MATH 1010 Sec. 3 Midterm 2
April 14, 2011
(Total: 100 points + extra credits: 5 points)
You have to show all the work on the space given. Please be neat! If you need more
space, you can use extra sheets of paper provided. (Please put your name on all the
sheets and label question numbers.) Not showing your work gets 0 point. Where
appropriate, clearly indicate your final answer by circling it. You are not allowed to
get help from your textbook, class notes, calculators, other students, or any other form
of outside aid. Do not forget to turn off your cell phone. If you have any question,
please ask the instructor. Good luck!
1. (10 points) Simplify the complex fraction
x+1
xy−3y
x+1
xy−3y
x2 −1
y
=
x2 −1
y
=
x+1
y
· 2
=
xy − 3y x − 1
y
1
x+1
·
=
y(x − 3) (x + 1)(x − 1)
(x − 3)(x − 1)
1
2. (10 points) Write the following quotient of complex numbers
4 + 3i
2−i
in standard form.
4 + 3i 2 + i
8 + 4i + 6i − 3
5 + 10i
·
=
=
= 1 + 2i
2−i 2+i
4+1
5
2
3. (10 points) Combine and simplify the following rational expression
33 − 18x
x + 2 2x − 10
÷
x2 − 16
x+4
x−4
33 − 18x x + 2 2x − 10 33 − 18x
x + 2 x − 4
+
÷
=
+
·
=
x2 − 16
x+4
x−4
(x + 4)(x − 4) x + 4
2(x − 5)
=
+
33 − 18x + (x + 2)(x + 4)
x−4
33 − 18x + x2 − 2x − 8
1
·
=
·
=
(x + 4)(x − 4)
2(x − 5)
(x + 4)
2(x − 5)
=
1
x2 − 20x + 25
x2 − 20x + 25
·
=
x+4
2(x − 5)
2(x − 5)(x + 4)
3
4. (10 points) Divide x4 − 2x3 − 23x2 − x + 2 by x + 4.
x4 − 2x3 − 23x2 − x + 2
22
= x3 − 6x2 + x − 5 +
x+4
x+4
4
5. (10 points) Solve 4x2 + x − 5 = 0.
x=
−1 ±
√
1 + 80
8
Solutions: x = − 54 , x = 1.
5
=
−1 ± 9
8
6. (Total: 10 points) Evaluate each radical function when x = 3.
√
(a) (5 points) f (x) = x + 6
√
√
f (3) = 3 + 6 = 9 = 3
(b) (5 points) g(x) =
√
3
15x2
g(3) =
√
3
15 · 9 =
√
3
√
3
27 · 5 = 3 5
7. (Total: 10 points) Find the domain of each rational function.
(a) (5 points) f (x) =
x+3
x−1
Domain = {x | x 6= 1}
(b) (5 points) g(x) =
x−2
x2 −9
Domain = {x | x 6= ±3}
6
8. (10 points) Solve
p
(
x2 + 1 =
√
2x − 5
p
√
x2 + 1)2 = ( 2x − 5)2
x2 + 1 = 2x − 5
x2 − 2x + 6 = 0
√
√
2 ± 4 − 24
2 ± −22
x=
=
2
2
No (real) solution.
7
9. (Total: 10 points) Perform each operation and write the result in standard form.
(a) (5 points) (6 + i) + (4 − 2i)
(6 + i) + (4 − 2i) = 10 − i
(b) (5 points) (5 − 3i)(3 − i)
(5 − 3i)(3 − i) = 15 − 9i − 5i − 3 = 12 − 14i
8
10. (5 points) Simplify the following radical expression
√
√
√
√
5
5
7 27 + 5 2 + 2 3 − 3 2
√
√
√
√
√
√
√
5
5
5
7 27 + 5 2 + 2 3 − 3 2 = 7 · 3 3 + (5 − 3) 2 + 2 3 =
√
√
√
√
5
5
= (21 + 2) 3 + 2 2 = 23 3 + 2 2
11. (5
√ points) Find the two consecutive positive integers such that their product is 132.
( 529 = 23)
n(n + 1) = 132
n2 + n − 132 = 0
√
−1 ± 1 + 528
−1 ± 23
n=
=
2
2
Therefore n = 11. The positive integers are 11 and 12.
9
12. (Extra problem: 5 points) Given f (x) = x + 4 and g(x) = 2x2 − 1, find (f ◦ g)(x).
(f ◦ g)(x) = f (g(x)) = f (2x2 − 1) = (2x2 − 1) + 4 = 2x2 + 3
–END–
10