cho 0<(a khác 1+căn 2 và các hàm f(x) = ( a^x +a^-x)/2

* Hướng dẫn giải

Đáp án D

$f^2\left(x\right)-g^2\left(x\right)$$={\left(\frac{a^x+a^{-x}}{2}\right)}^2-{\left(\frac{a^x-a^{-x}}{2}\right)}^2=1\Rightarrow$I đúng 

$\begin{array}{l}g\left(2x\right)=\frac{a^{2x}-a^{-2x}}{2}\\ =\frac{\left(a^x-a^{-x}\right)\left(a^x+a^{-x}\right)}{2}\\ =2.\frac{a^x-a^{-x}}{2}.\frac{a^x+a^{-x}}{2}\\ =2g\left(x\right).f\left(x\right)\end{array}$

=> II đúng

$\begin{array}{l}\left\{\begin{array}{l}f\left(g\left(0\right)\right)=f\left(0\right)=1\\ g\left(f\left(0\right)\right)=g\left(1\right)=\frac{a-\frac{1}{a}}{2}=\frac{a^2-1}{2a}\end{array}\right.\\ \Rightarrow f\left(g\left(0\right)\right)\ne g\left(f\left(0\right)\right)\end{array}$

=> III sai

do $g\left(2x\right)=2g\left(x\right)f\left(x\right)$ nên $g^{\text{'}}\left(2x\right)=2\left[g^{\text{'}}\left(x\right)f\left(x\right)-g\left(x\right)f^{\text{'}}\left(x\right)\right]\Rightarrow$IV sai