Question

If $p,q,r,s$ are in $A.P$ ad if $\left(x\right)=\left|\right|\begin{array}{ccc}p+sinx& q+sinx& q-r+sinx\\ q+sinx& r+sinx& -1+sinx\\ r+sinx& s+sinx& s-q+sinx\end{array}$ such that ${\int }_0^{2}f\left(x\right)dx=-4$ then the common difference of the $A.P$. cn

• A. $-1$
• B. $\frac{1}{2}$
• C. $1$
• D. none

Answer 1

The correct option is D none

Given $f\left(x\right)=\left|\begin{array}{ccc}9+\sin x& q+\sin x& q-r+\sin x\\ q+\sin x& r+\sin x& -1+\sin x\\ r+\sin x& s+\sin x& s-q+\sin x\end{array}\right|$ & $p,q,r,s$ are in $AP$

$\Rightarrow \left|\begin{array}{ccc}p-q& q-r& q-r+1\\ q+\sin x& r+\sin x& -1+\sin x\\ r-q& s-r& s-q+1\end{array}\right|\begin{array}{c}{R}_1\to R_1-R_2\\ R_{3\to }{R}_3-R_2\end{array}$

Let the common difference be $d$

$\Rightarrow f\left(x\right)=\left|\begin{array}{ccc}-d& -d& -d+1\\ q+\sin x& r+\sin x& -1+\sin x\\ d& +d& 2d+1\end{array}\right|R_1\to R_1+R_3$

$=\left|\begin{array}{ccc}0& 0& d+2\\ q+\sin x& r+\sin x& \sin x-1\\ d& d& 2d+1\end{array}\right|$ (expanding)

$=(d+2)-d(q-r)$

$=-d^2(d+2)$

Now, ${\int }_0^{2}f\left(x\right)dx=-4$

$\Rightarrow {\int }_0^2-d^2(d+2)dx=-4\Rightarrow d^2(d+2)=2---\left(1\right)$

Now of the option satisfy $\left(1\right)$