电大计算机专科软件数学基础形成性考核册答案(附题目)

0+(一)单项选择题A.奇函数B.偶函数C.非奇非偶函数D.既奇又偶函数2xxxx2x32x2EQ\*jc3\*hps35\o\al(\s\up11(s),s)EQ\*jc3\*hps35\o\al(\s\up11(i),i)EQ\*jc3\*hps35\o\al(\s\up11(nmx),n2x)xx喻0x──xx喻0x12.若函数f(x)在点x连续，则一定在点x处可微.2))))解：原式＝解：原式＝limx喻011x喻x喻01=42•••1-——)1-——)x喻伪-4]EQ\*jc3\*hps21\o\al(\s\up18(l),x)-4xxEQ\*jc3\*hps21\o\al(\s\up14(1),x)++1xxEQ\*jc3\*hps20\o\al(\s\up11(1),x)xEQ\*jc3\*hps20\o\al(\s\up11(1),x)+（x),1=sinx=根sinEQ\*jc3\*hps20\o\al(\s\up9(11=sinx=根（x)（x2（x)（x2)xy确定的隐函数,求y,.xyxyxyx+y确定的隐函数,求dy.,,x+y,,x+y一EQ\*jc3\*hps21\o\al(\s\up6(1),2)x)x)xx2x47=2EQ\*jc3\*hps35\o\al(\s\up2(2),EQ\*jc3\*hps35\o\al(\s\up2(2),3)EQ\*jc3\*hps35\o\al(\s\up2(1),4)2-EQ\*jc3\*hps21\o\al(\s\up10(1),x)EQ\*jc3\*hps21\o\al(\s\up10(1),x)1x21（x)1EQ\*jc3\*hps21\o\al(\s\up13(π),2)EQ\*jc3\*hps21\o\al(\s\up13(π),2)EQ\*jc3\*hps21\o\al(\s\up11(π),2)EQ\*jc3\*hps21\o\al(\s\up13(π),2)EQ\*jc3\*hps21\o\al(\s\up11(π),2)=xlnxe-e单项选择题单项选择题填空题EQ\*jc3\*hps35\o\al(\s\up12(2),0)EQ\*jc3\*hps35\o\al(\s\up12(0),2)EQ\*jc3\*hps35\o\al(\s\up12(2),5)EQ\*jc3\*hps35\o\al(\s\up12(3),0)EQ\*jc3\*hps35\o\al(\s\up12(0),0)EQ\*jc3\*hps35\o\al(\s\up12(0),0)EQ\*jc3\*hps35\o\al(\s\up12(3),3)EQ\*jc3\*hps35\o\al(\s\up12(2),2)EQ\*jc3\*hps35\o\al(\s\up12(0),0)EQ\*jc3\*hps35\o\al(\s\up12(0),0)EQ\*jc3\*hps35\o\al(\s\up12(3),0)EQ\*jc3\*hps35\o\al(\s\up12(2),0)EQ\*jc3\*hps36\o\al(\s\up5(4),0)EQ\*jc3\*hps36\o\al(\s\up5(5),3)EQ\*jc3\*hps36\o\al(\s\up5(7),2)EQ\*jc3\*hps36\o\al(\s\up5(0),3)EQ\*jc3\*hps36\o\al(\s\up5(0),0)EQ\*jc3\*hps36\o\al(\s\up5(35),13)EQ\*jc3\*hps36\o\al(\s\up5(22),2)|EQ\*jc3\*hps36\o\al(\s\up9(0),0)EQ\*jc3\*hps36\o\al(\s\up9(020,X=A-1B=|0]EQ\*jc3\*hps35\o\al(\s\up11(0),1)EQ\*jc3\*hps35\o\al(\s\up13(0),0)0]EQ\*jc3\*hps35\o\al(\s\up11(0),1)EQ\*jc3\*hps35\o\al(\s\up13(0),0)EQ\*jc3\*hps35\o\al(\s\up10(0),1)12-52-2102-1EQ\*jc3\*hps35\o\al(\s\up13(0),0)-1-310EQ\*jc3\*hps35\o\al(\s\up13(1),2)EQ\*jc3\*hps35\o\al(\s\up12(2),1)EQ\*jc3\*hps35\o\al(\s\up12(1),a)EQ\*jc3\*hps35\o\al(\s\up12(0),b)-1|「1EQ\*jc3\*hps35\o\al(\s\up13(0),0)EQ\*jc3\*hps35\o\al(\s\up8(2),3)EQ\*jc3\*hps35\o\al(\s\up8(1),4)|「-6|L-5EQ\*jc3\*hps35\o\al(\s\up13(2),3)EQ\*jc3\*hps35\o\al(\s\up13(5),8)EQ\*jc3\*hps35\o\al(\s\up13(3),λ)EQ\*jc3\*hps35\o\al(\s\up13(0),0)EQ\*jc3\*hps35\o\al(\s\up12(1),0)-32]「1EQ\*jc3\*hps35\o\al(\s\up14(1),1)EQ\*jc3\*hps35\o\al(\s\up14(0),0)||lx23(x3为任意实数)2-2|EQ\*jc3\*hps35\o\al(\s\up14(0|EQ\*jc3\*hps35\o\al(\s\up14(0),0)0-a-1-2(xlx226，当λ=4时有非零解，lx227，证：因为（E＋AE－AE－A2＝0AA＝E，（B-1AB)T=BTAT(B-1)T=B-1AB31,11,51,93,13,53,94,（1）M=|EQ\*jc3\*hps36\o\al(\s\up9(1),0)EQ\*jc3\*hps36\o\al(\s\up1(0),1)||61）R1,12,23,3EQ\*jc3\*hps35\o\al(\s\up1(1),0EQ\*jc3\*hps35\o\al(\s\up1(1),0)EQ\*jc3\*hps35\o\al(\s\up18(0),0)|EQ\*jc3\*hps35\o\al(\s\up1(0),0)EQ\*jc3\*hps35\o\al(\s\up18(0),0)设（A－B）∪（A－CA由德﹒摩根律得（A－B）∪（A－CA根据（12）可知（A－B）∪（A－CAA∩B∩C＝Ф2，用两边互相包含来证等式证：设xA(BC)则xA且x(BC)即xA且xB或者xA且xC从而有xAB,或xAC所以x(AB)(AC)即A(BC)(AB)(AC)即xA且xB或者xA且xC从而有xA且所以xA(BC)即(AB)(AC)A(BC)112345678）21）(FR)M（2）C(HP)（3）ME（4）MR(C(FS))31,11,51,93,13,53,910，(AB)AB，A(AB)A，A(AB)A4,（1）M=|EQ\*jc3\*hps36\o\al(\s\up9(1),0)EQ\*jc3\*hps36\o\al(\s\up1(0),1)|EQ\*jc3\*hps35\o\al(\s\up8(0),0)EQ\*jc3\*hps35\o\al(\s\up8(0),1)EQ\*jc3\*hps35\o\al(\s\up8(0),0)EQ\*jc3\*hps35\o\al(\s\up8(0),0)EQ\*jc3\*hps35\o\al(\s\up8(0),0)EQ\*jc3\*hps35\o\al(\s\up16(0),0)EQ\*jc3\*hps35\o\al(\s\up16(1),0)EQ\*jc3\*hps35\o\al(\s\up16(0),0)EQ\*jc3\*hps35\o\al(\s\up16(0),0)EQ\*jc3\*hps35\o\al(\s\up16(0),0)61）R1,12,23,3EQ\*jc3\*hps35\o\al(\s\up1(1),0EQ\*jc3\*hps35\o\al(\s\up1(1),0)EQ\*jc3\*hps35\o\al(\s\up18(0),0)|EQ\*jc3\*hps35\o\al(\s\up