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Q.
Consider, $f ( x )=( x -1) \tan ^{-1} x -\frac{1}{2} \ln \left(1+ x ^2\right), x \geq 1$.
Identify which of the following statement(s) is(are) correct?
Application of Derivatives
Solution:
$ f ( x )=( x -1) \tan ^{-1} x -\frac{1}{2} \ln \left(1+ x ^2\right)$
$f ^{\prime}( x )=( x -1) \frac{1}{1+ x ^2}+\tan ^{-1} x -\frac{ x }{1+ x ^2}=\tan ^{-1} x -\frac{1}{1+ x ^2}$
$f ^{\prime}( x )>0 \forall x \geq 1 \Rightarrow f ( x )$ is $\uparrow$
$f ^{\prime \prime}( x )=\frac{1}{1+ x ^2}+\frac{2 x }{\left(1+ x ^2\right)^2}=\frac{(1+ x )^2}{\left(1+ x ^2\right)^2}$
$f^{\prime \prime}(x)>0 \Rightarrow f^{\prime}(x)$ is $\uparrow \forall x \geq 1$
$f ^{\prime}(\infty) \rightarrow \frac{\pi}{2} \Rightarrow f ^{\prime}( x )<\frac{\pi}{2} \forall x \geq 1$
Using LMVT in $[x, x+4], x \geq 1$
$\frac{f(x+4)-f(x)}{x+4-x}=f^{\prime}(c) \text { for } c \in(x, x+4) $
$\therefore f(x+4)-f(x)<2 \pi \forall x \in[1, \infty)$