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

Statement$-1:$ $\underset{x\to 0}{lim} \left[\frac{{\mathrm{tan}}^{-1}x}{x}\right]=0$ where $\left[x\right]$ representsgreatest integer  $\le .x$Statement$-2:$

1. A

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

2. B

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

3. C

STATEMENT-1 is True, STATEMENT-2 is False

4. D

STATEMENT-1 is False, STATEMENT-2 is True

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Solution:

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

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

$\varphi {\mathrm{tan}}^{-1}x>x\varphi t\frac{{\mathrm{tan}}^{-1}x}{x}<1\left(x<0\right)$

Thus $\underset{x\to 0+}{lim} \left[\frac{{\mathrm{tan}}^{-1}x}{x}\right]=0$ and

$\underset{x\to {0}^{-}}{lim} \left[\frac{{\mathrm{tan}}^{-1}x}{x}\right]=0$

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