ａ）
求ln x = (x/e) - ∫ √(1-cos2xdx) ∫由π到0
於(0, +∞)內只有兩個唔同既實根
ｂ）
e^x + 2 , x<0
f(x) ={
a + x , x≧0
於x = 0 , a = ?
數學問題疑難～
2007-03-03 9:03 pm
回答 (1)
2007-03-03 9:32 pm
✔ 最佳答案
a) differniate both sides wrt x :1/x=1/e -sqrt (1- cos2x)(put x=0)
By change of variable in fundemental law of calculus
1/x=1/e-0
x=e
b)It is very easy !!!
e^x + 2=a+x
a= e^x-2-x= 1-2-0 =-1
收錄日期: 2021-04-13 13:40:31
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070303000051KK01826