Question

If ${\sin }^{-1}\left(\sin p\right)=3\pi -p$ and the point of intersection of the lines $x+y=6$ and $px-y=3$ will have integral co-ordinates (both abscissa and ordinate), then the number of values of $p$ is

Answer 1

Given : ${\sin }^{-1}\left(\sin p\right)=3\pi -p$

We know that ${\sin }^{-1}\left(\sin x\right)=3\pi -x$ iff$\frac{5\pi }{2}\le x\le \frac{7\pi }{2}$
$\Rightarrow p\in [\frac{5\pi }{2},\frac{7\pi }{2}]$

Now given lines are : $x+y=6$ and $px-y=3$ (solving both)
$\Rightarrow$ Point of intersection of lines is $(\frac{9}{1+p},\frac{6p-3}{1+p})$
$\because$ Co-ordinates are integral,
$\Rightarrow \frac{9}{1+p}\in Z$
$\Rightarrow 1+p=\pm 1$ (or) $1+p=\pm 3$ (or) $1+p=\pm 9$
$\Rightarrow p\in \{-10,-4,-2,0,2,8\}$
Also, $\frac{6p-3}{1+p}\in Z$ for $p\in \{-10,-4,-2,0,2,8\}$
But $p\in [\frac{5\pi }{2},\frac{7\pi }{2}]$
$\therefore p=8$
$\therefore$ Number of possible value of $p$ is $1$.