Question

Let $f\left(x\right)=\left|\begin{array}{ccc}{\omega }^3& {\omega }^4& {\omega }^5\\ \sin (m-1)x& \sin mx& \sin (m+1)x\\ \cos (m-1)x& \cos mx& \cos (m+1)x\end{array}\right|$,

where $m\in N$ and $\omega$ is the cube root of unity.
If $\underset{0}{\overset{\pi /2}{\int }}f\left(x\right)dx=a\omega +b{\omega }^2,$ then $(a,b)=$

• A. $(2,1)$
• B. $(2,0)$
• C. $(1,2)$
• D. $(1,-2)$

Answer 1

The correct option is B $(2,0)$

$f\left(x\right)=\left|\begin{array}{ccc}1& \omega & {\omega }^2\\ \sin (m-1)x& \sin mx& \sin (m+1)x\\ \cos (m-1)x& \cos mx& \cos (m+1)x\end{array}\right|$

$C_1\to C_1+C_3-2\cos x{C}_2$

$f\left(x\right)=\left|\begin{array}{ccc}1+{\omega }^2-2\omega \cos x& \omega & {\omega }^2\\ \sin (m-1)x+\sin (m+1)x-2\sin mx\cos x& \sin mx& \sin (m+1)x\\ \cos (m-1)x+\cos (m+1)x-2\cos mx\cos x& \cos mx& \cos (m+1)x\end{array}\right|$

$=\left|\begin{array}{ccc}1+{\omega }^2-2\omega \cos x& \omega & {\omega }^2\\ 0& \sin mx& \sin (m+1)x\\ 0& \cos mx& \cos (m+1)x\end{array}\right|$

$=(1+{\omega }^2-2\omega \cos x)\left[\sin mx\cos \right(m+1)x-\cos mx\sin (m+1\left)x\right]$
$=(-\omega -2\omega \cos x)(-\sin x)$
$=\omega (1+2\cos x)\left(\sin x\right)$

$\Rightarrow f\left(x\right)=\omega (\sin x+\sin 2x)$
So $\underset{0}{\overset{\pi /2}{\int }}f\left(x\right)dx=\omega {[-\cos x-\frac{\cos 2x}{2}]}_0^{\pi /2}\Rightarrow \underset{0}{\overset{\pi /2}{\int }}f\left(x\right)dx=2\omega$
$\Rightarrow a\omega +b{\omega }^2=2\omega$
$\Rightarrow (a,b)=(2,0)$