Question

Let $\Delta (x)=\left|\begin{array}{ccc}3+2{\sin }^{4}x& 2{\cos }^{4}x& {\sin }^{2}2x\\ 2{\sin }^{4}x& 3+2{\cos }^{4}x& {\sin }^{2}2x\\ 2{\sin }^{4}x& 2{\cos }^{4}x& 3+{\sin }^{2}2x\end{array}\right|$ then $\underset{-\pi /2}{\overset{\pi /2}{\int }}x\Delta (x)dx$ equals

• A. ${\pi }^2$
• B. $\pi (\pi -1)$
• C. 1
• D. 0

Answer 1

Answer: (D)

Using $C_1\to C_1+C_2+C_3$, we get
$\Delta (x)=f(x)\left|\begin{array}{ccc}1& 2{\cos }^{4}x& {\sin }^{2}2x\\ 1& 3+2{\cos }^{4}x& {\sin }^{2}2x\\ 1& 2{\cos }^{4}x& 3+{\sin }^{2}2x\end{array}\right|$
where
$f(x)=3+2{\sin }^{4}x+2{\cos }^{4}x+{\sin }^{2}2x$
$=3+2\left({\sin }^{4}x+{\cos }^{4}x+2{\sin }^{2}x{\cos }^{2}x\right)$
$=3+2{\left({\cos }^{2}x+{\sin }^{2}x\right)}^2=5$
Applying $C_2\to C_2-\left(2{\cos }^{4}x\right)C_1$ and $C_3\to C_3-$ $\left({\sin }^{2}2x\right)C_1$, we get
$\Delta (x)=5\left|\begin{array}{ccc}1& 0& 0\\ 1& 3& 0\\ 1& 0& 3<br/>\end{array}\right|=45$
Thus, $\underset{-\pi /2}{\overset{\pi /2}{\int }}x\Delta (x)dx=45\underset{-\pi /2}{\overset{\pi /2}{\int }}xdx=0$