Question

$f\left(x\right)=2x{\tan }^{-1}x,g\left(x\right)=\log (1+x^2),h\left(x\right)=\sin x,u\left(x\right)=x-\frac{x^3}{6}+\frac{x^4}{120}$ then

• A. $f\left(x\right)<g\left(x\right)$
• B. $f\left(x\right)\ge g\left(x\right)$
• C. $h\left(x\right)>u\left(x\right),x>0$
• D. $h\left(x\right)\ge u\left(x\right),x>0$

Answer 1

The correct option is B $f\left(x\right)\ge g\left(x\right)$

Let
$h\left(x\right)=2x{\tan }^{-1}\left(x\right)-log(1+x^2)$
Hence
$h^{\prime }\left(x\right)=2{\tan }^{-1}\left(x\right)+\frac{2x}{1+x^2}-\frac{2x}{1+x^2}$
$=2{\tan }^{-1}\left(x\right)$.
Now
For
$x>0$ ${\tan }^{-1}\left(x\right)>0$.
Hence for all x>0,
$h^{\prime }\left(x\right)>0$
Or
$h\left(x\right)\ge 0$
Or
$f\left(x\right)\ge g\left(x\right)$. where $f\left(x\right)=g\left(x\right)$ at x=0.
Now for $x<0$.
$h^{\prime }\left(x\right)=m\left(x\right)=2{\tan }^{-1}\left(x\right)$.
Now
$m^{\prime }\left(x\right)=\frac{2}{1+x^2}>0$
Hence
$m^{\prime }\left(x\right)>0$ for all x.
Thus $m\left(x\right)>0$ for all x.
Or
$h^{\prime }\left(x\right)>0$ for all x.
Hence
$h\left(x\right)>0$ for all x.
Hence
$f\left(x\right)\ge g\left(x\right)$ for all x.