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PRE-CALCULUS TEAM QUESTIONS Sponsors' Copy : February 27, 2016 Sickles Invitational 1. Arc-functions have the traditional restricted range. 7 5 A = Arc sin sin Arc cos cos . Write A in fraction form. 6 6 3 3 B = sin(2 x) for sin x and x . Write B in fraction form. 5 2 2 7 C = sin for sin and 0 . Write C in fraction form. 2 4 2 h k 1 3 when sin , 0 and sin , . 2 2 5 2 10 ----------------------------------------------------------------------------------------------------------------------------- D = h k for sin( ) = 2. f ( x) a cos(bx) 2 for a 0 . g ( x) sin(2 x c ) d for c 0 . Ferris Wheel, for parts C and D A = a b if the minimum value of f is 14 , and the period of the graph of f is . 4 to the right. 8 C = 2m n if d (t ) m cos(4t ) n gives the distance above ground in feet that a rider of a Ferris wheel is, if the rider's seat ranges from 2 feet to 200 feet above ground. m 0 . d (t ) is measured in feet above ground. D = p q if D(t ) p cos(qt ) 3 gives the distance above ground in feet that a rider of a Ferris wheel is, if the wheel makes a complete rotation in 30 seconds, t is measured in minutes, and the rider has a maximum height of 20 feet above ground. D(t ) is measured in feet above ground. p 0 . B = c d if the minimum value of g is 12, and the phase shift of g is PRE-CALCULUS TEAM QUESTIONS Sponsors' Copy : February 27, 2016 Sickles Invitational ---------------------------------------------------------------------------------------------------------------------------R U 3. 12 10 42 30o o S 10 T V 12 W Acute RST and obtuse UVW have SR=VW=12, ST=UV=10, mS 42o , and mW 30o . mT mR mS . Triangles may not be drawn to scale. A = the value of p q for ( RT )2 p q , in RST , using for sin 42o . B = the length of the shortest altitude of RST , using 2 as an approximation 3 2 as an approximation for sin 42o . 3 Write B as a reduced fraction. C = the value of h k , given that UW= h k . V is the obtuse angle of UVW . D = the y-coordinate for vertex U, if V is at the origin on the xy-coordinate plane, and W is at the point (12, 0) on the x-axis. 4. E1 : 9 x 2 y 2 18 x 6 y 9 0 E2 : x 2 4 y 2 40 y 104 0 . A = the larger y-coordinate of the two foci of the conic determined by E1 above. B = the coordinates of the endpoint of the transverse axis of the conic determined by E2 above, written as (a, b) with a 0 . Put your answer as a coordinate pair. C = the sum of the focal radii of the conic determined by E1 above. D = the product of the eccentricities of the two conics defined by E1 and E2 above. Write D as a reduced fraction. ---------------------------------------------------------------------------------------------------------------------------5. E1 : y 2 4 x 6 y 1 0 . E2 : ( x 2) 2 8 y 40 . E3 : x 2 y 2 2 x 2 y 9 0 A = the sum of the x- and y-coordinates of the focus of the parabola defined by E1 . B = the sum of the x- and y-coordinates of the focus of the parabola defined by E2 . C = the distance between the vertex and focus of the graph of the parabola with 3 directrix equation y x 5 and focus (10, 5). 4 D = the area of the circle with equation E3 . PRE-CALCULUS TEAM QUESTIONS Sponsors' Copy : February 27, 2016 Sickles Invitational ---------------------------------------------------------------------------------------------------------------------x 6. x 1 f ( x) 3 . 1 1 g ( x ) . h( x ) . log 3 x 9 A = the real solution to the equation 81 f ( x) g ( x) . B = the real solution to the equation h( f ( x)) g (1) , written in reduced fraction form. C = f 1 (3 3) g 1 ( 3) , for f 1 and g 1 the inverses of the functions f and g respectively. D: h1 ( x) 3m( x ) for h 1 the inverse of h . Give an expression for m( x) in terms of x . --------------------------------------------------------------------------------------------------------------------7. f ( x) 4 x3 14 x 2 6 x . g ( x) x3 13x 2 112 x 100 . h( x) a4 x 4 a3 x 3 a2 x 2 a1 x a0 A = the coefficient of the quadratic term, a2 , of the function h above, given that a4 1 and the roots of h are 3 2i, 3 2i, i and i . B = the value of r2 given that the roots of f are r1 , r2 and r3 , and r1 r2 r3 . C = the sum of absolute values of the roots of g . D = sum of the roots of the functions y f ( x 1) , y g ( x 1) and y h( x 1) . --------------------------------------------------------------------------------------------------------------------- 8. Consider the function y f ( x) defined by parametric equations x t 1 and y t 2 2t 2 , for t 0 . A = the minimum y-value of the graph of y f ( x) for the function defined above. B = the change in x-coordinates as the y-coordinate changes from 2 (at t 0 ) to 10. C = the value of k if the directrix of the graph of y f ( x) has equation y k . (Note: consider all real values of t for the graph of f for this part.) D: the graph of y f ( x) for t 0 is reflected over the y-axis , and the corresponding parametric equations for the reflected function are x2 a1t a2 and y2 a3t 2 a4t a5 and t 0 . D = a1 a2 a3 a4 a5 . ---------------------------------------------------------------------------------------------------------------------5 ab 9. A = the value of for 1 i 3 a bi and i 1 . 16 B = the sum of the distinct y-intercepts when the polar graph r 12 4sin is graphed on the xy-plane. C = the area of the graph of r 8cos . D = the distance between points P and Q on the graph of r 2 9sin(2 ) , given that PRE-CALCULUS TEAM QUESTIONS Sponsors' Copy : February 27, 2016 Sickles Invitational the maximum radius of the graph occurs at points P and Q for 0 2 . ------------------------------------------------------------------------------------------------------------------------10. A = the sum of p and q for the unit vector p, q which has the same direction as 6,8 . B = the angle between the vectors a 2.5 3, 2.5 and b 3, 3 3 . C = m n so that vector 3, 4 is perpendicular to vector m, n and the length of m, n is twice that of 3, 4 . 7 and magnitude 4. 12 -------------------------------------------------------------------------------------------------------------------------ax 2 bx 40 11. f ( x) for a and c relatively prime integers. cx 2 d 2 The graph of y f ( x) has asymptotes at x 4 , y , and 3 a removable discontinuity at x 4 . D = the sum r s , of the vector r , s with direction A = a b for the function f described above. B = c d for the function f described above. C = the x-intercept of the graph of f described above. D = the y-intercept of the graph of f described above. Write D in reduced fraction form. -------------------------------------------------------------------------------------------------------------------------12. A is the sum of the values of a for 0 a 2 so that 2 cos(a ) 1 0 . B is the sum of the values of b for 0 b 2 so that 1 cos 2 b 3 cos(b) . C is sum of the values of c for 0 b 2 so that sin(2c) tan c . D is the product of all values of cos x that correspond to each value of x in 16 16 . That is, if each solution to sin 2 x is xn ,cos( xn ) 0 x 2 for which sin 2 x 25 25 for 0 xn 2 , give the product of all values of cos( xn ) . ----------------------------------------------------------------------------------------------------------------------------- 13. A = sin cos . 12 12 B = the angle of inclination (with the positive x-axis: 0 B ), in radians, of the line with equation 3 x y 3 . 1 C = the value of cos 2 sin 2 for so that 0 and sin cos . 2 3 pq 2 14 4 D = p q for sin cos if tan 2 , and 0 . 9 4 7 PRE-CALCULUS TEAM QUESTIONS Sponsors' Copy : February 27, 2016 14. Diagram 1 shows a semi-elliptical arch with base width 600 feet. The center of the arch is 100 feet high. Diagram 2 shows a parabolic arch with base width 600 feet. The center of the arch is 100 feet high. Sickles Invitational 100 ft 100 ft 600 600ftft Diagram 1 600 ft Diagram 2 Both diagrams have vertical symmetry in the center of the diagram. A = the height in feet of the arch in diagram 1, ten feet to the right of the center. "Ground level" is the major axis of the ellipse. B = the height in feet of the arch in diagram 2, ten feet to the right of the center, (All heights are of outer arch shown.) C = the distance x in feet horizontally from the center of diagram 1, if the height at that distance x is 50 feet above ground. D = the distance x in feet horizontally from the center of diagram 2, if the height at that distance x is 50 feet above ground. 15. E1 : x 2 4 y 2 2 x 4 y 1 0 . E2 : x 2 y 2 2 x 4 y 5 0 E3 : y 2 x 2 y 4 0 E4 : x 2 y 2 10 x 2 y 30 0 Statement S: The equation produces a graph on the xy-plane = 2 points. Statement T: The equation produces no graph on the xy-plane = 3 points. Statement U: The equation represents a parabola with a vertical axis of symmetry = 4 points. Statement V: The equation produces a parabola which opens to the right = 5 points. A = the number of points that would be assigned to E1 based on the four statements S, T, U and V. More than 1 statement may apply to E1 . B = the number of points that would be assigned to E2 based on the four statements S, T, U and V. More than 1 statement may apply to E2 . C = the number of points that would be assigned to E3 based on the four statements S, T, U and V. More than 1 statement may apply to E3 . D = the number of points that would be assigned to E4 based on the four statements S, T, U and V. More than 1 statement may apply to E4 . PRE-CALCULUS TEAM QUESTIONS Sponsors' Copy : February 27, 2016 Sickles Invitational ANSWERS: (fraction) means answer must be in reduced fraction form. 2. A= 24 2 (fraction) 3 24 53 B= (fraction) B= or 13.25 25 4 C= 299 2 C= (fraction) 4 D = 23 D = 21 Thrown out 3. A = 32244 ------------------------5. A = 2 B = 9 ------------------------7. A = 14 1 B= 2 C = 21 1. A = C= 3 D = 11 ------------------------9. A= 1 3 (no i ) B = 8 C = 16 D=6 ------------------------2 13. A= 2 2 B= 3 17 C= 9 D = 17 20 (fraction) 3 C = 116 B= D = 43 3 ------------------------6. A = 1 10 B= (fraction) 9 9 C= or 2.25 4 1 or x 1 D= x ------------------------1 10. A = 0.2 or 5 B= 6 C=2 D= 2 2 ------------------------10 899 14. A = 3 899 8 B= or 99 9 9 C = 150 3 1 1. A: Arc sin . 6 2 D = 150 2 Arc cos( 3 / 2) 25 or 12.5 2 ------------------------11. A = 0 D= B = 45 C=5 5 D= (fraction) 6 ------------------------15. A=2 B=2 C=2 D=3 5 2 . Sum = 6 3 4. A = 3 2 2 B = (2, 5 ) C=6 10 3 ------------------------8. A = 1 B=4 D= 3 or 0.75 4 D = 1 C= ------------------------12. A = 2 B = 2 C = 5 81 D= 625 ------------------------- PRE-CALCULUS TEAM QUESTIONS Sponsors' Copy : February 27, 2016 B = 2sin x cos x 2 Sickles Invitational 3 4 24 for the angle x with terminal side in Quadrant II. 5 5 25 3 4 4 3 2 . Note: Use cos x 1 2sin 2 x . Find cosx= 3 . 4 2 8 4 2 1 C= 1 4 3 3 4 3 3 h k = . 4 27 h k . h+k= 23. 2 5 2 5 10 10 10 D thrown out: not specified as integers 2 . b 8 . a b = 24 . 2. A: the graph will go from a 2 to a 2 . a 2 14 . a 16 . b 4 c 1 53 so c . B = 13.25 or B: The function has range 1 d , so 1 d 12 . d=13. . 2 8 4 4 C: m n =2 and m n 200 . Solve to get n=101. m=99. 2m+n=299. 2 1 D: We want a period of 1/2 minute so . q 4 . p+3=20. p=17. p+q=21 q 2 D = sin( ) 3. A: Use the law of cosines to get ( RT )2 144 100 2(120) cos 42o . Using 2/3 for sin 42o 5 . ( RT )2 244 80 5 p q p=244 and q=6400(5)=32000 3 and p+q=32244. 1 2 B: Area = (12)(10) 40 . Since mT mR mS , T is the largest angle so 12 is the 2 3 1 20 largest side, and the shortest altitude h gives (12)h 40 . h 2 3 C: Use the two right triangles shown to the U right to get UW=8+ 6 3 . h k = 8 36(3) 10 6 h+k = 8+108 = 116. D: Drop the altitude on the diagram to the right 30o and use the results from part C to get the height V W from U is half of 8+ 6 3 or 4 3 3 . ( x 1)2 ( y 3) 2 4. E1 : 9 x 2 y 2 18 x 6 y 9 0 . 9( x 2 2 x 1) ( y 2 6 y 9) 9 9 9 . 1. 1 9 x 2 ( y 5)2 E2 : x 2 4 y 2 40 y 104 0 . x 2 4( y 2 10 y 25) 104 100 . 1. 4 1 A: Foci will be 9 1 2 2 from the vertex, vertically. Largest y-value is 3 2 2 . B: The transverse axis is horizontal with endpoints (2, -5) and (-2, -5). Answer = (2, -5). C: The sum of the focal radii of an ellipse is 2a which is 2(3)=6. 2 2 5 10 D: eccentricity is c/a so for E1 this gives . For E2 this gives . D= . 3 2 3 gives cos 42o PRE-CALCULUS TEAM QUESTIONS Sponsors' Copy : February 27, 2016 Sickles Invitational 5. E1 : y 2 4 x 6 y 1 0 . ( y 3)2 4 x 1 9 . ( y 3)2 4( x 2) . V( -2, 3), opens right. E2 : ( x 2)2 8 y 40 . E2 : ( x 2)2 8( y 5) . V(-2, -5). Opens down. E3 : x 2 y 2 2 x 2 y 9 0 . ( x 1)2 ( y 1)2 11 . A circle with C(-1, 1), r= 11 A: focus is 1 unit to the right of V(-2, 3) which gives (-1, 3). Sum = 2. B: focus is 2 units down from (-2, -5) which gives (-2, -7). Sum = -9. C: Distance from a point to a line using the formula with line 3x-4y= -20 and point 3(10) 4(5) 20 (10, 5). 6 4 4 6 . Half of this is vertex to focus, which is 3. 9 16 D= 11 2 11 1 . Base 3 gives powers 4 x 1 2x gives x= -1. 9x 1 1 1 10 9 . 9x 9 1 . x= . 9. B: h( f ( x)) . g(-1)=9. x 1 x1 x 1 9 log 3 3 log 3 3 6. A: 81 3x 1 x 3 5 1 1 C: 3 3 3 . Powers of 3: x 1 . f 1 (3 3) = . 3 . Powers of 3: 2 x 2 2 9 2 1 9 so g 1 ( 3) = . Sum = . 4 4 1 1 1 D: x . x log3 y 1 . log3 y . y 31/ x . D= . x x log 3 y x1 7. A: Sum of 3 2i is 6= -b/a. Product is 13=c/a. So a quadratic factor is x 2 6 x 13 . Similarly the other two roots give x 2 1 . Multiply to get a quadratic factor of 1+13=14. 1 1 B 2 x(2 x 2 7 x 3) 2 x( x 3)(2 x 1) so r1 r2 r3 gives r1 0, r2 , r3 =3. B= r2 = . 2 2 2 C: Since g(1)=0, divide synthetically by x-1 to get x 12 x 100 . Use the Quadratic Formula 12 144 400 to get and divide by 2 to get 6 36 100 6 8i . Absolute values are 2 1, 62 82 10 and 62 82 10 for a sum of 21. D: Since the (x+1) makes each root 1 less (corresponding to moving the graph to the left 1), making each root 1 less, we just add -1 to each root. (Note: you can check that the non-real roots also are made less by 1 by comparing x 2 1 to ( x 1)2 1 and comparing the roots i to 1 i . Back to the problem: Sum of the roots is –b/a. f sum is 7/2. g sum is 13. h sum is 6. Subtract 10 for the number of roots, to get 7/2 + 13 + 6 -10 = 12.5 or 25/2. 8. A: y ( x 1)2 2( x 1) 2 x 2 4 x 5 . y is min at x= -b/(2a)=2. y= 4 8 5 1. Check that time is in the given interval: (2, 1) gives time 1. So A=1. B: time changes from 0 to 4, since t 2 2t 2 10 , (t 4)(t 2) gives t=4. x changes from 1 to 5. B=4. C: y x 2 4 x 5 ( x 2)2 1 when you complete the square. Directrix is 1/4 unit down from PRE-CALCULUS TEAM QUESTIONS Sponsors' Copy : February 27, 2016 Sickles Invitational vertex (2, 1) to get y=0.75. D: replace x with (-x) to reflect over the y-axis. So we get x t 1 and the y-values remain the same at y t 2 2t 2 . D= -1+-1 + (1-2+2)= -1. 5 5 9. A: Using DeMoivre's Theorem, 1 i 3 = 2cis = 32cis 3 3 1 ab 3 32 1 3 32 cis 32 i , this is . Since we want 1 3 16 3 2 16 2 2 2 3 B: y-intercept is where or or coterminal angles with them. r= 8 or 16. Since at 2 2 3 the graph is at y= -16, the sum is 8-16= -8. To verify we can change the equation 2 5 to rectangular x 2 y 2 12 x 2 y 2 4 y . The y-intercept occurs at x=0 so y 2 12 y 4 y . Either y 2 8 y 0 or y 2 16 y 0 . y 0 so y = 8 or -16, for a sum of -8. C: This is a circle with diameter 8, so area is 16 . D: This is a lemniscate and a max radius is at where sine is at its maximum. 4 Points P and Q must have a radius of 9 3 so distance is 6. 1 6 8 , 10. A: magnitude is 10 so the vector is for a sum of . 5 10 10 7.5 3 7.5 3 3 = 3 . B= 56 2 6 C: Reverse the components and negate one of them to get a perpendicular. Since the magnitude of the original vector is 5, we want magnitude 10. So 8, 6 or 8,6 both B: cos give a sum of components with absolute value 2. 7 D: For cosine and sine of , you can use the half angle identities or addition of and 12 4 7 6 2 7 6 2 , or memorization. cos . sin . 4 times the sum is 2 2 12 4 12 4 3 11. a=2 and c=3 since the horizontal asymptote is y=2/3. The denominator must be 3( x 4)( x 4) since there is a vertical asymptote x=4 and a removable discontinuity at x= -4. So d= -48. The numerator is 2( x 4)( x ___) due to the removable discontinuity and since we need to multiply to get constant -40 so the gives 2( x 4)( x 5) to give an x-intercept of 5. That makes the numerator 2 x 2 2 x 40 and b= -2. 40 40 5 . A = a b 2 2 0 . B = c d 3 48 45 . C = 5. D = d 48 6 PRE-CALCULUS TEAM QUESTIONS Sponsors' Copy : February 27, 2016 Sickles Invitational 1 2 4 2 . 12. A: cos a . 2 3 3 sin b B: sin b 3 cos b . 3 . cosb must be positive but sinb can be either sign. So in cos b 5 QI or QIV, tan b 3 . b or for a sum of 2 . 3 3 2 sin c 3 5 7 C: 2sin c cos c . sin c 0 or 2 cos 2 c 1 . cos c . 0, ; , , , . Sum of 5 . cos c 4 4 4 4 2 16 4 3 D: sin 2 x sin x which occurs four times in 0 x 2 and cos x is twice and 25 5 5 3 81 twice for a product of . 5 625 2 3 2 1 2 6 2 6 2 13. A = sin . and cos ; the difference . 2 4 4 3 4 2 2 2 2 3 4 2 B: slope is 3 = tan . The angle of inclination for a negative is . 3 3 1 1 8 C: Square both sides of sin cos to get 1 sin(2 ) . sin(2 ) . 3 9 9 1 Since cos 2 sin 2 = (cos sin )(cos sin ) = (cos sin ) and since is in 3 17 8 Use sin(2 ) to get cos(2 ) . Due to the absolute value, we don't have to worry 9 9 about the sign of the radical. 2 14 7 in QI, sin(2 ) (using the Pythagorean Th.) and cos(2 ) 3 7 3 2 2 2 2 11 2 2 p q 2 (sin cos ) 2 1 sin(2) 1 . Square: 1 = . p q 17 . 3 9 3 9 3 9 x2 y2 14. A: Set center ground at (0, 0). 1 is the entire ellipse. x=10 gives 3002 1002 10 899 100 y2 899 y2 1 . . y 3 300 300 100 100 900 100 100 2 B: Set center ground at (0, 0). y ax 100 . Use (300, 0) to get 300(300)a 100 1 899 (100) 100 . 99 8/9 feet or a= -1/900. x=10 gives y . 900 9 300 3 x2 50 50 x2 3 150 3 1 C: See part A. . .x= 2 2 2 300 100 300(300) 4 D: tan 2 PRE-CALCULUS TEAM QUESTIONS Sponsors' Copy : February 27, 2016 Sickles Invitational 1 2 x2 x 100 . y=50. 50 100 . y 50 900 150 2 900 900 E1 : x 2 4 y 2 2 x 4 y 1 0 . ( x 1)2 4( y 0.5)2 1.75 . hyperbola with a graph. D: See part B. y 15. E2 : x 2 y 2 2 x 4 y 5 0 . ( x 1)2 ( y 2)2 0 . Graph of a single point. E3 : y 2 x 2 y 4 0 . ( y 1)2 1( x 3) . Parabola, open to the left. E4 : x 2 y 2 10 x 2 y 30 0 . ( x 5)2 ( y 1)2 4 . No graph. A : E1 has 2 for statement S. A=2. B : E2 has 2 points for statement S. B=2. C : E3 has 2 points for statement S. C=2. D : E4 .has 3 points for statement T. D=3.