Question

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

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

Answer 1

Answer: (D)

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

Answer 2

xy > 0 ̃ x & y are of same sign

x + $\frac{1}{x}$ ³ 2 or £ – 2

sec^–1$\left( x + \frac{1}{x} \right)$ Î $\left\lbrack \frac{\pi}{3},\frac{\pi}{2} \right)$ È $\left( \frac{\pi}{2},\frac{2\pi}{3} \right\rbrack$

\ z Î $\left\lbrack \frac{2\pi}{3},\pi \right)$ È $\left( \pi,\frac{4\pi}{3} \right\rbrack$

Hence, value of z (among the given) which does not lie in the

set is $\frac{5\pi}{3}$.