Exam 1 (A)
Calculus I: MATH 1351-030
Fall 2011
Name:
Instructions. (Part 1: Multiple Choice) Solve each of the following problems. Choose the
best solution to each problem and clearly mark your choice on the scantron form.
1. Solve the following equation for x:
|7x − 1| = 6
(a) x = −5/7
(c)
4 x = −5/7 and 1
(e) x = 5/7
(b) x = ±5/7
(d) x = 5/7 and −1
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2. Solve the following inequality: |6x − 1| > 7
S
(a)
4 (−∞, −1) S 34 , ∞
(b) −1, 34
S 4
(c) −∞, − 43
[1, ∞)
(d) (−∞, −1]
,∞
3
(e) (−∞, ∞)
c)
20
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3. Solve the following inequality: −2 ≤ 4 − 3x < 2
S
2
(a) 23 , 2
(b)
−∞,
−
(2, ∞)
3
S
2
(c) −∞,
[2, ∞)
(d) (−∞, ∞)
−3
2
(e)
4 3, 2
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4. Express the given function h(x) as a composition of two functions f (x) and g(x) such
that h(x) = f (g(x)):
h(x) = √
6
7x + 6
(a) f (x) = x6 , g(x) = 7x + 6
(c) f (x) = 7x + 6, g(x) = x6
(e)
4 f (x) = √6x , g(x) = 7x + 6
1
(b) f (x) = 6, g(x) = √7x+6
1
(d) f (x) = 7x
, g(x) = x + 6
5. Which of the following functions are equal?
(I) f (x) =
2x2 +7x−4
2x−1
(II) g(x) = x + 4
(III) h(x) = x + 4, x 6=
(a) II and III
(c) none are equal
(e) I, II, and III
1
2
(b) I and II
(d)
4 I and III
Calculus I: MATH 1351-030/Exam 1 (A) – Page 2 of 6 – Name:
6. Find an equation for the line passing through the point (3, −2) and parallel to the line
2x + 5y − 5 = 0.
(a)
4 2x + 5y + 4 = 0
(b) 5x + 2y − 16 = 0
(c) 2x + 5y − 4 = 0
(d) 5x + 2y + 16 = 0
(e) 2x − 5y − 4 = 0
7. What are the exact values of tan−1 (−1)
(a) π2
(c) π4 , 3π
4
(e)
4 − π4
(b) π4
(d) − π2
8. Let f be a function with domain D and range R. Which of the following defines “the
inverse of f ”?
(a) The function f −1 with domain D and range R is the inverse of f if f −1 (f (x)) = x
and f (f −1 (y)) = y for all real numbers.
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(b) The function f −1 with domain R and range D is the inverse of f if f −1 (f (x)) = x
for all x in D and all x in R.
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(c)
4 The function f −1 with domain R and range D is the inverse of f if f −1 (f (x)) = x
for all x in D and f (f −1 (y)) = y for all y in R.
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(d) The function f −1 with domain R and range D is the inverse of f if f −1 (x) = −x.
ht
(
c)
20
(e) The function f −1 with domain D and range R is the inverse of f if f −1 (f (x)) = x
for all x in R and f (f −1 (y)) = y for all y in D.
(I) f (x) =
1
x2
(II) g(x) = x5
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9. Which of the following are odd functions (i.e., “symmetric with respect to the origin”)?
(III) h(x) = sin(x)
(IV) s(x) = sin(x) + cos(x)
(a) none are odd
(c) I and III
(e)
4 II and III
(b) only II
(d) I, II, and III
10. Which of the following best describes the curve represented by y − k = m(x − h)?
(a) The equation of a line with slope m that passes through the point (−h, −k).
(b) The equation of a line with slope
h
k
and x-intercept (m, 0).
(c) The equation of a circle with center C = (h, k) and radius R =
√
m.
(d) The equation of a line with slope m and y-intercept (h, k).
(e)
4 The equation of a line with slope m that passes through the point (h, k).
Calculus I: MATH 1351-030/Exam 1 (A) – Page 3 of 6 – Name:
11. Find the center and radius of the circle defined by x2 − 4x + y 2 + 2y + 3 = 0.
√
(a) Center is C = (−2, 1) and radius is R = 2
(b) Center is C = (−2, 1) and radius is R = 2
√
(c)
4 Center is C = (2, −1) and radius is R = 2
(d) Center is C = (2, −1) and radius is R = 2
(e) Center is C = (−2, −1) and radius is R = 4
12. Find the slope m, x-intercept, and y-intercept of the line given by the equation
5x + 3y − 15 = 0.
(a)
4 m = − 35 , x-intercept = (3, 0), y-intercept = (0, 5)
(b) m = 53 , x-intercept = (−3, 0), y-intercept = (0, −5)
e
(c) m = 3, x-intercept = (3, 0), no y-intercept
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(d) m = − 53 , x-intercept = (−3, 0), y-intercept = (0, 5)
11
(b) f −1 (x) = 2x − 3
(d) f −1 (x) = 2(x + 3)
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c)
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13. Let f (x) = 2x + 3. Find f −1 if it exists.
(a) The inverse does not exist.
(c)
4 f −1 (x) = 21 (x − 3)
(e) f −1 (x) = 21 x − 3
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(e) m = 3, x-intercept = (3, 0), y-intercept = (0, 5)
14. If f (x) = −2x + 3 and g(x) = 5x + 5, what is g(f (x))?
(a) 10x + 13
(b) 10x + 20
(c) −10x − 7
(d)
4 −10x + 20
(e) −10x − 13
15. Which of the following statements best describes The Horizontal Line Test?
(a) A curve in the plane is the graph of a function if and only if it intersects no horizontal
line more than once.
(b) A function f has an inverse if and only if its graph intersects a horizontal line at two
or more points.
(c) The y-intercepts of the graph of f are the points (x, y) that intersect a horizontal
line.
(d) A curve in the plane is the graph of a function if and only if it intersects any horizontal
line at least once.
(e)
4 A function f has an inverse if and only if no horizontal line intersects the graph of
y = f (x) at more than one point.
Calculus I: MATH 1351-030/Exam 1 (A) – Page 4 of 6 – Name:
16. Find the domain of the function f defined by
√
f (x) = x2 + 2x
and compute the values f (−1) and f (1) if the x-values are in the domain.
√
S
(a) domain = (−∞, −2) (0, ∞), f (−1) is undefined, f (1) = 3.
S
(b) domain = (−∞, 0] [2, ∞), f (−1) = 1, f (1) is undefined.
√
S
(c) domain = (−∞, −2) (0, ∞), f (−1) = 0 f (1) = 3.
√
S
(d)
4 domain = (−∞, −2] [0, ∞), f (−1) is undefined, f (1) = 3.
√
√
(e) domain = (−∞, ∞), f (−1) = − 3, f (1) = 3.
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(b) √
1/2
(d)
4 2
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what is the distance between them?
(b) 2√
(d)
4 4 2
20
17. Given the two points (−1, 2) and (3, −2),
(a) 8
(c) −2
√
(e) 2 2
18. Find the exact value of sec π4 .
√
(a) 2√ 3/3
(c) √2/2
(e) 3
ht
(
c)
19. Which of the following lines are perpendicular to each other?
(II) y = − 13 x − 5
(III) y = 3x − 4
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(I) y = 13 x + 2
(IV) y = 13 x + 4
(a) I, II, and III
(c)
4 II and III
(e) I and III
(b) I and II
(d) none are perpendicular
20. Let f (x) = tan(x) on the interval − π2 , π2 . Find the inverse f −1 if it exists and state its
domain and range.
(a) f −1 (x) = tan−1 (x) with domain [0, π] and range (−∞, ∞).
(b) f −1 (x) = tan−1 (x) with domain (−1, 1) and range − π2 , π2 .
(c) f −1 (x) = tan−1 (x) and domain − π2 , π2 with range (−∞, ∞).
(d) The inverse does not exist.
(e)
4 f −1 (x) = tan−1 (x) with domain (−∞, ∞) and range − π2 , π2 .
Calculus I: MATH 1351-030/Exam 1 (A) – Page 5 of 6 – Name:
Instructions. (Part 2: Written Response) Solve each of the following problems. Show your
work clearly. You must write out all relevant steps. Simply having the correct
answer does not give you credit.
1. Plot the two points (−2, 3) and (4, 1) on a Cartesian plane, calculate the distance between
them, find the coordinates of the midpoint of the line segment between them, and plot
the midpoint on the same graph. Be sure to label all points clearly on your plot.
√
Solution: Distance is 2 10, midpoint is M = (1, 2).
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c)
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2. Find an equation for the line passing through the point (4, 10) and perpendicular to the
line 6x − 3y − 5 = 0.
Solution: Slope of original line is 2, so slope of perpendicular line is m = −1/2. Use
point-slope or another line form to get: x + 2y − 24 = 0
Calculus I: MATH 1351-030/Exam 1 (A) – Page 6 of 6 – Name:
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3. Sketch a graph of f (x) = cos(x). Clearly indicate a restricted interval on which an
inverse exists. Sketch a graph of cos−1 (x).
Solution: If we restrict cos(x) to the interval [0, π], it is monotonic and the inverse exists.
4. (Bonus) Simplify the expression tan (cos−1 x) Your answer should be completely algebraic; there should be no trigonometric functions in your answer.
Solution: Let α = cos−1 (x). Then cos(α) = x.√ Use reference triangle with sides
√
2
x, 1 − x2 , 1 to see that tan (cos−1 x) = tan(α) = 1−x
x