例 53、设函数 =a(x)^2 与 =ln x 相切,则a的值等于 ()-|||-A、 dfrac (1)(2e) B、 dfrac (1)(e) C、e D
[题目]设 (x)=sqrt (1+{ln )^2x} 则 (e)=()-|||-A、 dfrac (sqrt {2)}(4)-|||-B、 dfrac (sq
[ dfrac {{e)^2}({(z-1))^2},1] =-|||-A e-|||-B dfrac (1)(e)-|||-C dfrac (e)(2)-||
(3)曲线 =(x)^2 与曲线 =aln x(aneq 0) 相切,则a等于 ()-|||-(A)4e (B) 3e (C) 2e (D) e
dfrac (1)(2)pi (R)^2E-|||-D. sqrt (2)(R)^2E-|||-E. dfrac (pi {R)^2E}(sqrt {2)}
(2)极限 lim _(xarrow infty )((1-dfrac {1)(2x))}^dfrac (x{2)}= () .-|||-(A)e (B) ^-
设函数 (x)=(e)^dfrac (1{x-1)}dfrac (ln |x+2|)({x)^2+x-6}求(x)=(e)^dfrac (1{x-1)}dfra
2 , 1 B . 1 , 2 C . (x)=dfrac ({e)^x-1}(x), 1D . (x)=dfrac ({e)^x-1}(x), 2设函数,若当
f(x)= ({e)^4-dfrac (1)(3))-|||-dfrac (1)(2)(e)^4-|||-dfrac (1)(2)((e)^2-dfrac
dfrac (pi {R)^2E}(2)-|||-C.2πR^2E-|||-D.0