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

Let f(x)=tan−1⁡x−xf′(x)=11+x2−1=−x21+x2 0,tan−1⁡x f(0)=0ϕtan−1⁡x>xϕttan−1⁡xx<1(x<0)Thus limx→0+ tan−1⁡xx=0 and limx→0− tan−1⁡xx=0

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

Let f(x)=tan−1⁡x−xf′(x)=11+x2−1=−x21+x2 0,tan−1⁡x f(0)=0ϕtan−1⁡x>xϕttan−1⁡xx<1(x<0)Thus limx→0+ tan−1⁡xx=0 and limx→0− tan−1⁡xx=0

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

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STATEMENT-1 is True, STATEMENT-2 is False

STATEMENT-1 is False, STATEMENT-2 is True

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for
STATEMENT-1

answer is A.

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Detailed 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

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

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

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