Question

已知x∈R,向量=(acos²x,1),=(2,asin2x-a),f(x)=·,且a≠0.

(1)求函數(shù)f(x)解析式,并求當(dāng)a＞0時(shí)函數(shù)f(x)的單調(diào)增區(qū)間;

(2)當(dāng)x∈［0,］時(shí),f(x)的最大值為5,求a的值.

Answer 1

解:(1)f(x)=2acos²x+asin2x-a

=asin2x+acos2x=2asin(2x+).

由a＞0，當(dāng)2kπ≤2x+≤2kπ+(k∈Z）時(shí),

即kπ-≤x≤kπ+(k∈Z)時(shí),

f(x)為增函數(shù),即f(x)的增區(qū)間為［kπ-,kπ+］(k∈Z).

(2)f(x)=2asin(2x+),當(dāng)x∈［0,］時(shí),

2x+∈［,］.

若a＞0,當(dāng)2x+=時(shí),f(x)最大值為2a=5,則a=.

若a＜0,當(dāng)2x+=時(shí),f(x)的最大值為-a=5,則a=-5.