已知函數(shù)f(x)是定義在R上的單調(diào)增函數(shù)且為奇函數(shù),數(shù)列{an}是等差數(shù)列,a1007>0,則f(a1)+f(a2)+f(a3)+…+f(a2012)+f(a2013)的值( )
A.恒為正數(shù)
B.恒為負(fù)數(shù)
C.恒為0
D.可正可負(fù)
【答案】分析:由題意可得f(0)=0,且當(dāng)x>0,f(0)>0; 當(dāng)x<0,f(0)<0.由數(shù)列{an}是等差數(shù)列,a1007>0,可得f(a1007)>0 可得 a1+a2013=2a1007>0,故f(a1)+f(a2013)>0,同理可得,f(a2)+f(a2012)>0,f(a3)+f(a2011)>0,…,從而得到所求式子的符號(hào).
解答:解:∵函數(shù)f(x)是R上的奇函數(shù)且是增函數(shù)數(shù)列,∴f(0)=0,且當(dāng)x>0,f(0)>0; 當(dāng)x<0,f(0)<0.
∵數(shù)列{an}是等差數(shù)列,a1007>0,故f(a1007)>0.
再根據(jù) a1+a2013=2a1007>0,∴f(a1)+f(a2013)>0.
同理可得,f(a2)+f(a2012)>0,f(a3)+f(a2011)>0,…,
∴f(a1)+f(a2)+f(a3)+…+f(a2012)+f(a2013)>0,
故選A.
點(diǎn)評(píng):本題主要考查等差數(shù)列的性質(zhì),函數(shù)的奇偶性和單調(diào)性的綜合應(yīng)用,屬于中檔題.