Giải bài 5.51 trang 208 - SBT Đại số và Giải tích lớp 11
Chứng minh rằng \(f'(x)=0\forall x\in \mathbb{R}\), nếu:
\(a)f\left( x \right)=3\left( {{\sin }^{4}}x+{{\cos }^{4}}x \right)-2\left( {{\sin }^{6}}x+{{\cos }^{6}}x \right);\)
\(b)f\left( x \right)={{\cos }^{6}}x+2{{\sin }^{4}}x{{\cos }^{2}}x+3{{\sin }^{2}}x{{\cos }^{4}}x+{{\sin }^{4}}x\);
\(c)f\left( x \right)=\cos \left( x-\dfrac{\pi }{3} \right)\cos \left( x+\dfrac{\pi }{4} \right)+\cos \left( x+\dfrac{\pi }{6} \right)\cos \left( x+\dfrac{3\pi }{4} \right)\);
\(d)f\left( x \right)={{\cos }^{2}}x+{{\cos }^{2}}\left( \dfrac{2\pi }{3}+x \right)+{{\cos }^{2}}\left( \dfrac{2\pi }{3}-x \right)\)
Gợi ý:
Biến đổi các biểu thức đã cho không phụ thuộc vào x
a)
\( \begin{align} & f\left( x \right)=3{{\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)}^{2}}-6{{\sin }^{2}}x{{\cos }^{2}}x-2\left( {{\sin }^{4}}x-{{\sin }^{2}}x{{\cos }^{2}}x+{{\cos }^{4}}x \right) \\ & =3-6{{\sin }^{2}}x{{\cos }^{2}}x-2\left[ \left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)-3{{\sin }^{2}}x{{\cos }^{2}}x \right] \\ & =3-6{{\sin }^{2}}x{{\cos }^{2}}x-2+6{{\sin }^{2}}x{{\cos }^{2}}x=1 \\ & \Rightarrow f'\left( x \right)=0 \\ \end{align}\)
b)
\(\begin{align} & f\left( x \right)=\left( {{\cos }^{6}}x+{{\sin }^{2}}x{{\cos }^{4}}x \right)+\left( 2{{\sin }^{2}}x{{\cos }^{4}}x+2{{\sin }^{4}}x{{\cos }^{2}}x \right)+{{\sin }^{4}}x \\ & ={{\cos }^{4}}x+2{{\sin }^{2}}x{{\cos }^{2}}x+{{\sin }^{4}}x \\ & ={{\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)}^{2}}=1 \\ & \Rightarrow f'\left( x \right)=0 \\ \end{align}\)
c)
\(\begin{align} & f\left( x \right)=\cos \left( x-\frac{\pi }{3} \right)\cos \left( x+\frac{\pi }{4} \right)+\sin \left( \frac{\pi }{3}-x \right)\sin \left( -\frac{\pi }{4}-x \right) \\ & =\cos \left( x-\frac{\pi }{3} \right)\cos \left( x+\frac{\pi }{4} \right)+\sin \left( x-\frac{\pi }{3} \right)\sin \left( x+\frac{\pi }{4} \right) \\ & =\cos \left[ \left( x-\frac{\pi }{3} \right)-\left( x+\frac{\pi }{4} \right) \right]=\cos \frac{7\pi }{12} \\ & \Rightarrow f'\left( x \right)=0 \\ \end{align}\)
d)
\(\begin{align} & f\left( x \right)=\frac{1}{2}\left[ 1+\cos 2x+1+\cos \left( \frac{4\pi }{3}+2x \right)+1+\cos \left( \frac{4\pi }{3}-2x \right) \right] \\ & =\frac{1}{2}\left[ 3+\cos 2x+2.\cos \frac{4\pi }{3}\cos 2x \right] \\ & =\frac{3}{2} \\ & \Rightarrow f'\left( x \right)=0 \\ \end{align}\)