Let f : R to R,f(x) = max{|tan^{-1}x|, cot^{- | vedclass

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(B) Let $g(x) = |	an^{-1}x|$ and $h(x) = \cot^{-1}x$.We need to find $f(x) = \max\{g(x), h(x)\}$.For $x < 0$,$g(x) = |	an^{-1}x| = -	an^{-1}x$ and

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Exactly one of the above statements is correct.

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(B) Let $g(x) = |	an^{-1}x|$ and $h(x) = \cot^{-1}x$.We need to find $f(x) = \max\{g(x), h(x)\}$.For $x  -	an^{-1}x$,$f(x) = \cot^{-1}x$ for $x For $x \ge 0$,$g(x) = 	an^{-1}x$ and $h(x) = \cot^{-1}x$. They intersect where $	an^{-1}x = \cot^{-1}x$,i.e.,$x = 1$. At $x=1$,$f(1) = \frac{\pi}{4}$.For $x \in [0, 1]$,$\cot^{-1}x \ge 	an^{-1}x$,so $f(x) = \cot^{-1}x$. For $x > 1$,$	an^{-1}x > \cot^{-1}x$,so $f(x) = 	an^{-1}x$.Statement $I$: The function is continuous everywhere but not differentiable at $x=1$ (where the functions switch) and potentially at $x=0$ (where $|	an^{-1}x|$ has a corner). Thus,$I$ is wrong.Statement $II$: The range is $[\frac{\pi}{4}, \pi)$. As $x 	o -\infty$,$f(x) 	o \pi$. As $x 	o \infty$,$f(x) 	o \frac{\pi}{2}$. The minimum value is $\frac{\pi}{4}$ at $x=1$. Thus,$II$ is wrong.Statement $III$: Since the function is not monotonic and does not cover the entire codomain $R$,it is many-one into. Thus,$III$ is correct.Only one statement is correct.

Let $f : R \to R$,$f(x) = \max\{|\tan^{-1}x|, \cot^{-1}x\}$. Consider the following statements:
$I.$ The function is continuous and derivable $\forall x \in R$.
$II.$ The range of the function is $\left[ \frac{\pi}{4}, \pi \right]$.
$III.$ $f(x)$ is many-one into.
Identify the correct option.

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Let $f : R \to R$,$f(x) = \max\{|\tan^{-1}x|, \cot^{-1}x\}$. Consider the following statements:$I.$ The function is continuous and derivable $\forall x \in R$.$II.$ The range of the function is $\left[ \frac{\pi}{4}, \pi \right]$.$III.$ $f(x)$ is many-one into.Identify the correct option.