Question

If $\alpha ,\beta ,\gamma ,\delta$ are in $A.P.$ and $\underset{0}{\overset{2}{\int }}f\left(x\right)dx=-4$, where $f\left(x\right)=\left|\begin{array}{ccc}x+\alpha & x+\beta & x+\alpha -\gamma \\ x+\beta & x+\gamma & x-1\\ x+\gamma & x+\delta & x-\beta +\delta \end{array}\right|$ then common difference $d$ is:

• A. $1$
• B. $-1$
• C. $2$
• D. $-2$

Answer 1

Answer: (A), (B)

The correct options are
A $1$
B $-1$

As $\alpha ,\beta ,\gamma ,\delta$ are in $A.P.$ so,
$\beta =\alpha +d\gamma =\alpha +2d\delta =\alpha +3d$
Given:
$f\left(x\right)=\left|\begin{array}{ccc}x+\alpha & x+\beta & x+\alpha -\gamma \\ x+\beta & x+\gamma & x-1\\ x+\gamma & x+\delta & x-\beta +\delta \end{array}\right|$
Using row operation $R_3\to R_3-R_2$
$\Rightarrow f\left(x\right)=\left|\begin{array}{ccc}x+\alpha & x+\beta & x+\alpha -\gamma \\ x+\beta & x+\gamma & x-1\\ d& d& 1+2d\end{array}\right|$

Using row operation $R_2\to R_2-R_1$
$\Rightarrow f\left(x\right)=\left|\begin{array}{ccc}x+\alpha & x+\beta & x+\alpha -\gamma \\ d& d& -1+2d\\ d& d& 1+2d\end{array}\right|$

Using row operation $C_2\to C_2-C_1$
$\Rightarrow f\left(x\right)=\left|\begin{array}{ccc}x+\alpha & d& x+\alpha -\gamma \\ d& 0& -1+2d\\ d& 0& 1+2d\end{array}\right|\Rightarrow f\left(x\right)=-d(d+2d^2+d-2d^2)\Rightarrow f\left(x\right)=-2d^2$
We now that,
$\underset{0}{\overset{2}{\int }}f\left(x\right)dx=-4\Rightarrow -2d^2\times 2=-4\Rightarrow d^2=1\Rightarrow d=\pm 1$