下列各选项中，函数相同的是（ ）。
A: \(
f(x) = \ln {x^2},g(x) = 2\ln x \)
B: \(
f(x) = x,g(x) = \sqrt { { x^2}} \)
C: \(
f(x) = \sqrt { { x^2}} ,g(x) = \left| x \right| \)
D: \(
f(x) = { { {x^2} - 1} \over {x - 1}},g(x) = x + 1 \)
A: \(
f(x) = \ln {x^2},g(x) = 2\ln x \)
B: \(
f(x) = x,g(x) = \sqrt { { x^2}} \)
C: \(
f(x) = \sqrt { { x^2}} ,g(x) = \left| x \right| \)
D: \(
f(x) = { { {x^2} - 1} \over {x - 1}},g(x) = x + 1 \)
举一反三
- 下列函数相等的是( )。 A: \( f(x) = \ln {x^2},g(x) = 2\ln x \) B: \( f(x) = x,g(x) = \sqrt { { x^2}} \) C: \( f(x) = \sqrt { { x^2}} ,g(x) = \left| x \right| \) D: \( f(x) = { { {x^2} - 1} \over {x - 1}},g(x) = x + 1 \)
- 3. 下列各对函数$y=f(u), u=g(x)$中, 可以复合成复合函数$y=f(g(x))$的是( ). A: $f(u) = \sqrt {{u^2} + 1} ,\quad g(x) = {{\rm{e}}^x}<br/>$ B: $<br/>f(u) = \arccos (1 + 2u),\quad g(x) = 1 + {x^2}<br/>$ C: $f(u) = \sqrt {u + 1} ,\quad g(x) = \sin x - 3<br/>$ D: $<br/>f(u) = {\ln ^2}u,\quad g(x) = \arcsin x<br/>$
- \( \int {({1 \over x} - {2 \over {\sqrt {1 - {x^2}} }})dx} = \)( ) A: \( \ln \left| x \right| + 2\arcsin x + C \) B: \( \ln \left| x \right| - 2\arcsin x + C \) C: \(- \ln \left| x \right| - 2\arcsin x + C \) D: \(- \ln \left| x \right| +2\arcsin x + C \)
- 求函数$y = \root 3 \of {x + \sqrt x } $的导数$y' = $( ) A: ${{1 + 2\sqrt x } \over {\root 3 \of {{{\left( {x + \sqrt x } \right)}^2}} }}$ B: $ {{1 + 2\sqrt x } \over {6\root 3 \of {{{\left( {x + \sqrt x } \right)}^2}} }}$ C: $ {{1 + 2\sqrt x } \over {6\sqrt x \cdot \root 3 \of {{{\left( {x + \sqrt x } \right)}^2}} }}$ D: $ {{1 + 2\sqrt x } \over {\sqrt x \cdot \root 3 \of {{{\left( {x + \sqrt x } \right)}^2}} }}$
- 【单选题】若 f ( x ) = ( x − 1 ) x 2 − 1 2 , g ( x ) = x − 1 x + 1 ,则? A. f ( x ) = g ( x ) "> f ( x ) = g ( x ) B. lim x → 1 f ( x ) = g ( x ) "> lim x → 1 f ( x ) = g ( x ) C. lim x → 1 f ( x ) = lim x → 1 g ( x ) "> lim x → 1 f ( x ) = lim x → 1 g ( x ) D. 以上等式均不成立