30 câu trắc nghiệm Toán 11 Chân trời sáng tạo Đạo hàm có đáp án

TXĐ: \[{\rm{D}} = [ - 1; + \infty )\]

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

\[ = \mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{1}{{\sqrt {{\rm{x}} + 1} + \sqrt 2 }} = \frac{1}{{2\sqrt 2 }} = \frac{{\sqrt 2 }}{4}\]

Đáp án cần chọn là: A

Giả sử \[\Delta x\] là số gia của đối số tại x₀.

Ta có \[{\rm{\Delta y = f}}\left( {{{\rm{x}}_{\rm{0}}}{\rm{ + \Delta x}}} \right) - {\rm{f}}\left( {{{\rm{x}}_{\rm{0}}}} \right){\rm{ = }}{\left( {{{\rm{x}}_{\rm{0}}}{\rm{ + \Delta x}}} \right)^{\rm{2}}} - {\rm{x}}_{\rm{0}}^{\rm{2}}{\rm{ = \Delta x}}\left( {{\rm{2}}{{\rm{x}}_{\rm{0}}}{\rm{ + \Delta x}}} \right)\]

\[\mathop {\lim }\limits_{{\rm{\Delta x}} \to 0} \frac{{{\rm{\Delta y}}}}{{{\rm{\Delta x}}}} = \mathop {\lim }\limits_{{\rm{\Delta x}} \to 0} \left( {{\rm{2}}{{\rm{x}}_{\rm{0}}}{\rm{ + \Delta x}}} \right){\rm{ = 2}}{{\rm{x}}_{\rm{0}}}\].

Vậy \[{\rm{f'}}\left( {{{\rm{x}}_{\rm{0}}}} \right){\rm{ = 2}}{{\rm{x}}_0}\]

Đáp án cần chọn là: C

Bài giải trên hoàn toàn đúng.

Đáp án cần chọn là: D

Định nghĩa \[{\rm{f'}}\left( {{{\rm{x}}_{\rm{0}}}} \right) = \mathop {\lim }\limits_{{\rm{\Delta x}} \to 0} \frac{{{\rm{f(}}{{\rm{x}}_{\rm{0}}}{\rm{ + \Delta x)}} - {\rm{f(}}{{\rm{x}}_{\rm{0}}}{\rm{)}}}}{{{\rm{\Delta x}}}}\] hay \[{\rm{f'}}\left( {{{\rm{x}}_{\rm{0}}}} \right) = \mathop {\lim }\limits_{{\rm{h}} \to 0} \frac{{{\rm{f(}}{{\rm{x}}_{\rm{0}}}{\rm{ + h)}} - {\rm{f(}}{{\rm{x}}_{\rm{0}}}{\rm{)}}}}{{\rm{h}}}\] (nếu tồn tại giới hạn).

Đáp án cần chọn là: C

Lời giải

\[{\rm{f'}}\left( 0 \right) = \mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{{\rm{f}}\left( {\rm{x}} \right) - f\left( 0 \right)}}{{{\rm{x}} - 0}} = \mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{3 - \sqrt {4 - {\rm{x}}} - 1}}{{\rm{x}}} = \mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{2 - \sqrt {4 - {\rm{x}}} }}{{\rm{x}}} = \mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{4 - 4 + {\rm{x}}}}{{{\rm{x}}\left( {2 + \sqrt {4 - {\rm{x}}} } \right)}}\]

\[ = \mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{1}{{2 + \sqrt {4 - {\rm{x}}} }} = \frac{1}{4}\]

Xét \[\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( {\rm{0}} \right)}}{{{\rm{x}} - 0}} = \mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{\frac{{3 - \sqrt {4 - {\rm{x}}} }}{4} - \frac{1}{4}}}{{\rm{x}}} = \mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{2 - \sqrt {4 - {\rm{x}}} }}{{4{\rm{x}}}}\]

\[ = \mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{\left( {2 - \sqrt {4 - {\rm{x}}} } \right)\left( {2 + \sqrt {4 - {\rm{x}}} } \right)}}{{4{\rm{x}}\left( {2 + \sqrt {4 - {\rm{x}}} } \right)}} = \mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{\rm{x}}}{{4{\rm{x}}\left( {2 + \sqrt {4 - {\rm{x}}} } \right)}} = \mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{1}{{4\left( {2 + \sqrt {4 - {\rm{x}}} } \right)}} = \frac{1}{{16}}.\]

Đáp án cần chọn là: B

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

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

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

Vậy không tồn tại đạo hàm của hàm số tại x = 1.

Đáp án cần chọn là: D

Ta có \[{\rm{\Delta y = f}}\left( {{\rm{x + \Delta x}}} \right) - {\rm{f}}\left( {\rm{x}} \right){\rm{ = 2}}{\left( {{\rm{x + \Delta x}}} \right)^{\rm{3}}} - {\rm{2}}{{\rm{x}}^{\rm{3}}}{\rm{ = 6}}{{\rm{x}}^{\rm{2}}}{\rm{\Delta x + 6x}}{\left( {{\rm{\Delta x}}} \right)^{\rm{2}}}{\rm{ + 2}}{\left( {{\rm{\Delta x}}} \right)^{\rm{3}}}\]

\[ \Rightarrow \frac{{{\rm{\Delta y}}}}{{{\rm{\Delta x}}}}{\rm{ = 6}}{{\rm{x}}^{\rm{2}}}{\rm{ + 6x\Delta x + 2}}{\left( {{\rm{\Delta x}}} \right)^{\rm{2}}}{\rm{.}}\]

Đáp án cần chọn là: C

Lời giải

Ta có \[{\rm{\Delta y = f}}\left( {{\rm{x + \Delta x}}} \right) - {\rm{f}}\left( {\rm{x}} \right){\rm{ = }}\frac{{\rm{1}}}{{{\rm{x + \Delta x}}}} - \frac{{\rm{1}}}{{\rm{x}}}{\rm{ = }} - \frac{{{\rm{\Delta x}}}}{{{\rm{x}}\left( {{\rm{x + \Delta x}}} \right)}}\]

\[ \Rightarrow \frac{{{\rm{\Delta y}}}}{{{\rm{\Delta x}}}}{\rm{ = }} - \frac{{\rm{1}}}{{{\rm{x}}\left( {{\rm{x + \Delta x}}} \right)}}{\rm{.}}\]

\[\mathop {\lim }\limits_{{\rm{x}} \to {1^ + }} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( {\rm{1}} \right)}}{{{\rm{x}} - 1}} = \mathop {\lim }\limits_{{\rm{x}} \to {1^ + }} \frac{{2{\rm{x}} + 3 - 5}}{{{\rm{x}} - 1}} = \mathop {\lim }\limits_{{\rm{x}} \to {1^ + }} \frac{{2{\rm{x}} - 2}}{{{\rm{x}} - 1}} = 2\]

\[\mathop {\lim }\limits_{{\rm{x}} \to {1^ - }} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( {\rm{1}} \right)}}{{{\rm{x}} - 1}} = \mathop {\lim }\limits_{{\rm{x}} \to {1^ - }} \frac{{\frac{{{{\rm{x}}^3} + 2{{\rm{x}}^2} - 7{\rm{x}} + 4}}{{{\rm{x}} - 1}} - 5}}{{{\rm{x}} - 1}}\]

\[ = \mathop {\lim }\limits_{{\rm{x}} \to {1^ - }} \frac{{{{\rm{x}}^3} + 2{{\rm{x}}^2} - 12{\rm{x}} + 9}}{{{{\left( {{\rm{x}} - 1} \right)}^2}}} = \mathop {\lim }\limits_{{\rm{x}} \to {1^ - }} \frac{{\left( {{\rm{x}} - 1} \right)\left( {{{\rm{x}}^2} + 3{\rm{x}} - 9} \right)}}{{{{\left( {{\rm{x}} - 1} \right)}^2}}} = \mathop {\lim }\limits_{{\rm{x}} \to {1^ - }} \frac{{{{\rm{x}}^2} + 3{\rm{x}} - 9}}{{{\rm{x}} - 1}} = + \infty \]

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

Vậy hàm số không tồn tại đạo hàm tại x = 1.

Đáp án cần chọn là: D

Dễ thấy \[{\rm{f}}\left( {\rm{x}} \right){\rm{ = }}{{\rm{x}}^2} - 1\]khi \[{\rm{x}} \ge 0\] là hàm đa thức nên nó liên tục tại x = 2.

Ngoài ra \[\mathop {\lim }\limits_{{\rm{x}} \to 2} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( {\rm{2}} \right)}}{{{\rm{x}} - 2}} = \mathop {\lim }\limits_{{\rm{x}} \to 2} \frac{{\left( {{{\rm{x}}^2} - 1} \right) - \left( {{2^2} - 1} \right)}}{{{\rm{x}} - 2}} = \mathop {\lim }\limits_{{\rm{x}} \to 2} \frac{{{{\rm{x}}^2} - 4}}{{{\rm{x}} - 2}} = \mathop {\lim }\limits_{{\rm{x}} \to 2} \left( {{\rm{x}} + 2} \right) = 4\]

Do đó hàm số liên tục và có đạo hàm tại x = 2.

Xét các giới hạn \(\left\{ {\begin{array}{*{20}{c}}{\mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} \left( {{x^2} - 1} \right) = - 1}\\{\mathop {\lim }\limits_{{\rm{x}} \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{{\rm{x}} \to {0^ - }} \left( { - {x^2}} \right) = 0}\end{array}} \right.\)

Do \[\mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} {\rm{f}}\left( {\rm{x}} \right) \ne \mathop {\lim }\limits_{{\rm{x}} \to {0^ - }} {\rm{f}}\left( {\rm{x}} \right)\] nên hàm số không liên tục tại x = 0.

Do đó, hàm số không có đạo hàm tại x = 0.

Đáp án cần chọn là: D

Giả sử \[{\rm{\Delta x}}\] là số gia của đối số tại \[{x_0}\]

Ta có \[{\rm{\Delta y = f}}\left( {{{\rm{x}}_{\rm{0}}}{\rm{ + \Delta x}}} \right) - {\rm{f}}\left( {{{\rm{x}}_{\rm{0}}}} \right){\rm{ = }}\frac{{\rm{1}}}{{{{\rm{x}}_{\rm{0}}}{\rm{ + \Delta x}}}} - \frac{{\rm{1}}}{{{{\rm{x}}_{\rm{0}}}}}{\rm{ = }} - \frac{{{\rm{\Delta x}}}}{{{{\rm{x}}_{\rm{0}}}\left( {{{\rm{x}}_{\rm{0}}}{\rm{ + \Delta x}}} \right)}}\]

\[\mathop {\lim }\limits_{{\rm{\Delta x}} \to 0} \frac{{{\rm{\Delta y}}}}{{{\rm{\Delta x}}}} = \mathop {\lim }\limits_{{\rm{\Delta x}} \to 0} \left( { - \frac{1}{{{{\rm{x}}_0}\left( {{{\rm{x}}_0} + {\rm{\Delta x}}} \right)}}} \right) = - \frac{1}{{{\rm{x}}_0^2}}\]

Vậy \[{\rm{f'}}\left( {{{\rm{x}}_{\rm{0}}}} \right) = - \frac{1}{{{\rm{x}}_0^2}} \Rightarrow {\rm{f'}}\left( {\sqrt 2 } \right) = - \frac{1}{2}\]

Ta có:

\[{\rm{f'}}\left( {\rm{0}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( {\rm{0}} \right)}}{{\rm{x}}} = \mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{\frac{{\sqrt {{{\rm{x}}^2} + 1} - 1}}{{\rm{x}}}}}{{\rm{x}}} = \mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{\sqrt {{{\rm{x}}^2} + 1} - 1}}{{{{\rm{x}}^2}}} = \mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{{{\rm{x}}^2} + 1 - 1}}{{{{\rm{x}}^2}\left( {\sqrt {{{\rm{x}}^2} + 1} + 1} \right)}}\]

\[ = \mathop {\lim }\limits_{x \to 0} \frac{1}{{\sqrt {{{\rm{x}}^2} + 1} + 1}} = \frac{1}{2}\]

Đáp án cần chọn là: A

Ta có: \[\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( {\rm{0}} \right)}}{{{\rm{x}} - 0}} = \mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{\sqrt {\rm{x}} }}{{{{\rm{x}}^2}}} = \mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{1}{{{\rm{x}}\sqrt {\rm{x}} }} = + \infty \Rightarrow \] Hàm số không có đạo hàm tại x = 0.

Đáp án cần chọn là: B

(I) hiển nhiên đúng.

(II) sai.

Ví dụ: Xét hàm số \[{\rm{f}}\left( {\rm{x}} \right){\rm{ = }}\left| {\rm{x}} \right|\] ta có

\[\mathop {\lim }\limits_{{\rm{x}} \to {{\rm{x}}_0}} {\rm{ = }}\left| {{{\rm{x}}_{\rm{0}}}} \right|{\rm{ = f}}\left( {{{\rm{x}}_{\rm{0}}}} \right) \Rightarrow \] Hàm số liên tục tại trên R. Tuy nhiên hàm số không có đạo hàm tại x = 0

\(f'\left( 0 \right) = \mathop {\lim }\limits_{x \to {0^{}}} \frac{{\left| x \right| - 0}}{{x - 0}} = \mathop {\lim }\limits_{x \to {0^{}}} \frac{{\left| x \right|}}{x}\)

\(\left\{ {\begin{array}{*{20}{c}}{\mathop {\lim }\limits_{x \to {0^ + }} \frac{{\left| x \right|}}{x} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{x}{x} = 1}\\{\mathop {\lim }\limits_{x \to {0^ - }} \frac{{\left| x \right|}}{x} = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{ - x}}{x} = - 1}\end{array}} \right. \Rightarrow \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\left| x \right|}}{x} \ne \mathop {\lim }\limits_{x \to {0^ - }} \frac{{\left| x \right|}}{x}\)

Không tồn tại đạo hàm của hàm số tại x = 0.  

Đáp án cần chọn là: A

Dựa vào nhận xét: Hàm số có đạo hàm tại x = x₀ thì liên tục tại x = x₀. Điều ngược lại không đúng.

Ta thấy đáp án C đúng.

Đáp án cần chọn là: C

\(\left\{ {\begin{array}{*{20}{c}}{\mathop {\lim }\limits_{x \to {0^ + }} \frac{{f\left( x \right) - f\left( 0 \right)}}{{x - 0}} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{{x^2}}}{x} = \mathop {\lim }\limits_{x \to {0^ + }} x = 0}\\{\mathop {\lim }\limits_{x \to {0^ - }} \frac{{f\left( x \right) - f\left( 0 \right)}}{{x - 0}} = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{ - {x^2}}}{x} = \mathop {\lim }\limits_{x \to {0^ - }} \left( { - x} \right) = 0}\end{array}} \right.\)

\[ \Rightarrow \mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( {\rm{0}} \right)}}{{{\rm{x}} - 0}} = \mathop {\lim }\limits_{{\rm{x}} \to {0^ - }} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( {\rm{0}} \right)}}{{{\rm{x}} - 0}} = 0\]

⇒ Hàm số y = f(x) có đạo hàm tại x = 0.

\[\mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} \frac{{{\rm{g}}\left( {\rm{x}} \right) - {\rm{g}}\left( {\rm{0}} \right)}}{{{\rm{x}} - 0}} = \mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} \frac{{\sqrt {\rm{x}} }}{{\rm{x}}} = \mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} \frac{1}{{\sqrt {\rm{x}} }} = + \infty \]

⇒ Hàm số y = g(x) không có đạo hàm tại x = 0.

Đáp án cần chọn là: A

Ta có:

\[f'\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} \frac{{f\left( x \right) - f\left( 0 \right)}}{{x - 0}} = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[3]{{4{x^2} + 8}} - \sqrt {8{x^2} + 4} }}{{{x^2}}}\]

\[ = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[3]{{4{x^2} + 8}} - 2}}{{{x^2}}} - \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[{}]{{8{x^2} + 4}} - 2}}{{{x^2}}}\]

\[ = \mathop {\lim }\limits_{x \to 0} \frac{{4{x^2}}}{{\left( {{{\sqrt[3]{{4{x^2} + 8}}}^2} + 2\sqrt[3]{{4{x^2} + 8}} + 4} \right)}} - \mathop {\lim }\limits_{x \to 0} \frac{{8{x^2}}}{{{x^2}\left( {{{\sqrt[{}]{{4{x^2} + 8}}}^2} + 2} \right)}}\]

\[ = \mathop {\lim }\limits_{x \to 0} \frac{4}{{{{\sqrt[3]{{4{x^2} + 8}}}^2} + 2\sqrt[3]{{4{x^2} + 8}} + 4}} - \mathop {\lim }\limits_{x \to 0} \frac{8}{{\sqrt[{}]{{8{x^2} + 4}} + 2}} = \frac{1}{3} - 2 = \frac{{ - 5}}{3}\]

Dựa vào đồ thị hàm số ta thấy \[\mathop {\lim }\limits_{{\rm{x}} \to {1^ - }} {\rm{f}}\left( {\rm{x}} \right) = 1,\,\,\mathop {\lim }\limits_{{\rm{x}} \to {1^ + }} {\rm{f}}\left( {\rm{x}} \right) = 0 \Rightarrow \mathop {\lim }\limits_{{\rm{x}} \to {1^ - }} {\rm{f}}\left( {\rm{x}} \right) \ne \mathop {\lim }\limits_{{\rm{x}} \to {1^ + }} {\rm{f}}\left( {\rm{x}} \right) \Rightarrow \] Không tồn tại \[\mathop {\lim }\limits_{{\rm{x}} \to 1} {\rm{f}}\left( {\rm{x}} \right)\], hàm số không liên tục tại x = 1.

Ngoài ra tại các điểm x = 0, x = 2, x = 3  thì hàm số đều có đạo hàm.

Vậy hàm số không có đạo hàm tại x = 1.

Đáp án cần chọn là: B

Ta có:\[\mathop {\lim }\limits_{{\rm{x}} \to {1^ + }} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( {\rm{1}} \right)}}{{{\rm{x}} - {\rm{1}}}} = \mathop {\lim }\limits_{{\rm{x}} \to {1^ + }} \frac{{\frac{{{{\rm{x}}^{\rm{3}}} - {\rm{4}}{{\rm{x}}^{\rm{2}}}{\rm{ + 3x}}}}{{{{\rm{x}}^{\rm{2}}} - {\rm{3x + 2}}}} - {\rm{0}}}}{{{\rm{x}} - {\rm{1}}}} = \mathop {\lim }\limits_{{\rm{x}} \to {1^ + }} \frac{{{\rm{x}}\left( {{\rm{x}} - {\rm{3}}} \right)\left( {{\rm{x}} - {\rm{1}}} \right)}}{{{{\left( {{\rm{x}} - {\rm{1}}} \right)}^{\rm{2}}}\left( {{\rm{x}} - {\rm{2}}} \right)}}\]

\[ = \mathop {\lim }\limits_{{\rm{x}} \to {1^ + }} \frac{{{\rm{x}}\left( {{\rm{x}} - 3} \right)}}{{\left( {{\rm{x}} - 1} \right)\left( {{\rm{x}} - 2} \right)}} = + \infty \]

\[\mathop {\lim }\limits_{{\rm{x}} \to {1^ - }} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( {\rm{1}} \right)}}{{{\rm{x}} - 1}} = \mathop {\lim }\limits_{{\rm{x}} \to {1^ - }} \frac{{\frac{{{{\rm{x}}^{\rm{3}}} - {\rm{4}}{{\rm{x}}^{\rm{2}}}{\rm{ + 3x}}}}{{{{\rm{x}}^{\rm{2}}} - {\rm{3x + 2}}}} - {\rm{0}}}}{{{\rm{x}} - {\rm{1}}}} = \mathop {\lim }\limits_{{\rm{x}} \to {1^ - }} \frac{{{\rm{x}}\left( {{\rm{x}} - {\rm{3}}} \right)\left( {{\rm{x}} - {\rm{1}}} \right)}}{{{{\left( {{\rm{x}} - {\rm{1}}} \right)}^{\rm{2}}}\left( {{\rm{x}} - {\rm{2}}} \right)}}\]

\[ = \mathop {\lim }\limits_{{\rm{x}} \to {1^ - }} \frac{{{\rm{x}}\left( {{\rm{x}} - 3} \right)}}{{\left( {{\rm{x}} - 1} \right)\left( {{\rm{x}} - 2} \right)}} = - \infty \]

Do đó không tồn tại giới hạn \[\mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( 1 \right)}}{{{\rm{x}} - 1}}\]

Vậy hàm số không có đạo hàm tại x = 1

Đáp án cần chọn là: D

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

Ta có:

\[\mathop {\lim }\limits_{{\rm{x}} \to {{\left( { - 1} \right)}^ + }} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( { - 1} \right)}}{{{\rm{x}} + 1}} = \mathop {\lim }\limits_{{\rm{x}} \to {{\left( { - 1} \right)}^ + }} \frac{{\frac{{{x^2} + x + 1}}{x} + 1}}{{x + 1}} = \mathop {\lim }\limits_{{\rm{x}} \to {{\left( { - 1} \right)}^ + }} \frac{{{x^2} + 2x + 1}}{{x\left( {x + 1} \right)}} = \mathop {\lim }\limits_{{\rm{x}} \to {{\left( { - 1} \right)}^ + }} \frac{{x + 1}}{x} = 0\]

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

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

Do đó không tồn tại  \[\mathop {\lim }\limits_{{\rm{x}} \to \left( { - 1} \right)} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( { - 1} \right)}}{{{\rm{x}} + 1}}\], vậy không tồn tại đạo hàm của hàm số tại \[{{\rm{x}}_0} = - 1\].

Đáp án cần chọn là: D

Ta có: \[{\rm{y = }}\frac{{\left| {\rm{x}} \right|}}{{{\rm{x + 1}}}} = \left\{ {\begin{array}{*{20}{c}}{\frac{x}{{x + 1}}\,\,khi\,\,x \ge 0}\\{\frac{{ - x}}{{x + 1}}\,\,khi\,\,x < 0}\end{array}} \right.\]

Ta có \(\left\{ {\begin{array}{*{20}{c}}{\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \frac{x}{{x + 1}} = 0 = f\left( 0 \right)}\\{\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{ - x}}{{x + 1}} = 0}\end{array}} \right. \Rightarrow \mathop {\lim }\limits_{x \to {x_0}^ + } f\left( x \right) = \mathop {\lim }\limits_{x \to {x_0}^ - } f\left( x \right) = f\left( 0 \right) = 0 \Rightarrow \)Hàm số liên tục tại x = 0.

\[{\rm{f'}}\left( 0 \right) = \mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( {\rm{0}} \right)}}{{{\rm{x}} - {\rm{0}}}}\]

Ta có:

\(\left\{ {\begin{array}{*{20}{c}}{\mathop {\lim }\limits_{x \to {0^ + }} \frac{{f\left( x \right) - f\left( 0 \right)}}{{x - 0}} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\frac{x}{{x + 1}} - 0}}{x} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{1}{{x + 1}} = 1}\\{\mathop {\lim }\limits_{x \to {0^ - }} \frac{{f\left( x \right) - f\left( 0 \right)}}{{x - 0}} = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{\frac{{ - x}}{{x + 1}} - 0}}{x} = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{ - 1}}{{x + 1}} = - 1}\end{array}} \right. \Rightarrow x = {x_0}\)

\( \Rightarrow \mathop {\lim }\limits_{x \to {x_0}^ + } {\frac{{f\left( x \right) - f\left( {{x_0}} \right)}}{{x - {x_0}}}_{}} \ne \mathop {\lim }\limits_{x \to {x_0}^ - } \frac{{f\left( x \right) - f\left( {{x_0}} \right)}}{{x - {x_0}}}\)

Hàm số không tồn tại đạo hàm tại x = 0.

Đáp án cần chọn là: B

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

Trước tiên hàm số phải liên tục tại x = 0.

Ta có:

\[\begin{array}{*{20}{l}}{\mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} \frac{{{{\rm{x}}^2} + 1}}{{{\rm{x}} + 1}} = 1 = {\rm{f}}\left( 0 \right)}\\{\mathop {\lim }\limits_{{\rm{x}} \to {0^ - }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {0^ - }} \left( {{\rm{ax + b}}} \right) = b}\end{array}\]

Để hàm số liên tục tại x = 0 thì \[\mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {0^ - }} {\rm{f}}\left( {\rm{x}} \right){\rm{ = f}}\left( {\rm{0}} \right) \Leftrightarrow {\rm{b}} = 1\]

Khi đó ta có \[{\rm{f'}}\left( 0 \right) = \mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( {\rm{0}} \right)}}{{\rm{x}}}\]

Ta có

\[\mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( 0 \right)}}{{\rm{x}}} = \mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} \frac{{\frac{{{{\rm{x}}^2} + 1}}{{{\rm{x}} + 1}} - 1}}{{\rm{x}}} = \mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} \frac{{{{\rm{x}}^2} - {\rm{x}}}}{{{\rm{x}}\left( {{\rm{x}} + 1} \right)}} = \mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} \frac{{{\rm{x}} - 1}}{{{\rm{x}} + 1}} = - 1\]

\[\mathop {\lim }\limits_{{\rm{x}} \to {0^ - }} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( 0 \right)}}{{\rm{x}}} = \mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} \frac{{\left( {{\rm{ax}} + 1} \right) - 1}}{{\rm{x}}} = \mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} {\rm{a = a}}\]

Để hàm số có đạo hàm tại x = 0 thì  \[\mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( {\rm{0}} \right)}}{{\rm{x}}} = \mathop {\lim }\limits_{{\rm{x}} \to {0^ - }} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( {\rm{0}} \right)}}{{\rm{x}}} \Leftrightarrow {\rm{a}} = - 1\]

Vậy a = −1, b = 1.

Đáp án cần chọn là: D

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

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

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

Hàm số có đạo hàm tại x = 1, do đó hàm số liên tục tại x = 1.

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

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

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

Hàm số có đạo hàm tại \[{\rm{x}} = 1 \Leftrightarrow \mathop {\lim }\limits_{{\rm{x}} \to {1^ + }} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( {\rm{1}} \right)}}{{{\rm{x}} - {\rm{1}}}} = \mathop {\lim }\limits_{{\rm{x}} \to {1^ - }} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( {\rm{1}} \right)}}{{{\rm{x}} - {\rm{1}}}} \Leftrightarrow {\rm{a}} = 1\] (2)

Từ (1) và (2), ta có \[{\rm{a}} = 1,\;{\rm{b}} = - \frac{1}{2}\]

Đáp án cần chọn là: A

Một hàm số liên tục tại x₀ chưa chắc có đạo hàm tại điểm đó, hơn nữa

\[\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( {\rm{0}} \right)}}{{{\rm{x}} - {\rm{0}}}} = \mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{{\rm{xsin}}\frac{{\rm{\pi }}}{{\rm{x}}} - {\rm{0}}}}{{\rm{x}}} = \mathop {\lim }\limits_{{\rm{x}} \to 0} \sin \frac{{\rm{\pi }}}{{\rm{x}}} = + \infty \]hàm số không có đạo hàm tại x = 0.

Lập luận trên sai từ bước 4.

Đáp án cần chọn là: D

\[{\rm{f'}}\left( {\rm{0}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( {\rm{0}} \right)}}{{{\rm{x}} - 0}} = \mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{{\rm{x}}\left( {{\rm{x}} - {\rm{1}}} \right)\left( {{\rm{x}} - {\rm{2}}} \right)...\left( {{\rm{x}} - {\rm{1000}}} \right) - {\rm{0}}}}{{\rm{x}}}\]

\[ = \mathop {\lim }\limits_{{\rm{x}} \to 0} \left( {{\rm{x}} - 1} \right)\left( {{\rm{x}} - 2} \right)...\left( {{\rm{x}} - 1000} \right) = \mathop {\lim }\limits_{{\rm{x}} \to 0} \left( { - 1} \right)\left( { - 2} \right)\left( { - 3} \right)...\left( { - 1000} \right) = {\left( { - 1} \right)^{1000}}.1000! = 1000!\]

Đáp án cần chọn là: B

Để hàm số có đạo hàm tại x = 1 thì trước hết hàm số phải liên tục tại x = 1.

Ta có: f(0) = 1

\[\mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} \left( {{\rm{a}}{{\rm{x}}^{\rm{2}}}{\rm{ + bx + 1}}} \right){\rm{ = 1 = f}}\left( {\rm{0}} \right)\]

\[\mathop {\lim }\limits_{{\rm{x}} \to {0^ - }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {0^ - }} \left( {asinx + bcosx} \right) = b\]

Để hàm số liên tục tại x = 1 thì \[\mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {0^ - }} {\rm{f}}\left( {\rm{x}} \right){\rm{ = f}}\left( {\rm{0}} \right) \Leftrightarrow {\rm{b}} = 1\]

Khi đó ta có: \[{\rm{f'}}\left( 0 \right) = \mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( 0 \right)}}{{{\rm{x}} - 0}}\]

\[\mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( {\rm{0}} \right)}}{{{\rm{x}} - {\rm{0}}}} = \mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} \frac{{{\rm{a}}{{\rm{x}}^{\rm{2}}}{\rm{ + x + 1}} - {\rm{1}}}}{{\rm{x}}} = \mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} \left( {{\rm{ax + 1}}} \right) = 1\]

\[\mathop {\lim }\limits_{{\rm{x}} \to {0^ - }} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( {\rm{0}} \right)}}{{{\rm{x}} - {\rm{0}}}} = \mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} \frac{{{\rm{asinx + cosx}} - {\rm{1}}}}{{\rm{x}}}\]

\[ = \mathop {\lim }\limits_{{\rm{x}} \to {0^ - }} \frac{{{\rm{2asin}}\frac{{\rm{x}}}{{\rm{2}}}{\rm{cos}}\frac{{\rm{x}}}{{\rm{2}}} - {\rm{2si}}{{\rm{n}}^{\rm{2}}}\frac{{\rm{x}}}{{\rm{2}}}}}{{\rm{x}}}\]

\[ = \mathop {\lim }\limits_{{\rm{x}} \to {0^ - }} \frac{{\sin \frac{{\rm{x}}}{2}}}{{\frac{{\rm{x}}}{2}}}\mathop {\lim }\limits_{{\rm{x}} \to {0^ - }} \left( {{\rm{acos}}\frac{{\rm{x}}}{{\rm{2}}} - {\rm{2sin}}\frac{{\rm{x}}}{{\rm{2}}}} \right) = {\rm{a}}\]

Để tồn tại \[{\rm{f'}}\left( 0 \right) \Leftrightarrow \mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( {\rm{0}} \right)}}{{{\rm{x}} - {\rm{0}}}} = \mathop {\lim }\limits_{{\rm{x}} \to {0^ - }} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( {\rm{0}} \right)}}{{{\rm{x}} - {\rm{0}}}} \Leftrightarrow {\rm{a}} = 1.\]

Đáp án cần chọn là: A

Bước 1:

Ta có: \[\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{\sin {\rm{x}} - \sin 3{\rm{x}}}}{{\rm{x}}} = \mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{\sin {\rm{x}}}}{{\rm{x}}} - \mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{\sin 3{\rm{x}}}}{{\rm{x}}}\]

Bước 2:

\[ = \mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{\sin {\rm{x}}}}{{\rm{x}}} - \mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{3.\sin 3{\rm{x}}}}{{3{\rm{x}}}} = 1 - 3 = - 2\]

Đáp án cần chọn là: C