Let f x = 2 x + tan - 1 x and g x = log e 1 + x 2 + x ,   x ∈ 0 , 3 . Then

Given:

$f\left(x\right)=2x+{\tan }^{-1}x$

$\Rightarrow f^{\text{'}}\left(x\right)=2+\frac{1}{1+x^2}$

And,

$g\left(x\right)=\ln \left(\sqrt{1+x^2}+x\right)$

$\Rightarrow g^{\text{'}}\left(x\right)=\frac{1}{\sqrt{1+x^2}}$

Now,

$0\le x\le 3$

$\Rightarrow 0\le x^2\le 9$

$\Rightarrow 1\le 1+x^2\le 10$

So,

$\frac{1}{10}\le \frac{1}{1+x^2}\le 1$

$\Rightarrow 2+\frac{1}{10}\le 2+\frac{1}{1+x^2}\le 3$

$\Rightarrow 2+\frac{1}{10}\le f^{\text{'}}\left(x\right)\le 3$

$\Rightarrow \frac{21}{10}\le f^{\text{'}}\left(x\right)\le 3$

And,

$\frac{1}{\sqrt{10}}\le g^{\text{'}}\left(x\right)\le 1$

So, $\min f^{\text{'}}\left(x\right)=\frac{21}{10}\ne \left[1+\max g^{\text{'}}\left(x\right)\right]$

Option $\left(4\right)$ is incorrect

From above,

$g^{\text{'}}\left(x\right)<f^{\text{'}}\left(x\right)\forall x\in \left[0,3\right]$

Option $\left(1\right)$ is incorrect.

Since, $f^{\text{'}}\left(x\right)$ and $g^{\text{'}}\left(x\right)$ are both positive so $f\left(x\right)$ and $g\left(x\right)$ both are increasing

So,

$\max \left(f\left(x\right)\right)$ at $x=3$ is $6+{\tan }^{-1}3$

$\max \left(g\left(\mathrm{x}\right)\right)$ at $x=3$ is $\ln \left(3+\sqrt{10}\right)$ 

And, $6+{\tan }^{-1}3>\ln \left(3+\sqrt{10}\right)$ 

Option $\left(2\right)$ is correct