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Câu hỏi: Chứng minh rằng \(f'\left( x \right) = 0\forall x \in R,\) nếu:
Phương pháp giải:
Chứng minh các biểu thức đã cho không phụ thuộc vào x.
Từ đó suy ra \(f'\left( x \right) = 0.\)
Lời giải chi tiết:
\(\begin{array}{l}
f\left(x \right) = 3\left[ {{{\left({{{\sin }^2}x + {{\cos }^2}x} \right)}^2} - 2{{\sin }^2}x{{\cos }^2}x} \right]\\
- 2\left[ {{{\left({{{\sin }^2}x + {{\cos }^2}x} \right)}^3} - 3{{\sin }^2}x{{\cos }^2}x\left({{{\sin }^2}x + {{\cos }^2}x} \right)} \right]\\
= 3\left[ {1 - 2{{\sin }^2}x{{\cos }^2}x} \right] - 2\left[ {1 - 3{{\sin }^2}x{{\cos }^2}x} \right]\\
= 3 - 6{\sin ^2}x{\cos ^2}x - 2 + 6{\sin ^2}x{\cos ^2}x\\
= 1\\
\Rightarrow f'\left(x \right) = 0
\end{array}\)
Lời giải chi tiết:
\(\begin{array}{l}
f\left(x \right) = \left({{{\cos }^6}x + 3{{\sin }^2}x{{\cos }^4}x} \right)\\
+ \left({2{{\sin }^4}x{{\cos }^2}x + {{\sin }^4}x} \right)\\
= {\cos ^4}x\left({{{\cos }^2}x + 3{{\sin }^2}x} \right)\\
+ {\sin ^4}x\left({2{{\cos }^2}x + 1} \right)\\
= {\cos ^4}x\left({1 + 2{{\sin }^2}x} \right)\\
+ {\sin ^4}x\left({2{{\cos }^2}x + 1} \right)\\
= {\cos ^4}x + 2{\sin ^2}x{\cos ^4}x\\
+ 2{\sin ^4}x{\cos ^2}x + {\sin ^4}x\\
= \left({{{\cos }^4}x + {{\sin }^4}x} \right)\\
+ 2{\sin ^2}x{\cos ^2}x\left({{{\cos }^2}x + {{\sin }^2}x} \right)\\
= {\left({{{\sin }^2}x + {{\cos }^2}x} \right)^2} - 2{\sin ^2}x{\cos ^2}x\\
+ 2{\sin ^2}x{\cos ^2}x\\
= 1\\
\Rightarrow f'\left(x \right) = 0
\end{array}\)
Lời giải chi tiết:
\(\begin{array}{l}
f\left(x \right) = \left({\cos x\cos \frac{\pi }{3} + \sin x\sin \frac{\pi }{3}} \right)\left({\cos x\cos \frac{\pi }{4} - \sin x\sin \frac{\pi }{4}} \right)\\
+ \left({\cos x\cos \frac{\pi }{6} - \sin x\sin \frac{\pi }{6}} \right)\left({\cos x\cos \frac{{3\pi }}{4} - \sin x\sin \frac{{3\pi }}{4}} \right)\\
= \left({\frac{1}{2}\cos x + \frac{{\sqrt 3 }}{2}\sin x} \right)\left({\frac{{\sqrt 2 }}{2}\cos x - \frac{{\sqrt 2 }}{2}\sin x} \right)\\
+ \left({\frac{{\sqrt 3 }}{2}\cos x - \frac{1}{2}\sin x} \right)\left({ - \frac{{\sqrt 2 }}{2}\cos x - \frac{{\sqrt 2 }}{2}\sin x} \right)\\
= \frac{{\sqrt 2 }}{4}{\cos ^2}x + \frac{{\sqrt 6 }}{4}\sin x\cos x - \frac{{\sqrt 2 }}{4}\sin x\cos x - \frac{{\sqrt 6 }}{4}{\sin ^2}x\\
- \frac{{\sqrt 6 }}{4}{\cos ^2}x + \frac{{\sqrt 2 }}{4}\sin x\cos x - \frac{{\sqrt 6 }}{4}\sin x\cos x + \frac{{\sqrt 2 }}{4}{\sin ^2}x\\
= \frac{{\sqrt 2 - \sqrt 6 }}{4}{\cos ^2}x + \frac{{\sqrt 2 - \sqrt 6 }}{4}{\sin ^2}x\\
= \frac{{\sqrt 2 - \sqrt 6 }}{4}\left({{{\cos }^2}x + {{\sin }^2}x} \right)\\
= \frac{{\sqrt 2 - \sqrt 6 }}{4}\\
\Rightarrow f'\left(x \right) = 0
\end{array}\)
Lời giải chi tiết:
\(\begin{array}{l}
f\left(x \right) = {\cos ^2}x + {\cos ^2}\left({\frac{{2\pi }}{3} + x} \right) + {\cos ^2}\left({\frac{{2\pi }}{3} - x} \right)\\
= {\cos ^2}x + {\left({\cos \frac{{2\pi }}{3}\cos x - \sin \frac{{2\pi }}{3}\sin x} \right)^2}\\
+ {\left({\cos \frac{{2\pi }}{3}\cos x + \sin \frac{{2\pi }}{3}\sin x} \right)^2}\\
= {\cos ^2}x + {\left({ - \frac{1}{2}\cos x - \frac{{\sqrt 3 }}{2}\sin x} \right)^2}\\
+ {\left({ - \frac{1}{2}\cos x + \frac{{\sqrt 3 }}{2}\sin x} \right)^2}\\
= {\cos ^2}x + \left({\frac{1}{4}{{\cos }^2}x + \frac{{\sqrt 3 }}{2}\sin x\cos x + \frac{3}{4}{{\sin }^2}x} \right)\\
+ \left({\frac{1}{4}{{\cos }^2}x - \frac{{\sqrt 3 }}{2}\sin x\cos x + \frac{3}{4}{{\sin }^2}x} \right)\\
= {\cos ^2}x + \frac{1}{2}{\cos ^2}x + \frac{3}{2}{\sin ^2}x\\
= \frac{3}{2}{\cos ^2}x + \frac{3}{2}{\sin ^2}x\\
= \frac{3}{2}\left({{{\cos }^2}x + {{\sin }^2}x} \right)\\
= \frac{3}{2}\\
\Rightarrow f'\left(x \right) = 0
\end{array}\)
Câu a
\(f\left( x \right) = 3\left({{{\sin }^4}x + {{\cos }^4}x} \right) \) \(- 2\left( {{{\sin }^6}x + {{\cos }^6}x} \right)\)Phương pháp giải:
Chứng minh các biểu thức đã cho không phụ thuộc vào x.
Từ đó suy ra \(f'\left( x \right) = 0.\)
Lời giải chi tiết:
\(\begin{array}{l}
f\left(x \right) = 3\left[ {{{\left({{{\sin }^2}x + {{\cos }^2}x} \right)}^2} - 2{{\sin }^2}x{{\cos }^2}x} \right]\\
- 2\left[ {{{\left({{{\sin }^2}x + {{\cos }^2}x} \right)}^3} - 3{{\sin }^2}x{{\cos }^2}x\left({{{\sin }^2}x + {{\cos }^2}x} \right)} \right]\\
= 3\left[ {1 - 2{{\sin }^2}x{{\cos }^2}x} \right] - 2\left[ {1 - 3{{\sin }^2}x{{\cos }^2}x} \right]\\
= 3 - 6{\sin ^2}x{\cos ^2}x - 2 + 6{\sin ^2}x{\cos ^2}x\\
= 1\\
\Rightarrow f'\left(x \right) = 0
\end{array}\)
Câu b
\(f\left( x \right) = {\cos ^6}x + 2{\sin ^4}x{\cos ^2}x \) \(+ 3{\sin ^2}x{\cos ^4}x + {\sin ^4}x\)Lời giải chi tiết:
\(\begin{array}{l}
f\left(x \right) = \left({{{\cos }^6}x + 3{{\sin }^2}x{{\cos }^4}x} \right)\\
+ \left({2{{\sin }^4}x{{\cos }^2}x + {{\sin }^4}x} \right)\\
= {\cos ^4}x\left({{{\cos }^2}x + 3{{\sin }^2}x} \right)\\
+ {\sin ^4}x\left({2{{\cos }^2}x + 1} \right)\\
= {\cos ^4}x\left({1 + 2{{\sin }^2}x} \right)\\
+ {\sin ^4}x\left({2{{\cos }^2}x + 1} \right)\\
= {\cos ^4}x + 2{\sin ^2}x{\cos ^4}x\\
+ 2{\sin ^4}x{\cos ^2}x + {\sin ^4}x\\
= \left({{{\cos }^4}x + {{\sin }^4}x} \right)\\
+ 2{\sin ^2}x{\cos ^2}x\left({{{\cos }^2}x + {{\sin }^2}x} \right)\\
= {\left({{{\sin }^2}x + {{\cos }^2}x} \right)^2} - 2{\sin ^2}x{\cos ^2}x\\
+ 2{\sin ^2}x{\cos ^2}x\\
= 1\\
\Rightarrow f'\left(x \right) = 0
\end{array}\)
Câu c
\(f\left( x \right) = \cos \left({x - {\pi \over 3}} \right)\cos \left({x + {\pi \over 4}} \right) \) \(+ \cos \left( {x + {\pi \over 6}} \right)\cos \left({x + {{3\pi } \over 4}} \right)\)Lời giải chi tiết:
\(\begin{array}{l}
f\left(x \right) = \left({\cos x\cos \frac{\pi }{3} + \sin x\sin \frac{\pi }{3}} \right)\left({\cos x\cos \frac{\pi }{4} - \sin x\sin \frac{\pi }{4}} \right)\\
+ \left({\cos x\cos \frac{\pi }{6} - \sin x\sin \frac{\pi }{6}} \right)\left({\cos x\cos \frac{{3\pi }}{4} - \sin x\sin \frac{{3\pi }}{4}} \right)\\
= \left({\frac{1}{2}\cos x + \frac{{\sqrt 3 }}{2}\sin x} \right)\left({\frac{{\sqrt 2 }}{2}\cos x - \frac{{\sqrt 2 }}{2}\sin x} \right)\\
+ \left({\frac{{\sqrt 3 }}{2}\cos x - \frac{1}{2}\sin x} \right)\left({ - \frac{{\sqrt 2 }}{2}\cos x - \frac{{\sqrt 2 }}{2}\sin x} \right)\\
= \frac{{\sqrt 2 }}{4}{\cos ^2}x + \frac{{\sqrt 6 }}{4}\sin x\cos x - \frac{{\sqrt 2 }}{4}\sin x\cos x - \frac{{\sqrt 6 }}{4}{\sin ^2}x\\
- \frac{{\sqrt 6 }}{4}{\cos ^2}x + \frac{{\sqrt 2 }}{4}\sin x\cos x - \frac{{\sqrt 6 }}{4}\sin x\cos x + \frac{{\sqrt 2 }}{4}{\sin ^2}x\\
= \frac{{\sqrt 2 - \sqrt 6 }}{4}{\cos ^2}x + \frac{{\sqrt 2 - \sqrt 6 }}{4}{\sin ^2}x\\
= \frac{{\sqrt 2 - \sqrt 6 }}{4}\left({{{\cos }^2}x + {{\sin }^2}x} \right)\\
= \frac{{\sqrt 2 - \sqrt 6 }}{4}\\
\Rightarrow f'\left(x \right) = 0
\end{array}\)
Câu d
\(f\left( x \right) = {\cos ^2}x + {\cos ^2}\left({{{2\pi } \over 3} + x} \right) + {\cos ^2}\left({{{2\pi } \over 3} - x} \right).\)Lời giải chi tiết:
\(\begin{array}{l}
f\left(x \right) = {\cos ^2}x + {\cos ^2}\left({\frac{{2\pi }}{3} + x} \right) + {\cos ^2}\left({\frac{{2\pi }}{3} - x} \right)\\
= {\cos ^2}x + {\left({\cos \frac{{2\pi }}{3}\cos x - \sin \frac{{2\pi }}{3}\sin x} \right)^2}\\
+ {\left({\cos \frac{{2\pi }}{3}\cos x + \sin \frac{{2\pi }}{3}\sin x} \right)^2}\\
= {\cos ^2}x + {\left({ - \frac{1}{2}\cos x - \frac{{\sqrt 3 }}{2}\sin x} \right)^2}\\
+ {\left({ - \frac{1}{2}\cos x + \frac{{\sqrt 3 }}{2}\sin x} \right)^2}\\
= {\cos ^2}x + \left({\frac{1}{4}{{\cos }^2}x + \frac{{\sqrt 3 }}{2}\sin x\cos x + \frac{3}{4}{{\sin }^2}x} \right)\\
+ \left({\frac{1}{4}{{\cos }^2}x - \frac{{\sqrt 3 }}{2}\sin x\cos x + \frac{3}{4}{{\sin }^2}x} \right)\\
= {\cos ^2}x + \frac{1}{2}{\cos ^2}x + \frac{3}{2}{\sin ^2}x\\
= \frac{3}{2}{\cos ^2}x + \frac{3}{2}{\sin ^2}x\\
= \frac{3}{2}\left({{{\cos }^2}x + {{\sin }^2}x} \right)\\
= \frac{3}{2}\\
\Rightarrow f'\left(x \right) = 0
\end{array}\)
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