Question

Let $\Delta \left(x\right)=\left|\begin{array}{ccc}x+a& x+b& x+a-c\\ x+b& x+c& x-1\\ x+c& x+d& x-b+d\end{array}\right|$ and ${\int }_0^2\Delta \left(x\right)dx=-16$ where a, b, c, d are in A.P.,

then the common difference of the A.P. is equal to

• A. $\pm$ 1
• B. $\pm$ 2
• C. $\pm$ 3
• D. $\pm$ 4

Answer 1

The correct option is B $\pm$ 2

Given, a,b,c,d are in A.P.
$\Rightarrow b-a=c-b=d-c=D$
Now, $\Delta \left(x\right)=\left|\begin{array}{ccc}x+a& x+b& x+a-c\\ x+b& x+c& x-1\\ x+c& x+d& x-b+d\end{array}\right|$
$R_3\to R_3-R_2,R_2\to R_2-R_1$
$\Delta \left(x\right)=\left|\begin{array}{ccc}x+a& x+b& x+a-c\\ b-a& c-b& -1-a+c\\ c-b& d-c& -b+d+1\end{array}\right|$
$=\left|\begin{array}{ccc}x+a& x+b& x+a-c\\ D& D& -1-a+c\\ D& D& -b+d+1\end{array}\right|$
$C_1\to C_1-C_2$
$=\left|\begin{array}{ccc}-D& x+b& x+a-c\\ 0& D& -1-a+c\\ 0& D& -b+d+1\end{array}\right|$
$=-D\left|\begin{array}{ccc}-1& x+b& x+a-c\\ 0& D& -1-a+c\\ 0& D& -b+d+1\end{array}\right|$
$\Delta \left(x\right)=-2D^2$
Given, ${\int }_0^2\Delta \left(x\right)dx=-16$

$\Rightarrow -2D^2{\int }_0^{2}1dx=-16$

$\Rightarrow 4D^2=16$

$\Rightarrow D=\pm 2$