最新消息：期末考试的答案

2006–2007I
1. ( √ ) 2. (×) 3. (×) 4. (×) 5. ( √ ) 6. ( √ )
1. (B) 2. (C) 3. (D) 4. (B) 5. (D) 6. (C) 7. (C) 8. (D) 9. (A) 10. (B)
3
2 + √ 1 + x 2 + c
1. 1
2π 2. 12 ln 3
2 + √ 1 + x
3
− 12 2 + √ 1 + x + 3
2
3. 0 4. 1
5. √ x 2 − 1 − arctan 1
√ x 2 −1 + c 6. 25π 7. 1
2
5− arctan 1
√ x 2 −1 sign (x) arcsec (x)
1. (3) √ ×
(1)f(x)0U(0),f(0) = 0limx→0 f(x)
x 2 =
1,x = 0
(2)f(x, y)(x0, y0),(x0, y0)
.
(3)
(4)f(x)[a, b],f ′ (x)[a, b]
(5)f(x)x = 1,limx→0+ f(x x ) = f(1).
(6)f(x)[a, b],(a, b)ξ ∈ (a, b)
f(b) − f(a) = f ′ (ξ)(b − a).
2. (10).
1
n√ ( )
n
(A) 0; (B) 1; (C) 2; (D).
(1){xn} =
(2)x → 0( )
(A) sin 1 1
; (B) e x; (C) x x+1
x2 1 ; (D) arctan −1 x
(3)y = f(−x) ,y ′ = ( )
(A) f ′ (x); (B) −f ′ (x); (C) f ′ (−x); (D) −f ′ (−x).
1

(4)f(x) = e −x , f ′ (ln x)
x dx = ( )
(A) − 1
1
+ C; (B) x x
+ C; (C) − ln x + C; (D) ln x + C.
(5)f(x, y)f(0, 0) = 0,ax + by = 0
(0, 0)( )
(A); (B);
(C); (D).
(6)y = x 2
0 sin tdt( )
(A) y ′ = sin x; (B) y ′ = − cos x; (C) y = 2x sin x 2 ; (D) y ′ = −2x cosx 2
x (7) limx→0
3
(tan x−sinx) √ 1+x2 = ( )
(A) 0; (B) 1; (C) 2; (D) 3.
(8)f(x)(a, b)f ′′ (x) < 0( )
(A)(a, b)f(x) > 0 (B)(a, b)f(x)
(C)(a, b)f(x) (D)(a, b)Lf(x)
(9)f(x, y)U(x0, y0)f(x, y)(x0, y0)
(x0, y0)f(x)x0( )
(A) (B)
(C); (D)
√ √ 3
x3 − 3x2 − x2 + 2x = ( )
(10) limx→+∞
(A) −1; (B) −2; (C) −3; (D) 1.
3.7
(1)D = {(x, y) |x2 + y2 ≤ 1, y ≥ 0}
dxdy = ;
D
(2)
1
2+ 3√ dx = ;
1+x
(3)D = {(x, y) | |x| + |y| ≤ 1}
D x2ydxdy = ;
(4)Dy = x 2 y = x2
2
;
(5)
(6) 10
0
dx = ;
√
20x − x2dx = ;
x2 +1
x √ x2−1 sin x x = πD
x2 dxdy =
(7)f(x) 1
0 f(x)dx = 1, 1
0 dx x
f(x)f(y)dy = .
0
4. (8)()
(1)
1
1+ √ 1−x2dx (2)
√ arcsin x
(1−x2 ) 3dx (3) π
0 x sin2 xdx (4) 1
−1 x ln(1 + ex )dx
2

(1) x = sin θ
1
1+ √ 1−x2dx = cos θdθ
1+cos θ = 1+cos θ−1
1+cos θ
= θ −
1
2cos2 θ dθ = θ +
2
2 θ sec 2dθ 2
= arcsin x − tan arcsinx
2
= arcsin x −
= arcsin x −
(2)x = sin θ
sinarcsin x
1+cos arcsin x
+ c
x
1+ √ 1−x 2 + c
√ arcsin x
(1−x2 ) 3dx = θ cos θ
cos3 θ
dθ =
dθ = θ −
θ = θ − tan + c 2
1
1+cos θdθ θ
cos 2 θ dθ = θd tanθ = θ tanθ − tan θdθ
= θ tan θ + ln (cosθ) + c = x
√ 1−x 2 arcsin x + ln √ 1 − x 2 + c
(3) π
0 x sin2 xdx = π 2x
x1−cos
0
= π2
4
+ π
2
− π
2
π
π t + cos 2tdt = 2
2
4
dx = 2 x2
2
π
0
− π
0
(4)f(x) = x ln(1+ex ), 1
−1 x ln(1+ex )dx = 1
= 1
= 1
2
f(x)+f(−x)
−1 2
dx = 1
1
−1 x ln ex dx = 1
2
2
1
x cos 2xdx
−1
f(x)+f(−x)
2
−1 x ln(1 + ex ) − x ln (1 + e −x )dx = 1
2
1
−1 x2dx = 1
3
dx+ 1
−1
f(x)−f(−x)
dx 2
1 1+ex x ln −1 1+e−xdx 5. (6)z = 1 − x 2 − y 2 z = 0.
(0, 0, 1)zx 2 + y 2 = 1 − z
π(1 − z)z 1
0
π(1 − z)dz = π
2 .
6. (6)x 2 + y 2 + z 2 = 1x + 2y + 3z + 9 = 0
x 2 0 + y2 0 + z2 0 = 1(x0, y0, z0)x + 2y +
3z + 9 = 0
d = |x0 + 2y0 + 3z0 + 9|
√ 1 2 + 2 2 + 3 2
f(x, y, z, λ) = 1
14 (x0 + 2y0 + 3z0 + 9) 2 + λ x 2 0 + y2 0 + z2 0 − 1
3

fx = fy = fz = fλ = 0,
⎧
⎪⎨
⎪⎩
2
14 (x0 + 2y0 + 3z0 + 9) + 2λx0 = 0
4
14 (x0 + 2y0 + 3z0 + 9) + 2λy0 = 0
6
14 (x0 + 2y0 + 3z0 + 9) + 2λz0 = 0
x2 0 + y2 0 + z2 0 − 1 = 0
x0 = √ 14
14 , y0 = √ 14
7 , z0 = 3√14 14 ;x0 = − √ 14
14 , y0 = − √ 14
7 , z0 = −3√14 14
x0 = − √ 14
14 , y0 = − √ 14
7 , z0 = − 3√ 14
14
7. (6)x > 0, nf(x) = x
0 (t − t2 )sin 2n tdt. : f(x) ≤
1
(2n+2)(2n+3) .
: f ′ (x) = (x − x 2 ) sin 2n x.0 < x < 1f ′ (x) > 0x > 1f ′ (x) ≤
0
f (x)x = 1f(1) ≤
f(1) = 1
0 (t − t2 ) sin 2n tdt ≤ 1
0 (t − t2 )t2ndt = 1
1
(2n+2)(2n+3) .
1
(2n+2)(2n+3) .
2n+2t2n+2 − 1
2n+3t2n+3 1
8. (7)f(x)[a, b],f ′ (a) < f ′ (b). : c ∈
(f ′ (a), f ′ (b)),ξ ∈ (a, b),f ′ (ξ) = c.
g(x) = f(x) − cxg ′ (a) < 0, g ′ (b) > 0.
ξ ∈ (a, b),f ′ (ξ) = 0.g ′ (a) < 0a
≥ 0g ′ (b) > 0b
ξ ∈ (a, b),g ′ (ξ) = 0.
9. (7): 1
2
1
2
1
2
x + 1
2
x 1
1 x + ≤ 1, (x > 0).
2
x 1
1 x + ≤ 1, (0 < x ≤ 1).
2
1
x 1
≤ 1ln 2 ≤ x ln 1 − 1
2
x
g (x) = x ln (1 − 0.5 x ) − ln 0.5g(1) = 0
g ′ (x) = ln (1 − 0.5 x ) + x −0.5x ln 0.5
1−0.5 x .g ′ (1) = 0
g ′′ (x) = −0.5x ln 0.5
1−0.5x + −0.5x ln 0.5
1−0.5x + x −0.5x (ln0.5) 2 (1−0.5x )+0.5x ln 0.5(−0.5x ln 0.5)
(1−0.5x ) 2
= −0.5x ln 0.5
1−0.5 x
2 + x ln 0.5 + x ln 0.5 0.5 x
1−0.5 x
4
0 =

= −0.5x ln 0.5
1−0.5x
1 2 + x ln 0.51−0.5x
−0.5 ≥ x ln 0.5
1−0.5x (2 + 2 ln0.5) > 0
g ′ (x)(0, 1],g ′ (x) ≤ g ′ (1) = 0g(x)(0, 1]
g(x) ≥ g(1) = 0
5