Question

Let $△\left(x\right)=\left|\begin{array}{ccc}x+a& x+b& x+a-c\\ x+b& x+c& x-1\\ x+c& x+d& x-b+d\end{array}\right|$and ${\int }_0^2△\left(x\right)dx=-16$,there $a,b,c,d$ are in $AP$, then the common difference of the $AP$ is

• A. $1$
• B. $2$
• C. $-2$
• D. none of these

Answer 1

Answer: (B), (C)

The correct options are
B $2$
C $-2$

$a,b,c$ and $d$ are in AP,
so,
$a=a,b=a+d,c=a+2d,d=a+3d$
$=\left|\begin{array}{ccc}x+a& x+a+d& x-2d\\ x+a+d& x+a+2d& x-1\\ x+a+2d& x+a+3d& x+2d\end{array}\right|$
By, $C_1=C_1-C_2$,
$=\left|\begin{array}{ccc}-d& x+a+d& x-2d\\ -d& x+a+2d& x-1\\ -d& x+a+3d& x+2d\end{array}\right|$
By, $R_1=R_1-R_2$ and $R_2=R_2-R_3$,
$=\left|\begin{array}{ccc}0& -d& -2d+1\\ 0& -d& -1-2d\\ -d& x+a+3d& x+2d\end{array}\right|$
$=d^2(-1-2d+2d-1)$
$=-2d^2$
${\int }_0^2-2d^{2}dx=-2d^{2}x{|}_0^2=-4d^2=-16$
$d^2=4$ so $d=2$ or $-2$