14. If

,

, and

are positive numbers satisfying

x + \frac{1}{y} = 4

,

y + \frac{1}{z} = 1

, and

z + \frac{1}{x} = \frac{7}{3}

, then find

xyz

.

(A)

\frac{2}{3}

(B)

(C)

\frac{4}{3}

(D)

(E)

\frac{7}{3}

Solution: $(x+\frac{1}{y})(y+\frac{1}{z}) (z+\frac{1}{x})$
$=xyz+x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{xyz}$
$=\frac{28}{3}$

Since
$x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$
$=(x+\frac{1}{y})+ (y+\frac{1}{z})+(z+\frac{1}{x})$
$=\frac{22}{3}$
Sub that in, and let $xyz=k$
$k+\frac{1}{k}+\frac{22}{3}=\frac{28}{3}$
$3k^2-6k+3=0$
$(k-1)^2=0$
$k=1$ , or B.

15. If

\frac{\pi}{2} < \alpha < \frac{3\pi}{4}

,

\cos{\alpha} + \sin{\alpha} = \frac{1}{3}

, and

\left|\cos{2\alpha} + \sin{2\alpha}\right|= \frac{n + \sqrt{p}}{q}

, where

,

, and

are relatively prime positive integers and

is not divisible by the square of any prime, find

n+p+q

.

(A) 34 (B) 35 (C) 36 (D) 37 (E) 38

Solution: $(\sin{\alpha}+\cos{\alpha})^2$
$=(\sin^2{\alpha}+\cos^2{\alpha})+2\sin{\alpha}\cos{\alpha}$
$=1+\sin{2\alpha}$
$=\frac{1}{9}$
Solve for $\sin{2\alpha}$
to get $\sin{2\alpha}=-\frac{8}{9}$
$\cos{2\alpha}=\pm \sqrt{1-(\frac{8}{9})^2}$
$=\pm \frac{\sqrt{17}}{9}$
Because of our restrictions, $\cos{2\alpha}$ is negative
$=-\frac{\sqrt{17}}{9}$
absolute value of $\cos{2\alpha}+\sin{2\alpha}=\frac{\sqrt{17}+8}{9}$
17+8+9=34, or A.

16. If

r_1

,

r_2

, and

r_3

are the solutions to the equation

x^3 - 4x^2 - 4x + 17 = 0

, find

r_1^3 + r_2^3 + r_3^3

.

(A) -68 (B) -4 (C) 0 (D) 61 (E) 68

Solution:First off, rename $r_1=a; r_2=b; r_3=c$ (so that you don't get all the r's confused). Since they are solutions to the equation, the equation $x^3-4x^2-4x+17=(x-a)(x-b)(x-c)$ . Expanding the right side gives $x^3-(a+b+c)x^2+(ab+bc+ac)x-abc=0$ . Therefore, $a+b+c=4; abc=-17$ ; $ab+bc+ac=-4$ . Now take $a+b+c=4$ and multiply both sides by (a+b+c), giving $a^2+b^2+c^2+ab+ac+bc=16$ . Replace $ab+ac+bc$ with -8, swing the -8 to the other side, then multiply both sides by $(a+b+c)$ again. Now you have (with a little bit of simplification):
$a^3+b^3+c^3+a(ab+ac)+b(ab+bc)+c(ac+bc)=96$
Change the equation we already had to ab+ac=-4-bc and put that into the equation we have, then repeat for the other two with the other variables. It now looks like this:
$a^3+b^3+c^3+a(-4-bc)+b(-4-ac)+c(-4-ab)=96$ which is equivalent to this:
$a^3+b^3+c^3-4(a+b+c)-3abc=96$ Substituting from our equations for a, b, and c, gives us $a^3+b^3+c^3-16+51=96$ ; therefore, $a^3+b^3+c^3=61$ and the answer is (D).

17. Find

\sum_{k=0}^{49} (-1)^k  {99 \choose 2k}

.

(A)

(B)

-2^{49}

(C)

2^{49}

(D)

-2^{50}

(E)

2^{50}

Solution:Note that
$(1-i)^{99}=1-{99 \choose 1}i-{99 \choose 2}+ {99 \choose 3}i+{99 \choose 4}.....$
$=\displaystyle\sum_{k=0}^{49} (-1)^k {99 \choose 2k}+i\sum_{k=0}^{49} (-1)^k {99 \choose {2k+1}}$
And $1=-(\sqrt2)(\cos\frac{3\pi}{4}), -1=-(\sqrt2)(\sin\frac{3\pi}{4})$
By DeMorvie's Theorem
$(1-i)^{99}=-((\sqrt2)(\cos\frac{3\pi}{4})+i(\sqrt2)(\sin\frac{3\pi}{4}))^{99} =-(\sqrt2)^{99}(\cos\frac{(99)(3)\pi}{4}+i(\sin\frac{(99)(3)\pi}{4}))$
Thus
$\displaystyle\sum_{k=0}^{49} (-1)^k {99 \choose 2k}$
$=-(\sqrt2)^{99}(\cos\frac{297\pi}{4})$
$=-2^{49}(\sqrt2)(\cos\frac{\pi}{4})$
$=-2^{49}$ , or B.

That took A LOT of Copy and Paste.