Question

If $x^2-4x+5\sin y=0,y\in (0,2\pi )$ then x = ................ & y = ................ .

• A. $x=2$ & $y=-\pi /2$
• B. $x=-2$ & $y=\pi /2$
• C. $x=5$ & $y=3\pi /2$
• D. $x=-2$ & $y=-\pi /2$

Answer 1

Answer: (C)

$x=5$ & $y=3\pi /2$

$x^2-4x+5siny=0$
$siny=\frac{4x-x^2}{5}$
Now $-1\le siny\le 1$
Hence
$-1\le \frac{4x-x^2}{5}\le 1$
$-5\le 4x-x^2\le 5$
Hence
$4x-x^2\le 5$
Or
$x^2-4x+5\ge 0$
Since discriminant is less than 0, hence the above inequality is always true for any real value of x.
Now consider
$-5\le 4x-x^2$
$x^2-4x-5\le 0$
$(x-5)(x+1)\le 0$
$x\ge -1$ and $x\le 5$
Therefore
$xϵ[-1,5]$
Hence the integral solutions are
$x=-1$ $y=\frac{3\pi }{2}$
$x=0$, $y=0,2\pi$
$x=4$, $y=0,2\pi$
$x=5$, $y=\frac{3\pi }{2}$