MATH 121  Final Exam Solution (version II)  Radford 12/03/02

1. Let y = f(x) = 6 + \sqrt{5x − 7}. The calculation

y = 6 + \sqrt{5x − 7},
y − 6 = \sqrt{5x − 7},
(y − 6)^{2} = 5x − 7,
(y − 6)^{2} + 7 = 5x,
\frac{(y − 6)^{2} + 7}{5} = x (15 points)

shows that the inverse of f(x) is given by f^{−1}(x) = \frac{(x − 6)^{2} + 7}{5} (5 points).

2. Since f(x) = x^{3} + 2x − 3 has integer coefficients, the rational roots of f(x) are among ±1, ±3. Since f(1) = 0 it follows that x = 1 is a root of f(x). Therefore x − 1 divides f(x). Dividing x − 1 into f(x) gives the quotient of x^{2} + x + 3 and remainder 0. Thus

x^{3} + 2x − 3 = (x − 1)(x^{2} + x + 3)

By the quadratic formula the roots of x^{2} + x + 3 are \frac{−1 ± \sqrt{11}ı}{2}. Thus the roots of f(x) are

1 (4 points), \frac{−1 + \sqrt{11}ı}{2} (4 points), \frac{−1 − \sqrt{11}ı}{2} (4 points).

As a product of linear factors,

f(x) = (x − 1) (x − \frac{−1 + \sqrt{11}ı}{2}) (x − \frac{−1 − \sqrt{11}ı}{2}) (8 points).

3. The height function is given by h(t) = −16t^{2} + 10t + 500 feet.

(a) Rewriting h(t) as

−16t^{2} + 10t + 500 = −16(t^{2} − \frac{10}{16}t) + 500 = −16(t − \frac{10}{32})^{2} + (500 + 16 (\frac{10}{32})^{2})

we see that the maximal altitude is ≈ 501.563 feet (10 points).

(b) We want to solve h(t) = 0, where t ≥ 0, or −16t^{2} + 10t + 500 = 0, where t ≥ 0.

By the quadratic formula t = \frac{−10 ± \sqrt{10^{2} − (4)(−16)(500)}}{2(−16)}, so t ≈ 5.911 seconds. (10

points).

4. (a) The calculation

2 sin^{2}(x) + cos(x) = 2,

2(1 − cos^{2}(x)) + cos(x) = 2,
−2 cos^{2}(x) + cos(x) = 0,
cos(x) (−2 cos(x) + 1) = 0,

shows that the set of solutions to the original equation is the set of solutions to cos(x) = 0

together with the set of solutions to cos(x) = \frac{1}{2} (3 points). Thus all solutions the original equation are described by

x = \frac{π}{2} + πn (3 points), \frac{π}{3} + 2πn (3 points), or \frac{5π}{3} + 2πn (3 points),

where n is any integer.

(b) The calculation

log \sqrt{x^{2} − 10} = 3,
\sqrt{x^{2} − 10} = 10^{3},
x^{2} − 10 = (10^{3})^{2},
x^{2} = 10^{6} + 10,
x = ±\sqrt{10^{6} + 10},

the steps of which are reversible, shows that x = \sqrt{10^{6} + 10} (4 points), −\sqrt{10^{6} + 10} (4 points).

5. (a) 600 (1 + \frac{0.08}{4})^{4·9} (5 points) ≈ 1044.615 dollars (5 points).

(b) Let P be the initial amount invested at 100r% per year compounded continuously. Then 2P = P (15) = P e^{r15} which means 2 = e^{r15} (3 points) since P > 0. Therefore

r = \frac{ln 2}{15} (3 points) ≈ .046 (4 points) which can be expressed as 4.6%.

6. a) All real numbers except for x = 9/17 (3 points). b) y = 5/17 (3 points). c) x = −9/17 (3 points). d) At x = −3/5 (3 points). e) At y = −1/3 (3 points). f) Labeling (3 points) and shape of graph (2 points).

7. (a) no (1 points). Take x = \frac{π}{4} for example. In this case cot(x + π) = cot(\frac{π}{4} + π) = 1

and − cot(x) = − cot(\frac{π}{4} ) = −1 (3 points).

(b) no (1 points). Take x = 0 for example. In this case e^{x+5} = e^{5} and e^{x} + e^{5} = e^{0} + e^{5} = 1 + e^{5} (3 points).

(c) yes (4 points).

(d) no (1 points). Take x = 1 for example. In this case ln(x + 10) = ln(11) ≠ 0 and (ln 1)(ln 10)) = (0) ln(10) = 0 (3 points).

(e) no (1 points). Take x = 2π for example. In this case arccos(cos(2π)) = arccos(cos(0)) = 0, since 0 ≤ x ≤ π, which is not x = 2π (3 points).

8. Since tan(t) = −\frac{7}{6} and π < t < 2π, it follows that the terminal side of t is in the fourth quadrant. Therefore (6, −7) is a point on the terminal side of the angle t. Since r = \sqrt{6^{2} + (−7)^{2}} = \sqrt{85} we conclude that

cos(t) = \frac{6}{\sqrt{85}},  sin(t) = −\frac{7}{\sqrt{85}},  tan(t) = \frac{sin(t)}{cos(t)} = −\frac{7}{6}.

The last calculation is not necessary, of course, but does provide a consistency check. Thus

(a) cos(t) = \frac{6}{\sqrt{85}} (5 points), (b) csc(t) = −\frac{\sqrt{85}}{7} (5 points),

and since \frac{π}{2} < \frac{t}{2} < π,

(c) sin(t/2) = \sqrt{\frac{1 − cos(t)}{2}} = \sqrt{\frac{1 − \frac{6}{\sqrt{85}}}{2}} = \sqrt{\frac{\sqrt{85} − 6}{2\sqrt{85}}} (5 points),

and finally

(d) cos(2t) = cos^{2}(t) − sin^{2}(t) = (\frac{6}{\sqrt{85}})^{2} − (\frac{−7}{\sqrt{85}})^{2} = \frac{36}{85} − \frac{49}{85} = −\frac{13}{85} (5 points).

9. (a) Let A be the angle formed by the line segments from George to the top of the streetlight to Georgette, let B = 61^{o}, and let C = 27^{o}. Then A = 180^{o} − 27^{o} − 61^{o} = 92^{o}. Below is a picture which is not meant to be to scale.

[figure: C = 27^{o}; B = 61^{o}; A = 92^{o}; b; a = 120; c; h; d; George; Georgette]

By the Law of Sines

\frac{120}{sin 92^{o}} = \frac{b}{sin 61^{o}} .

Therefore

b = (\frac{120}{sin 92^{o}} ) sin 61^{o} ≈ 105.018 feet (4 points).

Consequently the height of the streetlight is h = b sin 27^{o} ≈ 47.677 feet (8 points).

(b) The distance from Georgette to the streetlight is

d = h cot 61^{o} = \frac{h}{tan 61^{o}} ≈ 26.428 feet (8 points).

10. (a) The modulus of 1 + \sqrt{3}ı is |1 + \sqrt{3}ı| = \sqrt{1^{2} + (\sqrt{3})^{2}} = 2. Therefore 1 + \sqrt{3}ı = 2 (\frac{1}{2} + \frac{\sqrt{2}}{2}ı) = 2 (cos(\frac{π}{3} ) + ı sin(\frac{π}{3} )) (10 points).

(b) Using the polar form of 1 + ı we calculate

(1 + \sqrt{3}ı)^{7} = (2 (cos(\frac{π}{3} ) + ı sin(\frac{π}{3} )))^{7}
= 2^{7} (cos(\frac{π}{3} ) + ı sin(\frac{π}{3} ))^{7}
= 2^{7} (cos(\frac{7π}{3} ) + ı sin(\frac{7π}{3} ))
= 128 (cos(\frac{π}{3} ) + ı sin(\frac{π}{3} )) (10 points).

Note that (1 + \sqrt{3}ı)^{7} = 64(1 + \sqrt{3}ı).