Question

Consider, $f(x)=(x-1){\tan }^{-1}x-\frac{1}{2}\ln \left(1+x^2\right),x\ge 1$. Identify which of the following statement(s) is(are) correct?

• A. $\underset{x\to \infty }{\mathrm{Lim}}{f}^{\prime }(x)=\frac{\pi }{2}$
• B. $f^{\prime }(x)$ is increasing in $[1,\infty )$
• C. $f(x+4)-f(x)>2\pi \forall x\in [1,\infty )$
• D. $f(x+4)-f(x)<2\pi \forall x\in [1,\infty )$

Answer 1

Answer: (A), (B), (D)

$f(x)=(x-1){\tan }^{-1}x-\frac{1}{2}\ln \left(1+x^2\right)$
$f^{\prime }(x)=(x-1)\frac{1}{1+x^2}+{\tan }^{-1}x-\frac{x}{1+x^2}={\tan }^{-1}x-\frac{1}{1+x^2}$
$f^{\prime }(x)>0\forall x\ge 1\Rightarrow f(x)$ is $\uparrow$
$f^{\prime \prime }(x)=\frac{1}{1+x^2}+\frac{2x}{{\left(1+x^2\right)}^2}=\frac{(1+x{)}^2}{{\left(1+x^2\right)}^2}$
$f^{\prime \prime }(x)>0\Rightarrow f^{\prime }(x)$ is $\uparrow \forall x\ge 1$
$f^{\prime }(\infty )\to \frac{\pi }{2}\Rightarrow f^{\prime }(x)<\frac{\pi }{2}\forall x\ge 1$
Using LMVT in $[x,x+4],x\ge 1$
$\frac{f(x+4)-f(x)}{x+4-x}=f^{\prime }(c)\text{ for }c\in (x,x+4)$
$\therefore f(x+4)-f(x)<2\pi \forall x\in [1,\infty )$