Question

Let $f\left(x\right)$ be an invertible function such that $f^{\prime }\left(x\right)>0$ and $f^{\prime \prime }\left(x\right)>0$ for all $x\in R,$ then which of the following is/are correct ?
(where $x_1,x_2,\cdots ,x_n$ are different points)

• A. $\frac{\sum _{r=1}^{n}f\left(x_r\right)}{n}>f\left(\frac{\sum _{r=1}^n{x}_r}{n}\right)$
• B. $\frac{\sum _{r=1}^{n}f\left(x_r\right)}{n}<f\left(\frac{\sum _{r=1}^n{x}_r}{n}\right)$
• C. $\frac{\sum _{r=1}^n{f}^{-1}\left(x_r\right)}{n}<f^{-1}\left(\frac{\sum _{r=1}^n{x}_r}{n}\right)$
• D. $\frac{\sum _{r=1}^n{f}^{-1}\left(x_r\right)}{n}>f^{-1}\left(\frac{\sum _{r=1}^n{x}_r}{n}\right)$

Answer 1

Answer: (A), (C)

The correct options are
A $\frac{\sum _{r=1}^{n}f\left(x_r\right)}{n}>f\left(\frac{\sum _{r=1}^n{x}_r}{n}\right)$
C $\frac{\sum _{r=1}^n{f}^{-1}\left(x_r\right)}{n}<f^{-1}\left(\frac{\sum _{r=1}^n{x}_r}{n}\right)$

The nature of the graph of $f\left(x\right)$ is shown,


So, the $n$ points form polygon, therefore centre of the polygon is,
$(\frac{x_1+x_2+\cdots +x_n}{n},\frac{f\left(x_1\right)+f\left(x_2\right)+\cdots +f\left(x_n\right)}{n})=(\frac{\sum _{r=1}^n{x}_r}{n},\frac{\sum _{r=1}^{n}f\left(x_r\right)}{n})$
Correspondingly there will be a point on the curve,
$(\frac{\sum _{r=1}^n{x}_r}{n},\sum _{r=1}^{n}f\left(\frac{\sum _{r=1}^n{x}_r}{n}\right))$
This point will be below the centre of the polygon so the ordinate will follow the relation,
$f\left(\frac{\sum _{r=1}^n{x}_r}{n}\right)<\frac{\sum _{r=1}^{n}f\left(x_r\right)}{n}$


Let $g\left(x\right)=f^{-1}\left(x\right)$
$f\left(g\right(x\left)\right)=x\Rightarrow f^{\prime }\left(g\right(x\left)\right)g^{\prime }\left(x\right)=1\Rightarrow g^{\prime }\left(x\right)>0[\because f^{\prime }(g\left(x\right))>0]$

Now,
$g^{\prime }\left(x\right)=\frac{1}{f^{\prime }\left(g\right(x\left)\right)}\Rightarrow g^{\prime \prime }\left(x\right)=-\frac{1}{\left(f^{\prime }\right(g\left(x\right)){)}^2}{f}^{\prime \prime }\left(g\right(x\left)\right)g^{\prime }\left(x\right)\Rightarrow g^{\prime \prime }\left(x\right)<0$
The graph of $f^{-1}\left(x\right)$ is,


So, the $n$ points form polygon, therefore centre of the polygon is,
$(\frac{x_1+x_2+\cdots +x_n}{n},\frac{f^{-1}\left(x_1\right)+f^{-1}\left(x_2\right)+\cdots +f^{-1}\left(x_n\right)}{n})=(\frac{\sum _{r=1}^n{x}_r}{n},\frac{\sum _{r=1}^n{f}^{-1}\left(x_r\right)}{n})$
Correspondingly there will be a point on the curve,
$(\frac{\sum _{r=1}^n{x}_r}{n},\sum _{r=1}^n{f}^{-1}\left(\frac{\sum _{r=1}^n{x}_r}{n}\right))$
This point will be above the centre of the polygon, so the ordinate will follow the relation,
$\frac{\sum _{r=1}^n{f}^{-1}\left(x_r\right)}{n}<f^{-1}\left(\frac{\sum _{r=1}^n{x}_r}{n}\right)$

Alternate Solution:
Let $f\left(x\right)=e^x$
$f^{\prime }\left(x\right)=e^x,f^{\prime }\left(x\right)>0$
$f^{\prime \prime }\left(x\right)=e^x,f^{\prime \prime }\left(x\right)>0$
$f\left(\frac{\sum _{r=1}^n{x}_r}{n}\right)=e^{(\sum x_i)/n}$
$\frac{\sum _{r=1}^{n}f\left(x_r\right)}{n}=\frac{e^{x_1}+e^{x_2}+e^{x_3}+...+e^{x_n}}{n}$

Using $\text{AM-GM}$ inequality,
$\frac{e^{x_1}+e^{x_2}+e^{x_3}+...+e^{x_n}}{n}>e^{(\sum x_i)/n}$

$\Rightarrow \frac{\sum _{r=1}^{n}f\left(x_r\right)}{n}>f\left(\frac{\sum _{r=1}^n{x}_r}{n}\right)...\left(1\right)$

Now, inverse of $e^x$ is $\ln x$ i.e., $f^{-1}\left(x\right)=\ln x$
Since, $\ln x$ is increasing function with decreasing slope, therefore if we take inverse of eqn $\left(1\right)$, then the inquality sign will change. Hence, we obtained
$\frac{\sum _{r=1}^n{f}^{-1}\left(x_r\right)}{n}<f^{-1}\left(\frac{\sum _{r=1}^n{x}_r}{n}\right)$