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fx ()0 = fx ()0 = fxxxx ()2030121 =−+− =−−+−+ (42)(551)21 xxxx 5510 xx −+= x 5510 ±= fx () fx ()0 = 20301210 xxx −+−=

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シェア "fx ()0 = fx ()0 = fxxxx ()2030121 =−+− =−−+−+ (42)(551)21 xxxx 5510 xx −+= x 5510 ±= fx () fx ()0 = 20301210 xxx −+−="

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[  東京工業大学  1959 年  数学Ⅱ  2  ] 

三次方程式 20x3−30x2+12x− =1 0 3つの実根をもつことを証明せよ。

    f x( )=20x3−30x2+12x−1 とおく。

( ) 0

f x = 3つの実根を持つための条件は,

    ( f x( )の極大値)≧0 かつ f x( )の極小値)≦0   が成り立つときである。

次に, f x( )の極値を調べると

    f´( )x =60x2−60x+12 であり, f´( )x =0 となるのは

5x2−5x+ =1 0 より 5 5

x= ±10 のとき。

  ここで, f x( )=20x3−30x2+12x−1=(4x−2)(5x2−5x+ −1) 2x+1 と変形できることから

  5 5 5 5

2 1

10 10

f ⎛⎜⎜⎝ + ⎞⎟⎟⎠= − ⋅ + +

5 0

= − 5 <

5 5 5 5

2 1

10 10

f ⎛⎜⎜ − ⎞⎟⎟= − ⋅ − +

⎝ ⎠

5 0

= 5 >

である。 したがって f x( )=0 3つの(相異なる)実根をもつ。

  [別解] f x( )=20x3−30x2+12x−1 (2x 1)(10x2 10x 1)

= − − +

3 3

1 1

5 5

10(2 1)

2 2

x x x

⎛ ⎞⎛ ⎞

+ −

⎜ ⎟⎜ ⎟

⎜ ⎟⎜ ⎟

= − − −

⎜ ⎟⎜ ⎟

⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

    と因数分解できるので, f x( )=0 3つの(相異なる)実根をもつ。

参照

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