Question

f(x) is an increasing, concave upward function for all $X\in [a,b]$ , then for any $\lambda \in R^{+}$ (f is invertible)

• A. $\left[\frac{b+a\lambda }{1+\lambda }\right]>f^{-1}\left[\frac{f\left(b\right)+\lambda f\left(a\right)}{1+\lambda }\right]$
• B. $\left[\frac{b+a\lambda }{1+\lambda }\right]\ge f^{-1}\left[\frac{f\left(b\right)+\lambda f\left(a\right)}{1+\lambda }\right]$
• C. $\left[\frac{b+a\lambda }{1+\lambda }\right]<f^{-1}\left[\frac{f\left(b\right)+\lambda f\left(a\right)}{1+\lambda }\right]$
• D. $\left[\frac{b+a\lambda }{1+\lambda }\right]\le f^{-1}\left[\frac{f\left(b\right)+\lambda f\left(a\right)}{1+\lambda }\right]$

Answer 1

The correct option is C $\left[\frac{b+a\lambda }{1+\lambda }\right]<f^{-1}\left[\frac{f\left(b\right)+\lambda f\left(a\right)}{1+\lambda }\right]$

$f\left[\frac{a\lambda +b}{\lambda +1}\right]<\left[\frac{f\left(b\right)+\lambda f\left(a\right)}{1+\lambda }\right]$