Nếu ${x_0}$ là điểm cực đại của hàm số thì $\left( {{x_0};f\left( {{x_0}} \right)} \right)$ là:

$f\left( 0 \right) < 5$

$f\left( 2 \right) \geqslant 5$ 

$f\left( 1 \right) = 5$      

$f\left( 0 \right) = 5$

\(a > 1\)

\(0 < a < 1\)

\(a \ge 1\)        

\(a > 0\) 

\(\mathop {\lim }\limits_{x \to 0} \dfrac{{\ln \left( {1 + x} \right)}}{x} = 1\)

\(\mathop {\lim }\limits_{x \to 0} \dfrac{{\ln \left( {1 - x} \right)}}{x} = 1\)

\(\mathop {\lim }\limits_{x \to 0} \dfrac{{\ln x}}{x} = 1\)

\(\mathop {\lim }\limits_{x \to 0} \dfrac{{\ln \left( {1 + x} \right)}}{{1 + x}} = 1\)

$R$ 

$R\backslash \left\{ 0 \right\}$ 

$\left( { - \infty ;0} \right)$ 

$\left( {0; + \infty } \right)$ 

\(\dfrac{{3{a^3}\sqrt 3 }}{8}\)

\(\dfrac{{{a^3}\sqrt 3 }}{8}\)           

\(\dfrac{{3{a^3}}}{8}\)        

\(\dfrac{{{a^3}}}{8}\)

\(2\log a + \log b\)

$\log a + 2\log b$

$2\left( {\log a + \log b} \right)$

$\log a + \dfrac{1}{2}\log b$

$a > 0$ 

$a = 0$

$a < 0$ 

$a \ne 0$