Suggested Solutions Test MaC1NVC10 K1 Algebra & Functions NV-College©
[email protected] Not for sale. Free to use for educational purposes 1 Test MaC1NVCO10 on Algebra and Functions: Quadratic Equations, Exponential Equations, Logarithms, Simultaneous Equations Instructions Test period Oct 6, 2011; 13:00-15:35 Part I : You are not allowed to use calculator in part I. Part II: You may use calculator in part II. You are not allowed to share your calculator during test with another student. Phones are not allowed to be used as a calculator. Resources Formula sheet, your personalised formula booklet, ruler, protractor and a graphic calculator. The test For all items a single answer is not enough. It is also expected • that you write down what you do • that you explain/motivate your reasoning • that you draw any necessary illustrations. Score and The maximum score is (30G/36VG/37VG¤¤¤¤) points depending on the mark levels level of problems selected to solve) The maximum number of points you can receive for each solution is indicated after each problem. If a problem can give 2 ”Pass”-points and 1 ”Pass with distinction”-point this is written [2/1]. Some problems are marked with ¤, which means that they more than other problems offer opportunities to show knowledge that can be related to the criteria for ”Pass with Special Distinction” in Assessment Criteria 2000. Lower limit for the mark on the test G: Pass: 20 points VG: Pass with distinction: 30 points (minimum of 15 VG points) or 25 VG. MVG: Pass with special distinction: 30 points (minimum of 25 VG points) and ¤¤. Prob. 1a 1b 1c 1d 1e 1f 2a 2b 2c 2d 2e 2c 2d 2e G 2 3 1 2 1 2 2 VG 2 3 1 1 3 2 2 2 2 MVG¤ ¤ ¤ Prob. 2f 2g 2h 3a 3b 4a 4b 5 6 7 8a 8b 9 10 G 2 2 2 4 1 2 VG 3 3 2 2 4 4 3 4 MVG¤ ¤ ¤ ¤ Prob. 11a 11b 11c 12 13a 13b 13c 13d 14 sum limits G 1 1 2 30 20 VG 4 1 1 1 1 4 36 30/(V15) MVG¤ ¤ ¤ ¤ ¤ 37 30/(V25) Prob. 1a 1b 1c 1d 1e 1f 2a 2b 2c 2d 2e 2c 2d 2e G 2 3 1 2 1 2 2 VG 2 3 1 1 3 2 2 2 2 MVG¤ ¤ ¤ Prob. 2f 2g 2h 3a 3b 4a 4b 5 6 7 8a 8b 9 10 G 2 2 2 4 1 2 VG 3 3 2 2 4 4 3 4 MVG¤ ¤ ¤ ¤ Prob. 11a 11b 11c 12 13a 13b 13c 13d 14 sum limits Grade G 1 1 2 VG 4 1 1 1 1 4 MVG¤ ¤ ¤ ¤ ¤ Suggested Solutions Test MaC1NVC10 K1 Algebra & Functions NV-College ©
[email protected] Not for sale. Free to use for educational purposes 2 Note that most of problems have alternative. You will solve only one of those problems. In case you solve both alternatives, you will get best of two alternative grades, but not both. Part I: No calculator may be used in this part. - The solution for the part I must be written on a separate paper. - You may start working on part II without the access to any calculator. - Only after submission of the solution for the part I you may use your calculator. 1. Simplify the following expressions as far as possible: a) (G-alternative to 1b) 36 36 12 2 2 ÷ + ÷ x x x [2/0] or b) (VG-alternative to 1a) 2 2 2 2 1 1 1 2 1 y x y xy x ÷ + + [0/2] Solution 1a: (G alternative) ( ) ( )( ) ( )( ) ( )( ) 6 6 6 6 6 6 6 6 6 36 36 12 2 2 2 + ÷ = + ÷ ÷ ÷ = + ÷ ÷ = ÷ + ÷ x x x x x x x x x x x x Answer: 6 6 36 36 12 2 2 + ÷ = ÷ + ÷ x x x x x Solution 1b: (VG alternative) Answer: x y x y y x y xy x ÷ + = ÷ + + 2 2 2 2 1 1 1 2 1 x y x y xy x y xy x y x x y y y x x x y y y x y x y x y x y x y x y x y xy x ÷ + = ÷ + = · ÷ · · + · = ÷ + = | | . | \ | + | | . | \ | ÷ | | . | \ | + = ÷ + + 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 2 2 2 2 2 c) (G-Alternative to 1d) x x x x 2 2 ÷ ÷ ÷ [3/0] d) (VG-Alternative to 1c) y x x y y x x y y x [0/3] Solution 1c: (G-alternative) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x 2 2 2 2 2 4 2 2 4 2 2 4 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 = ÷ ÷ = ÷ ÷ = ÷ ÷ + ÷ = = ÷ + ÷ ÷ = ÷ ÷ ÷ = ÷ ÷ · ÷ ÷ · ÷ = ÷ ÷ ÷ Answer: x x x x x 2 2 2 = ÷ ÷ ÷ Suggested Solutions Test MaC1NVC10 K1 Algebra & Functions NV-College ©
[email protected] Not for sale. Free to use for educational purposes 3 Solution 1d: (VG-alternative) First method: 32 11 2 1 16 5 2 1 2 1 8 3 2 1 2 1 2 1 4 1 2 1 2 1 2 1 2 1 2 1 | | . | \ | = | | | . | \ | | . | \ | = | | | | . | \ | | | | . | \ | | | . | \ | = = | | | | | . | \ | | | | | . | \ | | | | . | \ | | . | \ | = | | | | | | | . | \ | | | | | | | . | \ | | | | | . | \ | | | | . | \ | | | . | \ | = y x x y y x y x x y y x x y y x x y y x y x x y y x x y y x y x x y y x x y y x Answer: 32 11 | | . | \ | = y x y x x y y x x y y x Using: 4 1 2 1 2 1 2 1 2 1 | . | \ | = | | | . | \ | | . | \ | = | | | . | \ | | | . | \ | x y x y y x x y 8 3 2 1 4 3 2 1 4 1 2 1 2 1 2 1 | | . | \ | = | | | . | \ | | | . | \ | = | | | . | \ | | . | \ | = | | | | . | \ | | | | . | \ | | | . | \ | y x y x x y y x y x x y y x 16 5 2 1 8 5 2 1 8 3 2 1 2 1 2 1 2 1 | . | \ | = | | | . | \ | | . | \ | = | | | . | \ | | | . | \ | = | | | | | | . | \ | | | | | . | \ | | | | . | \ | | | . | \ | x y x y y x x y y x x y y x x y 32 11 2 1 16 11 2 1 16 5 2 1 2 1 2 1 2 1 2 1 | | . | \ | = | | | . | \ | | | . | \ | = | | | . | \ | | . | \ | = | | | | | | | . | \ | | | | | | | . | \ | | | | | . | \ | | | | . | \ | | | . | \ | y x y x x y y x y x x y y x x y y x Suggested Solutions Test MaC1NVC10 K1 Algebra & Functions NV-College ©
[email protected] Not for sale. Free to use for educational purposes 4 Second method: 32 11 32 10 32 21 16 5 32 21 16 1 4 32 1 4 16 16 1 4 1 32 1 8 1 2 1 32 1 16 1 8 1 4 1 2 1 | | . | \ | = | . | \ | | | . | \ | = | . | \ | | | . | \ | = | . | \ | | | . | \ | = | . | \ | | | . | \ | = | | . | \ | | . | \ | | | . | \ | | . | \ | | | . | \ | = + + + + + + y x x y y x x y y x x y y x x y y x y x x y y x x y y x y x x y y x x y y x Answer: 32 11 | | . | \ | = y x y x x y y x x y y x e) 125 log 81 log 2 9 log 5 5 9 3 + ÷ [1/1] Suggested solution 1e: 9 3 4 10 5 log 3 9 log 2 2 3 log 2 5 5 log 9 log 2 3 log 5 125 log 81 log 2 9 log 5 5 9 3 3 5 2 9 2 3 5 9 3 = + ÷ = · + · ÷ · · = = + ÷ = + ÷ Using 1 log = a a , and a n a b n b log log · = . Answer: 9 125 log 81 log 2 9 log 5 5 9 3 = + ÷ f) ( ) | | . | \ | ÷ | . | \ | + 10000 log 1000 log 100 log 4 2 2 x x x [2/1] Suggested solution 1f: ( ) ( ) ( ) ( ) x x x x x x x x x x log 4 9 4 log 4 log 2 3 log 2 2 10 log log log 10 log log 10 log 10000 log 1000 log 100 log 4 4 2 3 2 2 4 2 2 ÷ = + ÷ ÷ + · + = = ÷ ÷ ÷ + + = | | . | \ | ÷ | . | \ | + Using 1 10 log = , x n x n log log · = , ( ) b a b a log log log + = · and b a b a log log log ÷ = | . | \ | . Answer: ( ) x x x x log 4 9 10000 log 1000 log 100 log 4 2 2 ÷ = | | . | \ | ÷ | . | \ | + Suggested Solutions Test MaC1NVC10 K1 Algebra & Functions NV-College ©
[email protected] Not for sale. Free to use for educational purposes 5 2. Solve the following equations. Show the detail of your solutions. Give the answers in the surd (exact) form. (a) (G-alternative to 2b) 0 6 5 2 = + ÷ x x [1/0] (b) (VG-alternative to 2a) 3 10 1 = + x x 0/3] Solution 2a: (G-alternative) ( )( ) ¹ ´ ¦ = · = ÷ = · = ÷ · = ÷ ÷ · = + ÷ 3 0 3 2 0 2 0 3 2 0 6 5 2 x x x x x x x x Answer: 3 , 2 2 1 = = x x Solution 2b: (VG-alternative) 0 1 3 10 3 10 1 3 10 1 3 10 1 2 2 = + · ÷ · · · = + · · · = | . | \ | + · · = + x x x x x x x x x x 0 1 36 100 6 10 0 1 6 10 6 10 0 1 3 10 2 2 2 2 = + ÷ | . | \ | ÷ · = + | . | \ | ÷ | . | \ | ÷ · = + · ÷ x x x x · = | . | \ | ÷ · = ÷ ÷ | . | \ | ÷ · = + ÷ | . | \ | ÷ 36 64 6 10 0 36 36 100 6 10 0 1 36 100 6 10 2 2 2 x x x ¦ ¦ ¹ ¦ ¦ ´ ¦ = = + = · = ÷ = = ÷ = · ÷ = ÷ · ± = ÷ · ± = ÷ · = | . | \ | ÷ 3 6 18 6 8 6 10 6 8 6 10 3 1 6 2 6 8 6 10 6 8 6 10 6 8 6 10 36 64 6 10 36 64 6 10 2 x x x x x x x Answer: 3 , 3 1 2 1 = = x x (c) (G-Alternative to 2d, 2e) 000 1 500 5 1 | . | \ | = x [2/0] (d) (VG-Alternative to 2c, 2e) x x 7 5 3 2 · = · [0/2] (e) (MVG-Alternative to 2c, 2d) ( ) 0 49 log 3 log log 7 5 7 2 7 = + ÷ x x [0/2/¤] Solution 2c: (G-alternative) ( ) 04 . 0 25 1 5 1 5 1 5 1 5 1 2 500 000 1 500 1 000 1 500 1 500 000 1 500 = = · | . | \ | = · | . | \ | = · | | . | \ | | . | \ | = · | . | \ | = x x x x x Answer: 04 . 0 25 1 = = x Solution 2d (VG-alternative): | . | \ | = | . | \ | · = | . | \ | · = · = · · · = · · · · = · 5 2 log 3 7 log 5 2 3 7 5 2 3 7 3 7 5 2 3 5 7 5 3 5 3 2 7 5 3 2 x x x x x x x x x x x x Suggested Solutions Test MaC1NVC10 K1 Algebra & Functions NV-College ©
[email protected] Not for sale. Free to use for educational purposes 6 3 log 7 log 5 log 2 log 3 7 log 5 2 log 5 2 log 3 7 log 5 2 log 3 7 log ÷ ÷ = · | . | \ | | . | \ | = · | . | \ | = | . | \ | · · | . | \ | = | . | \ | x x x x Answer: 3 log 7 log 5 log 2 log ÷ ÷ = x Solution 2e (MVG-alternative): ( ) 0 49 log 3 log log 7 5 7 2 7 = + ÷ x x ( ) ( ) 0 7 log 3 log 5 log 0 49 log 3 log log 2 7 7 2 7 7 5 7 2 7 = + · ÷ · = + ÷ x x x x Using 6 7 log 2 3 7 log 3 49 log 3 7 2 7 7 = · · = = , 1 7 log 7 = We may make change of variables to make the problem more clear: t x ÷ 7 log ( ) ( )( ) ¹ ´ ¦ = ÷ = ÷ · = ÷ ÷ · = + · ÷ · = · + · ÷ 0 3 0 2 0 3 2 0 6 5 0 7 log 2 3 log 5 log 2 7 7 2 7 t t t t t t x x ¦ ¹ ¦ ´ ¦ = = · = · = · = ÷ = = · = · = · = ÷ 343 7 3 log 3 0 3 49 7 2 log 2 0 2 3 7 2 7 x x t t x x t t Answer: ¹ ´ ¦ = = 343 49 2 1 x x (f) (G-alternative) 7 5 5 7 ÷ = ÷ x x [2/0] (g) (VG-alternative) ( ) ( ) 7 log 5 4 log 5 4 log = + + ÷ x x [0/3] (h) (MVG-alternative) 5 log 6 10 x x x = · [0/3/¤] Suggested solution 2f: 7 5 5 7 ÷ = ÷ x x ( )( ) ( ) 7 5 7 5 7 7 5 5 7 5 5 7 2 2 ± = ÷ · = ÷ · · = ÷ ÷ · ÷ = ÷ x x x x x x ¹ ´ ¦ = · + = · ± = ÷ ÷ = · + ÷ = · ÷ = ÷ 12 5 7 7 5 2 5 7 7 5 x x x x x x Answer: ¹ ´ ¦ = ÷ = 12 2 2 1 x x Suggested solution 2g: ( ) ( ) 7 log 5 4 log 5 4 log = + + ÷ x x Note the domain of x : - 5 4 4 5 5 4 0 5 4 < · < · > · > ÷ x x x x - 5 4 4 5 0 5 4 ÷ > · ÷ > · > + x x x - Answer: Domain of x is: 5 4 5 4 < < ÷ x ( ) ( ) ( ) ( ) ( ) 7 log 5 4 5 4 log 7 log 5 4 log 5 4 log = + · ÷ · = + + ÷ x x x x Suggested Solutions Test MaC1NVC10 K1 Algebra & Functions NV-College ©
[email protected] Not for sale. Free to use for educational purposes 7 ( ) 7 25 16 7 log 25 16 log 2 2 = ÷ · = ÷ x x 5 3 25 9 25 9 9 25 25 7 16 7 25 16 2 2 2 2 ± = ± = · = · = · = ÷ · = ÷ x x x x x Answer: ¦ ¦ ¹ ¦ ¦ ´ ¦ ÷ = = 5 3 5 3 2 1 x x Note that both answers ¦ ¦ ¹ ¦ ¦ ´ ¦ ÷ > ÷ = < = 5 4 5 3 5 4 5 3 2 1 x x are in the domain of x : 5 4 5 4 < < ÷ x Suggested Solution 2h: 5 log 6 10 x x x = · ( ) ( ) ( ) 0 6 5 0 6 log 5 log 6 log 5 log 10 log log log log 10 log log 10 10 2 2 2 6 5 6 5 log 6 5 log 5 log 6 = + · ÷ · = + · ÷ · ÷ · = ÷ = · · | | . | \ | = · = · = · t t x x x x x x x x x x x x x x x x ( )( ) ¦ ¹ ¦ ´ ¦ = = · = · = · = ÷ = = · = · = · = ÷ · = ÷ ÷ · = + · ÷ 1000 10 3 log 3 0 3 100 10 2 log 2 0 2 0 3 2 0 6 5 3 2 2 x x t t x x t t t t t t Using: x x log 3 log 3 · = , 2 100 log = , ( ) x n x n log log · = which implies ( ) x x x x log log log log · = . Lets change the variable and rename: t x ÷ log Answer: 100 1 = x and 1000 2 = x Check: ( ) ¦ ¦ ¹ ¦ ¦ ´ ¦ = = = = = · = · = · = · · ¹ ´ ¦ = = · 100 10 10 100 10 10 10 100 10 100 10 10 100 10 10 5 2 5 5 10 4 6 2 6 100 log 6 log 6 5 log 6 x x x x x x x x Checks. ( ) ¦ ¦ ¹ ¦ ¦ ´ ¦ = = = = = · = · = · = · · ¹ ´ ¦ = = · 1000 10 10 1000 10 10 10 1000 10 1000 10 10 1000 10 15 5 3 5 5 15 9 6 3 6 1000 log 6 log 6 5 log 6 x x x x x x x x OK! Suggested Solutions Test MaC1NVC10 K1 Algebra & Functions NV-College ©
[email protected] Not for sale. Free to use for educational purposes 8 Solve only one of problems#3 or #4. In case you solve both, you will receive grade only for one of them: 3. (G-Alternative to problem #4) The graph of a quadratic function ( ) x f is plotted in the figure. As illustrated in the figure The function crosses y-axis at ( ) 4 , 0 ÷ , and x- axis at ( ) 0 , 2 ÷ and ( ) 0 , 3 . a) Find analytically the equation of the quadratic function ( ) x f . [2/0] b) Find algebraically the coordinates of the minimum point of the function. Reading from the graph is not accepted. [2/0] -6 -4 -2 0 2 4 6 -4 -3 -2 -1 0 1 2 3 4 5 Suggested solution #3:Lets ( ) ( )( ) 2 1 x x x x A x f ÷ ÷ · = Function’s roots are ( ) 0 , 2 ÷ and ( ) 0 , 3 . Therefore, 2 1 ÷ = x and 3 2 = x : ( ) ( ) ( )( ) ( ) ( )( ) 3 2 3 2 ÷ + · = · ÷ ÷ ÷ · = x x A x f x x A x f The function passes ( ) 4 , 0 ÷ ( )( ) 3 2 6 4 4 6 3 0 2 0 4 = = · ÷ = ÷ · ÷ + · = ÷ A A A Answer: ( ) ( )( ) 3 2 3 2 ÷ + · = x x x f Minimum point of the function lies on the symmetry line of the function which is: 2 1 2 3 2 2 2 1 = + ÷ = + = x x x sym ( ) ( )( ) 6 25 2 6 1 2 4 1 3 2 3 2 1 2 2 1 3 2 2 1 3 2 3 2 ÷ = | . | \ | ÷ | . | \ | + · = | . | \ | ÷ | . | \ | + · = | . | \ | · ÷ + · = f x x x f Answer: The coordinates of the minimum point of the function is | . | \ | ÷ 6 25 , 2 1 . Suggested Solutions Test MaC1NVC10 K1 Algebra & Functions NV-College ©
[email protected] Not for sale. Free to use for educational purposes 9 4. (VG-Alternative to problem #3) The graph of a quadratic function ( ) x f is plotted in the figure. As illustrated in the figure The function crosses y-axis at ( ) 3 , 0 , and is tangent to the x-axis at ( ) 0 , 2 . a) Find analytically the equation of the quadratic function ( ) x f . [0/2] b) If ( ) c bx ax x f + + = 2 find conditions for the constants c b a & , , such that the function has: [0/2] i. two different solutions (roots.) ii. one double solution (root.) iii. no real solution. Note that function ( ) c bx ax x f + + = 2 has a ac b b x 2 4 2 ÷ ± ÷ = as its roots. -2 0 2 4 6 -1 0 1 2 3 4 5 Suggested solution #4:Lets ( ) ( )( ) ( ) 2 1 1 1 x x A x x x x A x f ÷ · = ÷ ÷ · = Function’s double root is ( ) 0 , 2 .Therefore, 2 2 1 = = x x : ( ) ( ) ( ) ( ) 2 2 1 2 ÷ · = · ÷ · = x A x f x x A x f The function passes ( ) 3 , 0 : ( ) ( ) ( ) 4 3 2 0 3 2 2 2 = · ÷ · = · ÷ · = a A x A x f Answer: ( ) ( ) 2 2 4 3 ÷ · = x x f Second method: ( ) c bx ax x f + + = 2 The function passes ( ) 3 , 0 : ( ) ( ) ( ) 3 3 0 0 0 2 = · = + + = c c b a f The function passes ( ) 0 , 2 : ( ) ( ) ( ) 0 3 2 4 0 3 2 2 2 2 = + + · = + + = b a b a f Due to the fact that the function has only one double root, 0 4 2 = ÷ ac b 12 3 4 4 4 0 4 2 2 2 2 2 b a b a c b a b ac ac b = · · = · = · = · = ÷ ( ) 0 3 0 9 6 0 3 2 3 0 3 2 12 4 0 3 2 4 2 2 2 2 = + · = + + · = + + · = + + · = + + b b b b b b b b a ( ) ( ) ( ) 2 2 2 2 2 4 3 3 3 4 3 4 3 12 3 12 3 ÷ · = · + ÷ = · = = = · ÷ = x x f x x x f b a b The function ( ) c bx ax x f + + = 2 has a ac b b x 2 4 2 ÷ ± ÷ = as its roots. Suggested Solutions Test MaC1NVC10 K1 Algebra & Functions NV-College ©
[email protected] Not for sale. Free to use for educational purposes 10 Depending on the value of ac b 4 2 ÷ the function has two, one or not any real roots: i. If 0 4 2 > ÷ ac b , the function has two different real roots. ii. If 0 4 2 = ÷ ac b , the function has one real double root. iii. If 0 4 2 < ÷ ac b , the function does not have any real root. Solve only one of problems#5, #6 or #7. In case you solve more than one problem, you will receive grade only for one of them: 5. (G-alternative to problems 6 and 7) Solve the following simultaneous equations: ¹ ´ ¦ = ÷ = + 12 log log 2 1 log 2 log y x y x [4/0] 6. (VG-Alternative to problems 5 and 7) Solve the following simultaneous equations: ( ) ¦ ¹ ¦ ´ ¦ = | | . | \ | = · 14 log 27 log 7 3 y x y x [0/4] 7. (MVG-Alternative to problems #5 & 6 V1) The equation ( ) ( ) 0 3 2 3 log 2 log > ¬ = x x x has a rational root. Find it in the form q p where p and q are integer numbers. (MVG-Alternative to problems #5 & 6 V2) The equation ( ) ( ) 0 11 7 11 log 7 log > ¬ = x x x has a rational root. Find it in the form q p where p and q are integer numbers. [0/4/¤] Suggested Solution #5: ¹ ´ ¦ = ÷ = + 12 log log 2 1 log 2 log y x y x We may multiply both sides of the second equation by 2 to eliminate y : 5 10 5 log 25 log 5 24 log 2 log 4 1 log 2 log 12 log log 2 1 log 2 log 2 = · = · = ÷ ÷÷ ÷ ¹ ´ ¦ = ÷ = + · ¹ ´ ¦ = ÷ = + x x x y x y x y x y x add Substitute 5 10 = x in the second equation to calculate y . 12 log 10 12 log 10 log 5 2 12 log 10 log 2 12 log log 2 10 5 5 = ÷ · = ÷ · · = ÷ · ¹ ´ ¦ = ÷ = y y y y x x 2 10 2 log log 12 10 12 log 10 ÷ = · ÷ = · = ÷ · = ÷ y y y y Answer: 5 10 = x , 2 10 ÷ = y Suggested Solutions Test MaC1NVC10 K1 Algebra & Functions NV-College ©
[email protected] Not for sale. Free to use for educational purposes 11 Suggested solutions # 6 Answers: 2 11 10 = x , 2 7 10 = y ;& 2 11 10 ÷ = x , 2 7 10 ÷ = y First Method: ( ) ( ) ( ) · ¦ ¹ ¦ ´ ¦ = = · · ¦ ¦ ¹ ¦ ¦ ´ ¦ = = | | . | \ | = = · · ¦ ¹ ¦ ´ ¦ = | | . | \ | · = · · · ¦ ¹ ¦ ´ ¦ = | | . | \ | = · 2 9 7 3 10 10 2 7 14 log 9 3 27 log 14 log 7 27 log 3 14 log 27 log y x y x y x y x y x y x y x y x ( ) 2 11 2 2 7 2 7 2 2 2 7 2 7 7 2 9 2 2 10 10 10 10 10 10 10 10 10 10 10 = = · = · ¦ ¹ ¦ ´ ¦ · = = · = · = · ¦ ¹ ¦ ´ ¦ = · · · = + x y x y y y y y y x Second Method Prob. #6: ( ) ( ) ( ) ¹ ´ ¦ = ÷ = + · ¦ ¦ ¹ ¦ ¦ ´ ¦ = = | | . | \ | = = · · ¦ ¹ ¦ ´ ¦ = | | . | \ | · = · · · ¦ ¹ ¦ ´ ¦ = | | . | \ | = · 2 log log 9 log log 2 7 14 log 9 3 27 log 14 log 7 27 log 3 14 log 27 log 7 3 y x y x y x y x y x y x y x y x 2 11 10 2 11 log 11 2 9 log 2 2 log log 9 log log = · = · = + = ÷ ÷÷ ÷ ¹ ´ ¦ = ÷ = + x x x y x y x add Substitute 4 log = x in 5 log log = + y x : 2 7 2 4 11 2 2 11 log 2 log 2 11 2 log 10 log 2 log log 10 2 11 2 11 = ÷ = ÷ = · = ÷ · = ÷ · ¦ ¹ ¦ ´ ¦ = ÷ = y y y y x x Answers: 2 11 10 = x , 2 7 10 = y Note: We may note that 2 11 10 ÷ = x , 2 7 10 ÷ = y are also simultaneous solutions of the simultaneous equations. This set of answers may be unnoticed and therefore missed in this method. Suggested Solutions Test MaC1NVC10 K1 Algebra & Functions NV-College ©
[email protected] Not for sale. Free to use for educational purposes 12 Suggested Solution Prob. #7V1:( ) ( ) 0 3 2 3 log 2 log > ¬ = x x x Take logarithm of both sides: ( ) ( ) 3 log 2 log 3 log 2 log x x = ( ) ( ) x x 3 log 3 log 2 log 2 log · = · ( ) ( ) x x log 3 log 3 log log 2 log 2 log + · = + · ( ) ( ) x x log 3 log 3 log log 2 log 2 log 2 2 · + = · + ( ) ( ) x x log 2 log log 3 log 3 log 2 log 2 2 · ÷ · = ÷ ( )( ) ( ) 2 log 3 log log 3 log 2 log 3 log 2 log ÷ · = + ÷ x ( )( ) ( ) 3 log 2 log log 3 log 2 log 3 log 2 log ÷ · ÷ = + ÷ x ( )( ) ( ) ( ) ( ) 3 log 2 log 3 log 2 log log 3 log 2 log 3 log 2 log 3 log 2 log ÷ ÷ · ÷ = ÷ + ÷ x ( ) ( ) | . | \ | = · | . | \ | = = ÷ = · ÷ = + ÷ = ÷ 6 1 log log 6 1 log 6 log 6 log 3 2 log 3 log 2 log log 1 x x Answer: 6 1 = x Suggested Solution Prob. #7V2:( ) ( ) 0 11 7 11 log 7 log > ¬ = x x x Take logarithm of both sides: ( ) ( ) 0 11 log 7 log 11 log 7 log > ¬ = x x x ( ) ( ) x x 11 log 11 log 7 log 7 log · = · ( ) ( ) x x log 11 log 11 log log 7 log 7 log + · = + · ( ) ( ) x x log 11 log 11 log log 7 log 7 log 2 2 · + = · + ( ) ( ) x x log 7 log log 11 log 11 log 7 log 2 2 · ÷ · = ÷ ( )( ) ( ) 7 log 11 log log 11 log 7 log 11 log 7 log ÷ · = + ÷ x ( )( ) ( ) 11 log 7 log log 11 log 7 log 11 log 7 log ÷ · ÷ = + ÷ x ( )( ) ( ) ( ) ( ) 11 log 7 log 11 log 7 log log 11 log 7 log 11 log 7 log 11 log 7 log ÷ ÷ · ÷ = ÷ + ÷ x ( ) ( ) ( ) ( ) | . | \ | = = ÷ = · ÷ = + ÷ = ÷ 77 1 log 77 log 77 log 11 7 log 11 log 7 log log 1 x Answer: 77 1 = x Suggested Solutions Test MaC1NVC10 K1 Algebra & Functions NV-College ©
[email protected] Not for sale. Free to use for educational purposes 13 Part II: You should submit your solutions to part I problem before having access to your calculator! Note: Most problems come in a group of two or three alternative problems. Choose only one of problems in the group to solve. Solve only one of problems#8, #9 or #10. In case you solve more than one problem, you will receive grade only for one of them: 8. (G-alternative to problems 9 and 10) ( ) ( ) x x x f x log 5 2 3 2 3 + · + = i. Find ( ) 1 f [1/0] ii. Find the domain and range of the function. Why? Explain. [2/0] 9. (VG-Alternative to problems 8 and 10) ( ) 5 3 2 + ÷ = x x x f . Find and simplify as far as possible: ( ) ( ) h f h f 2 2 ÷ + . [0/3] 10. (MVG-Alternative to problems #8 & 9) ( ) c ax x f + = 2 is given. Determine the constants a and c , if ( ) ( ) 15 24 8 2 4 + ÷ = x x x f f [0/4/¤] Suggested Solution prob. # 8: ( ) ( ) x x x f x log 5 2 3 2 3 + · + = ( ) ( ) ( ) ( ) 8 0 6 2 1 log 5 2 3 1 2 1 1 3 = + + = + · + = f The domain of the function is all positive real numbers, i.e. 0 > x . This is due to the fact that logarithm is defined only or positive real numbers. The range of the function is all real numbers larger than -492. Even though ( ) ( ) x x x f x log 5 2 3 2 3 + · + = is an increasing function of x , but for small 1 0 < < x , the logarithm term increases negatively. Suggested Solution prob. # 9: ( ) 5 3 2 + ÷ = x x x f ( ) ( ) ( ) ( ) ( ) ( ) ¦ ¹ ¦ ´ ¦ + + = + ÷ ÷ + + = + + ÷ + = + = + ÷ = + ÷ = 3 5 3 6 4 4 5 2 3 2 2 3 5 6 4 5 2 3 2 2 2 2 2 2 h h h h h h h h f f ( ) ( ) ( ) 1 1 3 3 2 2 2 + = + = / ÷ / + + = ÷ + h h h h h h h h f h f Suggested Solution prob. #10: ( ) c ax x f + = 2 , ( ) ( ) 15 24 8 2 4 + ÷ = x x x f f ( ) ( ) ( ) ( ) ( ) ( ) ( ) c c ax a x f f c x af x f f c ax x f + + · = · + = · + = 2 2 2 2 ( ) ( ) ( ) ( ) c ac cx a x a c c acx x a a c c ax a x f f + + + = + + + · = + + · = 2 2 2 4 3 2 2 4 2 2 2 2 2 ( ) ( ) ( ) ( ) ¦ ¦ ¹ ¦ ¦ ´ ¦ = ÷ · · = + · = + ÷ = ÷ = · ÷ = · · · ÷ = = · = · ¦ ¹ ¦ ´ ¦ + ÷ = + + + = 15 3 3 2 15 2 15 3 8 24 24 4 2 24 2 2 8 15 24 8 2 2 2 2 2 3 2 4 2 2 2 4 3 c c c ac c c c a a a x x x f f c ac cx a x a x f f Answer: 3 , 2 ÷ = = c a Suggested Solutions Test MaC1NVC10 K1 Algebra & Functions NV-College ©
[email protected] Not for sale. Free to use for educational purposes 14 Solve only one of problems#11, #12 or #13. In case you solve more than one problem, you will receive grade only for one of them: 11. (G-alternative to problems 9 and 10) The formula ( ) kr P t 82 . 0 000 375 · = describes the value of a car which decreases as a function of time (year). a) What is the price of the car as new? [1/0] b) What does 82 . 0 as the base represent? [1/0] Calculate after how many years the value of the car is reduced to kr 000 150 . [2/0] Suggested Solution #11: a) The car costs kr 000 375 as new. b) 82 . 0 as the base of the exponential means that the car’s value is decreased at the rate of % 18 per year. c) ( ) kr P t 82 . 0 000 375 · = ( ) kr t 82 . 0 000 375 000 150 · = ( ) t 82 . 0 375 150 · = ( ) 375 150 82 . 0 = t ( ) ( ) ( ) years years t t t 5 6 . 4 82 . 0 log 375 150 log 375 150 log 82 . 0 log 375 150 log 82 . 0 log ~ = | . | \ | = · | . | \ | = · · | . | \ | = Answer: The value of the car is dropped to 150 000 kr five years after its purchase at 375 000 kr. Suggested Solutions Test MaC1NVC10 K1 Algebra & Functions NV-College ©
[email protected] Not for sale. Free to use for educational purposes 15 12. (VG-Alternative to problems 11 and 12) An archeologist believes that he has found the remnant of Noah’s Ark. According to bile Noah’s Ark was build 4000 BC. How many percent of the normal amount of C-14 should be expected in the ark? Half-life of C-14 decay is years 5730 . [0/4] Suggested Solution prob. # 12 730 5 0 730 5 1 1 730 5 730 5 0 0 0 2 2 2 2 1 2 t t N N a a a N N a N N ÷ ÷ ÷ · = · = · = = · · = · · = 730 5 0 2 t N N ÷ · = 4000 BC is about 6000 years ago, i.e. years t 000 6 ~ 0 730 5 0 48 . 0 2 N N N t = · = ÷ Answer: Only 48% of the normal value of C-14 must be left in the ark! Suggested Solutions Test MaC1NVC10 K1 Algebra & Functions NV-College ©
[email protected] Not for sale. Free to use for educational purposes 16 13. (MVG-Alternative to problems #11 and 12) [0/4/¤] In the begging of year 1900 a laboratory got mg 0 . 65 Radium-226 which is radioactive with half-life of 1600 years. a) Set up a mathematical which demonstrates how Radium decays as a function of time. [0/1/¤] b) Show that the amount of Radium mg y left over after years x may be expressed as T x y y ÷ · = 2 0 [0/1/¤] where 0 y is initial amount of the Radium , and T is the half-life of Radium. c) Calculate the amount of the Radium year 2000. [0/1] d) Which year the amount of Radium in the laboratory is decreased by 25%? [0/1/¤] Suggested Solution prob. .# 13 a) x a C y · = where mg C 0 . 65 = is the initial amount of the Uranium (It is the value of y at 0 = x : i.e. mg C a C 0 . 65 0 . 65 0 = · · = . The half- life Uranium 226 is 1600 years. This means that after years x 600 1 = , mg C y 5 . 32 2 0 . 65 2 = = = 600 1 0 . 65 5 . 32 a · = 1600 1 600 1 2 1 2 1 | . | \ | = · = a a Answer: mg y x 1600 2 1 0 . 65 | . | \ | · = T x x x y y mg y mg y ÷ ÷ · = · · = · | . | \ | · = 2 2 0 . 65 2 1 0 . 65 0 1600 1600 QED b) Year 2000, years x 100 1900 2000 = ÷ = mg y mg y 2 . 62 2 1 0 . 65 1600 100 ~ · | . | \ | · = Answer: mg y 2 . 62 ~ c) ( ) 75 . 0 log 2 1 log 2 1 75 . 0 2 1 0 . 65 1600 1600 1600 = | . | \ | · | . | \ | · / = / · · | . | \ | · = x x x C C mg y ( ) ( ) ( ) | . | \ | · = · · = | . | \ | · · = | . | \ | 2 1 log 75 . 0 log 1600 75 . 0 log 1600 2 1 log 75 . 0 log 2 1 log 1600 x x x ( ) years x 664 2 1 log 75 . 0 log 1600 ~ | . | \ | · = Answer: years x 664 ~ , Year 2675. Suggested Solutions Test MaC1NVC10 K1 Algebra & Functions NV-College ©
[email protected] Not for sale. Free to use for educational purposes 17 14. Newton's Law of Cooling states that the rate of change of the temperature of an object is proportional to the difference between its own temperature and the ambient temperature (i.e. the temperature of its surroundings). Let us assume that T is the temperature of the body and 0 T is the temperature of the surrounding such that T T < 0 . If the surrounding temperature 0 T is constant and the ventilation is good, the cooling takes place such that the temperature difference 0 T T D ÷ = is a decreasing exponential function of time. A high ranked secret police clerk is found death in her well air-conditioned office. The body temperature of the death secret police was C ° 5 . 31 at the time of discover of the body at 00 : 14 . Two hours and fifty minutes later, at 50 : 16 the temperature of the body was dropped to C ° 5 . 27 . The temperature of the office was C ° 0 . 21 . Estimate the time of the murder. Normal body-temperature is C ° 0 . 37 . [0/4/¤] Suggested Solution prob. #16. 0 T T D ÷ = t a C T D · = ° ÷ = 0 . 21 ¦ ¹ ¦ ´ ¦ · = ° ÷ ° · = ° ÷ ° 170 0 . 21 5 . 27 0 . 21 5 . 31 a C a C t Where t is the time in minutes counted from 00 : 14 .i.e.: we may set 0 = t at 14:00. Therefore: ° = · · = ° ÷ ° 5 . 10 0 . 21 5 . 31 C a C t 170 5 . 10 0 . 21 5 . 27 a · ° = ° ÷ ° ° = · ° 5 . 6 5 . 10 170 a 21 13 105 65 5 . 10 5 . 6 5 . 10 5 . 10 170 = = ° ° = · ° ° a 170 1 170 21 13 21 13 | . | \ | = · = a a 170 1 21 13 | . | \ | = a Newton’s cooling temperature law applied to estimate the time of death of the secret police: 170 21 13 5 . 10 0 . 21 t T | . | \ | · ° = ° ÷ 21 32 105 160 5 . 10 0 . 16 21 13 0 . 16 21 13 5 . 10 21 13 5 . 10 0 . 21 0 . 37 170 170 170 = = = | . | \ | · ° = | . | \ | · ° · | . | \ | · ° = ° ÷ ° t t t min 149 min 21 13 log 21 32 log 170 21 32 log 21 13 log 170 21 32 log 21 13 log 170 ÷ = | . | \ | | . | \ | · = · | . | \ | = | . | \ | · · | . | \ | = | . | \ | t t t Answer: The secret police was murdered 149 minutes before 14:00. She was murdered at 11:31 AM.