Cho hai đa thức $f\left( x \right) = 4{x^4} - 2a{x^2} + \left( {a + 1} \right)x + 2$ và $g\left( x \right) = 2ax + 5.$ Tìm $a$ để $f\left( 1 \right) = g\left( 2 \right).$

$f\left( 1 \right) = 1011;f\left( { - 1} \right) =  - 1011$

$f\left( 1 \right) = 1011;f\left( { - 1} \right) = 1011$

$f\left( 1 \right) = 1011;f\left( { - 1} \right) =  - 1009$

$f\left( 1 \right) = 2021;f\left( { - 1} \right) = 2021$

$14$

$9$

$5$

$7$

$f\left( x \right) = 3x + 4.$

$f\left( x \right) = \dfrac{8}{3}x - 4.$

$f\left( x \right) = \dfrac{8}{3}x + 4.$

$f\left( x \right) =  - \dfrac{8}{3}x + 12.$

$f\left( x \right) = x + 2$

$f\left( x \right) =  - x + 4$

$f\left( x \right) =  - \dfrac{7}{3}x + \dfrac{{16}}{3}$

$f\left( x \right) = x - 2$

$f\left( { - 2} \right) = g\left( { - 2} \right)$

$f\left( { - 2} \right) = 3.g\left( { - 2} \right)$

$f\left( { - 2} \right) > g\left( { - 2} \right)$

$f\left( { - 2} \right) < g\left( { - 2} \right)$

$f\left( 0 \right) = g\left( 1 \right)$

$f\left( 0 \right) > g\left( 1 \right)$

$f\left( 0 \right) < g\left( 1 \right)$

$f\left( 0 \right) \ge g\left( 1 \right)$