第 167 回数学検定 (2009.04.12 実施) 1 級 1 次：計算技能検定 解答例 ...

167 (2009.04.12 ) 1 1
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(x + y + z)(−x 2 − y 2 − z 2 + 2xy + 2yz + 2zx) − 8xyz
x, y, z z = 0
→ (x + y)(−x 2 − y 2 + 2xy)
= (x + y){−(x − y) 2 } = (x + y)(x − y)(−x + y)
z 3 1
(x + y − z)(x − y + z)(−x + y + z)
= −x 3 + {(2y + 2z) − (y + z)}x 2 + {(−y 2 − z 2 + 2yz) + (y + z)(2y + 2z)}x
+ (y + z)(−y 2 − z 2 + 2yz) − 8xyz
= −x 3 + (y + z)x 2 + (y 2 + z 2 − 2yz)x − (y + z)(y 2 + z 2 − 2yz)
= −x 3 + (y + z)x 2 + (y − z) 2 x − (y + z)(y − z) 2
= −x 2 (x − y − z) + (y − z) 2 (x − y − z) = (x − y − z){−x 2 + (y − z) 2 }
= (x − y − z)(x + y − z)(−x + y − z)
= (−x + y + z)(x + y − z)(x − y + z)
x

lim
x→+0
( tan x
x
) 1
x 2
tan x x = 0
tan x = x + 1 3! · 2x3 + O(x 5 )
tan x
x
lim
x→+0
= 1 + 1 3 x2 + O(x 4 )
( tan x
x
) 1
x 2
(
= lim 1 + 1 ) 1
x→+0 3 x2 + O(x 4 x 2
)
= e 1/3
(1/x 2 )

z = f(x, y)
dz = ∂f ∂f
dx +
∂x ∂y dy
z = e 2x3 cos 4y 2
dz = (6x 2 e 2x3 cos 4y 2 )dx + (e 2x3 · 8y · (− sin 4y 2 ))dy
= (6x 2 e 2x3 cos 4y 2 )dx − (8e 2x3 y sin 4y 2 )dy

1 1 1 1 1
1 2 3 4 5
1 3 6 10 15
1 4 10 20 35
∣1 5 15 35 70∣
1
1 1 1 1 1
1 1 1 1 1
1 2 3 4 5
0 1 2 3 4
1 2 3 4
1 3 6 10 15
=
0 1 3 6 10
= 1 ·
0 1 3 6
1 4 10 20 35
0 1 4 10 20
0 1 4 10
∣
∣1 5 15 35 70∣
∣0 1 5 15 35∣
0 1 5 15∣
1 3 6
∣ ∣∣∣
= 1 ·
0 1 4
∣0 1 5∣ = 1 · 1 4
0 1∣ = 1

3 √6 +
1⃝
2⃝
√
√
√
980
980
27 + 3 6 −
27
1⃝
1⃝ x = 3 √6 +
x 3 =
(
6 +
√
√
√
980
980
27 + 3 6 −
27
√ ) (
980
+ 6 −
27
√
+ 3 3 √6 +
√
980
27
3
6 −
x 3
√ )
980
27
√
980
27
√
= 12 + 3 3 36 − 980
27 x
= 12 + 3√ 972 − 980 x = 12 − 2x
⎛
⎝ 3 √6 +
x 3 + 2x − 12 = 0
√
980
27 + 3 √
6 −
√
⎞
980
⎠
27
2⃝ 1⃝ x = 2
(x − 2)(x 2 + 2x + 6) = 0
x 2 + 2x + 6 = 0 x x = 2
3√ a 3√ b = 3√ ab
1

{f n } n=0,1,2,...
{
fn+1 = f n + f n−1 (n ≥ 1)
f 0 = 0 , f 1 = 1
∞∑
n=0
f n
2 n
S n S n =
S n+1 =
=
n+1
∑
k=0
n∑
k=1
n∑
k=0
n+1
f k
2 k = ∑
f k
2 k
k=2
f k
2 k + 1 n+1
2 = ∑
n−1
f k
2 k+1 + ∑ f k
2 k+2 + 1 2
k=0
S n+1 = 1 2 S n + 1 4 S n−1 + 1 2
k=2
lim
n→∞ S n = α n → ∞
f k−1 + f k−2
2 k + 1 2
α = 1 2 α + 1 4 α + 1 2
α = 2
2
S n
1 S n
3 OK

= θ (θ ≥ 0) O
l (θ = 0) O l A OA
O OA
S 1
S 1 =
=
∫ 2π
0
∫ 2π
0
= θ3
6
∣
2π
1
2 r2 dθ
1
2 θ2 dθ
θ=0
= (2π)3
6
O
OA (= 2π) S 2
S 2 = π(2π) 2 = (2π)3
2
S 1 : S 2 = 1 : 3
1 2 r2 dθ r , dθ
∫∫
f(r, θ) rdr dθ
D
f(r, θ) = 1