Question

The number of ordered triplets $(u,v,w)$ where $u,v,w\in [0,10]$ which satisfy the inequality $\frac{5^{{\sin }^{2}u+3{\cos }^{2}v}}{2^{{\cos }^{2}w}}\ge 625$ is

• A. No solution
• B. $32$
• C. $36$
• D. $48$

Answer 1

The correct option is C $36$

$2^{-{\cos }^{2}w}\times 5^{{\sin }^{2}u+3{\cos }^{2}v}\ge 2^0\times 5^4$
Comparing exponents from both sides
$-{\cos }^2\omega \ge 0\Rightarrow {\cos }^2\omega \le 0\Rightarrow \cos \omega =0\Rightarrow \omega =\frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{2}$

${\sin }^{2}u+3{\cos }^{2}v\ge 4$
$\Rightarrow 1-{\cos }^{2}u+3-{\sin }^{2}v\ge 4\Rightarrow {\cos }^{2}u+3{\sin }^{2}v\le 0\Rightarrow \cos u=0\&\sin v=0\Rightarrow u=\frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{2}$
and $v=0,\pi ,2\pi ,3\pi$

$\therefore$ Number of ordered triplets
$=3\times 4\times 3=36$