Question

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

$f ( x )=2 x +\tan ^{-1} x$ 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, $0 \leq x \leq 3$
$ 0 \leq x^2 \leq 9$
$ 1 \leq 1+x^2 \leq 10$
So $ 2+\frac{1}{10} \leq f^{\prime}(x) \leq 3 $
$ \frac{21}{10} \leq f^{\prime}(x) \leq 3 \text { 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 $\max \left( f ( x )\right.$ 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