Question

已知函數(shù)f(x)=x²-(-1)^K·2lnx(k∈N^*).

（1）討論函數(shù)f(x)的單調(diào)性；

（2）k是偶數(shù)時，正項數(shù)列{a_n}滿足a₁=1,f′(a_n)=，求{a_n}的通項公式；

（3）k是奇數(shù)，x＞0，n∈N^*時，求證：［f′(x)］ⁿ-2^n-1·f′(xⁿ)≥2ⁿ(2ⁿ-2).

Answer 1

解:（1）函數(shù)f(x)的定義域是(0,+∞)，k是奇數(shù)時,f(x)=x²+2lnx,f′(x)=2x+,x∈(0,+∞)時,f′(x)＞0.?

∴k是奇數(shù)時，f(x)在區(qū)間（0,+∞）內(nèi)是增函數(shù).                                                      ?

k是偶數(shù)時，f(x)=x²-2lnx,f′(x)=2x-,x∈(1,+∞)時，f′(x)＞0,x∈(0,1)時，f′(x)＜0.

∴k是偶數(shù)時，f(x)在區(qū)間（1，+∞）內(nèi)是增函數(shù)，f(x) 在區(qū)間（0，1）內(nèi)是減函數(shù).?

（2）k是偶數(shù)時，f′(x)=2x-,∵f′(a_n)=,

∴2a_n-=.?

化簡得2a_n²=a_n+1²-1,2(a_n²+1)²=a_n+1²+1,                                                                            ?

∴{a_n²+1}是以2為首項，公比q=2的等比數(shù)列，?

∴a_n²+1=2×2^n-1=2ⁿ.∴a_n＞0(n∈N^*).?

∴a_n=.                                                                                                        ?

（3）k是奇數(shù)時，f′(x)=2x+,x＞0,x∈N^*.?

［f′(x)］ⁿ-2^n-1·f′(xⁿ)=(2x+)ⁿ-2^n-1(2xⁿ+)?

=2ⁿ·［(x+)ⁿ-(xⁿ+)］?

=2ⁿ(C¹_nx^n-2+C²_nx^n-4+…+C^n-2_n·+C^n-1_n·).                                                       ?

令S=C¹_nx^n-2+C²_nx^n-4+…+C^n-2_n·+C^n-1_n·.

則2S=C¹_n（x^n-2+）+C²_n(x^n-4+)+…+C^n-1_n(x^n-2+)

≥2C¹_n+2C²_n+…+2C^n-1_n                                                                                                                                                                                          ?

=2·2ⁿ-4，?

∴S≥2ⁿ-2.      

∴［f′(n)］ⁿ-2^n-1·f′(xⁿ)≥2ⁿ(2ⁿ-2)(n∈N^*).