若函數(shù)f(x)=x+asin x在R上遞增,則實(shí)數(shù)a的取值范圍為 ________.
[-1,1]
分析:先對(duì)函數(shù)f(x)=x+asin x進(jìn)行求導(dǎo),根據(jù)原函數(shù)是R上的增函數(shù)一定有其導(dǎo)函數(shù)在R上大于等于0恒成立得到1+acosx≥0,再結(jié)合cosx的范圍可求出a的范圍.
解答:∵f′(x)=1+acosx,
∴要使函數(shù)f(x)=x+asinx在R上遞增,則1+acosx≥0對(duì)任意實(shí)數(shù)x都成立.
∵-1≤cosx≤1,
①當(dāng)a>0時(shí)-a≤acosx≤a,
∴-a≥-1,∴0<a≤1;
②當(dāng)a=0時(shí)適合;
③當(dāng)a<0時(shí),a≤acosx≤-a,
∴a≥-1,
∴-1≤a<0.
綜上,-1≤a≤1.
故答案為:[-1,1]
點(diǎn)評(píng):本題主要考查函數(shù)的單調(diào)性與其導(dǎo)函數(shù)的正負(fù)之間的關(guān)系,即當(dāng)導(dǎo)函數(shù)大于0時(shí)原函數(shù)單調(diào)遞增,當(dāng)導(dǎo)函數(shù)小于0時(shí)原函數(shù)單調(diào)遞減.