Question

The function $f(x)={\sec }^{-1}\frac{x}{\sqrt{x-[x]}}$ is defined for

• A. all real x
• B. R-[(-1, 1) U Z]
• C. $R^{+}$ - (0, 1)
• D. $R^{+}$ - Z

Answer 1

Answer: (B)

$se{c}^{-1}$ 1 x is defined for x $\ge 1$ or $x=\le -1$ i.e., for R-(-1. 1)
$\therefore$ f(x) is defined if $\frac{x}{\sqrt{x-[x]}}\le -1$ or $\ge 1$
Also x - [x] $\ne$ 0 $\therefore$ X is not an integer. [$\because$ [x] = n and n -[n]= 0 for all integer n].
Hence domain of $se{c}^{-1}\frac{x}{\sqrt{x-[x]}}$ = R - {(-1, 1) ($\cup$ Z} where Z is the set of all integers.