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 $\mathrm{h}(\mathrm{x})=\mathrm{f}(\mathrm{x})-\mathrm{g}(\mathrm{x})$
$h^{\prime}(x)>0 \forall x \in(0,3)$
$\therefore \mathrm{h}(\mathrm{x})$ is increasing function this implies $ max \ f(x) > max \ g(x)$
