giải các bpt sau
1. f(x) – (x-1).f'(x) = 0 với f(x) =$\frac{x-1}{2}$ .$cos^{2}x$
2. f'(x)=0 với f(x)=3cosx + 4sinx + 5x
giải các bpt sau
1. f(x) – (x-1).f'(x) = 0 với f(x) =$\frac{x-1}{2}$ .$cos^{2}x$
2. f'(x)=0 với f(x)=3cosx + 4sinx + 5x
Đáp án:
$\begin{array}{l}
f\left( x \right) = \dfrac{{x – 1}}{2}.{\cos ^2}x\\
f’\left( x \right) = \dfrac{1}{2}.{\cos ^2}x + \dfrac{{x – 1}}{2}.2.\left( { – \sin x} \right).\cos x\\
= \dfrac{1}{2}{\cos ^2}x – \dfrac{1}{2}\left( {x – 1} \right).\sin 2x\\
Pt:f\left( x \right) – \left( {x – 1} \right).f’\left( x \right) = 0\\
\Leftrightarrow \dfrac{{x – 1}}{2}.{\cos ^2}x – \left( {x – 1} \right).\left[ {\dfrac{1}{2}{{\cos }^2}x – \dfrac{1}{2}\left( {x – 1} \right).\sin 2x} \right] = 0\\
\Leftrightarrow \dfrac{{x – 1}}{2}.\left[ {{{\cos }^2}x – {{\cos }^2}x + \left( {x – 1} \right).\sin 2x} \right] = 0\\
\Leftrightarrow \left( {x – 1} \right).\left( {x – 1} \right).\sin 2x = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
2x = k\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = \dfrac{{k\pi }}{2}
\end{array} \right.\left( {k \in Z} \right)\\
Vậy\,x = 1;x = \dfrac{{k\pi }}{2}\\
2)\\
f\left( x \right) = 3\cos x + 4\sin x + 5x\\
\Leftrightarrow f’\left( x \right) = – 3\sin x + 4\cos x + 5 = 0\\
\Leftrightarrow 3\sin x – 4\cos x = 5\\
\Leftrightarrow \dfrac{3}{5}\sin x – \dfrac{4}{5}\cos x = 0\\
\Leftrightarrow \sin \left( {x – \arccos \dfrac{3}{5}} \right) = 0\\
\Leftrightarrow x – \arccos \dfrac{3}{5} = k\pi \\
\Leftrightarrow x = \arccos \dfrac{3}{5} + k\pi
\end{array}$