Question

Let $f\left(x\right)=\left|\begin{array}{ccc}1+{\sin }^{2}x& {\cos }^{2}x& 4\sin 2x\\ {\sin }^{2}x& 1+{\cos }^{2}x& 4\sin 2x\\ {\sin }^{2}x& {\cos }^{2}x& 1+4\sin 2x\end{array}\right|$, then the maximum value of $f\left(x\right)=$.

Answer 1

$=(1+{\sin }^{2}x)\left[\right(1+{\cos }^{2}x\left)\right(1+{\sin }^{2}x)-4\sin 2x{\cos }^{2}x]+{\cos }^{2}x[4\sin 2x{\sin }^{2}x-{\sin }^{2}x(1+4\sin 2x\left)\right]+4\sin 2x{\sin }^{2}x({\cos }^{2}x-1-{\cos }^{2}x)$

$=(1+{\sin }^{2}x)(1+4\sin 2x+{\cos }^{2}x)+{\cos }^{2}x(-{\sin }^{2}x)-4\sin 2x{\sin }^{2}x$

$=1+4\sin 2x+{\cos }^{2}x+{\sin }^{2}x+4\sin 2x{\sin }^{2}x-4\sin 2x{\sin }^{2}x+{\cos }^{2}x{\sin }^{2}x-{\cos }^{2}x{\sin }^{2}x$

$=2+4\sin 2x$ $({\sin }^{2}x+{\cos }^{2}x=1)$

$\therefore f^{\prime }\left(x\right)=\frac{d}{dx}(4\sin 2x+2)=8\cos 2x$

$f^{\prime }\left(x\right)=0\Rightarrow 8\cos 2x=0$

$\therefore \cos 2x=0$

$2x(2n-1)\frac{\pi }{2}$

$x=(2n-1)\frac{\pi }{4}$

For max value

$f^{\prime }\left(\alpha \right)<0$

$\therefore \frac{d}{dx}\left(8\cos 2x\right)=-16\sin 2x<0\Rightarrow \sin 2x>0$

When $\cos 2x=0\Rightarrow \sin 2x=\pm 1.$ Since $\sin 2x>0$

$\therefore n\sin 2x=1$

Max $f\left(a\right)=2+4\left(1\right)=6$