Question

Let a function $f$ is a strictly increasing and $f^{\prime \prime }\left(x\right)<0$, also $a,b$ and $c$ are three distinct real numbers in the domain of inverse of $f\left(x\right).$ If $A=\frac{f^{-1}\left(a\right)+f^{-1}\left(b\right)+f^{-1}\left(c\right)}{3}$ and $B=f^{-1}\left(\frac{a+b+c}{3}\right),$ then which of the following is correct

• A. $A<B$
• B. $A\le B$
• C. $A>B$
• D. $A\ge B$

Answer 1

The correct option is C $A>B$

Given : $f$ is increasing and $f^{\prime \prime }\left(x\right)<0,$ i.e. downward concave.
We know, inverse of $f$ will also be increasing and upward concave.
Now plotting the curve of $f^{-1}\left(x\right):$

Let points $A\equiv (a,f^{-1}(a\left)\right),B\equiv (b,f^{-1}(b\left)\right)$ and $C\equiv (c,f^{-1}(c\left)\right)$
then Coordinate of centroid, $G\equiv (\frac{a+b+c}{3},\frac{f^{-1}\left(a\right)+f^{-1}\left(b\right)+f^{-1}\left(c\right)}{3})$
As, $G$ lies above the curve :
$\Rightarrow \frac{f^{-1}\left(a\right)+f^{-1}\left(b\right)+f^{-1}\left(c\right)}{3}>f^{-1}\left(\frac{a+b+c}{3}\right)$
$\therefore A>B$