Let $f(x),f'(x),f''(x)$ be all positive for all $x\in[0,7]$. If $f^{-1}(x)$ exists then $f^{-1}(5)+4f^{-1}(6)-5f^{-1}(\frac{29}5)$

Question

Question:

Let $f(x)$ be continuous and twice differentiable function. Let $f(x),f'(x),f''(x)$ be all positive for all $x\in[0,7]$. If $f^{-1}(x)$ exists then $f^{-1}(5)+4f^{-1}(6)-5f^{-1}(\frac{29}5)$

• A) is always positive
• B) is always negative
• C) is zero for at least one $x\in(0,7)$
• D) is zero for all $x\in(0,7)$

My Attempt:

$f''(x)\gt0\implies f'(x)$ is increasing.

Also, $f^{-1}(f(x))=x$. So, to find $f^{-1}(5)$, I need that $x$ for which $f(x)$ is $5$.

Not able to think anything else.

(This question appeared in a JEE mock exam, held two days ago. Answer given is option B)

Answer 1

The function $f$ is strictly monotonously increasing $(f'>0)$ and strictly convex $(f''>0)$. If $f^{-1}$ exists, it is strictly monotonously increasing and strictly concave. Hence, \begin{align}\frac15 f^{-1}(5) +\frac{4}{5}f^{-1}(6)<f^{-1}\left( \frac15 \cdot 5 +\frac{4}{5}\cdot 6\right)=f^{-1}\left(\frac{29}{5}\right). \end{align}