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 $\underset{0}{\overset{2}{\int }}\Delta \left(x\right)dx=-16,$ where $a,b,c,d$ are in A.P. Then the common difference of the A.P. can be

• A. $1$
• B. $2$
• C. $-4$
• D. $-2$

Answer 1

Answer: (B), (D)

$-2$

$\Delta \left(x\right)=\left|\begin{array}{ccc}x+a& x+b& x+a-c\\ x+b& x+c& x-1\\ x+c& x+d& x-b+d\end{array}\right|$

Applying $R_2\to R_2-R_1$
$\Delta \left(x\right)=\left|\begin{array}{ccc}x+a& x+b& x+a-c\\ b-a& c-b& c-a-1\\ x+c& x+d& x-b+d\end{array}\right|$

Applying $R_3\to R_3-R_1$
$\Delta \left(x\right)=\left|\begin{array}{ccc}x+a& x+b& x+a-c\\ b-a& c-b& c-a-1\\ c-a& d-b& (d-b)+(c-a)\end{array}\right|$

$=\left|\begin{array}{ccc}x& x& x\\ b-a& c-b& c-a-1\\ c-a& d-b& (d-b)+(c-a)\end{array}\right|+\left|\begin{array}{ccc}a& b& a-c\\ b-a& c-b& c-a-1\\ c-a& d-b& (d-b)+(c-a)\end{array}\right|$

Let the common difference be $k.$
Then, $\Delta \left(x\right)=\left|\begin{array}{ccc}x& x& x\\ k& k& 2k-1\\ 2k& 2k& 4k\end{array}\right|+\left|\begin{array}{ccc}a& b& -2k\\ k& k& 2k-1\\ 2k& 2k& 4k\end{array}\right|$

$=0+\left|\begin{array}{ccc}a& b& -2k\\ k& k& 2k-1\\ 2k& 2k& 4k\end{array}\right|$

$\Rightarrow \Delta \left(x\right)=2\left|\begin{array}{ccc}a& b& -2k\\ k& k& 2k-1\\ k& k& 2k\end{array}\right|$

Again, applying $R_3\to R_3-R_2$
$\Delta \left(x\right)=2\left|\begin{array}{ccc}a& b& -2k\\ k& k& 2k-1\\ 0& 0& 1\end{array}\right|$

$\Rightarrow \Delta \left(x\right)=2k(a-b)=-2k^2$

Now, $\underset{0}{\overset{2}{\int }}\Delta \left(x\right)dx=-16$
$\Rightarrow 4k^2=16$
$\Rightarrow k=\pm 2$