Đáp án:
$\begin{array}{l}
a)A\left( {2;4} \right) \in f\left( x \right)\\
\Rightarrow 4 = \left( {2a - 1} \right).2\\
\Rightarrow 4 = 4a - 2\\
\Rightarrow 4a = 6\\
\Rightarrow a = \dfrac{3}{2}\\
\Rightarrow y = 2x\\
b)f\left( x \right) = y = 2x\\
+ Cho:x = 0 \Rightarrow y = 0
\end{array}$
=> đồ thị hàm số là đường thẳng đi qua gốc tọa độ O và điểm A
$\begin{array}{l}
c)f\left( x \right) = 1002\\
\Rightarrow 2x = 1002\\
\Rightarrow x = 501\\
d)f\left( {1002} \right) = 2.1002 = 2004\\
B6)\\
a)x = \dfrac{1}{2}\\
\Rightarrow \left\{ \begin{array}{l}
f\left( {\dfrac{1}{2}} \right) = 2.{\left( {\dfrac{1}{2}} \right)^2} = \dfrac{1}{2}\\
g\left( {\dfrac{1}{2}} \right) = - 3.\dfrac{1}{2} = - \dfrac{3}{2}\\
h\left( {\dfrac{1}{2}} \right) = {\left( {\dfrac{1}{2}} \right)^4} + {\left( {\dfrac{1}{2}} \right)^2} = \dfrac{5}{{16}}\\
t\left( {\dfrac{1}{2}} \right) = 2:\dfrac{1}{2} = 4
\end{array} \right.\\
+ Khi:x = - \dfrac{1}{2}\\
\Rightarrow \left\{ \begin{array}{l}
f\left( { - \dfrac{1}{2}} \right) = 2.{\left( { - \dfrac{1}{2}} \right)^2} = \dfrac{1}{2}\\
g\left( { - \dfrac{1}{2}} \right) = - 3.\left( {\dfrac{{ - 1}}{2}} \right) = \dfrac{3}{2}\\
h\left( {\dfrac{{ - 1}}{2}} \right) = {\left( {\dfrac{{ - 1}}{2}} \right)^4} + {\left( { - \dfrac{1}{2}} \right)^2} = \dfrac{5}{{16}}\\
t\left( { - \dfrac{1}{2}} \right) = 2:\dfrac{{ - 1}}{2} = - 4
\end{array} \right.\\
b)f\left( { - x} \right) = 2.{\left( { - x} \right)^2} = 2.{x^2}\\
\Rightarrow f\left( x \right) = f\left( { - x} \right)
\end{array}$
