Question

If $2^{\sqrt{{\sin }^{2}x-2\sin x+5}}\frac{1}{4^{{\sin }^{2}y}}\le 1,$ then the ordered pair (x, y) is equal to (m, n $\in$ I)

• A.
• B.
• C.
• D. x = n, y = m

Answer 1

The correct option is A

${\sin }^{2}x-2\sin x+5={\left(\sin x-1\right)}^2+4\ge 4$
$\therefore 2^{\sqrt{{\sin }^{2}x-2\sin x+5}}\ge 2^2=4$
and ${\sin }^{2}y\le 1\Rightarrow \frac{1}{4^{{\sin }^{2}y}}\ge \frac{1}{4}$
$\therefore LHS\ge$ 1 and according to question LHS $\le$ 1, so therefore, LHS = 1
for which
$\therefore si{n}^{2}x-2sinx+5=4$ and cosec${}^2$y = 1, sin${}^2$y = 1 or cos y = 0
(sin x - 1)${}^2$ = 0 y = (2m + 1)$\frac{\pi }{2}$
sin x = 1 ( x = (4n + 1)$\frac{\pi }{2}$