Precalculus and discrete mathematics answers?

Question

Hi,

I have uploaded a set of question here: http://krydom.com/math1.doc

Can anyone provide me with the answers to the questions? Also, if possible show your work or an explanation for numbers 12, 14, 17, 18, 22, and 25.

Thank you.

Answer 1

12)

csc x - cos x cot x
1/sin x - cos x(cos x/sin x)

1/sin x - cos² x/sin x

(1 - cos² x)/sin x

sin² x/sin x

sin x

14)

-21 mod 4 = (-21 + 24) mod 4 = 3 mod 4 so that's not it.

-5 mod 4 = (-5 + 8) mod 4 = 3 mod 4, so that's not it.

73 mod 4 = (18(4) + 1) mod 4 = 1 mod 4, so that's it... letter c.

17)

sin(7π/8) = sin(7π/4 /2) = √((1 – cos 7π/4) / 2) = √((1 - √2/2) / 2) = √(2 - √2) / 2

18)

a) g(x) = x³ - 8/|x² - 4| - 1

From the right, |x² - 4| = x² - 4, and the limit will be:

lim      (x³ - 8)/(x² - 4) - 1 =
x → 2+

lim [(x - 2)(x² + 2x + 4)]/[(x - 2)(x + 2)] - 1 =
x → 2+

lim (x² + 2x + 4)]/(x + 2) - 1 =
x → 2+

(2² + 2(2) + 4)/(2 + 2) - 1 = 12/4 - 1 = 3 - 1 = 2

From the left, |x² - 4| = -(x² - 4), and the limit will be:

lim -(x³ - 8)/(x² - 4) - 1 =
x → 2-

-(2² + 2(2) + 4)/(2 + 2) - 1 = -12/4 - 1 = -3 - 1 = -4

b) You probably should use L'Hopital's rule for this, but since the degree of the numerator (3) is 1 more than the degree of the denominator (2), for as x approaches infinity in both directions, the function will look more and more like f(x) = x. So the limit on the right is +∞ and on the left is -∞.

22)

(1 + tan x)/(1 + cot x) =
(1 + sin x/cos x)/(1 + cos x/sinx) =
((cos x + sin x)/cos x)/((sin x + cos x)/sin x) =
((cos x + sin x)/cos x) × (sin x/(sin x + cos x)) =
sin x/cos x =
tan x

25)

a) Every integer can be expressed as 3k, 3k + 1, or 3k + 2 because, given any integer, when dividing by three, the remainder can only be 0, 1, or 2. I'm not sure if that's formal enough for what your teacher is looking for, but there you have it.

b)

Case 1: x = 3k

Then x² = 9k², which must be a multiple of 3, so it's not 2 more than a multiple of 3.

Case 2: x = 3k + 1

Then x² = (3k + 1)² = 9k² + 6k + 1. 9k² and 6k are both multiples of 3, so x² is 1 more than a multiple of 3, which is not 2 more than a multiple of 3.

Case 3: x = 3k + 2

Then x² = (3k + 2)² = 9k² + 6k + 4. 9k² and 6k are both multiples of 3, so x² is 4 more than a multiple of 3...but that means that it's (4 - 3) = 1 more than the NEXT multiple of 3.

So all perfect squares are either a multiple of 3 or they are 1 more than a multiple of 3. Never 2.