2018-2019学年北师大版选修4-5 综合法放缩法 课时作业
1.已知a,b∈R+,A=√a+√b,B=√(a+b),则A,B的大小关系是( )
A.A=B B.A≤B
C.A>B D.A
解析:∵(√a+√b)2=a+2√ab+b,
∴A2-B2=2√ab>0.
又A>0,B>0,∴A>B.
答案:C
2.已知0A.logab+logba+2>0 B.logab+logba+2<0C.logab+logba+2≥0 D.logab+logba+2≤0解析:∵0 ∴logab<0, ∴-logab>0. ∴(-logab)+1/("-" log_a b)≥2("当且仅当" b=1/a "时取"" =""号" ), ∴-("-" log_a b+1/("-" log_a b))≤-2, 即logab+1/(log_a b)≤-2, ∴logab+logba≤-2. ∴logab+logba+2≤0.答案:D3.已知a,b∈R+,则下列各式成立的是( )A.cos2θ·lg a+sin2θ·lg bB.cos2θ·lg a+sin2θ·lg b>lg(a+b)C.a^(cos^2 θ)·b^(sin^2 θ)=a+bD.a^(cos^2 θ)·b^(sin^2 θ)>a+b解析:cos2θ·lg a+sin2θ·lg b答案:A4.已知S=1+1/(1×2)+1/(1×2×3)+...+1/(1×2×3×"..." ×n)(n是大于2的自然数),则有( )
A.logab+logba+2>0
B.logab+logba+2<0
C.logab+logba+2≥0
D.logab+logba+2≤0
解析:∵0 ∴logab<0, ∴-logab>0. ∴(-logab)+1/("-" log_a b)≥2("当且仅当" b=1/a "时取"" =""号" ), ∴-("-" log_a b+1/("-" log_a b))≤-2, 即logab+1/(log_a b)≤-2, ∴logab+logba≤-2. ∴logab+logba+2≤0.答案:D3.已知a,b∈R+,则下列各式成立的是( )A.cos2θ·lg a+sin2θ·lg bB.cos2θ·lg a+sin2θ·lg b>lg(a+b)C.a^(cos^2 θ)·b^(sin^2 θ)=a+bD.a^(cos^2 θ)·b^(sin^2 θ)>a+b解析:cos2θ·lg a+sin2θ·lg b答案:A4.已知S=1+1/(1×2)+1/(1×2×3)+...+1/(1×2×3×"..." ×n)(n是大于2的自然数),则有( )
∴logab<0,
∴-logab>0.
∴(-logab)+1/("-" log_a b)≥2("当且仅当" b=1/a "时取"" =""号" ),
∴-("-" log_a b+1/("-" log_a b))≤-2,
即logab+1/(log_a b)≤-2,
∴logab+logba≤-2.
∴logab+logba+2≤0.
答案:D
3.已知a,b∈R+,则下列各式成立的是( )
A.cos2θ·lg a+sin2θ·lg bB.cos2θ·lg a+sin2θ·lg b>lg(a+b)C.a^(cos^2 θ)·b^(sin^2 θ)=a+bD.a^(cos^2 θ)·b^(sin^2 θ)>a+b解析:cos2θ·lg a+sin2θ·lg b答案:A4.已知S=1+1/(1×2)+1/(1×2×3)+...+1/(1×2×3×"..." ×n)(n是大于2的自然数),则有( )
B.cos2θ·lg a+sin2θ·lg b>lg(a+b)
C.a^(cos^2 θ)·b^(sin^2 θ)=a+b
D.a^(cos^2 θ)·b^(sin^2 θ)>a+b
解析:cos2θ·lg a+sin2θ·lg b答案:A4.已知S=1+1/(1×2)+1/(1×2×3)+...+1/(1×2×3×"..." ×n)(n是大于2的自然数),则有( )
答案:A
4.已知S=1+1/(1×2)+1/(1×2×3)+...+1/(1×2×3×"..." ×n)(n是大于2的自然数),则有( )