### Application Of Derivatives Question 10

#### Question 10 - 01 February - Shift 1

Let $f(x)=2 x+\tan ^{-1} x$ and $g(x)=\log _e(\sqrt{1+x^{2}}+x)$,

$x \in[0,3]$. Then

(1) There exists $\hat{x} \in[0,3]$ such that $f^{\prime}(\hat{x})<g^{\prime}(\hat{x})$

(2) $\max f(x)>\max g(x)$

(3) There exist $0<x_1<x_2<3$ such that $f(x)<g(x)$, $\forall x \in(x_1, x_2)$

(4) $\min f^{\prime}(x)=1+\max g^{\prime}(x)$

## Show Answer

#### Answer: (2)

#### Solution:

#### Formula: Increasing and decreasing of a function, Maximum of function and minima of a function

$f(x)=2 x+\tan ^{-1} x$ and $g(x)=\ln (\sqrt{1+x^{2}}+x), x \in[0,3]$

$f^{\prime}(x)=2+\frac{1}{{1+x^{2}}}$ and $g^{\prime}(x)=\frac{1}{\sqrt{1+x^{2}}}$

Both does not have critical values

$f(0)=0, f(3)=6+\tan ^{-1}(3) $

$g(0)=0, g(3)=\log (3+\sqrt{10})$

Let $h(x)=f(x)-g(x)$

$h^{\prime}(x)>0 \forall x \in(0,3)$

$\therefore h(x)$ is increasing function this implies $ max \ f(x) > max \ g(x)$