Question

If $\omega ,{\omega }^2$ be the complex cube roots of unity, and $f\left(x\right)=\left|\begin{array}{ccc}x+1& \omega & {\omega }^2\\ \omega & x+{\omega }^2& 1\\ {\omega }^2& 1& x+\omega \end{array}\right|$

then ${\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}f\left(x\right)dx$ is equal to?

• A. $\frac{x^4}{4}$
• B. $\frac{3}{4}{x}^4$
• C. $0$
• D. None of the above

Answer 1

Answer: (C)

$0$

Since $w$ is the cube root of unity.$\therefore w^3=1\&1+w+w^2=0$where $w=\frac{-1+i\sqrt{3}}{2}\&w^2=\frac{-1-i\sqrt{3}}{2}$
$f\left(x\right)=\left|\begin{array}{ccc}x+1& w& w^2\\ w& x+w^2& 1\\ w^2& 1& x+w\end{array}\right|$
$R_1\to R_1+R_2+R_3$
$\Rightarrow f\left(x\right)=\left|\begin{array}{ccc}x+1+w+w^2& x+1+w+w^2& {x+1+w+w}^2\\ w& x+w^2& 1\\ w^2& 1& x+w\end{array}\right|=\left|\begin{array}{ccc}x& x& x\\ w& x+w^2& 1\\ w^2& 1& x+w\end{array}\right|$
$\Rightarrow f\left(x\right)=x\left|\begin{array}{ccc}1& 1& 1\\ w& x+w^2& 1\\ w^2& 1& x+w\end{array}\right|$
$C_1\to C_1-C_2{C}_2\to C_2-C_3$
$\Rightarrow f\left(x\right)=x\left|\begin{array}{ccc}0& 0& 1\\ w-x-w^2& x+w^2-1& 1\\ w^2-1& 1-x-w& x+w\end{array}\right|=x(w-x-w^2)(1-x-w)-(x+w^2-1)(w^2-1)$
$\Rightarrow f\left(x\right)=x(w-xw-w^2-x+x^2+xw-w^2+x{w}^2+w^3-x{w}^2+x-w^4+w^2+w^2-1)=x^3$
Since $f\left(x\right)$ is odd.
Therefore ${\int }_{-\frac{\Pi }{2}}^{\frac{\Pi }{2}}f\left(x\right)=0$