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Câu hỏi: Cho hàm số $y=f\left( x \right)$ xác định trên $\mathbb{R}$, có đạo hàm thỏa mãn ${f}'\left( 1 \right)=-10$. Tính $I=\underset{x\to 1}{\mathop{\lim }} \dfrac{f\left( \dfrac{x+1}{2} \right)-f\left( 1 \right)}{x-1}$.
A. $-5$.
B. $-20$.
C. $-10$.
D. $10$.
A. $-5$.
B. $-20$.
C. $-10$.
D. $10$.
$I=\underset{x\to 1}{\mathop{\lim }} \dfrac{f\left( \dfrac{x+1}{2} \right)-f\left( 1 \right)}{x-1}$.
Đặt $t=\dfrac{x+1}{2}\Rightarrow x-1=2\left( t-1 \right); $ Khi $x\to 1$ thì $t\to 1$.
Suy ra $I=\underset{x\to 1}{\mathop{\lim }} \dfrac{f\left( \dfrac{x+1}{2} \right)-f\left( 1 \right)}{x-1}=\underset{t\to 1}{\mathop{\lim }} \dfrac{f\left( t \right)-f\left( 1 \right)}{2\left( t-1 \right)}=\dfrac{1}{2}{f}'\left( 1 \right)=\dfrac{1}{2}.\left( -10 \right)=-5.$
Đặt $t=\dfrac{x+1}{2}\Rightarrow x-1=2\left( t-1 \right); $ Khi $x\to 1$ thì $t\to 1$.
Suy ra $I=\underset{x\to 1}{\mathop{\lim }} \dfrac{f\left( \dfrac{x+1}{2} \right)-f\left( 1 \right)}{x-1}=\underset{t\to 1}{\mathop{\lim }} \dfrac{f\left( t \right)-f\left( 1 \right)}{2\left( t-1 \right)}=\dfrac{1}{2}{f}'\left( 1 \right)=\dfrac{1}{2}.\left( -10 \right)=-5.$
Đáp án A.