Question

Equation $1+x^2+2x\sin \left({\cos }^{-1}y\right)=0$ is satisfied by

• A. exactly one value of $x$
• B. exactly two values of $x$
• C. exactly one value of $y$
• D. exactly two values of $y$

Answer 1

Answer: (A), (C)

exactly one value of $x$
exactly one value of $y$

$1+x^2+2x\sin \left({\cos }^{-1}y\right)=0$
For the above equation to have real solution, determinant has to be $\ge 0$.
Determinant, $D=4{\sin }^2\left({\cos }^{-1}y\right)-4\ge 0$
$\Rightarrow {\sin }^2\left({\cos }^{-1}y\right)\ge 1$
$\Rightarrow {\sin }^2\left({\cos }^{-1}y\right)=1(\because$ Range of $\sin x$ is $[-1,1])$
$\Rightarrow {\cos }^{-1}y=\frac{(2n+1)\pi }{2}$, $n\in I$
$\Rightarrow y=0$
For $y=0$, equation becomes
$1+x^2+2x=0$, which give $x=-1$
$\therefore 1+x^2+2x\sin \left({\cos }^{-1}y\right)=0$, is satisfied by one value of $x;x=-1$ and one value of $y;y=0$.