Question

(理)已知函數(shù)f(x)=x,g(x)=ln(1+x),h(x)=.

(1)證明當(dāng)x＞0時(shí),恒有f(x)＞g(x);

(2)當(dāng)x＞0時(shí),不等式g(x)＞(k≥0)恒成立,求實(shí)數(shù)k的取值范圍;

(3)在x軸正半軸上有一動(dòng)點(diǎn)D(x,0),過D作x軸的垂線依次交函數(shù)f(x)、g(x)、h(x)的圖象于點(diǎn)A、B、C,O為坐標(biāo)原點(diǎn).試將△AOB與△BOC的面積比表示為x的函數(shù)m(x),并判斷m(x)是否存在極值,若存在,求出極值;若不存在,請(qǐng)說明理由.

(文)已知函數(shù)f(x)=,x∈(0,+∞),數(shù)列{a_n}滿足a₁=1,a_n+1=f(a_n);數(shù)列{b_n}滿足b₁=1,b_n+1=,其中S_n為數(shù)列{b_n}的前n項(xiàng)和,n=1,2,3,….

(1)求數(shù)列{a_n}和數(shù)列{b_n}的通項(xiàng)公式;

(2)設(shè)T_n=,證明T_n＜3.

Answer 1

答案：(理)(1)證明:設(shè)F(x)=f(x)-g(x),則F′(x)=,

當(dāng)x＞0時(shí),F′(x)＞0,所以函數(shù)F(x)在(0,+∞)上單調(diào)遞增.又F(x)在x=0處連續(xù),所以F(x)＞F(0)=0,即f(x)-g(x)＞0.所以f(x)＞g(x).

(2)解：設(shè)G(x)=g(x),則G(x)在(0,+∞)上恒大于0,G(x)=ln(1+x)-k+,

G′(x)=,

x²+(2k-k²)x=0的根為0和k²-2k,即在區(qū)間(0,+∞)上,G′(x)=0的根為0和k²-2k,若k²-2k＞0,則G(x)在(0,k²-2k)上單調(diào)遞減,且G(0)=0,與G(x)在(0,+∞)上恒大于0矛盾;若k²-2k≤0,G(x)在(0,+∞)上單調(diào)遞增,且G(0)=0,滿足題設(shè)條件,所以k²-2k≤0.所以0≤k≤2.

(3)解：m(x)=,

m′(x)=,

其分母為正數(shù),其分子為ln(1+x)·.

由第(2)問,知ln(1+x)＞在(0,+∞)上恒成立,所以m′(x)＞0在(0,+∞)上恒成立,即m(x)在(0,+∞)上為單調(diào)遞增函數(shù),故m(x)無極值.

(文)(1)解：∵f(x)=,a_n+1=f(a_n),∴a_n+1=.∴.

∴為以=1為首項(xiàng)、以2為公差的等差數(shù)列.∴=1+(n-1)·2.∴a_n=.

又∵f(x)=,b_n+1=,∴b_n+1==2S_n+1.

∴b_n+2=2S_n+1+1.∴b_n+2-b_n+1=2(S_n+1-S_n).∴b_n+2=3b_n+1.∵b₁=1,b₂=2S₁+1=3,∴{b_n}為以b₁=1為首項(xiàng)、以q=3為公比的等比數(shù)列.b_n=3^n-1.

(2)證明:依題意T_n=1·1+3·+5·()²+7·()³+…+(2n-1)()^n-1,

T_n=1·+3·()²+5·()³+…+(2n-3)·()^n-1+(2n-1)·()ⁿ,

∴T_n=1+2［+()²+()³+…+()^n-1］-(2n-1)·()ⁿ=1+2·-(2n-1)()ⁿ.

∴T_n=2-()^n-1-(2n-1)()ⁿ.

∴T_n=3-·()^n-1-·(2n-1)()ⁿ＜3.