance.


My post shows and teaches a standard way  to avoid this trap.

Problem 5

To solve the problem correctly, we should reveal a context accurately.


The domain, i.e. the area where this expression is defined, is the set of points (x,y)
in the 1st quadrant or in the 3rd quadrant.  It means that the domain is the union of both quadrants.


In other words x >= 0, y >= 0 are allowed under the square root, or x <= 0, y <= 0 simultaneously.
But x and y of different signs are not allowed.


Second note is that when square root is considered as a function or as an expression,
by a commonly accepted agreement, it is considered as non-negative value or expression.


    Again, it is the common agreement for square root, when it is considered 
    as an expression of variables, that its output values are non-negative.



Next, the first move is to write   = .    (1)


But this expression is only half-right.  On the other half it is WRONG.



It is wrong in 3rd quadrant.  Numerical values under the square root are non-negative there. 
It is good, since we can extract the square root then.


But the values of the simplified expression (1) will be negative in 3rd quadrant, 
which is not consistent with the agreement mentioned above.


Therefore, we should write (1) in MODIFIED form

     =  = ,   (2)


placing x under the absolute value sign.


In this form (2),  it is consistent with (1) in the first quadrant and provides non-negative output
in the third quadrant.


Therefore, the commonly accepted ANSWER is THIS:

    }  is    in the 1st and 3rd quadrants.

Problem 6

The surface area of the cube is given by the polynomial expression  6x^2 - 36x + 54.


It is the combined area of 6 identical faces of the cube.  
The area of each face separately is

    (6x^2 - 36x + 54)/6 = x^2 - 12x + 9 = (x-3)^2.


Notice that the given polynomial is always non-negative, 
so its values make sense as the face area for all values of x =/= 3.


The length of the side of this cube is   = |x-3|.


Notice that I use the absolute value  |x-3|  here, 
as the length is always assumed to be non-negative.


The volume of the cube is then  .


This formula works for all values  x =/= 3, including  x < 3, 
while the formulas without the absolute value sign do not work at x < 3
giving negative edge length and negative volume.


So, my solution is MORE ACCURATE than the @mananth' solution and/or than the @Theo' solution.

Problem 7

This problem is a  ,  provoking you to answer "".


This answer is incorrect.


To answer correctly, you should remember that function arccos has values in interval  [,].


Therefore, you should transform    to the equivalent angle in interval  [,].


This equivalent angle is  ,  so the correct answer to the problem's question is  .

Problem 8

Regarding the amplitude, the question is trivial: I hope, 9 of 10 tutors will answer "amplitude is 2 units".


With the period, the question is trivial, too: I hope that many tutors will give a correct answer "the period is 2".


But regarding the phase shift, 9 of 10 tutors and 99 of 100 students will answer incorrectly, 
saying that the phase shift is  .


To answer correctly, we should remember that the parent function to compare with is cos(x).


When we consider y = , we should first write it equivalently with the positive coefficient 
before cosine

    y = ,


shifting right by half-period .  ONLY AFTER THAT we can rewrite the function, extracting the shift explicitly

    y =  = .


Now it becomes obvious that the shift is    units right, comparing with the parent function cos(x).


ANSWER.  The shift is    units right.