第9题函数y=xcosx+sinx的图象大致为().解析结合四个选项,会发现有三个选项均为奇函数,所以先考虑验证函数奇偶性,由f(-x)=-xcos(-x)+sin(-x)=-xcosx-sinx=-f(x),得该函数为奇函数,排除B选项;剩余的三个选项x<0时,符号有差异,所以验证符号:x∈(-π2,0)时,cosx>0,x<0,sinx<0,xcosx<0,所以x<0时,y<0,排除C选项;剩余两个选项当x>0时,符号不同,所以取特值x=π,由πcosπ=-π,sinπ=0,得x=π时,y=-π排除A选项,答案为D. Problem 9 The image of y = xcosx + sinx is approximately (). To parse the four options, we find that there are three options that are odd functions, so we first consider the parity of the function f (x) = - (-x) + sin (-x) = - xcosx-sinx = -f (x), the function is an odd function, excluding the B option; the remaining three options x <0, the symbols are different, : x ∈ (-π2,0), cosx> 0, x <0, sinx <0, xcosx <0, so x <0, y <0, excluding C options; the remaining two options when x> 0 , The symbol is different, so take the special value of x = π, by πcosπ = -π, sinπ = 0, when x = π, y = -π exclude A option, the answer is D.