Lösung zu "Kreuzzahlrätsel: Elementare Funktionen"

Kreuzzahlrätsel: Elementare Funktionen (Lösung)
von Dr. Roland Mildner (alpha 6/1983, S. 140)
Kurzlösung
Waagerecht: 3) 13; 4) 11; 6) 12; 8) 14; 9) 1111; 10) 10; 11) 13; 13) 52; 15) 15; 16) 30; 17) 41; 18) 1615; 19) 12; 21) 15; 22) 100; 23) 764; 25) 17; 26) 40; 27) 128; 29) 09; 30) 17; 32) 25; 33) 10; 34) 10; 35) 16; 37) 1251; 40) 6000; 41) 81; 43) 90; 44) 91; 46) 12; 47) 11; 49) 10; 50) 24; 53) 74; 54) 45; 56) 40; 58) 90; 59) 83; 61) 32; 62) 18; 63) 60.
Senkrecht: 1) 13; 2) 11; 3) 121; 5) 144; 7) 210; 8) 111; 12) 385; 14) 2301; 15) 1245; 20) 21; 21) 10; 23) 77; 24) 44; 27) 19500; 28) 8110; 29) 02; 31) 70; 34) 181; 36) 6450; 37) 16; 38) 2048; 39) 10; 42) 100; 43) 999; 45) 10; 46) 111112; 48) 10; 51) 54; 52) 14; 53) 7104; 55) 5100; 57) 01; 58) 99; 60) 362880; 61) 31.
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Ausführliche Lösung
Waagerecht:
3) f(4,5) = 3,5 ⋅ 4,5 - 2,75 = 13 ;
4) 0,5x - 5,5 = 0 → x = 11 ;
6) 5x + y - 12 = 0 und x = 0 → y = 12 ;
8) y = 14x + 3/2 → m = 14 ;
9) y = x - 1109 = 2x/1111 → x = 1111 ;
10) x ≥ 0 → 1,4x + 1 = x + 5 → x = 10 ;
11) 6x - 6 = 2x + 46 → x = 13 ;
13) (4x - 8)/20 = 10 → x = 52 ;
15) Gerade (Zweipunktegleichung): x - 15y - 15 = 0, für y = 0 ergibt sich x = 15 ;
16) x/2 + x/3 - x/5 = 19x/30 = 19 → x = 30 ;
17) (x - 5) ⋅ (x + 14) = (x + 3) ⋅ (x + 4) → 2x = 82 → x = 41 ;
18) x/19 - 17 = 68 → x = 1615 ;
19) x² - 11x - 12 = 0 → x₁ = 12 (x₂ = - 1) ;
21) y - 57 = x² - 14x = (x - 7)² - 49 → y - 8 = (x - 7)² → S = (7, 8) → Summe = 7 + 8 = 15 ;
22) x₀ = 2 → f(2) = 10² = 100 ;
23) x³ - 28x² + 259x - 792 = (x - 8)(x - 9)(x - 11) = 0 → x₁ = 8, x₂ = 9, x₃ = 11 → Differenz = x₁x₂x₃ - (x₁ + x₂ + x₃) = 792 - 28 = 764 (Einfacher: Nach dem Vietaschen Wurzelsatz Produkt und Summe der Nullstellen sofort aus dem Funktionsterm f(x) ablesen.) ;
25) x² - 17x + 72 = (x - 9)(x - 8) = 0 → x₁ = 9, x₂ = 8 → x₁ + x₂ = 17 (Einfacher: Nach dem Vietaschen Wurzelsatz Summe der Nullstellen sofort aus dem Funktionsterm f(x) ablesen.) ;
26) a = sin(&pi/2) + ln(e³) + (log_0,5(1/64))² = 1 + 3 + 36 = 40 ;
27) a = 2³⁰, b = 2²³ → c = a/b = 2⁷ = 128 ;
29) Setze e^x = a und log₃y = b: 2a + b = 4 und 5a + 6b = 17 → a = 1 = e^x, b = 2 = log₃y → x = 0, y = 9 → (x, y) = (0, 9), Schreibweise 09 ;
30) 2e^x-17 = lg(10²) = 2 → e^x-17 = 1 → x - 17 = 0 → x = 17 ;
32) log₃(3^{log₃(log₅(x^1/2))}) = log₃(log₅(x^1/2) = log₃1 = 0 → log₅(x^1/2) = 1 → log₅x = 2 → x = 5² = 25 ;
33) Setze lgx = y → y² + 3y - 4 = 0 → y₁ = lgx₁ = 1, y₂ = lgx₂ = - 4 → x₁ = 10, x₂ = 10^-4 ;
34) Quadrieren ergibt: x² + 21 = (x + 1)² → 2x = 20 → x = 10 ;
35) log₂x + log₄x + log₁₆x = log₂x + (1/2)log₂x + (1/4)log₂x = (7/4)log₂x = 7 → log₂x = 4 → x = 16 ;
37) a = 3, b = 3, c = f(12) = 12² - 5 = 139 → d = abc = 3 ⋅ 3 ⋅ 139 = 1251 ;
40) a₁ = 10, b₁ = 30, c₁ = 20 → V = abc = a₁a₂a₃ cm³ = 6000 cm³ ;
41) Mit sin(180° + α) = - sinα und sin(180° - α) = sinα ergibt sich a = (9 ⋅ sin270°)² = (9 ⋅ (- 1))² = 81 ;
43) → → → γ = 90° ;
44) a₁ = 7 ⋅ (- 1)² = 7, b₁ = 13 → A = 7 ⋅ 13 cm² = 91 cm² ;
46) g = 11 + 0 + 1 = 12 ;
47) f ' (x) = 9x² + 12x - 10 → f ' (1) = 9 + 12 - 10 = 11 ;
49) f ' (x) = 3 - (x³ - 3(x + 3)x²)/x⁶ = 3 + (2x + 9)/x⁴ → f ' (- 1) = 10 ;
50) → f ' (1/9) = 24 ;
53) y = f(x) = x³ + x² - x + 15 → f ' (x) = 3x² + 2x - 1, f '' (x) = 6x + 2 → f '' (12) = 74 ;
54) f ' (x) = cos²x - sin²x → f ' (0) = 1 = tanα → α = 45° ;
56) f ' (x) = 3x² - 42x + 120 = 0 → x² - 14x + 40 = 0 → x₁ = 10, x₂ = 4 → x₁x₂ = 40 (f '' (x) = 6x - 42 → f '' (10) = 18 > 0, f '' (4) = -18 < 0) ;
58) f(x) = ax³ + bx² + cx + d, f ' (x) = 3ax² + 2bx + c → f (0) = d = 2, f (1) = a + b + c + 2 = 6, f ' (2) = 12a + 4b + c = 18, f ' (3) = 27a + 6b + c = 35 → a = 1, b = 1, c = 2, f (x) = x³ + x² + 2x + 2 → f (4) = 90;
59) Ansatz y = f(x) = ax² + bx + c → f ' (x) = 2ax + b, f '' (x) = 2a → f (0) = c = - 2, f ' (1) = 2a + b = 2, f '' (2) = 2a = 10 → a = 5, b = -8, c = - 2, f (x) = 5x² - 8x - 2 → f (5) = 83 ;
61) Die beiden Rechteckseiten seien x (an der Mauer angrenzend) und y → 2x + y = 16, Fläche A = xy = x(16 - 2x), A ' = dA/dx = 16 - 4x = 0 → x = 4, y = 8 → A_max = 4 ⋅ 8 = 32 (A '' = - 4 < 0 → Maximum) ;
62) H/(r - R) = h/r, H = h(r - R)/r (R Grundkreisradius, H Höhe des Kreiszylinders) → V = V_Zyl = π R²H = πR²h(r - R)/r, V ' = dV/dR = 2 π hR - 3πhR²/r = 0 → R(3R - 2r) = 0 → R = 2r/3 = 3 cm, H = 3 cm (da R > 0) → Längsquerschnitt 2HR = 18 cm² ;
63) Querschnittsfläche A = (x + a)h/2 (x und a Parallelseiten - dabei x Breite eines Brettes - und h Höhe des Trapezes), h = x ⋅ sinα, b = x ⋅ cosα, a = x + 2b = x + 2xcosα → A = (x + a)h/2 = (1 + cosα)sinα ⋅ x², A ' = dA/dα = (cosα + cos²α - sin²α) ⋅ x² = 0 → 2cos²α + cosα - 1 = 0 → cosα = - 1 (entfällt: 0< α <90°) und cosα = 1/2 → α = 60° .
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Senkrecht:
1) V = Alter des Vaters, P = Alter von Peter: V + t = 2(P + t) → 33 + t = 2(10 + t) → t = 13 (Jahre) ;
2) (31 - x)/20 = 1 → x = 11 ;
3) Zweipunktegleichung: (y - 0)/(x - 0) = (11 - 0)/(1/11 - 0) → y = 121x → m = 121 ;
5) y = - x/12 → Ansatz für g: y = - x/12 + n, P(0, 12) einsetzen: n = 12 → y = - x/12 + 12 = 0 → x = 144 ;
7) s = v ⋅ t = 126 km/h ⋅ 6s = 126 ⋅ 10³m ⋅ 6s/3600s = 210 m ;
8) m = ρ ⋅ V = (3000 kg/m³) ⋅ (37 cm³) = 3 ⋅ 10⁶ ⋅ 37 g cm³/10⁶cm³ = 111 g ;
12) x/5 + x/7 + x/11 = (77 + 55 + 35)x/385 = 167x/385 = 167 → x = 385 ;
14) 6903/x - 1 = 2301/x + 1 → (6903 - 2301)/x = 2 → x = 4602/2 = 2301 ;
15) x/83 + 68 = x/15 → x/15 - x/83 = (83 - 15)x/1245 = 68 → x = 1245 ;
20) x² - 10x + 21 = 0 → x₁ = 7, x₂ = 3 → P = 7 ⋅ 3 = 21 ;
21) f(10) = 1,5 ⋅ 10² - 15 ⋅ 10 + 10 = 10 ;
23) x² - 4x + 4 = (x - 2)² = 0 → a = b = 2, c = 7, y = f(x) = 2x² - 2x - 7 → f(7) = 2 ⋅ 7² - 2 ⋅ 7 - 7 = 77 ;
24) a - b = 46, 46a + b = 1 → a = 1, b = - 45, f(x) = x² - 45x + 44 = 0 → x₁ = 1, x₂ = 44 ;
27) h = gt²/2 = 9,81 ⋅ 6,3²/2 m ≈ 194,68 m ≈ 195 m = 19500 cm ;
28) 0 = f(10) = 10⁴ - 189 ⋅ 10 - c → c = 10000 - 1890 = 8110 ;
29) 2n(n² + 1) = 5n² → n₁ = 0, für n ≠ 0 gilt 2(n² + 1) = 5n bzw. n² - 5n/2 + 1 = 0 → n₂ = 2, n₃ = 1/2 Lösung (ganzzahlig, in geordneter Reihenfolge): n₁ = 0, n₂ = 2 (Schreibweise: 02) ;
31) 1 = f(0) = lg( c ⋅ (1 - 0)/(7 - 0)) = lg( c/7 ) → 10¹ = c/7 → c = 70 ;
34) 3x - 2y = 7, x + 2y = 29 → x = 9, y = 10 → a = x² + y² = 9² + 10² = 181 ;
36) f(8) = 3⁸ - 111 = 6450 ;
37) Durch Quadrieren ergibt sich:
38) x₀ = 11 → f(11) = 2¹¹ = 2048 ;
39) Quadrieren von abermaliges Quadrieren liefert x² - 8x -20 = 0 → x₁ = -2, x₂ = 10 ;
42) 1 + lgx ≥ 0 (x ≥ 1/10) → | 1 + lgx | = 1 + lgx = 3 → lgx = 2 → x = 10² = 100 (Der Fall 1 + lgx < 0 bzw. 0 < x < 1/10 liefert x = 10^-4.) ;
43) 2x - 99 = | x + 900 | ≥ 0 → x ≥ 99/2 → x + 900 ≥ 0 → | x + 900 | = x + 900 = 2x - 99 → x = 999 ;
45) 2^x+1 - 2^x - 2^x-1 = 4 ⋅ 2^x-1 - 2 ⋅ 2^x-1 - 2^x-1 = (4 - 2 - 1)2^x-1 = 2^x-1 = 512 → x - 1 = 9 → x = 10 ;
46)
48) a = tan(arccos0,1)/sin(arccos0,1) = sin(arccos0,1)/(sin(arccos0,1) ⋅ cos(arccos0,1)) = 1/cos(arccos0,1) = 1/0,1 = 10 ;
51) x = - 3 ist Nullstelle von f(x) in (-4, -2) → Stammfunktion F von f: F(x) = x⁵/5 - 10x³/3 + 9x → A = ( F(-3) - F(-4) ) + ( F (-3) - F(-2) ) = 2 ⋅ F(-3) - F(-2) - F(-4) = 2 ⋅ 72/5 - 34/15 - (- 412/15) = 810/15 = 54 (LE²) ;
52)
53) Grenzen: y = 111 = 37x²/768 → x₁ = - 48, x₂ = 48 → ;
55) ;
57) f(x) = x⁴/12 - x³/6 + x + 1, f ' (x) = x³/3 - x²/2 + 1, f '' (x) = x² - x, f ''' (x) = 2x - 1 → f '' (x) = x² - x = x(x - 1) = 0 → x₁ = 0, x₂ = 1, f ''' (0) = - 1 ≠ 0, f ''' (1) = 1 ≠ 0 → Ergebnis (Wendepunkt-Abszissen in geordneter Reihenfolge): 01 ;
58) ;
60) f⁽⁹⁾(x) = 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 9! = 362880 ;
61) .
Kontrollmöglichkeit:
Bei richtiger Auflösung des Kreuzzahlrätsels ergeben sich in den Kästchen mit den zweistelligen Primzahlnummern 11-13-17-19-23-29-31-37-41-43-47-53-59-61 die Ziffern 1-5-4-1-7-0-7-1-8-9-1-7-8-3 und damit mittels geeigneter Punktierung die Lebensdaten von Leonhard Euler: 15.4.1707 bis 18.9.1783.
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