Question

(理)已知函數(shù)f(x)=e^x-k-x,其中x∈R.

(1)當(dāng)k=0時,若g(x)=的定義域為R,求實數(shù)m的取值范圍;

(2)給出定理:若函數(shù)f(x)在［a,b］上連續(xù),且f(a)·f(b)＜0,則函數(shù)y=f(x)在區(qū)間(a,b)內(nèi)有零點,即存在x₀∈(a,b),使f(x₀)=0.運用此定理,試判斷當(dāng)k＞1時,函數(shù)f(x)在［k,2k］內(nèi)是否存在零點.

(文)已知數(shù)列{a_n}的前n項和為S_n,a₁=2,且na_n+1=S_n+n(n+1)(n∈N^*).

(1)求a_n;

(2)設(shè)b_n=,求{b_n}的最大項.

Answer 1

解：(理)(1)當(dāng)k=0時,g(x)=,要使g(x)的定義域為R,則m≠x-e^x在R上恒成立.令h(x)=x-e^x,h′(x)=1-e^x,∴h(x)在(-∞,0)上遞增,在[0,+∞)上遞減.∴h(x)_max=h(0)=-1.

∴h(x)≤-1.∴m的取值范圍為(-1,+∞).

(2)設(shè)函數(shù)f(x)在R上連續(xù),∴f(x)=e^x-k-x在[k,2k]上連續(xù),而f(2k)=e^k-2k.令g(k)=e^k-2k,∴g′(k)=e^k-2.∵k＞1,g′(k)=e^k-2＞0,∴g(k)在(1,+∞)上遞增.由g(1)=e-2＞0,得g(k)＞g(1)＞0,∴f(2k)＞0.又∵f(k)=1-k＜0,∴f(k)·f(2k)＜0.綜上所述,可以判斷函數(shù)f(x)在區(qū)間[k,2k]上存在零點.

(文)(1)na_n+1=S_n+n(n+1),∴(n-1)a_n=S_n-1+n(n-1)(n≥2).兩式相減,得na_n+1-(n-1)a_n=a_n+2n,即a_n+1-a_n=2(n≥2).當(dāng)n=1時,a₂=a₁+1×2=4,此時a₂-a₁也符合.∴a_n+1-a_n=2(n∈N^*).故a_n=2+(n-1)·2=2n.

(2)S_n=n(n+1).

∵=,∴當(dāng)n≥2時,b_n+1≤b_n.

又b₁==1＜b₂==,∴b₁＜b₂=b₃＞b₄＞b₅＞…＞b_n.

∴{b_n}的最大項為第二項或第三項,即a₂=a₃=.