平成24年度 愛媛大学 解答例
1.
(ⅰ)
(a) x
2= u, cos u = v とおくと, √
cos x
2= √ v . d √
cos x
2dx = d √ v dv · dv
du · du dx = 1
2 √
v · ( − sin u) · 2x = − x sin x
2√ cos x
2. (b) { log(x + 2 + p
x
2+ 4x + 5) }
0= 1 +
2x+42√ x2+4x+5
x + 2 + √
x
2+ 4x + 5
=
√ x
2+ 4x + 5 + x + 2
√ x
2+ 4x + 5(x + 2 + √
x
2+ 4x + 5) = 1
√ x
2+ 4x + 5 . (ⅱ)
0 + π 2 = 1
1 − c
2から, 1 − c
2= 4
π
2, c
2= 1 − 4 π
2.
− 1 < c < 0 より, c = − r
1 − 4 π
22.
( ⅰ )
Z 1 + t 1 + t
2dt =
Z 1
1 + t
2dt + 1 2
Z 2t
1 + t
2dt = tan
−1t + 1
2 log(1 + t
2) + C .
(ⅱ) e
x= t とおくと,dx = 1
t dt, x −∞ → log √ 3
t 0 → √
3
, Z
12log 3−∞
e
x+ e
2x1 + e
2xdx =
Z
√3 0t(1 + t) 1 + t
2· 1
t dt
= h
tan
−1t + 1
2 log(1 + t
2) i
√30
= tan
−1√ 3 + 1
2 log 4 = π
3 + log 2 . ( ⅲ ) ロピタルの定理を用いる.
x
lim
→01 x
Z
x 0√ 3 + 5 sin t + e
tdt = lim
x→0
√ 3 + 5 sin x + e
x= 2 .
3.
( ⅰ )
∂f
∂x = log 2 · 2
yx·
³
− y x
2´
= − y log 2
x
2· 2
yx, ∂f
∂y = log 2 · 2
xy· 1
x = log 2 x · 2
yx. ( ⅱ )
(a) ZZ
D1
sin x
2dxdy = Z √
π2
0
Z
x 0sin x
2dydx = Z √
π2
0
[y sin x
2]
x0dx = Z √
π2
0
x sin x
2dx
=
h − cos x
22
i √
π 20
= 1
2 . (b)
x = r cos θ, y = r sin θ (e
−1≤ r ≤ 1) とおくと,
1
ヤコビアンは ¯ ¯ ¯ ∂(x, y)
∂(r, θ)
¯ ¯
¯ = abs
¯ ¯
¯ ¯
¯
cos θ − r sin θ sin θ r cos θ
¯ ¯
¯ ¯
¯ = r . ZZ
D2
r log r
2drdθ = ZZ
D2
2r log r drdθ = Z
2π0
Z
1 e−12r log r drdθ
= 2π n
[r
2log r]
1e−1− Z
1e−1
r dr o
= 2π n
− e
−2( − 1) − h r
22
i
1 e−1dr o
= 2π n
e
−2− ³ 1 2 − e
−22
´o
= π(3e
−2− 1).
4.
( ⅰ )
| A − λE | =
¯ ¯
¯ ¯
¯ ¯
¯ ¯
¯ ¯
a + 2b − λ 0 0 a
0 b − λ b 0
0 b b − λ 0
a 0 0 a + 2b − λ
¯ ¯
¯ ¯
¯ ¯
¯ ¯
¯ ¯
= (a + 2b − λ)
2{ (b − λ)
2− b
2} − a { a(b − λ)
2− ab
2} = (a + 2b − λ)
2λ(λ − 2b) − a
2λ(λ − 2b)
= λ(λ − 2b) { (a + 2b − λ)
2− a
2} = λ(λ − 2b) { (a + 2b − λ) + a }{ (a + 2b − λ) − a }
= − λ(λ − 2b)
2(2a + 2b − λ) = 0 より, λ = 0, 2b (2重解) , 2a + 2b . ( ⅱ )
最小の固有値は λ = 0 .
λ = 0 のとき,
⎧ ⎪
⎨
⎪ ⎩
(a + 2b)x
1+ ax
4= 0 bx
2+ bx
3= 0 (a + 2b)x
4= 0
から, x
4= 0, x
1= 0, x
2= − x
3.
よって,
⎛
⎜ ⎜
⎜ ⎜
⎝ x
1x
2x
3x
4⎞
⎟ ⎟
⎟ ⎟
⎠ = t
⎛
⎜ ⎜
⎜ ⎜
⎝ 0
− 1 1 0
⎞
⎟ ⎟
⎟ ⎟
⎠ (t は任意の実数 ).
( ⅲ )
a + b > 0, b > 0, x
26 = − x
3を用いて,
t