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

Question

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

$f(x)=\left|\begin{array}{ccc} 1+\sin ^2 x & \cos ^2 x & 4 \sin 4 x \ \sin ^2 x & 1+\cos ^2 x & 4 \sin 4 x \ \sin ^2 x & \cos ^2 x & 1+4 \sin 4 x \end{array}ight|, x \in R$

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

Accepted Answer

$$\begin{aligned}
& \left|\[\begin{array}{ccc} 1+\sin ^2 x & \cos ^2 x & 4 \sin 4 x \ \sin ^2 x & 1+\cos ^2 x & 4 \sin 4 x \ \sin ^2 x & \cos ^2 x & 1+4 \sin 4 x \end{array}\end{aligned}$$\right|, x \in R \
\& R\_2 \rightarrow R\_2-R\_1 \& R\_3 \rightarrow R\_3-R\_1 \
\& f(x)\left|$$\begin{array}{ccc} 1+\sin ^2 x & \cos ^2 x & 4 \sin 4 x \ -1 & 1 & 0 \ -1 & 0 & 1 \end{array}$$\right|
\]

Expand about $\mathrm{R}_1$, we get

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

$\therefore M=\max$ value of $f(x)=6$

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

$$\therefore \mathrm{M}^4-\mathrm{m}^4=1280$$

Solutions · 1