Đáp án:
1A 3A
Giải thích các bước giải:
\(\begin{array}{l}
C1:\\
f'\left( x \right) = 3.{\cos ^2}x.\left( { - \sin x} \right).{\sin ^2}\dfrac{x}{3} + 2.\dfrac{1}{3}.\sin \dfrac{x}{3}.\cos \dfrac{x}{3}.{\cos ^3}x\\
= - 3{\cos ^2}x.\sin x.{\sin ^2}\dfrac{x}{3} + \dfrac{2}{3}.\sin \dfrac{x}{3}.\cos \dfrac{x}{3}.{\cos ^3}x\\
f'\left( \pi \right) = - \dfrac{{\sqrt 3 }}{6}\\
\to A\\
C3:\\
f'\left( x \right) = \dfrac{{2x + 6 - 2x + 1}}{{{{\left( {x + 3} \right)}^2}}}\\
= \dfrac{7}{{{{\left( {x + 3} \right)}^2}}}\\
\to A
\end{array}\)
Câu 2:
\(\begin{array}{l}
y'\left( {{x_0}} \right) = \dfrac{{2\left( {{x_0} - 1} \right) - 2{x_0} - 1}}{{{{\left( {{x_0} - 1} \right)}^2}}}\\
= \dfrac{{2{x_0} - 2 - 2{x_0} - 1}}{{{{\left( {{x_0} - 1} \right)}^2}}}\\
= \dfrac{{ - 3}}{{{{\left( {{x_0} - 1} \right)}^2}}}
\end{array}\)
\( \to PTTT:y = \dfrac{{ - 3}}{{{{\left( {{x_0} - 1} \right)}^2}}}\left( {x - {x_0}} \right) + {y_0}\)
\( \to y = \dfrac{{ - 3}}{{{{\left( {{x_0} - 1} \right)}^2}}}\left( {x - {x_0}} \right) + \dfrac{{2{x_0} + 1}}{{{x_0} - 1}}\)
Do phương trình tiếp tuyến đi qua điểm (-7;5)
\(\begin{array}{l}
\to 5 = \dfrac{{ - 3}}{{{{\left( {{x_0} - 1} \right)}^2}}}\left( { - 7 - {x_0}} \right) + \dfrac{{2{x_0} + 1}}{{{x_0} - 1}}\\
\to 5\left( {{x_0}^2 - 2{x_0} + 1} \right) = 21 + 3{x_0} + 2{x_0}^2 - {x_0} - 1\\
\to 3{x_0}^2 - 12{x_0} - 15 = 0\\
\to \left[ \begin{array}{l}
{x_0} = 5\\
{x_0} = - 1
\end{array} \right. \to \left[ \begin{array}{l}
{y_0} = \dfrac{{11}}{4}\\
{y_0} = \dfrac{1}{2}
\end{array} \right.\\
\to \left[ \begin{array}{l}
k = - \dfrac{3}{{16}}\\
k = - \dfrac{3}{4}
\end{array} \right.\\
\to PTTT:\left[ \begin{array}{l}
y = - \dfrac{3}{{16}}\left( {x - 5} \right) + \dfrac{{11}}{4}\\
y = - \dfrac{3}{4}\left( {x + 1} \right) + \dfrac{1}{2}
\end{array} \right.\\
\to \left[ \begin{array}{l}
y = - \dfrac{3}{{16}}x + \dfrac{{59}}{{16}}\\
y = - \dfrac{3}{4}x - \dfrac{1}{4}
\end{array} \right.
\end{array}\)
