Question

If $z=se{c}^{-1}(x+\frac{1}{x})+se{c}^{-1}(y+\frac{1}{y})$ , where xy > 0, then the value of z (among the given values) which is not possible is

• A. $\frac{5\pi }{6}$
• B. $\frac{7\pi }{10}$
• C. $\frac{9\pi }{10}$
• D. $\frac{5\pi }{3}$

Answer 1

The correct option is D

$\frac{5\pi }{3}$

$xy>0\Rightarrow x\&y$ are of same sign
$x+\frac{1}{x}\ge 2$ or $\le -2$

$se{c}^{-1}(x+\frac{1}{x})ϵ[\frac{\pi }{3},\frac{\pi }{2})\cup (\frac{\pi }{2},\frac{2\pi }{3}]\therefore zϵ[\frac{2\pi }{3},\pi )\cup (\pi ,\frac{4\pi }{3}]$

Hence, value of z (among the given) which does not lie in the set is $\frac{5\pi }{3}$.