Question

If $p,q,r,s$ are in arithmetic progression and $f(x)=\left|\begin{array}{ccc}p+\sin x& q+\sin x& p-r+\sin x\\ q+\sin x& r+\sin x& -1+\sin x\\ r+\sin x& s+\sin x& s-q+\sin x\end{array}\right|$ such that ${\int }_0^{2}f(x)dx=-4$, then the common difference of the progession is

• A. $\pm 1$
• B. $\frac{1}{2}$
• C. $\pm 2$
• D. None of these

Answer 1

Answer: (A)

Let $q=p+d,r=p+2d,s=p+3d$
$\therefore f(x)=\left|\begin{array}{ccc}p+\sin x& p+d+\sin x& -2d+\sin x\\ p+d+\sin x& p+2d+\sin x& -1+\sin x\\ p+2d+\sin x& p+3d+\sin x& 2d+\sin x\end{array}\right|$
Applying $R\to R_1+R_3-2R_2$, we get
$f(x)=\left|\begin{array}{ccc}0& 0& 2\\ p+d+\sin x& p+2d+\sin x& -1+\sin x\\ p+2d+\sin x& p+3d+\sin x& 2d+\sin x\end{array}\right|$
$=2[(p+d+\sin x)(p+3d+\sin x)$
$-(p+2d+\sin x{)}^2=-2d^2$
Given ${\int }_0^{2}f(x)dx=-4$
$<br/>\begin{array}{l}<br/>\Rightarrow {\int }_0^2\left(-2d^2\right)dx=-4\\ <br/>\Rightarrow d^2=1\Rightarrow d=\pm 1<br/>\end{array}<br/>$