Statement-1: limx→0 tan−1⁡xx=0 where x representsgreatest integer ≤.xStatement-2: tan−1⁡xx

A
STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1
B
STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for
STATEMENT-1
C
STATEMENT-1 is True, STATEMENT-2 is False
D
STATEMENT-1 is False, STATEMENT-2 is True

Solution:

Let $f (x) = \tan^{- 1} x - x$

$f^{'} (x) = \frac{1}{1 + x^{2}} - 1 = \frac{- x^{2}}{1 + x^{2}} < 0$ . Hence $f$ is decreasing function
so for $x > 0, \tan^{- 1} x < x$ and for $x < 0, f (x) > f (0) = 0$

$ϕ \tan^{- 1} x > x ϕ t \frac{\tan^{- 1} x}{x} < 1 (x < 0)$

Thus $lim_{x \to 0 +} [\frac{\tan^{- 1} x}{x}] = 0$ and

$lim_{x \to 0^{-}} [\frac{\tan^{- 1} x}{x}] = 0$

Statement $- 1 :$ $lim_{x \to 0} [\frac{\tan^{- 1} x}{x}] = 0$ where $[x]$ represents
greatest integer $\leq . x$
Statement $- 2 :$ $\frac{\tan^{- 1} x}{x} < 1 for all x \neq 0$

Solution:

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