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 $\int_{0}^{2} f(x) d x=-4$, 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 $q=p+ d, r=p+2 d, s=p+3 d$
$\therefore f(x)= \begin{vmatrix}p+\sin x & p +d+\sin x & -2 d+\sin x \\ p +d+\sin x & p+2 d+\sin x & -1+\sin x \\ p+2 d+\sin x & p+3 d+\sin x & 2 d+\sin x\end{vmatrix}$
Applying $R \rightarrow R_{1}+R_{3}-2 R_{2}$, we get
$f(x)= \begin{vmatrix}0 & 0 & 2 \\ p +d+\sin x & p+2 d+\sin x & -1+\sin x \\ p+2 d+\sin x & p+3 d+\sin x & 2 d+\sin x\end{vmatrix}$
$=2[(p +d+\sin x)(p+3 d+\sin x)$
$-(p+2 d+\sin x)^{2}=- 2 d^{2}$
Given $\int_{0}^{2} f(x) d x=-4$
$
\begin{array}{l}
\Rightarrow \int_{0}^{2}\left(-2 d^{2}\right) d x=-4 \\
\Rightarrow d^{2}=1 \Rightarrow d=\pm 1
\end{array}
$