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

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(B) We have $f(x) = \max \{ |	an^{-1} x|, \cot^{-1} x \}$.$1$. Analysis of $I$: The function $g(x) = |	an^{-1} x|$ and $h(x) = \cot^{-1} x$ are cont

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Exactly one of the above statements is correct.

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(B) We have $f(x) = \max \{ |	an^{-1} x|, \cot^{-1} x \}$.$1$. Analysis of $I$: The function $g(x) = |	an^{-1} x|$ and $h(x) = \cot^{-1} x$ are continuous everywhere. The maximum of two continuous functions is continuous. However,they intersect where $|	an^{-1} x| = \cot^{-1} x$. Since $\cot^{-1} x > 0$ for all $x$,we solve $	an^{-1} x = \cot^{-1} x$ (for $x > 0$) $\Rightarrow 	an^{-1} x = \frac{\pi}{2} - 	an^{-1} x \Rightarrow 2 	an^{-1} x = \frac{\pi}{2} \Rightarrow 	an^{-1} x = \frac{\pi}{4} \Rightarrow x = 1$. At $x=1$,the derivatives of the two functions are different,so it is not derivable at $x=1$. Thus,statement $I$ is wrong.$2$. Analysis of $II$: As $x 	o -\infty$,$f(x) = \max \{|-\frac{\pi}{2}|, \pi\} = \pi$. As $x 	o \infty$,$f(x) = \max \{\frac{\pi}{2}, 0\} = \frac{\pi}{2}$. The minimum value occurs at the intersection $x=1$,where $f(1) = \frac{\pi}{4}$. The range is $[\frac{\pi}{4}, \pi)$. Thus,statement $II$ is wrong.$3$. Analysis of $III$: Since the function is not surjective (range is not $R$) and it is not monotonic (it decreases then increases),it is many-one into. Thus,statement $III$ is correct.Conclusion: Only statement $III$ is correct. Therefore,exactly 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 $[\frac{\pi}{4}, \pi]$.
$III$. $f(x)$ 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 $[\frac{\pi}{4}, \pi]$.$III$. $f(x)$ is many-one into.Identify the correct option.