解ï¼f(x)=e^x-1-x-ax^2; f'(x)=e^x-1-2ax>0; a<(e^x-1)/2x;
å 为å½æ°å¨x>0æ¶ï¼æ¯å¢å½æ°ï¼å æ¤ï¼f(0+)æ¯å½æ°å¨å¨xâ0+æ¶,å½æ°å
·æçæå°å¼ï¼å æ¤ï¼æï¼ a<(e^x-1)/2x<=lim(xâ0+)(e^x-1)/2x=lim(xâ0+)e^x/2=e^0/2=1/2, a<1/2;açåå¼èå´:(-â,1/2)ã
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