shestakov-all-gdz-2004 (546287), страница 39
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Ответ: y = –9–x + 8ex + 26.4.5.В06.а) y ( x) =(1)3 ⋅ 36 x + 4 ⋅ 6 x +1 − 30 x ln 63 ⋅ ln 36 ⋅ 36 x + 4 ln 6 ⋅ 6 x +1 − 30 ln 6. y′( x) ==0;6 ln 66 ln 66ln6 ⋅ 62x + 24ln6 ⋅ 6x – 30ln6 = 0; 62x + 4 ⋅ 6x – 5 = 0; 6x = t, t > 0; t2 + 4t – 5= 0;⎡t = 1x⎢t = −5 ⇒ t = 1; 6 = 1 ⇒ x = 0. Ответ: x = 0.⎣б) y ( x) =2 ⋅16 x + 4 x +1 − 80 x ln 42 ⋅ ln16 ⋅16 x + ln 4 ⋅ 4 x +1 − 80 ln 4. y′( x) ==0;4 ln 44 ln 44ln4 ⋅ 42x + 4ln4 ⋅ 4x – 80ln4 =0; 42x + 4x – 20 = 0;⎡4x = 4⇒ 4x = 4 ⇒ x = 1. Ответ: x = 1.⎢ x⎢⎣ 4 = −54.5.В07.
f(x) = 14e15x + 5, F(x) ∩ f′(x) = т. M(0;y0).∫ (14e15 x298)+ 5 dx =14 15 xe + 5x + C ;15f ′(x) = 210e15x;143136143136+ С = 210 ⇒ С =⇒ y = e15 x + 5 x +.15151515143136.Ответ: y = e15 x + 5 x +1515x = 0 ⇒ y = 210 ⇒б) f(x) = 6e7x + 13; F(x) ∩ f ′(x) = т. M(0;y0).∫(6x7x + 13)dx =f ′(x) = 42e7x;6 7xe + 13x = C;762886288+ C = ⇒ 42 ⇒ C =⇒ y = e7x + 13x +.77776288.Ответ: y = e7x + 13x +77x = 0 ⇒ y = 42 ⇒4.5.В08.ln 5а) S = ∫ 2e3 x dx =ln 22 3ln 5 t2 t∫ e dt = e3 3ln 233 ln 5=3 ln 22 3 3( 5 − 2 ) = 23 (125 − 8) =32= ⋅117 = 78 ;3ln 7б) S = ∫ 3e2 x dx =ln 33 2 ln 7 t3 2 23∫ e dt = ( 7 − 3 ) = ⋅ 40 = 60 .2 2 ln 3224.5.В09.а) f(x) = 6x – 36xln6 + 5.
yкас ⊥ x = –19; x0 — ?f ′(x0) = 0 ⇒ yкас || Oy;f ′(x0) = ln6 ⋅ 6 x0 – 36ln6 = 0;6 x0 = 36 ⇒ x0 = 2. Ответ: x0 = 2.б) f(x) = 18x – 18xln18 + 29; x = –7; x0 — ?yкас || Oy ⇒ f ′(x0) = 0;f ′(x0) = ln18 ⋅ 18x0 – 18ln18 = 0 ⇒ x0 = 1. Ответ: x0 = 1.4.5.В10.а) f ( x) =26 x– 28x – 2; yкас || y = –2; x0 — ?ln 26Из условия следует, что f′(x0) = 0; f′(x0) = 26 x0 – 28 = 0; x0 = log26 28.Ответ: x0 = log2628.б) f ( x) =19 x– 25x + 7; yкас || y = –7; x0 — ?ln19f ′(x0) = 0; f′(x0) = 19 x0 – 25; ⇒ x0 = log19 25.Ответ: x0 = log1925.4.5.В11.а) f(x) = cos3x + ex; т. (0; 0) ∈ F(x); F(x) — ?x∫ (cos3x + e )dx =sin 3 x x+e +C = y ;3299⇒ 1 + C = 0 ⇒ C = –1; ⇒ y =sin 3 x x+ e −1 .313Ответ: y = sin3x + ex – 1.б) f(x) = sin4x + ex; т.
(0; 0) ∈ F(x); F(x) — ?− cos 4 x x+e +C ;413− cos 4 x x 3⇒1– +C=0⇒C=− ;⇒ y =+e − .444413xОтвет: y = − cos4x + e − .44x∫ (sin 4 x + e )dx =4.5.В12.а) f ( x) = 3x + x + 4 + 1 ; т. M(0; y0).yкас = f′(x0)(x – x0) + f(x0); f(x0) = 4; f ′( x0 ) = 3x ln 3x0 +⎛112 x0 + 4⎞⇒ yкас = ⎜ + ln 3 ⎟ x + 4 .⎝4⎠⎛1⎝4⎞⎠Ответ: y = ⎜ + ln 3 ⎟ x + 4 .б) f ( x) = e x + 2 x + 1 + 1 ; т. M(1; y0).f(x0) = e + 2 2 + 1; f′(x0) = e x0 +⎛1x0 + 1= e+121 ⎞31 ⎞3⎛. Ответ: y = ⎜ e +.⎟ x +1+⎟ x +1+2⎠22⎠2⎝⇒ yкас = ⎜ e +⎝Уровень С.4.5.С01.а) f(x)=2xln2+6x–5F(x)=2x+3x2–5x+CF'(x)=f(x)=0=2xln2+6x–5<0 на [–5; 0]Значит F(x) убывает на [–5; 0] xmax=–5 xmin=0.F(–5)–F(0)=2–5–1+3·25+25=99+2–5= 99Ответ: 991.321.32б) f(x)=5xln 5+4x–6, F(x)=5x+2x2–6x+CF'(x)=f(x)<0 при x ∈ [–2; 0].Значит F(x) убывает на [–2; 0]xmax=–2 xmin=0.F(–2)–F(0)=11− 1 + 8 + 12 = 19 += 19, 042525Ответ: 19,04.300;= ln 3 +1;4301б) f(x) =2 ln 5– 6x – 24;5x[–1; 2].–2 ln 2 5− 6 < 0 при ∀x; ⇒ f(–1) = 10ln5 – 18 — max;5x22248ln5 – 36 — min; ⇒ max – min = 10ln5 – 18 –ln5 + 36 =ln5 + 18.f(2) =252525248Ответ:ln 5 + 18 .25f′(x) =4.5.С02.а) f ( x) =9 ⋅ 82 x +1 145 ⋅ 72 x 8 ⋅ 92 x +1.−+2 ln 8ln 722 ln 9yкас ⊥ Oy ⇒ yкас || Ox ⇒ f′(x0) = 0;f ′( x0 ) =9 ⋅ 82 x +1 ⋅ ln 8 ⋅ 2 145 ⋅ 72 x ⋅ ln 72 8 ⋅ 92 x +1 ⋅ 2 ⋅ ln 9+=−2 ln 8ln 722 ln 9= 9 ⋅ 82x+1 – 145 ⋅ 72x + 8 ⋅ 92x+1 = 0;72 ⋅ 82x – 145 ⋅ 8x ⋅ 9x + 72 ⋅ 92x = 0;2xxxx⎛8⎞⎛8⎞⎛8⎞ ⎛9⎞ ⎛8⎞ ⎛8⎞72 ⋅ ⎜ ⎟ − 145 ⎜ ⎟ + 72 = 0 ; ⎜ ⎟ = ⎜ ⎟ ; ⎜ ⎟ = ⎜ ⎟ ⇒ x = 1; x = –1.⎝9⎠⎝9⎠⎝9⎠ ⎝8⎠ ⎝9⎠ ⎝9⎠Ответ: x = 1; x = –1.б) f ( x) =4 ⋅ 92 x +1 97 ⋅ 36 x 9 ⋅ 42 x +1.−+2 ln 9ln 362ln 4f ′(x0) = 0; f′(x0) = 4 ⋅ 92x+1 – 97 ⋅ 36x + 9 ⋅ 42x+1 = 0;36 ⋅ 92x – 97 ⋅ 9x ⋅ 4x + 36 ⋅ 42x = 0;⎡⎛ 9 ⎞ x 9⎢⎜ ⎟ =4⎝4⎠⎛9⎞⎛9⎞⇒ x = 1; x = –1.36 ⋅ ⎜ ⎟ − 97 ⎜ ⎟ + 36 = 0 ; ⎢⎢x⎝4⎠⎝ 4⎠94⎛⎞⎢=⎢⎣⎜⎝ 4 ⎟⎠92xxОтвет: x = 1; x = –1.4.5.С03.а) f(x) = 3e8x – 3e7x + 2,т.
M(x0; 2).3e8 x0 − 3e7 x0 + 2 = 2 ; 3e8 x0 − 3e7 x0 = 0 ⇒ x0 = 0;yкас = f′(x0)(x – x0) + f(x0);f(x0) = 2; f′(x0) = 24e8 x0 − 21e7 x0 = 24 – 21 = 3 ;⇒ yкас = 3x + 2. Ответ: y = 3x + 2.б) f(x) = 4e6x – 4e5x – 3,т. M(x0; –3).4e6 x0 − 4e5 x0 − 3 = −3 ⇒ x0 = 0;yкас = f′(x0)(x – x0) + f(x0); f′(x0) = 24e6 x0 − 20e5 x0 = 4 ; ⇒ yкас = 4x – 3.Ответ: y = 4x – 3.4.5.С04. а) f(x) = (x2 – 8x + 16)ex + 2; yкас || Ox.yкас || Ox ⇒ f′(x0) = 0;302f ′(x0) = (2x – 8)ex + ex(x2 – 8x + 16) = 0;x2 – 8x + 16 + 2x – 8 = 0;⎡ x0 = 2; ⇒ f(2) = 4e2 + 2; y = 4e2 + 2; f(4) = 2; y = 2.⎢⎣ x0 = 4Ответ: y = 4e2 + 2; y = 2.б) f(x) = (x2 – 9x + 21)ex + 1, у нас || Ox.f ′(x) = 0; ex(x2 – 9x + 21 + 2x – 9) = 0;⎡x = 443⎢ x = 3 ⇒ f(4) = e + 1 = y1; f(3) = 3e + 1 = y2⎣Ответ: y = e4 + 1; y = 3e3 + 1.4.5.С05.а) f(x) = ex – 2 sinx – 4x + 3; F(x) ∩ f′(x) = т.
(0; y0).f ′(x) = ex – 2cosx – 4; x = 0 ⇒ f′(x) = 1 – 2 – 4 = –5;x2x∫ (e − 2sin x − 4 x + 3)dx = e + 2cosx – 2x + 3x + C = y;1 + 2 + C = –5 ⇒ C = –8; ⇒ y = ex + 2cosx – 2x2 + 3x – 8.Ответ: y = ex + 2cosx – 2x2 + 3x – 8.б) f(x) = 2ex + 3sinx + 6x – 1; F(x) ∩ f′(x) = (0; y0).f ′(x) = 2ex + 3cosx + 6; x = 0 ⇒ y = 2 + 3 + 6 = 11;x2x∫ (2e + 3sin x + 6 x − 1)dx = 2e – 3cosx + 3x – x + C;⇒ 2 – 3 + C = 11 ⇒ C = 12 ⇒y = 2ex – 3cosx + 3x2 – x + 12.Ответ: y = 2ex – 3cosx + 3x2 – x + 12.4.5.С06.а) f(x) = 16x2x – 35ex;F(x) ∩ Ox = т. A и т. B; A(0; 0); B(x1, 0); x1 — ?2xx2xx(1)∫ (16e − 35e )dx = 8e – 35e + C = y;т. (0; 0) ∈ (1) ⇒ 8 – 35 + C = 0 ⇒ C = 27;⇒ y = 8e2x – 35ex + 27;т. (x1; 0) ∈ (2) ⇒8e2x – 35ex + 27 = 0;(2)⎡x = 02727⎢⇒ x1 = ln .
Ответ: ln.⎢ x = ln 2788⎢⎣8б) f(x) = 2e2x – 29ex;(0; 0); (x0; 0) ∈ F(x); x0 — ?2xx2xx∫ (2e − 29e )dx = e – 29e + C = 0;(0; 0): 1 – 29 + C = 0 ⇒ C = 28 ⇒y = e2x – 29ex + 28;⎡x = 0⇒ x0 = ln28.(x0; 0): e2 x0 − 29e x0 + 28 = 0 ; ⎢⎣ x = ln 28Ответ: ln28.4.5.С07.а) f(x) = 4x – 4 ⋅ 2x – 14xln2; yкас || y1; y1 = xln4; x0 — ?Т.к. yкас || y1 ⇒ f′(x0) = ln4;303f ′(x0) = 4xln4 – 4ln2 ⋅ 2x – 14ln2 = ln4 = 2ln2;2 ⋅ 22x ⋅ ln2 – 4ln2 ⋅ 2x – 16ln2 = 0;22x – 2 ⋅ 2x – 8 = 0; 2x = 4; x0 = 2. Ответ: x = 2.б) f(x) = 9x – 14 ⋅ 3x – 34xln3; yкас || y1; y1 = xln9; x0 — ?f ′(x0) = ln9;f′(x0) = ln9 ⋅ 9x – 14ln3 ⋅ 3x – 34ln3 = 2ln3;2 ⋅ 33x – 14 ⋅ 3x – 36 = 0; 32x – 7 ⋅ 3x – 18 = 0;3x = 9; x0 = 2. Ответ: x0 = 2.4.5.С08.
а) f(x) = (5x + 2)e2x; g(x) = (17x – 4)e2x;12yкас( f ) || yкас ( g ) ; x0f = x0g;Найти y1 и y2.По условия получаем: f′(x0) = g′(x0);f ′( x0 ) = 5e2 x0 + 2e2 x0 (5x0 + 2)⇒5e 2 x + 10e 2 x x0 + 4e2 x − 17e2 x − 34e 2 x x0 + 8e 2 x = 0;0g′( x0 ) = 17e2 x0 + 2e2 x0 (17 x0 − 4)00000−24e 2 x0 х0=0; x0 = 0;yкас = f′(x0)(x – x0) + f(x0);f′(x0) = g′(x0) = 9; f(x0) = 2; g(x0) = –4; ⇒ y1 = 9x + 2; y2 = 9x – 4.Ответ: yкас (f) = 9x + 2; yкас (g) = 9x – 4.б) f(x) = (4x + 5)e3x;g(x) = (22x – 1)e3x; yкас (f) || yкас (g);f′(x0) = g′(x0); Найти yкас (f) и yкас (g).f′(x0) = 3e3 x (4 x0 + 5) + 4 ⋅ e3 x ;00g′(x0) = 3e (22 x0 − 1) + 22e3 x ; ⇒ 12 x0e3 x + 15e3 x +3 x0+4e3 x0− 22e3 x00+ 3e3 x0− 66 xe3 x000= 0;–54x0 ⋅ e3x = 0 ⇒ x = 0; f′(x0) = g′(x0) = 19; f(x0) = 5; g(x0) = –1;⇒ yкас (f) = 19x + 5; yкас (g) = 19x – 1.
Ответ: yкас (f) = 19x + 5; yкас (g) = 19x – 1.4.5.С09. а) f(x)=1–2x–4xln4, x ∈ [0; 5].F(x)=x–x2–4x+C F'(x)=f(x)<0 на [0; 5]⇒ F(x) убывает. xmax=0F(xmax)=–1+C=5 ⇒ C=6.Ответ: F(x)=x–x2–4x+6;б) f(x)=1–4x–3xln3 x ∈ [–3; 0].F(x)=x–2x2–3x+C F'(x)=f(x) – сначала положительна, потом отрицательна.11+ C = −21 −+C .272711F(0)=–1+C. xmin=–3 −21 + C = −3 C = 1827271Ответ: x − 2 x 2 − 3x + 18 .27Сравним F(–3) и F(0): F (−3) = −3 − 18 −4.5.С10. а) f ( x) =30411⋅ 36 x 6 x +1+; yкас || y = 17x + 5.2 ln 6ln 6⇒ f ′(x0) = 17;f ′( x0 ) =11⋅ 2 ln 6 ⋅ 36 x 6 ⋅ ln 6 ⋅ 6 x+= 17;2 ln 6ln 611 ⋅ 62x + 6 ⋅ 62x – 17 = 0;⇒ 6x = 1 ⇒ x0 = 0;yкас = f ′(x0)(x – x0) + f(x0);11122323+=; ⇒ yкас = 17x +.2 ln 6 2 ln 6 2 ln 62 ln 623Ответ: y = 17x +.2 ln 617 ⋅ 4 x 2 x +1б) f ( x) =+; yкас || y = 19x + 1.2ln 2 ln 217 ⋅ 2 ⋅ ln 2 ⋅ 22 x 2 ⋅ 2 x ⋅ ln 2f ′(x0) = 19; f ′( x0 ) =+= 19 ;2 ln 2ln 2f ( x0 ) =17 ⋅ 22x + 2 ⋅ 2x – 19 = 0; 2x = 1 ⇒ x0 = 0;1742121+=; ⇒ yкас = 19x +.2 ln 2 2ln 2 2 ln 22 ln 221Ответ: y = 19x +.2 ln 2f ( x0 ) =4.5.С11.а) x(t) = 3t + e9–t + 38; V > 2; t — ?x′(t) — это есть скорость точки.x′(t) = 3 – e9–t > 2 ⇒ e9–t < 1 ⇒ 9 – t < 0 ⇒ t > 9.Ответ: начиная с t = 9.б) x(t) = 5t + e7–t + 41; V > 4; t — ?x′(t) = 5 – e7–t > 4; e7–t < 1 ⇒ 7 – t < 0 ⇒ t > 7.Ответ: начиная с t = 7.4.5.С12.а) x(t) = t – e4–t + 41; V < 2; t — ?x′(t) — скорость ⇒x′(t) = 1 + e4–t < 2 ⇒ e4–t < 1;4 – t < 0 ⇒ t > 4.
Ответ: начиная с t = 4.б) x(t) = 2t – e1–t + 38; V < 3; t — ?x′(t) = 2 + e1–t < 3 ⇒ e1–t < 1⇒ 1 –t < 0 ⇒ t > 1.Ответ: начиная с t = 1.Уровень D.4.5.D01.а) f(x) = 16xln16 – 2 ⋅ 4x ⋅ ln4; F(x) ∩ Oy = (0; –9); F(x) ∩ Ox = (x0; 0) — ?xxxx∫ (16 ln16 − 2 ⋅ 4 ln 4)dx = 16 – 2 ⋅ 4 + C = y;т. (0; –9) ∈ данной прямой ⇒ 1 – 2 + C = –9 ⇒ C = –8;⇒ y = 16x – 2 ⋅ 4x – 8 = 0; 4x = 4 ⇒ x = 1 ⇒ т. (1; 0).Ответ: (1; 0).б) f(x) = 25xln25 – 5xln5; (0; –20).F ( x) = ∫ (25x ln 25 − 5 x ln 5)dx = 25x – 5x + C = y;305т.
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