$\text { and } x \in[0,3]$
$g ^{\prime}( x )=\frac{1}{\sqrt{1+x^2}}$
Now, $0 \leq x \leq 3$
$0 \leq x^2 \leq 9$
$1 \leq 1+x^2 \leq 10$
So, $\quad 2+\frac{1}{10} \leq f^{\prime}(x) \leq 3$
$\frac{21}{10} \leq f^{\prime}(x) \leq 3$ and $\frac{1}{\sqrt{10}} \leq g^{\prime}(x) \leq 1$ option $(4)$ is incorrect
From above, $g ^{\prime}( x )< f ^{\prime}( x ) \forall x \in[0,3]$ Option $(1)$ is incorrect. $f ^{\prime}( x ) \& g ^{\prime}( x )$ both positive so $f ( x ) \& g ( x )$ both are increasing
So, $\quad \max \left( f ( x )\right.$ at $x =3$ is $6+\tan ^{-1} 3$
$\operatorname{Max}(g(x)$ at $x=3$ is $\ln (3+\sqrt{10})$
And $6+\tan ^{-1} 3 > \ln (3+\sqrt{10})$
Option $(2)$ is correct
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