最新消息:期末考试的答案
最新消息:期末考试的答案
最新消息:期末考试的答案
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2006–2007I<br />
<br />
<br />
<br />
1. ( √ ) 2. (×) 3. (×) 4. (×) 5. ( √ ) 6. ( √ )<br />
1. (B) 2. (C) 3. (D) 4. (B) 5. (D) 6. (C) 7. (C) 8. (D) 9. (A) 10. (B)<br />
3<br />
2 + √ 1 + x 2 + c<br />
1. 1<br />
2π 2. 12 ln 3<br />
2 + √ 1 + x <br />
3<br />
− 12 2 + √ 1 + x + 3<br />
2<br />
3. 0 4. 1<br />
5. √ x 2 − 1 − arctan 1<br />
√ x 2 −1 + c 6. 25π 7. 1<br />
2<br />
5− arctan 1<br />
√ x 2 −1 sign (x) arcsec (x)<br />
1. (3) √ ×<br />
(1)f(x)0U(0),f(0) = 0limx→0 f(x)<br />
x 2 =<br />
1,x = 0<br />
(2)f(x, y)(x0, y0),(x0, y0)<br />
.<br />
(3)<br />
(4)f(x)[a, b],f ′ (x)[a, b]<br />
(5)f(x)x = 1,limx→0+ f(x x ) = f(1).<br />
(6)f(x)[a, b],(a, b)ξ ∈ (a, b)<br />
f(b) − f(a) = f ′ (ξ)(b − a).<br />
2. (10).<br />
<br />
1<br />
n√ ( )<br />
n<br />
(A) 0; (B) 1; (C) 2; (D).<br />
(1){xn} =<br />
(2)x → 0( )<br />
(A) sin 1 1<br />
; (B) e x; (C) x x+1<br />
x2 1 ; (D) arctan −1 x<br />
(3)y = f(−x) ,y ′ = ( )<br />
(A) f ′ (x); (B) −f ′ (x); (C) f ′ (−x); (D) −f ′ (−x).<br />
1
(4)f(x) = e −x , f ′ (ln x)<br />
x dx = ( )<br />
(A) − 1<br />
1<br />
+ C; (B) x x<br />
+ C; (C) − ln x + C; (D) ln x + C.<br />
(5)f(x, y)f(0, 0) = 0,ax + by = 0<br />
(0, 0)( )<br />
(A); (B);<br />
(C); (D).<br />
(6)y = x 2<br />
0 sin tdt( )<br />
(A) y ′ = sin x; (B) y ′ = − cos x; (C) y = 2x sin x 2 ; (D) y ′ = −2x cosx 2<br />
x (7) limx→0<br />
3<br />
(tan x−sinx) √ 1+x2 = ( )<br />
(A) 0; (B) 1; (C) 2; (D) 3.<br />
(8)f(x)(a, b)f ′′ (x) < 0( )<br />
(A)(a, b)f(x) > 0 (B)(a, b)f(x)<br />
(C)(a, b)f(x) (D)(a, b)Lf(x)<br />
(9)f(x, y)U(x0, y0)f(x, y)(x0, y0)<br />
(x0, y0)f(x)x0( )<br />
(A) (B)<br />
(C); (D)<br />
√ √ 3<br />
x3 − 3x2 − x2 + 2x = ( )<br />
(10) limx→+∞<br />
(A) −1; (B) −2; (C) −3; (D) 1.<br />
3.7<br />
(1)D = {(x, y) |x2 + y2 ≤ 1, y ≥ 0} <br />
dxdy = ;<br />
D<br />
(2) <br />
1<br />
2+ 3√ dx = ;<br />
1+x<br />
(3)D = {(x, y) | |x| + |y| ≤ 1} <br />
D x2ydxdy = ;<br />
(4)Dy = x 2 y = x2<br />
2<br />
;<br />
(5) <br />
(6) 10<br />
0<br />
dx = ;<br />
√<br />
20x − x2dx = ;<br />
x2 +1<br />
x √ x2−1 sin x x = πD<br />
x2 dxdy =<br />
(7)f(x) 1<br />
0 f(x)dx = 1, 1<br />
0 dx x<br />
f(x)f(y)dy = .<br />
0<br />
4. (8)()<br />
(1) <br />
1<br />
1+ √ 1−x2dx (2) <br />
√ arcsin x<br />
(1−x2 ) 3dx (3) π<br />
0 x sin2 xdx (4) 1<br />
−1 x ln(1 + ex )dx<br />
2
(1) x = sin θ<br />
<br />
1<br />
1+ √ 1−x2dx = cos θdθ<br />
1+cos θ = 1+cos θ−1<br />
1+cos θ<br />
= θ − <br />
1<br />
2cos2 θ dθ = θ +<br />
2<br />
2 θ sec 2dθ 2<br />
= arcsin x − tan arcsinx<br />
2<br />
= arcsin x −<br />
= arcsin x −<br />
(2)x = sin θ<br />
<br />
sinarcsin x<br />
1+cos arcsin x<br />
+ c<br />
x<br />
1+ √ 1−x 2 + c<br />
√ arcsin x<br />
(1−x2 ) 3dx = θ cos θ<br />
cos3 θ<br />
dθ = <br />
dθ = θ − <br />
θ = θ − tan + c 2<br />
1<br />
1+cos θdθ θ<br />
cos 2 θ dθ = θd tanθ = θ tanθ − tan θdθ<br />
= θ tan θ + ln (cosθ) + c = x<br />
√ 1−x 2 arcsin x + ln √ 1 − x 2 + c<br />
(3) π<br />
0 x sin2 xdx = π 2x<br />
x1−cos<br />
0<br />
= π2<br />
4<br />
+ π<br />
2<br />
− π<br />
2<br />
<br />
π<br />
π t + cos 2tdt = 2<br />
2<br />
4<br />
dx = 2 x2<br />
<br />
<br />
2<br />
π<br />
0<br />
− π<br />
0<br />
(4)f(x) = x ln(1+ex ), 1<br />
−1 x ln(1+ex )dx = 1<br />
= 1<br />
= 1<br />
2<br />
f(x)+f(−x)<br />
−1 2<br />
dx = 1<br />
1<br />
−1 x ln ex dx = 1<br />
2<br />
2<br />
1<br />
x cos 2xdx<br />
−1<br />
f(x)+f(−x)<br />
2<br />
−1 x ln(1 + ex ) − x ln (1 + e −x )dx = 1<br />
2<br />
1<br />
−1 x2dx = 1<br />
3<br />
dx+ 1<br />
−1<br />
f(x)−f(−x)<br />
dx 2<br />
1 1+ex x ln −1 1+e−xdx 5. (6)z = 1 − x 2 − y 2 z = 0.<br />
(0, 0, 1)zx 2 + y 2 = 1 − z<br />
π(1 − z)z 1<br />
0<br />
π(1 − z)dz = π<br />
2 .<br />
6. (6)x 2 + y 2 + z 2 = 1x + 2y + 3z + 9 = 0<br />
<br />
x 2 0 + y2 0 + z2 0 = 1(x0, y0, z0)x + 2y +<br />
3z + 9 = 0<br />
d = |x0 + 2y0 + 3z0 + 9|<br />
√ 1 2 + 2 2 + 3 2<br />
<br />
f(x, y, z, λ) = 1<br />
14 (x0 + 2y0 + 3z0 + 9) 2 + λ x 2 0 + y2 0 + z2 0 − 1<br />
3
fx = fy = fz = fλ = 0,<br />
⎧<br />
⎪⎨<br />
⎪⎩<br />
2<br />
14 (x0 + 2y0 + 3z0 + 9) + 2λx0 = 0<br />
4<br />
14 (x0 + 2y0 + 3z0 + 9) + 2λy0 = 0<br />
6<br />
14 (x0 + 2y0 + 3z0 + 9) + 2λz0 = 0<br />
x2 0 + y2 0 + z2 0 − 1 = 0<br />
x0 = √ 14<br />
14 , y0 = √ 14<br />
7 , z0 = 3√14 14 ;x0 = − √ 14<br />
14 , y0 = − √ 14<br />
7 , z0 = −3√14 14<br />
<br />
<br />
x0 = − √ 14<br />
14 , y0 = − √ 14<br />
7 , z0 = − 3√ 14<br />
14 <br />
<br />
7. (6)x > 0, nf(x) = x<br />
0 (t − t2 )sin 2n tdt. : f(x) ≤<br />
1<br />
(2n+2)(2n+3) .<br />
: f ′ (x) = (x − x 2 ) sin 2n x.0 < x < 1f ′ (x) > 0x > 1f ′ (x) ≤<br />
0<br />
f (x)x = 1f(1) ≤<br />
f(1) = 1<br />
0 (t − t2 ) sin 2n tdt ≤ 1<br />
0 (t − t2 )t2ndt = 1<br />
1<br />
(2n+2)(2n+3) .<br />
1<br />
(2n+2)(2n+3) .<br />
2n+2t2n+2 − 1<br />
2n+3t2n+3 1<br />
8. (7)f(x)[a, b],f ′ (a) < f ′ (b). : c ∈<br />
(f ′ (a), f ′ (b)),ξ ∈ (a, b),f ′ (ξ) = c.<br />
g(x) = f(x) − cxg ′ (a) < 0, g ′ (b) > 0.<br />
ξ ∈ (a, b),f ′ (ξ) = 0.g ′ (a) < 0a<br />
≥ 0g ′ (b) > 0b<br />
ξ ∈ (a, b),g ′ (ξ) = 0.<br />
9. (7): 1<br />
2<br />
1<br />
2<br />
1<br />
2<br />
x + 1<br />
2<br />
x 1<br />
1 x + ≤ 1, (x > 0).<br />
2<br />
x 1<br />
1 x + ≤ 1, (0 < x ≤ 1).<br />
2<br />
1<br />
x 1<br />
≤ 1ln 2 ≤ x ln 1 − 1<br />
2<br />
x <br />
g (x) = x ln (1 − 0.5 x ) − ln 0.5g(1) = 0<br />
g ′ (x) = ln (1 − 0.5 x ) + x −0.5x ln 0.5<br />
1−0.5 x .g ′ (1) = 0<br />
g ′′ (x) = −0.5x ln 0.5<br />
1−0.5x + −0.5x ln 0.5<br />
1−0.5x + x −0.5x (ln0.5) 2 (1−0.5x )+0.5x ln 0.5(−0.5x ln 0.5)<br />
(1−0.5x ) 2<br />
= −0.5x ln 0.5<br />
1−0.5 x<br />
2 + x ln 0.5 + x ln 0.5 0.5 x<br />
1−0.5 x<br />
4<br />
<br />
0 =
= −0.5x ln 0.5<br />
1−0.5x <br />
1 2 + x ln 0.51−0.5x <br />
−0.5 ≥ x ln 0.5<br />
1−0.5x (2 + 2 ln0.5) > 0<br />
g ′ (x)(0, 1],g ′ (x) ≤ g ′ (1) = 0g(x)(0, 1]<br />
g(x) ≥ g(1) = 0<br />
5