MAT 113 Test #2 Solutions

Transcription

1 MAT 11 Test # Solutions There were two forms of the test given, A and B. The letters next to the problems indicate which version they came from. A 1. Let P and Q be the points (, 1) and ( 1, ). a. [4 pts] Find the distance between P and Q. b. [4 pts] Find the point halfway between P and Q. Solution: (a): Use the distance formula: d (x 1 x ) + (y 1 y ) ( ( 1)) + (1 ) 0. Grading: + points for the formula, + points for the final answer. Grading for common mistakes: 1 point for an approximation. (b): Use the midpoint formula: ( x1 + x, y ) ( 1 + y + ( 1), 1 + ) (1, ). Grading: + points for the formula, + points for substituting values. B 1. Let P and Q be the points (, 5) and ( 5, ). a. [4 pts] Find the distance between P and Q. b. [4 pts] Find the point halfway between P and Q. Solution: (a): Use the distance formula: d (b): Use the midpoint formula: Grading: See above. (x 1 x ) + (y 1 y ) ( ( 5)) + (5 ) 58. ( x1 + x, y ) ( 1 + y + ( 5), 5 + ) (, 7 ). 1

2 A. Let L be the line whose equation is x + y 8. Let P be the point (1, 1). a. [4 pts] Find the equation of the line through P parallel to L. b. [4 pts] Find the equation of the line through P perpendicular to L. Solution: First, we need to find the slope of line L. To do this, we solve the equation for y: x + y 8 y 8 x y 8 x. The slope is the number in front of the x, namely. To find the answer to (a), we need to write down an equation for a line with the same slope as L and passing through the point (1, 1), using the point-slope equation: y y 0 m(x x 0 ) In this case, we get y 1 (x 1), or y x + 5. For (b), we need to find the slope of a line perpendicular to a line with a slope of. To do this, we take the reciprocal and multiply by 1:. Then we put the slope into the point-slope equation for a line: y 1 (x 1), or y x 1. Grading for both parts: + points for finding the slope of the desired line; + points for finding an equation. B. Let L be the line whose equation is x y 6. Let P be the point (1, 1). a. [4 pts] Find the equation of the line through P parallel to L. b. [4 pts] Find the equation of the line through P perpendicular to L. Solution: See above for the general procedure. We solve the equation for L for the variable y: x y 6 y 6 x y + x The slope of L is, so a line parallel to L has slope as well. Putting this into the point-slope equation for a line: y + 1 (x 1) or y x 5. For the line perpendicular to L, we need to find the inverse reciprocal of first, which is. Then the equation for part (b) is: Grading: See above. y + 1 (x 1) or y x 1.

3 A. Consider the graph of the equation x + y y. a. [5 pts] Find the x-intercept(s); if there are none, indicate this. b. [5 pts] Find the y-intercept(s); if there are none, indicate this. c. [5 pts] Does the graph of the equation have any symmetry? If so, which type(s)? Solution: (a): The x-intercept(s) are the points on the graph where y 0. Thus, we need to find the x s such that x + 0 0, that is, x 0. The only x-intercept is x 0. Grading: + points for setting y 0, + points for finding x. (b): The y-intercept(s) are the points on the graph where x 0. Thus, we need to find the y s such that 0 + y y, or y y. This equation is a bit harder to solve. After moving the y over to the right hand side and factoring, we get y(y ) 0, so the y-intercepts are y 0 and y and y. Grading: Same as for (a). (c): The three types of symmetry are x-axis (top-bottom), y-axis (left-right), and origin (rotational). To check for x-axis symmetry, we replace y with y and see if we end up with the same equation: x + ( y) ( y), or x y y Evidently not. Replacing x with x gives ( x) + y y, or x + y y, so the graph has y-axis symmetry. We do both substitutions to check for origin symmetry: ( x) + ( y) ( y), or x y y, so the graph does not have rotational symmetry, either. Grading: +1 point for knowing how to find each type of symmetry, + points for finding the answers. Grading for common mistakes: + points (total) if there was no work; +1 point (total) for no work and a wrong answer.

4 B. Consider the graph of the equation x y x. a. [5 pts] Find the x-intercept(s); if there are none, indicate this. b. [5 pts] Find the y-intercept(s); if there are none, indicate this. c. [5 pts] Does the graph of the equation have any symmetry? If so, which type(s)? Solution: (a): The x-intercept(s) are the points on the graph where y 0. Thus, we need to find the x s such that x x. To solve this equation, we need to move the x to the left-hand side and factor it. We get x x 0, or x(x + 1)(x 1) 0, so there are three x-intercepts: 0, 1, and 1. Grading: + points for setting y 0, + points for finding x. (b): The y-intercept(s) are the points on the graph where x 0. Thus, we need to find the y s such that y 0. The only y-intercept is y 0. Grading: Same as for (a). (c): The three types of symmetry are x-axis (top-bottom), y-axis (left-right), and origin (rotational). To check for x-axis symmetry, we replace y with y and see if we end up with the same equation: x ( y) x, or x + y x. Evidently not. Replacing x with x gives ( x) y x, or x y x, so the graph does not have y-axis symmetry, either. We do both substitutions to check for origin symmetry: ( x) ( y) ( x), or x + y x. If we divide both sides of this equation by 1, we get x y x, so the graph does have rotational symmetry. Grading: +1 point for knowing how to find each type of symmetry, + points for finding the answers. Grading for common mistakes: + points (total) if there was no work; +1 point (total) for no work and a wrong answer. 4

5 A 4. [6 pts] Find an equation for the circle which has its center at (, ) and which passes through the point (1, 4). What is the radius of this circle? Solution: The standard form of the equation for a circle is (x h) + (y k) r, where the center is (h, k), and the radius is r. We know that the center of the circle, but we need to find the radius. Since (h, k) (, ), we know that (x ) + (y ) r, for any point on the circle. In particular, if x 1 and y 4, then we get a true statement (since (1, 4) is on the circle). So: (x ) + (y ) r so r 8. Thus an equation for this circle is (1 ) + (4 ) r 8 r, (x ) + (y ) 8, and the radius is 8. Grading: + points for the standard form, + points for finding the radius, + points for substituting into the equation. B 4. [6 pts] Find an equation for the circle which has its center at (1, 4) and which passes through the point (4, 1). What is the radius of this circle? Solution: The standard form of the equation for a circle is (x h) + (y k) r, where the center is (h, k), and the radius is r. We know that the center of the circle, but we need to find the radius. Since (h, k) (1, 4), we know that (x 1) + (y 4) r, for any point on the circle. In particular, if x 4 and y 1, then we get a true statement (since (4, 1) is on the circle). So: (x 1) + (y 4) r so r 18. Thus an equation for this circle is (4 1) + (1 4) r 18 r, (x 1) + (y 4) 18, and the radius is 18. Grading: + points for the standard form, + points for finding the radius, + points for substituting into the equation. 5

6 A 5. Find all complex solutions to the following equations: a. [5 pts] x + 9 x b. [5 pts] x + x + x 0 c. [5 pts] x 4 5x Solution: (a): To get rid of the sign, we move the to the right-hand side and square both sides, then solve for x: x + 9 x x + 9 x + x + 9 (x + ) x + 9 (x + )(x + ) x + 9 x + x + x x x 5 4 x. It appears that x 5 is a solution, and in fact it checks out. 4 Grading: + points for moving the over, + points for the rest of solving for x. Grading for common mistakes: +1 point (total) plus an electric shock for anyone who wrote x + 9 x+. (b): This is a polynomial equation. The first thing to do is to check whether we can factor out a power of x. We can, and we get the equation: x(x + x + 1) 0, so x 0 or x + x To find the solutions to the second equation, we need to use the quadratic formula: x b ± b 4ac a 1 ± ± 1 ± i. So the answers are x 0 and x 1 ± i ; all of them check out. Grading: + points for factoring out the x; +1 point for noticing that x 0 is a solution; + points for the quadratic formula. (c): This can be done several different ways, but the basic idea is to let y x, since the exponents are all even. Then x 4 (x ) y, and so the equation becomes y 5y (y )(y ) 0, so y or y. However, you need to find the value of x. Remembering that x y, we need to solve the equations x and x, which give the four solutions x ±, ±. Grading: + points for subsituting y x or factoring x 4 5x + 6 as (x )(x ), + points for finding x. Grading for common mistakes: + points (total) for x (x 5 + 6) 0. 6

7 B 5. Find all complex solutions to the following equations: a. [5 pts] x x b. [5 pts] x + x + 4x 0 c. [5 pts] x 6 x Solution: (a): To get rid of the sign, we move the to the right-hand side and square both sides, then solve for x: x x x + 4 x x + 4 (x ) x + 4 (x )(x ) x + 4 x x x x x 5 6 x It appears that x 5 is a solution, but checking with your calculator shows it is not. This 4 equation has no solutions, real, complex, or otherwise. Grading: Same as A. 1 point if the answer wasn t checked. (b): This is a polynomial equation. The first thing to do is to check whether we can factor out a power of x. We can, and we get the equation: x(x + x + 4) 0, so x 0 or x + x To find the solutions to the second equation, we need to use the quadratic formula: x b ± b 4ac a ± ± 1 ± 1 i. So the answers are x 0 and x ± 1 i ; all of them check out. (Note that ± 1 i 1 ± i.) Grading: Same as A. (c): This can be done several different ways, but the basic idea is to let y x, since the exponents are all even. Then x 6 (x ) y, and so the equation becomes y y (y 1)(y 1) 0 so y 1. Now we need to find the solutions to x 1. This is tricker than I intended, and the answer x 1 got full credit. If you use the factoring rule for the difference of two cubes, you find out that: x 1 0 (x 1)(x + x + 1) 0, so x 1, or x + x (which was solved in A ). Grading: Same as A. 7

8 A 6. Write the following expressions in the form a + bi: a. [4 pts] (1 + )(1 ) b. [4 pts] + i 1 + i Solution: (a): (1 + )(1 ) (1 + i )(1 i ) 1 i + i i 6 (1 + 6) + ( )i. Grading: + points for evaluating X, + points for FOILing out and combining terms. (b): + i 1 + i + i 1 + i 1 i 1 i ( + i)(1 i) (1 + i)(1 i) 6i + i 9i 1 9i i i i Grading: + points for multiplying the numerator and denominator by the conjugate, + points for the rest of the arithmetic. B 6. Write the following expressions in the form a + bi: a. [4 pts] (1 + 5)( + ) b. [4 pts] 8 i 4 + 4i Solution: (a): (1 + 5)( + ) (1 + i 5)( + i ) + i + 5 i + i 15 ( 15) + ( + 5)i. Grading: + points for evaluating X, + points for FOILing out and combining terms. Grading for common mistakes: + points (total) for i. (b): 8 i 4 + 4i 8 i 4 + 4i 4 4i 4 4i i 4i + 4i 16 16i (8 i)(4 4i) (4 + 4i)(4 4i) 8 6i i. Grading: + points for multiplying the numerator and denominator by the conjugate, + points for the rest of the arithmetic. Grading for common mistakes: 1 point for

9 A 7. [5 pts] Find all real values of x such that (x + )(x ) x + 1 Solution: First, we need to see where the expression above changes sign (from positive to negative, or vice versa). This will happen only when the numerator is zero, or when the denominator is zero; when x,, or 1. This divides the real number line into four pieces: the x s less than, the x s between and 1, the x s between 1 and, and the x x bigger than. We choose a number inside each interval and test it: x : x 1.5 : x 0 : x 4 : 0. (( ) + )(( ) ) ( 1)( 6) < 0 ( ) + 1 (( 1.5) + )(( 1.5) ) (0.5)( 4.5) > 0 ( 1.5) ((0) + )((0) ) 6 < 0 (0) + 1 ((4) + )((4) ) (6)(1) > 0 (4) So we want the x s between and 1, and the x s bigger than. We need to determine which of the endpoints we want. Note that we cannot divide by 0, so x 1 is NOT a solution. The other two endpoints are. The answer is: all x such that x < 1 or x, or [, 1) [, + ) in interval notation. Grading: + points for finding where the expression changes sign, + points for finding the intervals, and +1 point for determining whether to include the endpoints. B 7. [5 pts] Find all real values of x such that (x 4)(x + 1) x + Solution: First, we need to see where the expression above changes sign (from positive to negative, or vice versa). This will happen only when the numerator is zero, or when the denominator is zero; when x 4, 1, or. This divides the real number line into four pieces: the x s less than, the x s between and 1, the x s between 1 and 4, and the x x bigger than 4. We choose a number inside each interval and test it: x 4 : x : x 0 : x 5 : (( 4) 4)(( 4) + 1) ( 4) + (( ) 4)(( ) + 1) ( ) + ((0) 4)((0) + 1) (0) + ((5) 4)((5) + 1) (5) + 0. ( 8)( ) ( 1) ( 6)( 1) 1 ( 4)(1) () (1)(6) (8) So we want the x s less than, and the x s between 1 and 4. We need to determine which of the endpoints we want. Note that we cannot divide by 0, so x is NOT a solution. The other two endpoints are. The answer is: all x such that x < or 1 x 4, or (, ) [ 1, 4] in interval notation. Grading: See A. < 0 > 0 < 0 > 0 9

10 A 8. [5 pts] Find all real solutions to the equation x x + 4 Solution: Note that y is either y or y, so we will replace x + with x + for one equation, and (x + ) for the other. x(x + ) 4 x + x 4 x + x 4 0 (x + 4)(x 1) 0 x 4, 1 or x( 1)(x + ) 4 x + x 4 x + x The second equation has no real solutions (the discriminant b 4ac < 0), so the only possible solutions are 4 and 1. We need to check to see whether these really are solutions. Only x 1 is a solution. Grading: + points for replacing with and to get two equations, + points for finding the solutions, +1 point for checking the answers. B 8. [5 pts] Find all real solutions to the equation x x 4 4 Solution: Note that y is either y or y, so we will replace x 4 with x 4 for one equation, and (x 4) for the other. x(x 4) 4 x 4x 4 x 4x 4 0 or x( 1)(x 4) 4 x + 4x 4 0 x 4x (x )(x ) x To solve the first equation, we need to use the quadratic formula: x b ± b 4ac a 4 ± ±. Now we need to check to see whether these three numbers are actually solutions. Substituting them for x in the original equation, we find that and 4 + are solutions, and 4 is not. Grading: + points for replacing with and to get two equations, + points for finding the solutions, +1 point for checking the answers. 10

11 A 9. [10 pts] Marcy has $.0 in nickels, dimes, and quarters. If she has three times as many dimes as quarters and four more nickels than dimes, how many of each coin does she have? Solution: Let d be the number of dimes that Marcy has. She has three times as many dimes as quarters, so the number of quarters she has is d. She has four more nickels than dimes, so she has d + 4 nickels. Now, the total amount of money she has is $.0, which is 0 cents. Since dimes are 10 cents, nickels are 5 cents, and quarters are 5 cents, 10(d) + 5 Now we need to solve the equation for d: ( ) d + 5(d + 4) 0. ( ) d 10d d d + 5d + 15d ( )d d 60 d 9 So she has 9 dimes, quarters, and 1 nickels. Grading: + points for writing down how many dimes, quarters, and nickels she has, + points for setting up the equation, + points for solving the equation for d, + points for the final answer. Grading for common mistakes: 1 point if the number of quarters and nickels were not given; points for the right answer with no work. B 9. [10 pts] A plumber and his assistant work together to replace the pipes in an old house. The plumber charges $50 per hour for his own labor and $0 per hour for his assistant s labor. The plumber works three times as long as his assistant on this job, and the labor charge on the final bill is $415. How long did the plumber and his assistant work on this job? Solution: Let h be the number of hours that the assistant worked. Then the plumber worked h hours. The assistant gets $0 per hour, so the assistant earned 0h dollars; the plumber worked h hours and earned $50 per hour, so the plumber earned 50(h) dollars. Since they earned $415 together, 0h + 150h 415, so 180h 415, or h Thus the plumber worked 70. hours, and the assistant 180 worked.4 hours. Grading: + points for writing down how much the plumber and his assistant earned, + points for setting up the equation, + points for solving the equation for h, + points for the final answer. Grading for common mistakes: 1 point if the amount of time was not given for both people; points for the right answer with no work. 11

12 A 10. [10 pts] A gardener has 160 ft of deer-rististant fence. She wants to enclose a rectangular vegetable garden in her backyard, and she wants the area enclosed to be at least 1500 ft. What range of values is possible for the length of her garden? Solution: If we let l be the length of the garden, and w the width, then the gardener wants to have lw There is a restriction that she can only use 160 ft of fence, so w + l 160. If we solve for w, we get: w + l 160 w 160 l w 160 l 80 l. Since we want to have lw 1500, this means that l(80 l) Now we need to solve this inequality for l. To do this, we will first find out where the inequality is actually equal: l(80 l) l l l 80l (l 0)(l 50), l 0, 50. These two values divide the real number line into three pieces: l s less than 0, l s between 0 and 50, and l s larger than 50. If we choose a number from each region, we find that: l 0 : (0)(80 (0)) 0 < 1500 l 40 : (40)(80 (40)) 1600 > 1500 l 100 : (100)(80 (100)) 000 < 1500, so we want the l s between 0 and 50 (including the endpoints), or [0, 50] in interval notation. Grading: +5 points for setting up the inequality (for one variable), +5 points for solving it. Grading for partial credit: +5 points (total) for at least one solution; +7 points (total) for various solutions. B 10. [10 pts] A riverboat theater offers bus tours to groups on the following basis. Hiring the bus costs the group $550, to be shared equally by the group members. Theater tickets are $0 each, minus 50c/ for every person in the group. How many members must be in the group so that the cost of the theater tour (bus fare plus tickets) is less than $45 per person? Solution: Let n be the number of people. The tickets will cost 0.5n dollars each, so the total for the tickets (for everyone) will be n(0.5n). The bus costs 550 dollars no matter how many people there are, so the total cost is n(0.5n). The average cost will be this divided by n, and we want this to be less than 45 dollars, so we want n(0.5n) n This is the inequality that we need to solve. < 45, or n(0.5n) < 45n. (Continued...) 1

13 To solve it, first we find out where it is equal: n(0.5n) 45n n.5n 45n 0.5n + 15n 550 n 15 ± (15) 4(.5)( 550) (.5) 15 ± , We can ignore the negative solution, because clearly n 0. We have two regions of the number line to look at: numbers between 0 and 1.401, and numbers bigger than If we take a test point from each: n 1 : (1)(0.5(1)) (1) n 0 : (0)(0.5(0)) < (0). So we want all the n s bigger than In fact, since n must be an integer, n must be at least. Grading: +5 points for setting up the inequality, +5 points for solving it. Grading for common mistakes: points for n 9.5 < 45n, 1 point for 1.4 people. Note: I originally got an answer of 51.4 (or 5) for this problem and graded it that way. I will add points to the grade of everyone who took B. A 11. [10 pts] A large pond is stocked with fish. The fish population P grows according to the formula P t + t + 0, where t is the number of days since the fish were first introduced into the pond. How many days will it take for the fish population to reach 80? Solution: We want to find out when the fish population (P ) is 80, so we need to solve the equation t + t : t + t t + t 60 0 t ± ( ) (1) ± , We can discard the negative solution because the context implies t > 0. Thus, the population is 80 after 6.898, and so will need 7 days to reach 80. Grading: +5 points for setting up the inequality, +5 points for solving it. B 11. [10 pts] A square plot of land has a building 0 ft long and 45 ft wide at one corner. The rest of the land outside the building forms a parking lot. If the parking lot has an area of 000 ft, what are the dimensions of the entire plot of land? (Give your answer to the nearest tenth of a foot.) Solution: The area of the land is x, where x is the length of one side. It is also equal to the area of building (a rectangle which is 0 ft by 45 ft) plus the area of the parking lot (000). Thus x , so x 450, which makes x ± 450 ±65.95 ft. We can discard the negative solution since measurements are postive. The plot of land is ft by ft. Grading: +5 points for setting up the inequality, +5 points for solving it. 1