"Let M and m respectively be the maximum and the minimum values of $f(x)=\left|\begin{array}{ccc}1+\sin ^2 x & \cos ^2 x & 4 \sin 4 x \\ \sin ^2 x & 1+\cos ^2 x & 4 \sin 4 x \\ \sin ^2 x & \cos ^2 x & 1+4 \sin 4 x\end{array}\right|, x \in R$ Then $ M^4 - m^4 $ is equal to :"

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"

$$\begin{aligned} & \left|\begin{array}{ccc} 1+\sin ^2 x & \cos ^2 x & 4 \sin 4 x \\ \sin ^2 x & 1+\cos ^2 x & 4 \sin 4 x \\ \sin ^2 x & \cos ^2 x & 1+4 \sin 4 x \end{array}\right|, x \in R \\ & R_2 \rightarrow R_2-R_1 \& R_3 \rightarrow R_3-R_1 \\ & f(x)\left|\begin{array}{ccc} 1+\sin ^2 x & \cos ^2 x & 4 \sin 4 x \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right| \end{aligned}$$

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 $\mathrm{f}(\mathrm{x})=-2$

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

"

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

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

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

Solution

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Let M and m respectively be the maximum and the minimum values of $f(x)=\left|\begin{array}{ccc}1+\sin ^2 x & \cos ^2 x & 4 \sin 4 x \\ \sin ^2 x & 1+\cos ^2 x & 4 \sin 4 x \\ \sin ^2 x & \cos ^2 x & 1+4 \sin 4 x\end{array}\right|, x \in R$ Then $ M^4 - m^4 $ is equal to :

Solution

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More Matrices and Determinants questions

Drill 25 more like these. Every day. Free.

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

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

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