Let M and m respectively be the maximum and the minimum values of f(x)=vmatrix 1+ ^2x ^2x …

Question

Let M and m respectively be the maximum and the minimum values of

$f(x)=\begin{vmatrix} 1+{\sin }^{2}x \quad  {\cos }^{2}x \quad  4\sin 4x \ {\sin }^{2}x \quad  1+{\cos }^{2}x \quad  4\sin 4x \ {\sin }^{2}x \quad  {\cos }^{2}x \quad  1+4\sin 4x \end{vmatrix},x\in R$

Then ${M}^{4}-{m}^{4}$ is equal to :

Accepted Answer

$\begin{gathered} |\begin{array}{ccc}1+{\sin }^{2}x \quad  {\cos }^{2}x \quad  4\sin 4x \ {\sin }^{2}x \quad  1+{\cos }^{2}x \quad  4\sin 4x \ {\sin }^{2}x \quad  {\cos }^{2}x \quad  1+4\sin 4x\end{array}|,x\in R \ {R}_{2}	o {R}_{2}-{R}_{1}\quad {R}_{3}	o {R}_{3}-{R}_{1} \ f(x)|\begin{array}{ccc}1+{\sin }^{2}x \quad  {\cos }^{2}x \quad  4\sin 4x \ -1 \quad  1 \quad  0 \ -1 \quad  0 \quad  1\end{array}| \end{gathered}$

Expand about ${R}_{1}$, we get

$f(x)=2+4\sin 4x$

$	herefore M=max$ value of $f(x)=6$

$m=\min$ value of $f(x)=-2$

$	herefore {M}^{4}-{m}^{4}=1280$

Solutions · 1