Let f(x) = 2x + ^ - 1 x and g(x) = _ e ( 1 + x ^ 2 + x), x ∈ [0, 3]...

Given:

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

$⇒ {f}^{'}\left(x\right) = 2 + \dfrac{1}{1 + {x}^{2}}$

And,

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

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

Now,

$0 \le x \le 3$

$⇒ 0 \le {x}^{2} \le 9$

$⇒ 1 \le 1 + {x}^{2} \le 10$

So,

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

$⇒ 2 + \dfrac{1}{10} \le 2 + \dfrac{1}{1 + {x}^{2}} \le 3$

$⇒ 2 + \dfrac{1}{10} \le {f}^{'}\left(x\right) \le 3$

$⇒ \dfrac{21}{10} \le {f}^{'}\left(x\right) \le 3$

And,

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

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

Option \left(4\right)$is incorrect</p><p>From above,</p><p>${g}^{'}\left(x\right) < {f}^{'}\left(x\right) \forall x \in \left[0, 3\right]$</p><p>Option \left(1\right)$ is incorrect.

Since, ${f}^{'}$\left(x\right)$and${g}^{'}$\left(x\right)$ 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$6 + \tan^{- 1}3$</p><p>$max$\left(g\left(x\right)\right)$ at $x = 3$ is $ln$\left(3 + \sqrt{10}\right)$</p><p>And,$6 + \tan^{- 1}3 > ln$\left(3 + \sqrt{10}\right)$

Option $\left(2\right)$ is correct