Question

Let $f(x)=2x+{\tan }^{-1}x$ and $g(x)={\log }_e\left(\sqrt{1+x^2}+x\right),x\in [0,3]$. Then

• A. $\min f^{\prime }(x)=1+\max g^{\prime }(x)$
• B. there exist $0<x_1<x_2<3$ such that $f(x)<g(x),\forall x\in \left(x_1,x_2\right)$
• C. $\max f(x)>\max g(x)$
• D. there exists $\hat{x}\in [0,3]$ such that $f^{\prime }(\hat{x})<g^{\prime }(\hat{x})$

Answer 1

Answer: (C)

$f(x)=2x+{\tan }^{-1}x$ and $g(x)=\ln \left(\sqrt{1+x^2}+x\right)$ and $x\in [0,3]$
$g^{\prime }(x)=\frac{1}{\sqrt{1+x^2}}$
Now, $0\le x\le 3$
$0\le x^2\le 9$
$1\le 1+x^2\le 10$
So $2+\frac{1}{10}\le f^{\prime }(x)\le 3$
$\frac{21}{10}\le f^{\prime }(x)\le 3\text{ and }\frac{1}{\sqrt{10}}\le g^{\prime }(x)\le 1$
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 $\max (f(x)$ at $x=3$ is $6+{\tan }^{-1}3$
$\mathrm{Max}(g(x)$ at $x=3$ is $\ln (3+\sqrt{10})$
And $6+{\tan }^{-1}3>\ln (3+\sqrt{10})$
Option (2) is correct