Question

Let $\Delta(x)=\begin{vmatrix}3+2 \sin ^4 x & 2 \cos ^4 x & \sin ^2 2 x \\ 2 \sin ^4 x & 3+2 \cos ^4 x & \sin ^2 2 x \\ 2 \sin ^4 x & 2 \cos ^4 x & 3+\sin ^2 2 x\end{vmatrix}$ then $\int\limits_{-\pi / 2}^{\pi / 2} x \Delta(x) d x$ equals

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

Answer 1

Answer: (D)

Using $C_1 \rightarrow C_1+C_2+C_3$, we get
$\Delta(x)=f(x)\begin{vmatrix}1 & 2 \cos ^4 x & \sin ^2 2 x \\1 & 3+2 \cos ^4 x & \sin ^2 2 x \\1 & 2 \cos ^4 x & 3+\sin ^2 2 x\end{vmatrix}$
where
$f(x) =3+2 \sin ^4 x+2 \cos ^4 x+\sin ^2 2 x $
$ =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 $C_2 \rightarrow C_2-\left(2 \cos ^4 x\right) C_1$ and $C_3 \rightarrow C_3-$ $\left(\sin ^2 2 x\right) C_1$, we get
$\Delta(x)=5\begin{vmatrix}1 & 0 & 0 \\1 & 3 & 0 \\1 & 0 & 3
\end{vmatrix}=45$
Thus, $\int\limits_{-\pi / 2}^{\pi / 2} x \Delta(x) d x=45 \int\limits_{-\pi / 2}^{\pi / 2} x d x=0$