Question

For all $yϵR$, $f\left(y\right)=\left|\begin{array}{ccc}1& y& y+1\\ 2y& y(y-1)& y(y+1)\\ 3y(y-1)& y(y-1)(y-2)& y(y^2-1)\end{array}\right|$, then ${\int }_{-\pi /2}^{\pi /2}f(y^2+2)dy$ equals to

• A. $\pi$
• B. $-\pi$
• C. $0$
• D. $2\pi$

Answer 1

The correct option is C $0$

Taking $y$ common from $R_2$ and $y(y+1)$ from $R_3$, we get

$f\left(y\right)=y^2(y-1)\left|\begin{array}{ccc}1& y& y+1\\ 2& y-1& y+1\\ 3& y-2& y+1\end{array}\right|$

Applying $C_3\to C_3-C_2$, we get

$f\left(y\right)=y^2(y-1)\left|\begin{array}{ccc}1& y& 1\\ 2& y-1& 2\\ 3& y-2& 3\end{array}\right|=0$

Therefore, ${\int }_{-\pi /2}^{\pi /2}f(y^2+2)dy=0$

Hence, option 'C' is correct.