MCV4U - Practice Mastery Test #9 & #10

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1 Name: Class: Date: ID: A MCV4U - Practice Mastery Test #9 & #10 Multiple Choice Identify the choice that best completes the statement or answers the question. 1. x + 2 is a factor of i) x 3 3x 2 + 6x 8 ii) x a. i only b. ii only c. i and ii d. neither 2. If f(x) = 2-5x and f(a) = -13 then a = a. -3 b. 3 c d Which of the following limits do not exist? i) lim 3 x ii) lim 3 x iii) lim 3 x x 3 + x 3 x 2 a. i) only b. ii) only c. iii) only d. i) and iii) x + h x 4. The expression lim h 0 h is most likely to be... a. the derivative of a function c. slope of a secant b. the value of a derivative d. none of the above 5. If s(t) = t 3 + 2t then v(t) = a. 3t b. 3t 2 c. 6t + 2 d t 6. The number lines for y,, and d 2 y are shown below. All zeros are shown. Assume the function is 2 continuous for all x ò. U means undefined y U d 2 y U The x-coordinates of all points of inflection are a. -3, -1 and 1 b. -3, -1 and 3 c. -3 and 1 d. -3 and 3 e. none 1

2 Name: ID: A 7. If y = 3x 2 ˆ ( x + 3), then = a. ( 6x) ( x + 3) + ( 1) 3x 2 ˆ c. ( 6x) ( 1) + ( x + 3) 3x 2 ˆ b. ( 6x)(1) 3x 2 ˆ ( x + 3) d. ( 6x ) 3x 2 ˆ + ( x + 3) 8. If y = ( 3x + 2) x 2 ˆ, then = a. ( 3x + 2) x 2 ˆ + ( 3) ( 2x + 3) c. ( 3) ( 2x + 3) + ( 3x + 2) x 2 ˆ b. ( 3) ( 2x + 3) ( 3x + 2) x 2 ˆ d. ( 3) x 2 ˆ + ( 2x + 3) ( 3x + 2) 9. Given f(x) = 4 x and g(x) = x + 2. If k(x) = f û g(x), then the domain of y=k(x) is... Ï a. Ô Ì x x 2 or x 2, x ò Ô Ï ÓÔ Ô c. Ô Ì x 2 x 2, x ò Ô ÓÔ Ô Ï b. Ô Ì x x 2, x ò Ô Ï ÓÔ Ô d. Ô Ì x x 2, x ò Ô ÓÔ Ô 10. If y = 3x 2 5 ˆ, then = a. 5 3x 2 4 ˆ 4 ( 6x) c. 5( 6x) 3x 2 ˆ b. 3x 2 4 ˆ 4 ( 6x) d. 5( 6x ) 11. Complete the identity tanx = a. b. 1 c. sin 2 x d. sinx cos 2 x 12. If y = cos 4x 2 ˆ then a. = 8x cos 4x 2 ˆ b. = 8x sin 4x 2 ˆ c. = 4x 2 ˆ cos( 8x) + 8x sin 4x 2 ˆ d. = 4x 2 ˆ sin( 8x) 8x cos 4x 2 ˆ 13. What is the next number in the sequence 10,30,90,270,...? a. 810 b. 450 c. 360 d

3 Name: ID: A 14. If y = 6ln(x), then = a. 6 x b. 6 6x (It is not necessary to state restrictions) c. 6 x d. 6 x 15. If y = 5e 2x, then = (It is not necessary to state restrictions) a. 10e 2x b. 10xe x 2 c. 2e 2x d. 10e 2x 16. In the prism shown, FA to... is equal a. a + b + c b. a + b c 17. Given a 18. Given a 19. Given a = ( 1,2, 4) then a = c. a a. 21 c. 7 b. 3 d. 21 = (2, 3, 2) then 3a b c d. a b + c is equal to a. 6, 9, 6 ˆ c. 6, 9,1 ˆ b. 5,0, 6 ˆ d. 5, 9,1 ˆ = (1,2, 3) and b = (0, 2,3) then 2a b is equal to ˆ ˆ a. 2,6, 9 ˆ c. 2,6,3 b. 2,2, 9 ˆ d. 2,6,3 3

4 Name: ID: A 20. Given the following vectors where aä = 10 and b ä = 9 then a b is approximately a b. 45 c d Given the following vectors where aä = 8 and b ä = 9 then a b is approximately a b. 36 c d Given the following vectors where aä = 6 and b ä = 8 then a b is approximately a b c d

5 Name: ID: A 23. In ò 3,the system of equations x y + z = 3 3x 3y + 3z = 9 defines a. a point b. a line c. a plane d. no points 24. Given eqtn #1: 2x + 2y + 2z = 2 and eqtn #2: r = 2, 2, 2 ˆ + t(0,0,1),t ò The point wth coordinates 2, 2, 1ˆ lies on the graphs of... a. both b. neither c. only eqtn #1 d. only eqtn #2 25. The equation r = (0,3, 1) + s(0,0,0) where s ò defines a. a point b. a line c. a plane d. no points 26. The equation r = (3, 2,1) defines a. a point b. a line c. a plane d. no points 27. The equation r = ( 3,2, 1) + s( 3,1,2) + t( 9,1,2) where s,t ò defines a. a point b. a line c. a plane d. no points 28. In ò 3,the system of equations 2x + 2y = 6 6x + 6y = 17 x + 3 = 0 defines a. a point b. a line c. a plane d. no points 29. Given eqtn #1: 2x + 2y + 2z = 5 and eqtn #2: r = 1,2,3 ˆ + t( 1,1, 1),t ò The point wth coordinates 2,3,1ˆ lies on the graphs of... a. both b. neither c. only eqtn #1 d. only eqtn #2 30. Given eqtn #1: 3x 2y + z = 9 and eqtn #2: r = 4,3, 1 ˆ + t(1, 2,2),t ò The point wth coordinates 3,1,1ˆ lies on the graphs of... a. both b. neither c. only eqtn #1 d. only eqtn #2 5

6 MCV4U - Practice Mastery Test #9 & #10 Answer Section MULTIPLE CHOICE 1. ANS: B if f(x) = x 3 3x 2 + 6x 8, then f( 2) = ( 2) 3 3( 2) 2 + 6( 2) 8 0 So...x + 2 is not a factor of x 3 3x 2 + 6x 8 If g(x) = x 3 + 8, then g( 2) = ( 2) = 0 So...x + 2 is a factor of x ANS: B The easiest way to solve these is probably to just guess and check. However, an algebraic solution follows: f(x) = 2-5x f(a) = 2-5a -13 = 2-5a 5a = 15 a = 3 3. ANS: A As x 3 +, x is greater than 3, so 3 x will negative, so exist. 3 x will be undefined, so lim x x does not As x 3, x is less than 3, so 3 x will positive, so lim 3 x = 0 x 3 3 x will be defined and will approach 0, so As x 2, from either side, x is less than 3, so 3 x will positive, 1, so lim x 3 3 x = 1 3 x will be defined, and will approach 1

7 4. ANS: A If f(x) = x, then the expression is of the form lim h 0 f(x + h) f(x) h which is the derivative. 5. ANS: A if s(t) = t 3 + 2t then v(t) = s (t) = 3t ANS: E At a point of inflection, the second derivative will change sign, (it changes from concave up/down to concave down/up), so there are none. There is a cusp at x= ANS: A y = 3x 2 ˆ ( x + 3), so = d dx 3x 2 ˆ ˆ ( x + 3) + d dx ( x + 3) ˆ ˆ 3x2 = ( 6x) ( x + 3) + ( 1) 3x 2 ˆ 8. ANS: D y = ( 3x + 2) x 2 ˆ, so = d dx ( 3x + 2) ˆ x 2 ˆ + d dx x 2 ˆ ˆ ( 3x + 2) = ( 3) x 2 ˆ + ( 2x + 3) ( 3x + 2) 2

8 9. ANS: D k(x) = f û g(x), then k(x) = f g(x) ˆ so k(x) = f g(x) ˆ = f( x + 2) = 4 ( x + 2) = 2 x For k(x) to be defined, 2 x 0 x 2 Ï The domain of y=k(x) is Ô Ì x x 2, x ò Ô ÓÔ Ô 10. ANS: A y = 3x 2 5 ˆ, so d 3x 2 5 ˆ = d 3x 2 ˆ d 3x 2 ˆ = 5 3x 2 4 ˆ ( 6x) 3

9 11. ANS: D We know that tanx = sinx tanx = = = = sinx sinx 1 sinx 1 sinx cos 2 x for all x. Substituting this in, we get Another method is to rearrange tanx = sinx (by multiplying both sides by 1 ) tanx = sinx 1 (tanx) = 1 tanx cosx = sinx cos 2 x sinx ˆ 12. ANS: B y = cos 4x 2 ˆ, so d cos 4x 2 ˆ = d cos 4x 2 ˆ d 4x 2 = d 4x 2 ˆ = sin 4x 2 ˆ 8x = 8x sin 4x 2 ˆ (chain rule) 4

10 13. ANS: A Each term is 3 times the previous term (check by dividing consecutive terms), so if we take 270 and multiply it by 3, we get ANS: C d ln(x) = 6 ˆ + d 1 = 6 1 x + 0 = 6 x 15. ANS: D 2xˆ d e = 5 d (2x) d ( 2x) = 5e 2x 2 = 10e 2x 16. ANS: C In the prism shown, to get from F to A, follow a path along the edges. Go from F to E, then E to H, and then H to A, so... FA = FE + EH + HA = a b c because FE is in the opposite direction to a, EH is in the opposite direction to b and HA is in the opposite direction to c 5

11 17. ANS: A Starting at the origin, the 1st component of the vector comes out towards us (if positive) or goes back into the page (if negative). In this case the 1 is shown in green. The 2nd component of the vector moves left or right from there. In this case the 2 is shown in blue. These two vectors form a right triangle with hypotenuse of length ( 1) 2 + ( 2) 2 When we move up or down using the 3rd component, the former hypotenuse forms a right triangle with the red vector. The new hypotenuse has length ( 1) 2 + ( 2) 2 + ( 4) 2 = 21 So the length of the vector in simplest form is ANS: A 3a + b = 3 2, 3, 2ˆ = 3(2),3( 3),3( 2) ˆ = 6, 9, 6ˆ 6

12 19. ANS: A 2a b = 2 1,2, 3ˆ 0, 2,3ˆ = 2(1),2(2),2( 3) ˆ + (0), ( 2), (3) ˆ = 2,6, 9ˆ 20. ANS: A Because we know the angle between the vectors, to find a a b = a b cos(θ) = (10)(9) cos(30 ) ANS: A Because we know the angle between the vectors, to find a a b = a b cos(θ) = (8)(9)cos(150 ) 62.4 b b, use the formula..., use the formula ANS: B Because we know the angle between the vectors, to find a b = a b sin(θ) = (6)(8) sin(35 ) 27.5 a b use the formula... 7

13 23. ANS: C Think of equations as restrictions on the freedom variables naturally have. Both of these equations define a plane in ò 3, so they would normally intersect in a line. However, in this case the 2nd equation is exactly 3 the first, so any point that satisfies the 1st equation will also satisfy the 2nd. This means that the planes intersect in a plane - with equation x y + z = 3 (or 3x 3y + 3z = 9). 24. ANS: A By substitution, we can see that 2, 2, 1ˆ satisfies eqtn#1. I.e., 2(2) + 2( 2) + 2( 1) = 2. If we use t=1, we see that the vector eqtn gives r = 2, 2, 2 ˆ + (1)(0,0,1) or r = 2, 2, 1 ˆ. So the vector eqtn also generates the point at 2, 2, 1ˆ. So... it lies on both. 25. ANS: A The equation r = (0,3, 1) + s(0,0,0) takes us from the origin out to the point (0,3, 1) and then travels from there along multiples of (0,0,0)... in other words it doesn t move from the point at (0,3, 1). So... this equation defines a point. 26. ANS: A The equation r = (3, 2,1) takes us from the origin out to the point (3, 2,1) and then stays there! So... this equation defines a point. 27. ANS: C The equation r = ( 3,2, 1) + s( 3,1,2) + t( 9,1,2) takes us from the origin out to the point ( 3,2, 1) and then travels from there along multiples of ( 3,1,2) and along multiples of ( 9,1,2). Note that the direction vectors in this equation are NOT parallel because ( 9,1,2) is not a multiple of ( 3,1,2). This means that adding multiples of ( 9,1,2) moves us off the line already formed by travelling along multiples of ( 3,1,2).... in other words the equation defines all the points on the plane through ( 3,2, 1) with direction vectors ( 3,1,2) and ( 9,1,2). 8

14 28. ANS: D Think of equations as restrictions on the freedom variables naturally have. All of these equations define a plane in ò 3, so they would normally intersect in a point. However, in this case the left side of the 2nd equation is exactly 3 the left side of the first, so any point that satisfies the 1st equation would satisfy 3 2x + 2yˆ = 3( 6), or 6x + 6y = 18. This means that any point on the 1st plane could NOT be on the second plane. Although the third equation defines a plane that does intersect each of the other two planes, this means that this system defines a set of NO points, because there are no points in common to all three planes. 29. ANS: B By substitution, we can see that 2,3,1ˆ does NOT satisfy eqtn#1. I.e., 2( 2) + 2(3) + 2(1) = 4 NOT 5. If we use t=1, we see that the vector eqtn gives r = 1,2,3 ˆ + (1)( 1,1, 1) or r = 2,3,2 ˆ. So the vector eqtn gives us the point at 2,3,2ˆ, but not 2,3,1ˆ. So... 2,3,1ˆ does not lie on the graphs of either eqtn. 30. ANS: D By substitution, we can see that 3,1,1ˆ does NOT satisfy eqtn#1. I.e., 3( 3) 2(1) + (1) = 8 NOT 9. If we use t=1, we see that the vector eqtn gives r = 4,3, 1 ˆ + (1)(1, 2,2) or r = 3,1,1 ˆ. So the vector eqtn does give us the point at 3,1,1ˆ. So... 3,1,1ˆ only lies on the graph of eqtn #2 9

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