Cho hàm số y = f(x) xác định: \(f\left( x \right)\left\{ {\begin{array}{*{20}{c}}{\frac{{\sqrt {{x^2} + 1} - 1}}{x}\,\,khi\,\,x \ne 0}\\{0\,\,khi\,\,x = 0}\end{array}} \right.\). Giá trị của f′(0) bằng:

\[\mathop {\lim }\limits_{{\rm{x}} \to {1^ + }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {1^ + }} \left( {{\rm{a}}{{\rm{x}}^{\rm{2}}}{\rm{ + bx}}} \right){\rm{ = a + b = f}}\left( {\rm{1}} \right)\]

\[\mathop {\lim }\limits_{{\rm{x}} \to {1^ - }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {1^ - }} \left( {{\rm{2x}} - 1} \right) = 1\]

Để hàm số liên tục tại x = 1 thì \[\mathop {\lim }\limits_{{\rm{x}} \to {1^ + }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {1^ - }} {\rm{f}}\left( {\rm{x}} \right){\rm{ = f}}\left( {\rm{1}} \right) \Leftrightarrow {\rm{a + b = 1}}\,\,\,\left( {\rm{1}} \right)\]

Khi đó ta có: \[{\rm{f'}}\left( 1 \right) = \mathop {\lim }\limits_{x \to 1} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( {\rm{1}} \right)}}{{{\rm{x}} - {\rm{1}}}}\]

\[\mathop {\lim }\limits_{{\rm{x}} \to {1^ + }} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( {\rm{1}} \right)}}{{{\rm{x}} - {\rm{1}}}} = \mathop {\lim }\limits_{{\rm{x}} \to {1^ + }} \frac{{{\rm{a}}{{\rm{x}}^{\rm{2}}}{\rm{ + bx}} - \left( {{\rm{a + b}}} \right)}}{{{\rm{x}} - {\rm{1}}}} = \mathop {\lim }\limits_{{\rm{x}} \to {1^ + }} \frac{{{\rm{a}}\left( {{{\rm{x}}^{\rm{2}}} - {\rm{1}}} \right){\rm{ + b}}\left( {{\rm{x}} - {\rm{1}}} \right)}}{{{\rm{x}} - {\rm{1}}}}\]

\[ = \mathop {\lim }\limits_{{\rm{x}} \to {1^ + }} \left[ {{\rm{a}}\left( {{\rm{x + 1}}} \right){\rm{ + b}}} \right] = {\rm{2a + b}}\]

\[\mathop {\lim }\limits_{{\rm{x}} \to {1^ - }} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( {\rm{1}} \right)}}{{{\rm{x}} - {\rm{1}}}} = \mathop {\lim }\limits_{x \to {1^ - }} \frac{{{\rm{2x}} - {\rm{1}} - \left( {{\rm{a + b}}} \right)}}{{{\rm{x}} - 1}} = \mathop {\lim }\limits_{{\rm{x}} \to {1^ - }} \frac{{2{\rm{x}} - 2}}{{{\rm{x}} - 1}} = 2\]

Để hàm số có đạo hàm tại x = 1 thì

\[{\rm{f'}}\left( 1 \right) = \mathop {\lim }\limits_{{\rm{x}} \to {1^ + }} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( 1 \right)}}{{{\rm{x}} - 1}} = \mathop {\lim }\limits_{{\rm{x}} \to {1^ - }} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( 1 \right)}}{{{\rm{x}} - 1}} \Leftrightarrow {\rm{2a + b}} = 2\,\,\,\left( 2 \right)\]

Từ (1) và (2) ta có hệ: \(\left\{ {\begin{array}{*{20}{c}}{a + b = 1}\\{2a + b = 2}\end{array} \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{a = 1}\\{b = 0}\end{array}} \right.} \right.\)

Đáp án cần chọn là: C

Để hàm số có đạo hàm của hàm số tại điểm x = 1 thì trước hết hàm số phải liên tục tại x = 1, tức là \[\mathop {\lim }\limits_{{\rm{x}} \to 1} {\rm{f}}\left( {\rm{x}} \right) = {\rm{f}}\left( 1 \right) \Leftrightarrow \mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{{{\rm{x}}^2} - 1}}{{{\rm{x}} - 1}} = {\rm{a}} \Leftrightarrow \mathop {\lim }\limits_{{\rm{x}} \to 1} \left( {{\rm{x + 1}}} \right){\rm{ = a}} \Leftrightarrow 2 = {\rm{a}}\]

Khi đó hàm số có dạng: \(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{\frac{{{x^2} - 1}}{{x - 1}}\,\,khi\,\,x \ne 1}\\{2\,\,khi\,\,x = 1}\end{array}} \right.\)

\[ \Rightarrow {\rm{f'}}\left( {\rm{1}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( {\rm{1}} \right)}}{{{\rm{x}} - 1}} = \mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{\frac{{{{\rm{x}}^2} - 1}}{{{\rm{x}} - 1}} - 2}}{{{\rm{x}} - 1}} = \mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{{\rm{x}} + 1 - 2}}{{{\rm{x}} - 1}} = 1\]

Vậy a = 2.

Đáp án cần chọn là: B