Question 63 of 67MediumMCQ
If \[f(x)=\left|\[2cos4x3+2cos4x2cos4x2sin4x2sin4x3+2sin4x3+sin22xsin22xsin22x\]\right|,\] then \[\frac{1}{5} f^{\prime}(0)=\] is equal to : Solutions · 1
Official solution
\[\begin{aligned}
& \left|\[\begin{array}{|ccccccc|}2 \cos ^4 x & 2 \sin ^4 x & 3+\sin ^2 2 x \\ 3+2 \cos ^4 x & 2 \sin ^4 x & \sin ^2 2 x \\ 2 \cos ^4 x & 3+2 \sin ^2 4 x & \sin ^2 2 x\end{array}\end{aligned}\]\right| \\
\& \mathrm{R}\_2 \rightarrow \mathrm{R}\_2-\mathrm{R}\_1, \mathrm{R}\_3 \rightarrow \mathrm{R}\_3-\mathrm{R}\_1 \\
\& \left|\[2cos4x302sin4x033+sin22x−3−3\]\right| \\
\& \mathrm{f}(\mathrm{x})=45 \\
\& \mathrm{f}\^{\prime}(\mathrm{x})=0 \\
\&
\] Answer the question to unlock