Cho hàm số $f \left( x \right)$ liên tục trên $\left[\right. 1 ; 2 \left]\right.$ thoả mãn f \left( x \right) = 2 + \int_{1}^{2} \left( 6 x + 2 t \right) f \left( t \right) \text{d} t , \textrm{ } \forall x \in \left[ 1 ; 2 \left]\right.. Tính f \left(\right. \dfrac{1}{3} \right)bằng

A.  

−1.

B.  

$\dfrac{1}{3}$.

C.  

1.

D.  

$- \dfrac{1}{3}$.

Đáp án đúng là: C

Ta có: $f \left( x \right) = 2 + \int_{1}^{2} \left( 6 x + 2 t \right) f \left( t \right) d t$
$\Leftrightarrow f \left( x \right) = 6 x \int_{1}^{2} f \left( t \right) d t + 2 \int_{1}^{2} t f \left( t \right) d t + 2$
Đặt $a = \int_{1}^{2} f \left( t \right) \text{d} t \textrm{ } , \textrm{ } b = \int_{1}^{2} t f \left( t \right) \text{d} t$
$\Rightarrow f \left( x \right) = 6 a x + 2 b + 2 \textrm{ }\textrm{ }\textrm{ }\textrm{ } \left( 1 \right)$
$\Rightarrow x f \left( x \right) = 6 a x^{2} + \left( 2 b + 2 \right) x \textrm{ }\textrm{ } \left( 2 \right)$.
Từ đó ta có
$\left{\right. \begin{matrix}\int_{1}^{2} f \left( x \right) \text{d} x = \int_{1}^{2} \left( 6 x . a + 2 b + 2 \right) \text{d} x \\ \int_{1}^{2} x . f \left( x \right) \text{d} x = \int_{1}^{2} \left(\right. 6 a x^{2} + \left( 2 b + 2 \right) x \left.\right) \text{d} x\end{matrix} \Leftrightarrow \left{\right. \begin{matrix}a = \left( \left(\right. 3 a x^{2} + \left( 2 b + 2 \right) x \left.\right) \left|\right)_{1}^{2} \\ b = \left( \left(\right. 2 a x^{3} + \left(\right. b + 1 \right) x^{2} \left.\right) \left|\right.\right)_{1}^{2}\end{matrix} \\ \Leftrightarrow \left{\right. \begin{matrix}a = 9 a + 2 b + 2 \\ b = 14 a + 3 b + 3\end{matrix} \Leftrightarrow \left{\right. \begin{matrix}a = - \dfrac{1}{6} \\ b = - \dfrac{1}{3}\end{matrix} \Rightarrow f \left( x \right) = - x + \dfrac{4}{3} \Rightarrow f \left( \dfrac{1}{3} \right) = 1$