Download Theta Equations and Inequalities Solutions FAMAT State

Theta Equations and Inequalities Solutions
FAMAT State Convention 2015
𝑇
1. B- (8 + 2𝑖)(4 − 3𝑖) = 38 − 16𝑖 → 𝑇 = 38, 𝐵 = −16 → (2 − √−𝐵) = (19 − √16) =
(19 − 4) = 15 (Square root denotes a positive number only, in this case, 4.)
2. B- To find the average speed, take the harmonic mean of the speeds each way to and from
2𝑎𝑏
2(60)(80)
9600
480
the competition. 𝑎+𝑏 → 60+80 = 140 = 7
3. A- (𝑓 ∘ 𝑔)(𝑥) = 𝑓(𝑔(𝑥)) Therefore → 𝑔 (𝑓((𝑓 ∘ 𝑔)(𝑥))) = 𝑔 (𝑓 (𝑓(𝑔(𝑥)))) . 𝑔(−1) =
−1 → 𝑓(−1) = 2 → 𝑓(2) = 2 → 𝑔(2) = 5
4. D- 𝑥 + 2√𝑥 − 7 = 10 → 2√𝑥 − 7 = 10 − 𝑥 → 4(𝑥 − 7) = 100 − 20𝑥 + 𝑥 2 → 𝑥 2 − 24𝑥 + 128 = 0 →
(𝑥 − 8)(𝑥 − 16) → 𝑥 = 8, 16. Substituting these values into the original equation gives "8 +
2√8 − 7 = 10 ⋯ 8 + 2 = 10" and "16 + 2√16 − 7 = 10 ⋯ 16 + 6 ≠ 10". Therefore, 𝑥 = 8 is the only
solution.
5. C- 𝑁 = 𝑁0 ∙
𝑡
9.6
1 ℎ
(2)
1 1.2
(2)
→ 𝑁 = 320 ∙
1 8
320
5
→ 𝑁 = 320 ∙ (2) → 𝑁 = 256 = 4 = 1.25
6. B- We are given the following important equations, with Q, D, and N representing the number of
quarters, dimes, and nickels, respectively: 𝑄 + 𝐷 + 𝑁=105, 5Q=N, and D+5=N. Substituting the last
𝑁
11𝑁
two equations into the first in terms of N gives: 5 − 5 + 𝑁 + 𝑁 = 105 → 5 = 110 → 𝑁 = 50. If
there are 5 fewer dimes than nickels, there are 45 dimes.
𝑥−7
7. D- To find 𝑓 −1 (𝑥), solve for 𝑥 in 𝑓(𝑥) and then flip the variables. 𝑓(𝑥) = 𝑦 = 𝑥+15 → 𝑥𝑦 + 15𝑦 = 𝑥 −
7 → 𝑥𝑦 − 𝑥 = −15𝑦 − 7 → 𝑥(𝑦 − 1) = −15𝑦 − 7 → 𝑥 =
−15𝑦−7
𝑦−1
. Then, exchange, giving 𝑓 −1 (𝑥) =
−15𝑥−7
𝑥−1
8. E- 𝑥 =
𝑘𝑦
𝑧
→ 12 =
3𝑘
1
8
1
→ 𝑘 = 2. Using the new values for 𝑦 and 𝑧, 𝑥 =
1
(9)
2
1
4
= 18
9. A- First, we can multiply the entire equation by the quantity 6(𝑥 − 4)2 in order to clear all
denominators. From this, we are given 3𝑥(𝑥 − 4) − 48 = 2(𝑥 − 2)(𝑥 − 4). Simplifying gives 𝑥 2 = 64,
so 𝑥 = ±8. Thus, the largest solution is 8.
10. B- If the second term is 9 and the eighth term is 33, the common difference is 4, and the first term is
𝑁
5. The sum of an arithmetic sequence can be given by 2 (𝑎1 + 𝑎𝑁 ), where N is the number of terms
and 𝑎𝑁 is the n-th term. Thus, it is
101
2
(5 + (5 + 4 ∙ 100)) =
101
2
(410) = 20705
11. A- To find the quadratic equation, you must multiply the roots. The proper form is:
(𝑥 − 6 − 𝑖√3)(𝑥 − 6 + 𝑖√3) → 𝑥 2 − 6𝑥 + 𝑥𝑖√3 − 6𝑥 + 36 − 6𝑖√3 − 𝑥𝑖√3 + 6𝑖√3 − 3𝑖 2 → 𝑥 2 −
12𝑥 = −39
Theta Equations and Inequalities Solutions
FAMAT State Convention 2015
𝑟 𝑛𝑡
12. B- The formula for compound interest is 𝑃 = 𝑃0 (1 + ) . Thus, from Wayne Enterprises, Matt
makes 𝑃 = 1000 (1 +
𝑛
.04 2
.03 2
2
) = 1000(1.02)2 = 1040.40. Dave, with Stark Industries, makes 𝑃 =
1000 (1 + 1 ) = 1000(1.03)2 = 1060.90. Therefore, Stark Industries was the better investment
with 20.50 more profit.
13. C- This could occur when (𝑥 2 − 9𝑥 + 19) = 1 → 𝑥 2 − 9𝑥 + 18 = 0 → (𝑥 − 3)(𝑥 − 6) = 0 → 𝑥 = 3, 6.
−3
This could also occur when (2𝑥 2 − 𝑥 − 6) = 0 → (2𝑥 + 3)(𝑥 − 2) = 0 → 𝑥 = 2 , 2. Finally, it could
occur when (𝑥 2 − 9𝑥 + 19) = −1 and the exponent is even. So, (𝑥 2 − 9𝑥 + 19) = −1 → 𝑥 2 − 9𝑥 +
20 = 0 → (𝑥 − 4)(𝑥 − 5) = 0 → 𝑥 = 4, 5. However, the exponent is only even when 𝑥 = 4. Therefore,
−3
27
you must take the arithmetic mean of 2 , 2, 3, 4, and 6, which is 10.
−9
14. A- Both lines have a slope of and a y-intercept of −2, so the system is consistent and dependent,
2
meaning they are the same line.
15. A- When you graph this system of inequalities, you are given an irregular pentagon. By breaking the
31
pentagon into smaller rectangles and triangles, you can find that the area is equal to 2 .
16. D- First, you must put the ellipse into the form
(𝑥−ℎ)2
𝑎2
+
(𝑦−𝑘)2
𝑏2
= 1.
9𝑥 2 + 25𝑦 2 − 36𝑥 = 189 →
9(𝑥 2 − 4𝑥) + 25𝑦 2 = 189. Complete the square, giving 9(𝑥 − 2)2 + 25𝑦 2 = 225 →
1. The latus rectum is given by
2𝑏 2
𝑎
→
2(32 )
5
=
(𝑥−2)2
25
+
(𝑦−𝑘)2
9
=
18
5
17. E- A conic can be classified by the product of its C and A values in the general equation 𝐴𝑥 2 + 𝐶𝑦 2 +
𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0. In this case, the product of C and A is a negative value (2 ∙ −5) = −10. When AC<
0, the conic is a hyperbola.
18. E- Using synthetic division, you can determine that the roots to this equation are −2, 1, and 3. Using
a number line, you can determine that the only solutions to this inequality occur between
−2 and 1, as well as 3 to infinity → (−2, 1) ∪ (3, ∞). We cannot include − 2, 1, and 3 in the solution
set because the inequality is not greater than or equal to -6.
𝑦
𝑛
19. C- The fourth term of the expanded form of 𝑓(𝑥) = (3𝑥 − ) 6 can be found by (𝑝−1
) ∙ (𝑎)𝑛−𝑝+1 ∙
2
(𝑏)𝑝−1 , following basic binomial expansion. n is the exponent, and p is the p-th term you are
−𝑦 3
attempting to find, while a is the first term in the binomial and b is the second. (63) ∙ (3𝑥)3 ∙ ( 2 ) =
−𝑦 3
20(27𝑥 3 ) (
8
)=−
135
2
𝑥3𝑦3
20. B- The solution we are looking for, 𝐽2 𝐸𝑁 + 𝐽𝐸 2 𝑁 + 𝐽𝐸𝑁 2 , can be factored to (𝐽 + 𝐸 + 𝑁)(𝐽𝐸𝑁). This
−𝐵
is the sum of the roots multiplied by the product of the roots. The sum of roots is 𝐴 and the product
of roots is
−𝐾
𝐴
for any odd function in the form 𝐴𝑥 𝑛 + 𝐵𝑥 𝑛−1 + 𝐶𝑥 𝑛−2 + ⋯ + 𝐽𝑥 + 𝐾 = 0. Thus, the
sum of roots for this polynomial is
−𝐵
𝐴
and the product of roots is
−𝐷
𝐴
. So, 𝐽2 𝐸𝑁 + 𝐽𝐸 2 𝑁 + 𝐽𝐸𝑁 2 =
𝐵𝐷
𝐴2
Theta Equations and Inequalities Solutions
21. D- Factoring the expression
𝑥−4
𝑥 2 −4
≤ 0 gives
FAMAT State Convention 2015
(𝑥−4)
(𝑥−2)(𝑥+2)
≤ 0. −2, 2, and 4 are the critical numbers that
we must test on a number line. It can be found that the inequality works for numbers from negative
infinity to -2, and from 2 to 4. However, only 4 is included in the solution set, because the others (-2,
2, and negative infinity) are undefined. Therefore, the solution is (−∞,−2)∪(2,4].
22. A- 6𝑥 − |4𝑥 + 2| = 24 → |4𝑥 + 2| = 6𝑥 − 24. First, we will solve 4𝑥 + 2 = 6𝑥 − 24. With this, 𝑥 =
13. Plugging this value back into the original equation, we find that it is a solution. Next, we must
11
solve 4𝑥 + 2 = −6𝑥 + 24. With this, 𝑥 = 5 . Plugging this value back into the original equation, we
find that it is not a solution. Therefore, the sum of solutions is 13, as it is the only solution.
23. C- 𝑥 = √10 + 3√10 + 3√10 … → 𝑥 = √10 + 3𝑥 → 𝑥 2 − 3𝑥 − 10 = 0 → (𝑥 − 5)(𝑥 + 2) = 0 → 𝑥 = 5
24. C- When computing the determinant, we must set it equal to 36. This gives us: 2𝑥 − 48 = 0 after
combining like terms and simplifying. Therefore, 𝑥 = 24.
25. B- Using the remainder theorem, the remainder is given by 𝑓(−2) = 2𝑥 5 − 3𝑥 4 + 4𝑥 3 − 5𝑥 2 + 6𝑥 −
7 → −64 − 48 − 32 − 20 − 19 = −183.
26. C- Using Descartes’ Rule of Signs, first compute the equation 𝑓(−𝑥). 𝑓(𝑥) = 4𝑥 5 − 𝑥 4 − 10𝑥 3 +
4𝑥 2 − 𝑘𝑥 − 24 → 𝑓(−𝑥) = −4𝑥 5 − 𝑥 4 + 10𝑥 3 + 4𝑥 2 + 𝑘𝑥 − 24. Because there are two sign changes
in 𝑓(−𝑥), there are a maximum of 2 possible negative zeroes.
27. D- The sum of the reciprocals of the roots is given by
−𝐽
𝐾
for any function in the form 𝐴𝑥 𝑛 + 𝐵𝑥 𝑛−1 +
−0
𝐶𝑥 𝑛−2 + ⋯ + 𝐽𝑥 + 𝐾 = 0. Thus, for this equation, it is equal to 36 , or 0.
28. D- Because the volume and surface area are numerically equal given 𝑓(𝑑) = 𝑔(𝑑), we must start by
𝑑
4
𝑑
𝑑
setting the volume and surface area formulas equal to each other. 4𝜋( 2 )2 = 3 𝜋( 2 )3 → 1 = 6 → 𝑑 = 6
29. B- 42x − 4x+1 = −3 → 42𝑥 − 4𝑥+1 + 3 = 0 → (4𝑥 − 3)(4𝑥 − 1) = 0 → Because we know that 𝑥 ≠ 0,
1
we can say that 4𝑥 − 3 = 0. → 4𝑥 = 3 → 4−𝑥 = 3 → 2−𝑥 =
1
√3
=
√3
3
30. A- Each week, Cody loses a net amount of $20. We must set up the following inequality: 500 −
20𝑥 < 150 → 350 < 20𝑥 → 17.5 < 𝑥. Rounding to the fewest integer number of weeks, Cody will
have less than $150 after 18 weeks.