Question

Number of solution of pair $(x,y)$ of the equation $\sin x\sin y=min\{-1,{\alpha }^2-4\alpha +5\},\alpha \in R$ (where $0<x<\pi ,-\pi <y<0$) is

• A. $1$
• B. $2$
• C. $3$
• D. infinite

Answer 1

The correct option is A $1$

$\sin x\sin y=min\{-1,{\alpha }^2-4\alpha +5\}$
${\alpha }^2-4\alpha +5=(\alpha -2{)}^2+1\ge 1$
So,$min\{-1,{\alpha }^2-4\alpha +5\}=-1$
$\Rightarrow \sin x\sin y=-1$
$\Rightarrow \sin x=1,\sin y=-1or\sin x=-1,\sin y=1$
$\Rightarrow x=\frac{\pi }{2},y=-\frac{\pi }{2}orx=-\frac{\pi }{2},y=\frac{\pi }{2}$
So there is only one solution of pair $(x,y)$ satisfying the given equation.