Question

(理)設(shè)A={x|x≠kπ+,k∈Z},已知a=(2cos,sin),b=(cos,3sin),其中α、β∈A,

(1)若α+β=,且a=2b,求α,β的值;

(2)若a·b=,求tanαtanβ的值.

(文)已知函數(shù)f(x)=-x²+4,設(shè)函數(shù)F(x)=

(1)求F(x)的表達(dá)式;

(2)解不等式1≤F(x)≤2;

(3)設(shè)mn＜0,m+n＞0,判斷F(m)+F(n)能否小于0?

Answer 1

答案：(理)解:(1)∵α+β=,∴a=(1,sin(α)),b=(,3sin(α)).

由a=2b,得sin(α)=0,∴α=kπ+,β=-kπ+,k∈Z.

(2)∵a·b=2cos²+3sin²=1+cos(α+β)+3×

=+cos(α+β)-cos(α-β)=,

∴cos(α+β)=cos(α-β).展開,得2cosαcosβ-2sinαsinβ=3cosαcosβ+3sinαsinβ,

即-5sinαsinβ=cosαcosβ,∵α、β∈A,∴tanαtanβ=.

(文)解:(1)F(x)=

(2)當(dāng)x＞0時,解不等式1≤-x²+4≤2,得≤x≤;

當(dāng)x＜0時,解不等式1≤x²-4≤2,得≤x≤-5.

綜合上述不等式的解為≤x≤或≤x≤.

(3)∵mn＜0,不妨設(shè)m＞0,則n＜0,又m+n＞0,∴m＞-n＞0.∴|m|＞|n|.

∴F(m)+F(n)=-m²+4+n²-4=n²-m²＜0,即F(m)+F(n)能小于0.