Đáp án đúng: C
Giải chi tiết:Trước hết ta xét tính liên tục của hàm số tại
x = 1 x = 1 x = 1 .
Ta có:
lim x → 1 f ( x ) = lim x → 1 3 x + 1 − 2 x x − 1 = lim x → 1 ( 3 x + 1 − 2 x ) ( 3 x + 1 + 2 x ) ( x − 1 ) ( 3 x + 1 + 2 x ) = lim x → 1 3 x + 1 − 4 x 2 ( x − 1 ) ( 3 x + 1 + 2 x ) = lim x → 1 − ( x − 1 ) ( 4 x + 1 ) ( x − 1 ) ( 3 x + 1 + 2 x ) = lim x → 1 − 4 x − 1 3 x + 1 + 2 x = − 4 − 1 4 + 2 = − 5 4 = f ( 1 ) \begin{array}{l}\mathop {\lim }\limits_{x \to 1} f\left( x \right) = \mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {3x + 1} - 2x}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {\sqrt {3x + 1} - 2x} \right)\left( {\sqrt {3x + 1} + 2x} \right)}}{{\left( {x - 1} \right)\left( {\sqrt {3x + 1} + 2x} \right)}}\\ = \mathop {\lim }\limits_{x \to 1} \dfrac{{3x + 1 - 4{x^2}}}{{\left( {x - 1} \right)\left( {\sqrt {3x + 1} + 2x} \right)}} = \mathop {\lim }\limits_{x \to 1} \dfrac{{ - \left( {x - 1} \right)\left( {4x + 1} \right)}}{{\left( {x - 1} \right)\left( {\sqrt {3x + 1} + 2x} \right)}}\\ = \mathop {\lim }\limits_{x \to 1} \dfrac{{ - 4x - 1}}{{\sqrt {3x + 1} + 2x}} = \dfrac{{ - 4 - 1}}{{\sqrt 4 + 2}} = \dfrac{{ - 5}}{4} = f\left( 1 \right)\end{array} x → 1 lim f ( x ) = x → 1 lim x − 1 3 x + 1 − 2 x = x → 1 lim ( x − 1 ) ( 3 x + 1 + 2 x ) ( 3 x + 1 − 2 x ) ( 3 x + 1 + 2 x ) = x → 1 lim ( x − 1 ) ( 3 x + 1 + 2 x ) 3 x + 1 − 4 x 2 = x → 1 lim ( x − 1 ) ( 3 x + 1 + 2 x ) − ( x − 1 ) ( 4 x + 1 ) = x → 1 lim 3 x + 1 + 2 x − 4 x − 1 = 4 + 2 − 4 − 1 = 4 − 5 = f ( 1 ) ⇒ \Rightarrow ⇒ Hàm số liên tục tại
x = 1 x = 1 x = 1 .
Tính
f ′ ( 1 ) f'\left( 1 \right) f ′ ( 1 ) .
⇒ f ′ ( 1 ) = lim x → 1 f ( x ) − f ( 1 ) x − 1 = 3 x + 1 − 2 x x − 1 + 5 4 x − 1 = lim x → 1 4 3 x + 1 − 8 x + 5 x − 5 4 ( x − 1 ) 2 = lim x → 1 4 3 x + 1 − 3 x − 5 4 ( x − 1 ) 2 = lim x → 1 ( 4 3 x + 1 − 3 x − 5 ) ( 4 3 x + 1 + 3 x + 5 ) 4 ( x − 1 ) 2 ( 4 3 x + 1 + 3 x + 5 ) = lim x → 1 16 ( 3 x + 1 ) − ( 9 x 2 + 30 x + 25 ) 4 ( x − 1 ) 2 ( 4 3 x + 1 + 3 x + 5 ) = lim x → 1 − 9 x 2 + 18 x − 9 4 ( x − 1 ) 2 ( 4 3 x + 1 + 3 x + 5 ) = lim x → 1 − 9 ( x − 1 ) 2 4 ( x − 1 ) 2 ( 4 3 x + 1 + 3 x + 5 ) = lim x → 1 − 9 4 ( 4 3 x + 1 + 3 x + 5 ) = − 9 64 \begin{array}{l} \Rightarrow f'\left( 1 \right) = \mathop {\lim }\limits_{x \to 1} \dfrac{{f\left( x \right) - f\left( 1 \right)}}{{x - 1}} = \dfrac{{\dfrac{{\sqrt {3x + 1} - 2x}}{{x - 1}} + \dfrac{5}{4}}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \dfrac{{4\sqrt {3x + 1} - 8x + 5x - 5}}{{4{{\left( {x - 1} \right)}^2}}}\\ = \mathop {\lim }\limits_{x \to 1} \dfrac{{4\sqrt {3x + 1} - 3x - 5}}{{4{{\left( {x - 1} \right)}^2}}} = \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {4\sqrt {3x + 1} - 3x - 5} \right)\left( {4\sqrt {3x + 1} + 3x + 5} \right)}}{{4{{\left( {x - 1} \right)}^2}\left( {4\sqrt {3x + 1} + 3x + 5} \right)}}\\ = \mathop {\lim }\limits_{x \to 1} \dfrac{{16\left( {3x + 1} \right) - \left( {9{x^2} + 30x + 25} \right)}}{{4{{\left( {x - 1} \right)}^2}\left( {4\sqrt {3x + 1} + 3x + 5} \right)}} = \mathop {\lim }\limits_{x \to 1} \dfrac{{ - 9{x^2} + 18x - 9}}{{4{{\left( {x - 1} \right)}^2}\left( {4\sqrt {3x + 1} + 3x + 5} \right)}}\\ = \mathop {\lim }\limits_{x \to 1} \dfrac{{ - 9{{\left( {x - 1} \right)}^2}}}{{4{{\left( {x - 1} \right)}^2}\left( {4\sqrt {3x + 1} + 3x + 5} \right)}} = \mathop {\lim }\limits_{x \to 1} \dfrac{{ - 9}}{{4\left( {4\sqrt {3x + 1} + 3x + 5} \right)}} = \dfrac{{ - 9}}{{64}}\end{array} ⇒ f ′ ( 1 ) = x → 1 lim x − 1 f ( x ) − f ( 1 ) = x − 1 x − 1 3 x + 1 − 2 x + 4 5 = x → 1 lim 4 ( x − 1 ) 2 4 3 x + 1 − 8 x + 5 x − 5 = x → 1 lim 4 ( x − 1 ) 2 4 3 x + 1 − 3 x − 5 = x → 1 lim 4 ( x − 1 ) 2 ( 4 3 x + 1 + 3 x + 5 ) ( 4 3 x + 1 − 3 x − 5 ) ( 4 3 x + 1 + 3 x + 5 ) = x → 1 lim 4 ( x − 1 ) 2 ( 4 3 x + 1 + 3 x + 5 ) 1 6 ( 3 x + 1 ) − ( 9 x 2 + 3 0 x + 2 5 ) = x → 1 lim 4 ( x − 1 ) 2 ( 4 3 x + 1 + 3 x + 5 ) − 9 x 2 + 1 8 x − 9 = x → 1 lim 4 ( x − 1 ) 2 ( 4 3 x + 1 + 3 x + 5 ) − 9 ( x − 1 ) 2 = x → 1 lim 4 ( 4 3 x + 1 + 3 x + 5 ) − 9 = 6 4 − 9 Chọn C.
