Some Math Question

1. 3 2 n+1 + 6 n+1 + 1 n→∞ 7654321 + 2 ⋅ 6 n + 12 ⋅ 3 2 n −1 3 ⋅ 9 n + 6 n +1 + 1 = lim n →∞ 7654321 + 2 ⋅ 6 n + 4 ⋅ 9 n

lim

n

1 6 3+ 6⋅  + n 9 9 = lim n n →∞ 7654321 6 + 2⋅  + 4 9n 9 3 = 4

2. an 2 + 4n + 5 2 = n →∞ bn + 6 5 因為n趨近無窮，若分子與分母的最高次數不同

lim

結果必為0或正負無窮 2 但題目結果為 5 可知分子與分母的最高次數相同 (即a = 0，b ≠ 0) 且分子與分母最高次項的係數比為2 : 5 故4 : b = 2 : 5 b = 10 a + b = 0 + 10 = 10

3. lim n →∞

= lim

n →∞

= lim

n →∞

n2 + n − n n 2 + 3n − n 2 − n

(

(

n2 + n − n

)(

n 2 + 3n + n 2 − n

n + 3n − n − n 2

2

( n + n − n )( ( n + 3n − n 2

2

+n

)(

n2 + n + n

n + 3n + n − n 2

2

n 2 + 3n + n 2 − n

2

2

)(

)(

n2 + n + n

)

 3 1 n n 1 + + n 1 −  n n = lim  n →∞   1 4n n 1 + + n  n      1+ 3 + 1− 1   n n  1+1 1 = lim  = = n →∞ 4(1 + 1) 4  1  4 1 + + 1 n  

4. 4x   x lim − 2  x →2  x −2 x − 4  x( x + 2 ) −4 x = lim x →2 x 2 −4 x 2 −2 x = lim 2 x →2 x − 4 x( x − 2) = lim x →2 ( x − 2 )( x + 2 ) x 2 1 = lim = = x →2 x + 2 2 +2 2

5. 10 n lim n→∞ n! 10 10   10 10 lim × × × ... ×  n →∞ n n −1 n − 2 1  =0

)

)(

)

n +n+n 2

)

6.你好像寫錯了，應該是x → ∞ lim

(2x

( 3x

50

)

+ x +1

4

)

2

+ x 99 + x 88 + 100 這題可發現分子與分母的最高次數都為 200次方 x →∞

100

所以只要算出分子與分母最高次項的係數是多少就好了 =

34 81 = 4 22

7. lim

n →∞

1 1 2 n  + + ... +  2n  n n n

1 1 xdx 2 ∫0 1 1 = × x2 2 2 1 = 4 或者 =

1 0

1 1 2 n  + + ... +  2n  n n n 1 + 2 + ... + n = lim n →∞ 2n 2 n2 + n = lim n →∞ 4 n 2 1 = 4 lim

n →∞

n

 3x  8. <   > 為收斂數列，求x範圍  2x + 1 等比數列收斂的條件為 − 1 < r ≤ 1，r為公比 3x −1 < ≤1 2x + 1 2 ⇒ −( 2 x + 1) < 3 x(2 x + 1) ≤ (2 x + 1) 2 ⇒ −4 x 2 − 4 x − 1 < 6 x 2 + 3 x ≤ 4 x 2 + 4 x + 1 10 x 2 + 7 x + 1 > 0 ⇒ 2 2 x − x − 1 ≤ 0 ( 2 x + 1)( 5 x + 1) > 0 ⇒ ( 2 x + 1)( x − 1) ≤ 0 1 1   x > − 5 或x < − 2 ⇒ − 1 ≤ x ≤ 1  2 1 ⇒ − < x ≤1 5

9. f ( x) = f ′( x) =

x x +1 x2 + 1− x ⋅ 2x 2

(x

− x2 +1

( x + 1) ( − 2 x ) ( x + 1) − ( − x + 1) ⋅ 2( x f ′′( x) = ( x + 1) 2

)

=

+1

2

2

2

2

2

f ′′(1) =

2

2

4

2

)

+ 1 ⋅ 2x

− 2 ⋅ 22 − 0 1 =− 4 2 2

10.

(

)

f ( x) = ( x − 1) x 3 − x 2 − 1 = x 6 + ..... 因為題目要求f ( x )的六階導函數 3

又f ( x)為x的多項式 故f (6) ( x)中，原本次數小於6的項都會消失 剩下的x 6 微分六次之後 = 6 × 5 × 4 × 3 × 2 × 1 = 720 f (6) ( x) = 720

11. f ′( x ) − f ′(5) x →5 x−5 f ( x) − f (a ) 根據定義，f ′(a ) = lim x →a x−a f ′( x) − f ′(5) 故所求 lim = f ′′(5) x →5 x−5 f ′( x) = 3 x 2 − 4 x + 7 f ′′( x) = 6 x − 4 f ′′(5) = 26 f ( x) = x 3 − 2 x 2 + 7 x − 1，求 lim

12.

∫ (x

2x 2

)

+1

(

3

dx

)

−3

= ∫ 2 x x 2 + 1 dx

((

) ⇒ 2 x( x + 1)

因為 x 2 + 1 2

原式 = −

−2

−3

)′ = −2( x + 1) ⋅ 2x ′ 1 = − (( x + 1) ) 2

(

2

−3

2

−2

)

−2 1 2 x +1 + c 2

1 +c 2( x + 1) 2

=−

2

13. 1

∫

−2

x +1 dx −1

1

−2

−1

= ∫ x +1 dx + ∫ x +1dx −1

( − x −1) dx + ∫−1 ( x +1) dx −2

=∫

1

1  −1  1 1 = − x 2 + x  +  x 2 + x  2  −2  2  −1 5 = 2 這題其實畫出圖形 然後算出1到 - 2間x軸上方面積會快多了

14. 0

∫ (5x

3

4

− 4 x 2 + 3 x)dx

5 4 4 3 3 20 x − x + x 4 3 2 4 256 = −(320 − + 24) 3 776 =− 3 =

15.

∫

3

0

x+2 dx x +1

3 x +1 1  = ∫  + dx 0 x +1   x +1 1 1 3 −  = ∫ ( x + 1) 2 + ( x + 1) 2  dx 0   3

1 3

2 = ( x + 1) 2 + 2( x + 1) 2 3 3 2

1 2

0 3

1

2 2 = ( ⋅ 4 + 2 ⋅ 4 ) − ( ⋅1 2 + 2 ⋅1 2 ) 3 3 20 = 3

16.

∫ 2adx

= 2ax + c

17.

∫ ∫

5

2

1 5

1

5

f ( x) dx = ∫ f ( x)dx + ∫ f ( x)dx = 4 + (−2) = 2 1

2

3

5

g ( x)dx = ∫ g ( x)dx + ∫ g ( x )dx = (−1) + 3 = 2 1

3

求式 ∫ [ 2 f ( x) + 3 g ( x)] dx 1

5

5

5

= −2∫ f ( x )dx − 3∫ g ( x) dx 1

1

= −2 ⋅ 2 − 3 ⋅ 2 = −10

18.

∫

a

0

x( x − a)dx = a

4 3 4 3 4 = 3

⇒ ∫ ( x 2 − ax)dx = 0

1 a ⇒ x3 − x 2 3 2 3

a 0

3

a a 4 − = 3 2 3 3 ⇒ a = −8 ⇒ a = −2 ⇒

19. S為y = x 2 + 4 x − 5與x軸所圍封閉區域之面積 y = x 2 + 4 x − 5與x軸之交點的x座標 為x 2 + 4 x − 5 = 0的兩個解 ⇒ 交點之x座標為 − 5與1 ⇒S= =

1

∫

−5

( x 2 + 4 x − 5)dx

1 1 3 x + 2 x 2 − 5x 3 −5

1 125 = ( + 2 − 5) − (− + 50 + 25) 3 3 = 36

20. 求y = − x 2 與x + y + 6 = 0所圍區域面積 將拋物線代入直線中解出交點 : x2 − x − 6 = 0 x = 3, −2 所求面積 =

∫ [(− x

=

∫

3

−2 3

−2

2

]

) − (− x − 6) dx

(− x 2 + x + 6)dx

3 1 1 = − x3 + x 2 + 6x 3 2 −2

= ( −9 + =

125 6

9 8 + 18) − ( + 2 − 12) 2 3