ivlev-gdz-11-2001 (Дидактические материалы по алгебре и началам анализа для 10 - 11 класса - Ивлев), страница 19

Ответ: (0; f(e)]1= (0; e e ] .3. x(t) = Cx(t); x = C1eCt; 15 = C1e5C; 60 = C1e10C; 4 = e5C, тогда C1 =e5C = 4; 5C = ln4; e =15ln 4, тогда x = e45ln 4t515;4.ПРИМЕРНЫЕ КОНТРОЛЬНЫЕ РАБОТЫКонтрольная работа № 1Вариант 11. F ′ =1x2= f .157⎛π⎞⎝ ⎠2. F(x) = –4cosx + C; F ⎜ ⎟ = C − 0 = 0 ; C = 0; F = –4cosx.242dx = −2 x ;x3.

∫13102442∫ dx = −2 x = 8 − 4 = 4 .11 x3164. а) S = ∫ x 2 dx = x3 =211б) S1= ∫ x 2 dx = x326102=19;24 1 7− = ; S2=y(x2–x1)= 1 (2–1)= 1 ; S=S1–S2=223 6 67 1 2= − = .6 2 35. S = S1 + S2.2π32π380S1= ∫ 2sin xdx = − 2cos x2π32= + 2=3 ;S2= − ∫ − sin xdx = − cos x282π30=31; S= 4 .22Вариант 21.

F′ = −4x2= f ( x) .2. F = 8sinx + C.а) F = 8sinx; б) F(π) = 0 = C.93. ∫16xx39dx = 6∫ x−12 dx1= 12 x 2129= 36 − 12 = 24 .124. а) S = ∫ 2 x 2 dx = x3302016= ;321414 6 82б) S1 = ∫ 2 x 2 dx = . S2 = y(x2–x1)= 2(2–1)=2; S = S1 – S2= − = = 2 .33 3 3315. S=S1+S2;2π32π32π3000S= ∫ sin xdx − ∫ −2sin xdx =3 ∫ sin xdx = − 3cos x2π3019= − 3(− − 1)= .22Вариант 3121. F′(x) = +3x2= f ( x) .2.

а) F(x) = ∫ f ( x)dx = 2 ∫ sin 3 xdx = 2 3 ∫ sin 3xd (3 x) = − 2 3 cos3 x + C ;б) F(π) =1582+ C = 0 ; F(x) = − 2 cos3x − 2 .33343. ∫3x2,5x1dx = x34= 63 .12⎛⎝⎞⎠124. а) S = ∫ (4 − x 2 )dx = ⎜ 4 x − x3 ⎟3−21⎛ 1−2⎞8 ⎞ 32⎛;= 2⎜8 − ⎟ =3⎠ 3⎝1⎛1⎞22;б) S1 = ∫ (− x 2 + 4)dx = ⎜ − x3 + 4 x ⎟ = ⎜ 3 − ⎟ 2 =3⎠3⎝ 3⎠⎝−1−1S2 = y(x2–x1)= 3(1–(–1))=6; S = S1 – S2=π⎛0⎝5. S = ∫ ⎜ 2cos 2x ⎞+ 1⎟ dx = x2 ⎠π02241− 6 = =1 .333ππx+ 2 ∫ cos 2 dx = π + ∫ (cos 2 x + 1)dx = 2π ≈ 6, 28 .200Вариант 41 41. F′(x) = − 2 = f ( x) .3 x32322. а) F(x) = ∫ f ( x)dx = 3∫ cos 2 xdx = ∫ cos 2 xd (2 x) = sin 2 x + C ;⎛π⎞333б) F ⎜ ⎟ = + C = 0 ; F(x) = sin 2 x − .22⎝4⎠ 293.

∫ 6 x−12 dx1= 12 x 219= 36 − 12 = 24 .13⎛1⎞4. а) S = ∫ (3 − x )dx = ⎜ x3 + 3 x ⎟⎝3⎠− 3132⎛1⎞()= 3 3− 3 2=4 3 ;− 3116б) S1 = ∫ (3 − x 2 )dx = ⎜ 3 x − x3 ⎟ = . S2 = y(x2 – x1) = 2(1 –( –1)) = 4 ;3 ⎠3⎝−1−11641S = S1 – S2= − 4 = = 1 .333ππ005. S = ∫ (2sin 2 x 2 + 1)dx = ∫ (2 − cos x)dx = 2π .Контрольная работа № 2Вариант 11.449 − 33 = 4 16 = 2 .15912.1111(a 2 − b 2 )(a 2 + b 2 )1 111=1a2 − b21 1.a 2 b 2 (a 2 + b 2 )a 2b 2113.

а) x3 = ; x = ; б) 3x–2=16–8x+x2; x2–11x + 18 = 0; (x – 2)(x – 9) = 0;82x = 2, x = 9, т.к. 4 – 9 <0, то ответ: x = 2.⎧⎪ x + yx − y =8 ⎧ x + y =44. ⎨; ⎨; x = 3 , x = 9; y = 1 , y = 1.⎩ x− y =2⎪⎩ x − y = 2⎧ x ∈ [− π + 2πn; π + 2πn]22⎪⎧cos x ≥ 0⎪⎪⎪4; ⎨ x ∈ ⎡ π − arcsin 4 + 2πn; arcsin 4 + 2π( n + 1) ⎤ ;5. ⎨sin x ≤5⎢⎥⎪⎪552⎣⎦⎪ x ∈ (−1) n π + πn⎩2 − 2,5sin x = 1 − sin x⎪⎩6πx = + πn .6(()())Вариант 21.681 − 17 = 6 64 = 2 .12.11(a 2 − b)(a 2 + b)11a 2 (a 2=a2 + b1a2.− b)113. а) x = − ; x = − ; б) 3x+1 = x2 – 2x+1; x2 – 5x = 0; x = 0, x = 5, т.к.2730 – 1 < 0, то ответ: x = 5.⎧⎪ x − yx + y = 21 ⎧ x − y = 34.

⎨;⎨; x = 5 , x=25; y = 2 , y = 4.⎩ x+ y =7⎪⎩ x + y = 75. sin2x = 2 – 2,5cosx = 1 – cos2x; cos2x – 2,5cosx + 1 = 0;π(cosx – 2) (cos x − 1 2) = 0; cosx = 1 2 ; x = ± + 2πn , т.к. sin(− π 3 ) < 0 , то3πx=+ 2πn .33(()()44)Вариант 31. 95 − 14 = 81 = 3 .2. Применим формулу для разности кубов:a−b13a16013−b=13(a13− b )(a2313(a13 13+a b13−b )+b23)=a2313 13+a b+b23.11; x = ± ; б) 2x2 – 3x + 2 = 4x2 – 8x + 4; 2x2 – 5x + 2 = 0;1621(x – 2)(x –1 ) = 0, т.к. 2 ⋅ − 2 < 0 . Ответ: x = 2.2272xy=⎧⎪y = 13 ; x = ± 9; x = ± 4; y = ± 4; y = ± 9.4.

⎨( x + y )2 = 169 ; xx +1212− y=52⎪⎩( x − y ) = 253. а) x 4 ={5. x + 2 − x > 0 . Решим уравнение x + 2 − x = 0 ;x + 2 = x2; (x – 2)(x + 1) = 0; x ∈ [–2; 2).Вариант 41. 6 75 − 11 = 6 64 = 2 .2. Применим формулу для суммы кубов:a+b13a13+b=13(a13+ b )( a2313(a13 13−a b+b23)13+b )=a+[–223+–113 13−a b–2+b23x.3. а) x 6 = 1 64 ; x = ± 1 2 ; б) 2x2 + 5x + 4 = 4x2 + 8x + 4; 2x2 + 3x = 0;3x = 0, x = − 3 2 т.к. 2 ⋅ − + 2 < 0 . Ответ: x = 0.2⎧ x + y = 13⎪4. ⎨ x + y + 2 x y = 15 ;⎪⎩( x − y ) 2 = 25⎧2 x y = 12⎪2⎨( x + y ) = 25 ;2⎪( x − y ) = 1⎩⎡⎧⎢⎨⎢⎩⎢⎧⎢⎨⎢⎣ ⎩x+x−x+x−Ответ: (4, 9) и (9, 4).5. 2 – x > x2; x2 + x – 2 > 0; (x + 2)(x – 1) = 0;x ∈ (–∞; 1).Контрольная работа № 3Вариант 11. Отyyyy=5=4;=5= −1{xy == 94{xy == 49.⎡⎢⎢⎢⎣–++–21]21до 27.3()2.

а) 2x = 22 ⋅ 26 = 28; x = 8; б) 2 x 1 + 3 8 = 22 ; 2x = 16 = 24; x = 4.3. 3x2−4522≤ 243 = 3 ; x – 4 ≤ 5; x – 9 ≤ 0; x ∈ [–3; 3].1614. |sinx – 1| = 2; sinx = –1; x =3π+ 2πn ; n ∈ Z.2Вариант 21. Убывает от 3 до 1 27 .2. а) 32x = 34 ⋅ 33 = 37; x = 7 2 =3,5;б) 3x (1 + 1 9 ) = 57 ; 3x = 33; x = 3.23. 2 x −1 ≥ 8 ; x2 – 1 ≥ 3; x2 ≥ 4,x ∈ (–∞; –2] ∪ [2; +∞).4. |cosx –2| = 3; cosx = –1; x = π + 2πn .Вариант 311. Отдо 16.162.

а) 53x = 5–1 ⋅ 5−12=5−3232; 3x = − ;⎛ 13 ⎞1x = − ; б) 4 x ⎜ ⎟ = 52 ; 4x = 43; x = 3.2⎝ 16 ⎠23. (0,3) x − 2 x + 2 ≤ (0,3) 2 ; x2 – 2x + 2 ≤ 2; x(x–2) ≤ 0, x ∈ (–∞; 0] ∪ [2; +∞).4. |x – 1| = x – 1; x ≥ 1.Вариант 41.161. Убывает от 16 до−2. а) 32x = 3–2 ⋅ 3⎛⎝12−=35254; x =− ;7 ⎞⎠б) 5 x ⎜1 − ⎟ = 90 ; 5x = 52 ⋅ 5 = 53; x = 3.253. x2 – 4x + 2 ≤ 2; x ∈ [0; 4].4. 5|x+1| = 5x+1; x ≥ –1.Контрольная работа № 4Вариант 11.

Возрастает от –1 до 3.22. а)log x − 3 x2log 12= −2 ; x2 – 3x – 4 = 0;2(x – 4)(x + 1) = 0; x = 4, –1;б)1log x + log x = 3 ; log2x = 2; x = 4.22 21623. log4(x + 1) < –0,5; x + 1 < 4{=44. xy;y − 2x = 75.log (3 − x)2x−12= 2−1 =1 ⎧ x ≥ −11; ⎨1 ⇒ x ∈ [–1; – 2 ).2 ⎩x < − 2()⎪⎧( x + 4) x − 1 = 0⎧ y = 7 + 2x;x = 1 2 ; y = 8.2⎨7 x + 2 x 2 − 4 = 0 ; ⎨⎩⎪⎩ y = 7 + 2 x–+–≥ 0 ; x ∈ (0; 2].]x320Вариант 21.

Убывает от 1 до –3.2. а) x2 + 4x – 5 = 0; (x – 1)(x + 5) = 0;x = 1, x = –5;(1б) − log 3 x + log 3 x = −1 ; log3x − 1 22= –1; log3x = 2; x = 9.3. log0,5(x – 1) > –2; log2(x – 1) < 2;0 < x – 1 < 4; x ∈ (1; 5).{)={=3; x(8 + 3 x) = 3 ; x = 1 3 ; y = 9.4. xyy − 3 x = 8 xy = 35.log0,5( x + 3)x[≥ 0 ;x ∈ [–2; 0).+––2–3–0Вариант 31. Убывает от 2 до –3.2. а)log ( x 2 + 6 x)21log2 4= −2 ; x2 + 6x – 16 = 0;(x – 2)(x + 8) = 0; x = 2, x = –8;85 31= − log x = −2x x⋅ 2 2 2 233 − log x = 6 ; log x = 2 ; x = 4 ; log2x = 3; log2x = 2; x = 1.2223. lgx(lgx – 1) > 0; lgx ∈ (–∞; 0) ∪ (1; +∞), x ∈ (0; 1) ∪ (10; +∞).⎧⎪4 y 2 + 15 y − 4 = 01=4xy = 44. xy;;; y = ; x = 16.x = 15 + 4 y (15 + 4 y ) y = 4 ⎨ x = 44y⎪⎩б)log2{5.log{0,4( x − 2)x−6≤ 0 ;x ∈ (2; 3] ∪ (6; +∞).[2+–3–6Вариант 41.

Возрастает от –1 до 2.1632. а) log 1 (x2 + 8x) = –2; x2 + 8x = 9; (x+9)(x–1)=0; x1=–9; x2=1;35 1б) 2 − log 5 x + 1 2 + 1 2 log5 x = 2 ; − log5x = 2; log5x = 1; x = 5.2 2()3. lgx(lgx + 1) < 0; lgx ∈ (–1; 0); x ∈ 110 ; 1 .{{⎧ xy = 2=2; xy = 2;; y = 1 2 ; x = 4.4. xyx − 2 y = 3 y (3 + 2 y ) = 2 ⎨⎩3 y + 2 y 2 − 2 = 05.log (8 − x)34− x+–+≤ 0 ;x ∈ (4; 7].47]8Контрольная работа № 5Вариант 11. а) f′(x) = ex(cosx – sinx); f(0) = 1; б) ϕ′(x) = −22002. S1 = ∫ e x dx = e x1⎛ 1⎞ 4; ϕ′ ⎜ − ⎟ = .6x⎝ 8⎠ 3= e 2 − 1 ; S2 = y(x2–x1)= 1(2–0)=2; S=S1–S2=e2–1–2== e2–3 ≈ 4,4.3. f′(x) = 2lnx + 2; f′(x) = 0; lnx = –1; x = e–1; f убывает на (0; e–1]; возрастает на [e–1; +∞); xmin = e–1.1ln 24. f′=4tln4; ϕ′ = 2t+1ln2; 22t > 2 ⋅ 2t ln 4 =2 ⋅ 2t 2 =2t; 22t –2t > 0; 2t(2t –1) > 0,2t –1 > 0; t > 0.Вариант 21. а) f′(x) = ex(sinx + cosx); f(0) = 1; б) ϕ′(x) = 1 6x ; ϕ′(− 1 9 ) = − 3 2 .411 x42.

S = 3 − ∫ = 3 − ln x = 3 – ln4 ≈ 1,61.13. f′(x) = ex + xex = ex(x+1); f′ = 0 при x = –1; убывает при x ∈ (–∞; –1);возрастает при x ∈ [–1; +∞); xmin = –1.4. f′ = 2ln3 92t–1; ϕ′ = 2ln3 3t; 2t – 2 < t; t < 2, t ∈ (–∞; 2).Вариант 31. а) f′(x) = 2xln2cosx – 2xsinx = 2x(ln2 ⋅ cosx – sinx); f′(0) = ln2;164б) ϕ′(x) =6; ϕ′( 1 2) = 12 .x000−2−2−22. S = –2 + ∫ e− x dx = −2 − ∫ e − x d (− x) = −2 − e − x3.

f′ =2 − 2ln xx2=2(1 − ln x )x2= −3 + e 2 = e2 – 3 ≈ 4,4.; f = 0 при x = e; возрастает на (0; e]; убываетна [e; +∞); xmax = e.4. f′(x) =3x ln 3 − 3− x ln 32= 3x − 3− x ; f′ = 0 при x = 0; тогда fmin = f(0) =.ln 3ln 3Вариант 41. а) f′ = 3xln3sinx + 3xcosx = 3x(ln3 ⋅ sinx + cosx); f′(0) = 1;6⋅ 13 = 6 ; ϕ′ ⎛ 1 ⎞ = 18 .⎜ ⎟1 x x⎝ 3⎠33322. S = 4 – ∫ dx = 4 − 2ln x = 4 − ln 9 ≈ 1,8 .11 xб) ϕ′ =3. f′(x) =4e x − e x ⋅ 4 xe2 x=4(1 − x)ex; f = 0 при x = 1; возрастает на (–∞; 1];убывает на [1; +∞); x = 1 — максимум, f(1)= 4 e .4.

f′(x)=1 x х2–xx–x.3 (2 ln2 – 2 ln2) = 2 –2 ; f′=0 при x=0; тогда fmin=f(0) =ln 2ln 3Контрольная работа № 6Вариант 1π4π8π21. sin2x + cos2x = 0; tg2x + 1 = 0; tg2x = –1; 2x = − + πn ; x = − + n ;n ∈ Z.2216 32⎛1 ⎞.2. S = 16 – ∫ x dx = 16 − 2 ⎜ x3 ⎟ = 16 − =333⎝⎠−220⎧log ( y − x) = 1;3.

⎨ x +13 y⎩3 ⋅ 2 = 244.⎧y − x = 3;⎨ x +1 3+ x⎩3 ⋅ 2 = 24⎧y = 3 + x; x = 0; y = 3.⎨ x x⎩3 ⋅ 2 = 1x+5≥ 0 ; x ∈ [–5; –3] ∪ (3; +∞)( x − 3)( x + 3)[–5–+–3+3x5. f′(x) = e + cosx; f′(0) = 2; y = f′(x0)(x – x0) + f(x0); y = 2x + 1.Вариант 21. sin2x – cos2x = 0; tg2x = 1; 2x = π 4 + πn ; x = π 8 + π 2 n ; n ∈ Z.165x212. S1 = ∫ dx = x 26−2 222=8 8 8+ = ; S2 = y(x2–x1)= 2(2–(–2))=8;6 6 3−28 161S = S1 – S2= 8 − = = 5 .3 33x−y=2x=2+y⎧⎧; ⎨ y y; y = 1; x = 3.3. ⎨ y + 2 y +1237223⋅= −6⋅=⎩⎩4.x+6≤ 0 ; x ∈ [–6; –2) ∪ (2; +∞).(2 − x)(2 + x)[–6–22x5.

f′ = ex – sinx; y = x + 2.Вариант 31. sin2x + sinxcosx = 0; tg2x + tgx = 0; tgx = 0; tgx + 1 = 0; x = πn ;x=−π+ πk ; n, k ∈ Z.4(02. S = ∫ (1 − x 2 ) dx − 1 2 = x − 1 3 x3−1⎧x − y = 3)0−1−1 =2 −1 =1 .2326⎧x = 3 + y; ⎨ y y; y = 2; x = 5.3. ⎨ y +1 y −1⎩2 ⋅ 5 = 40 ⎩2 ⋅ 5 = 1004. f′ = ex+1 – e; f′ = 0; x = 0; f(–1) = 1 + e; f(0) = e; f(1) = e2 – e; fmax = e2 – e;fmin = e.125.

Т.к. 3x2 + 4 ≥ 0 для всех x, то 2sinx + 1 > 0; sinx ≥ − ;x ∈ [− π + 2πn; 7 π + 2πn] .66Вариант 41. cos2x–sinxcosx=0; cosx = 0; sinx = cosx; x =ππ+ πn ; x = + πn , n ∈ Z.240⎞1 ⎛ x311 1 12. S = ∫ (− x + 1)dx − = ⎜⎜ − + x ⎟⎟ − = 1 − − = .2 ⎝ 323 2 6−1⎠02−1⎧⎪ x + y = 2⎧x + y = 21 ; y = –2; x = 4.; ⎨ y y3. ⎨ y + 4 y + 3⎩3 ⋅ 4 = 36 ⎪⎩3 ⋅ 4 = 16 : 94. f′ = ex+2 – e; f′ = 0; x = –1; f(–1) = 2e; f(–2) = 1 + 2e; f(0) = e2; fmin = 2e;fmax = e2.125. Т.к.