a????1,即a??2时, f(sinx)在(?,)内是单调递增的; 222aaa2?b; 2)当???1,即?2?a?2时,函数f(sinx)在t?处取得极小值, f(sinx)极小??224a??3)当1?,即a?2 时, f(sinx)在(?,)内是单调递减的;
222另:f(sinx)?sin2x?asinx?b,f?(sinx)?2sinxcosx?acosx?cosx(2sinx?a),因为
??ax?(?,),所以cosx?0,由f?(sinx)?cosx(2sinx?a)?0得sinx?.
222aaa2?b; 1)当???1,即?2?a?2时,函数f(sinx)在sinx?处取得极小值, f(sinx)极小??2241)当
2)当a??2时, f(sinx)在(?3)当a?2时, f(sinx)在(?(2) 令t?sinx,x?[???,)内是单调递增的;
22??,)内是单调递减的; 22,],t?[?1,1], 22|f(sinx)?f0(sinx)|?|f(t)?f0(t)|?|?(a?a0)t?b?b0|?|(a?a0)t?(b?b0)|,
??所以|f(sinx)?f0(sinx)|在[???,]上的最大值为22D?max{|(a?a0)因为|(a?a0)??(b?b0)|,|(a?a0)(?)?(b?b0)|} . 22??(b?b0)|2?|(a?a0)(?)?(b?b0)|2??2?(a?a0)(b?b0) 22?a?a0?a?a0?a?a0?a?a0?所以当?或?时, D?|(a?a0)?(b?b0)|;当?或?时
b?bb?bb?bb?b20000????D?|(a?a0)???2?(b?b0)|.
?a?a??b|,|a(?)?b|}?max{|?b|,|?b|}; 2222?a?0?a?0?a?0?a?0a?a?a??b|2?|?b|2??2ab?;所以当??b|;当?由|或?时, D?|或?时222?b?0?b?0?b?0?b?0a?D?|?b|.
2?a?0?a?0?a?0?a?0a?a??b|?1;当??b|?1. 因此,当?或?时, 由D?|或?时D?|22?b?0?b?0?b?0?b?0(3)在(2)中, a0?b0?0时, D?max{|a
如图,当z?b?最大,
?a?a?a?b??1或?b?1相切时,z 与直线22422b6a25z=b-4432212ππ–4–3–2–1O1–1aπ-b=-12–2-–3–4aπ-b=12a?a?1?b代入z?b?得a2?2?a?4z?4?0,所以把24?2?4a22,因此z?b?满足条件4??4(4z?4)?0得z?44aπ+b=-12234aaπ+b=12D?1时的最大值为
?2?44.