If $\left\lbrack \cot^{- 1}x \right\rbrack + \left\lbrack \cos^{- 1}x \right\rbrack = 0,\$where $x$ is a non-negative real number and [.] denotes the greatest integer function, then complete set of values of $x$ is

Question

Dr. Abubacker · Accepted Answer

(b)
$\because 0 \leq \cos^{- 1}{x \leq \pi}$
And $0 < \cot^{- 1}{x < \pi}$
Given, $\lbrack\cot^{- 1}{xbrack + \lbrack\cot^{- 1}{xbrack = 0}}$
$\Rightarrow \lbrack\cot^{- 1}{xbrack = 0\ and\ \lbrack\cos^{- 1}{xbrack = 0}}$
$\Rightarrow 0 < \cot^{- 1}{x < 1\ and\ 0 \leq \cos^{- 1}{x < 1}}$
$	herefore x \in (\cot{1,\infty)\ and\ x \in (\cos{1,\ 1)}}$
$\Rightarrow x \in (\cot{1,\ 1})$

Q. If $\left\lbrack \cot^{- 1}x \right\rbrack + \left\lbrack \cos^{- 1}x \right\rbrack = 0,\$where $x$ is a non-negative real number and [.] denotes the greatest integer function, then complete set of values of $x$ is

Explanation

Reviewed by Dr. Abubacker

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