Answer Key (finals - Math 17)

MATHEMATICS 17:ANSWER KEY I. TRUE or FALSE 1. 2. 3. 4. 5.

(1 PT EACH)

FALSE FALSE FALSE TRUE TRUE

1.

II. Fill in the blanks. 1.

III. Do as indicated.

(2 PTS EACH)

Solution: 𝑚=

𝑘𝑛2 3

⇔

𝑙

3

𝑘𝑛2 𝑙= 𝑚

𝑙

=

𝑙𝑛𝑒𝑤

=

𝑘 2𝑛 2𝑚

=

𝑘𝑛2 8 𝑚

=

𝟖(𝒍𝒐𝒍𝒅 )

2.

𝑦 = 3𝑥 2 − 6𝑥 + 4 vertex : (𝟏, 𝟏)

3.

Solution:

=

𝒇−𝟏 (𝒙)

=

2𝑥 + 7 7𝑥 + 2 𝟕 − 𝟐𝒙 𝟕𝒙 − 𝟐 Rng𝒇 = ℝ − {𝟐/𝟕}

2𝑐𝑖𝑠 37.5°

4

−1 + 𝑖 3

4𝑐𝑖𝑠 150°

=

−𝑖(−1 + 𝑖 3)

=

4𝑐𝑖𝑠 150° 2𝑐𝑖𝑠 30°

=

2𝑐𝑖𝑠 120°

=

−𝟏 + 𝒊 𝟑

2 3

3

3.

Solution: RHS: cos 𝜃 sec 𝜃 − tan 𝜃

=

cos 𝜃 sec 𝜃 + tan 𝜃 ⋅ sec 𝜃 − tan 𝜃 sec 𝜃 + tan 𝜃

=

1 + sin 𝜃



IV. Find the solution set. log 4.5

4.

𝑓(𝑥)

Solution:

𝑖 −99

Then 𝑙 is multiplied by eight times. 2.

Solution:

Dom𝒇 = ℝ − {−𝟐/𝟕}

3

𝑘𝑛2 𝑚

(3 - 4 – 3 PTS)

Let sin 𝛼 =

−3 5

=

32 log 2

=

2 log 3 − log 2

=

𝟐𝒃 − 𝒂

, 𝛼 ∈ (𝜋,

3𝜋 2

)

By Pythagorean Identity, cos 𝛼 = tan

𝛼 2

1.

=

sin 𝛼 1 + cos 𝛼

=

−3 5 −4 1+ 5

=

−𝟑

−4 5

.

(5 PTS EACH)

Solution: CASE 1: C.N. = -6,-4 𝑥 𝑥+4 𝑥 − 3(𝑥 + 4) 𝑥+4 −2(𝑥 + 6) 𝑥+4

≤

3

≤

0

≤

0

By interval table, we have −∞

−6 − − − −

−2 𝑥+6 𝑥+4

−4 − + − +

+∞ − + + −

−3 − − + +

+∞ − + + −

𝑆𝑆1 = −∞, −6 ∪ (−4, +∞). 5.

Let 𝛼 = 𝐴𝑟𝑐𝑠𝑖𝑛 8 cos 2𝐴𝑟𝑐𝑠𝑖𝑛 15

8 15

, then sin 𝛼 = =

8 15

.

cos 2𝛼

=

1−

=

97 225

2 sin2

𝛼

CASE 2: C.N. = -4, -3 −𝑥 𝑥+4 −𝑥 − 3(𝑥 + 4) 𝑥+4 −4(𝑥 + 3) 𝑥+4

≤

3

≤

0

≤

0

By interval table, we have −∞ −4 𝑥+3 𝑥+4

−4 − − − −

𝑆𝑆2 = −∞, −4 ∪ [−3, +∞).

2.

𝑺𝑺𝟏 ∩ 𝑺𝑺𝟐 = (−∞, −𝟔] ∪ [−𝟑, +∞). 2.

Solution: Let 𝑚 = 2 𝑥 , then 22𝑥+1 − 5 ⋅ 2 𝑥 = 2

3

2𝑚 − 5𝑚 − 3

=

0

2𝑚 + 1 (𝑚 − 3)

=

0

Then, 2𝑚 + 1 = 0 or 2 𝑥 = −1/2. (no solution) Also, 𝑚 − 3 = 0 or 2 𝑥 = 3 or 𝒙 = 𝐥𝐨𝐠 𝟐 𝟑. 3.

log

4.

3

=

4 + log 3 (𝑥 + 6)

4𝑥 + 55 − log 3 (𝑥 + 6)

=

4

4𝑥 + 55 𝑥+6

=

9

𝒙

=

𝟏 𝟓

Solution:

3 sin 𝑥 + cos 2𝑥 3 sin 𝑥 + (1 − 2 sin2 𝑥) 2 sin2 𝑥 − 3 sin 𝑥 − 2 2 sin 𝑥 + 1 (sin 𝑥 − 2)

= = = =

= −8𝑖 3 Then 𝑧 = −8𝑖, 𝑎 = 0, 𝑏 = −8, 𝑟 = 8 and 𝜃 = 3𝜋/2. By De Moivre’s Theorem,

𝑧

=

3

8𝑐𝑖𝑠

For 𝑘 = 0,1,2, 3

3

3

𝑧

𝑧

𝑧

3𝜋 + 2𝜋𝑘 2 3

2𝑐𝑖𝑠

=

𝟐𝒊

=

2𝑐𝑖𝑠

=

− 𝟑−𝒊

=

2𝑐𝑖𝑠

7𝜋 6

0

=

𝑛2 + 9𝑛 − 1240

0

=

𝑛 − 31 (𝑛 + 40)

Let 𝑥1 be the horizontal distance from the helicopter to the headquarters. Let 𝑥2 be the horizontal distance from the helicopter to the evacuation site.

𝑥2

11𝜋 6

=

600 tan 75°

=

600(2 − 3)

=

1200 − 600 3 𝑓𝑡

=

600 tan 60°

=

200 3 𝑓𝑡

Thus the distance from the headquarters to the evacuation site is 𝒙𝟏 + 𝒙𝟐 = 𝟏𝟐𝟎𝟎 − 𝟒𝟎𝟎 𝟑 ft.

𝟑−𝒊 (5 PTS EACH)

rate 30 𝑟

=

𝑥1

Let 𝑟 be Sarah’s rate. Let 𝑡 be the time in hours when Sarah and Chuck met (prior to Sarah).

Chuck Sarah

1240

Then, we have two right triangles with equations:

V. Problem Solving 1.

=

𝜋 2

=

=

1240

𝑛 2𝑎1 + (𝑛 − 1)𝑑 2 𝑛 [2 10 + (𝑛 − 1)2] 2 2 𝑛 + 9𝑛

−1 −1 0 0

𝑧3

3

=

3.

Then sin 𝑥 = −1/2 or 𝒙 = 𝟕𝝅/𝟔 and 𝟏𝟏𝝅/𝟔. Also, sin 𝑥 = 2 or 𝑥 = arcsin 2. (no solution) 5.

𝑆𝑛

Then 𝒏 = 𝟑𝟏days, and Bella started saving money on September 30.

Solution: log 3 (4𝑥 + 55)

The saving follows an arithmetic sequence with 𝑎1 = 10 𝑑=2 𝑆𝑛 = 1240.

time 𝑡 + 1/2 𝑡

From Chuck’s equation for distance, 30 𝑡 +

distance 105 105 1 2

= 105km.

Thus, 𝑡 = 3hrs. From Sarah’s equation of distance, 𝑟𝑡 = 105km. Then since 𝑡 = 3hrs, 𝒓 = 𝟑𝟓kph.