Question

Let $f$ be any function continuous on $\left[a,b\right]$ and twice differentiable on $\left(a,b\right)$. If for all $x\in \left(a,b\right)$,$f\text{'}\left(x\right)>0$ and $f"\left(x\right)<0$, then for any $c\in \left(a,b\right)$ ,$\left(f\left(c\right)-f\left(a\right)]\right)÷\left(f\left(b\right)-f\left(c\right)]\right)$ is greater than

• A. $\frac{\left(b-c\right)}{\left(c-a\right)}$
• B. $1$
• C. $\frac{\left(c-a\right)}{\left(b-c\right)}$
• D. $\frac{\left(b+a\right)}{\left(b-a\right)}$

Answer 1

The correct option is C

$\frac{\left(c-a\right)}{\left(b-c\right)}$

Explanation for correct option(s)

Given:

$\left(f\left(c\right)-f\left(a\right)]\right)-\left(f\left(b\right)-f\left(c\right)]\right)$ where $f$ is a twice differentiable and continuous function and $c\in \left(a,b\right)$

To find:

Whether the value of the given function.

Explanation:

Let us use the Lagrange's mean value theorem

For $p\in \left(a,c\right)$

$f\text{'}\left(p\right)=\frac{\left(f\left(c\right)-f\left(a\right)\right)}{\left(c-a\right)}$

For $q\in \left(c,b\right)$

$f\text{'}\left(q\right)=\frac{\left(f\left(b\right)-f\left(c\right)\right)}{\left(b-c\right)}$

Since we know that, for a function, "if the value of $y$ is increasing on increasing the value of $x$, then the given function is called an increasing function, whereas if the value of $y$ is decreasing on increasing of the value $x$, then the function is called as a decreasing function".

By using the definition of decreasing function we get,

$f"\left(x\right)<0$,$f\text{'}\left(x\right)$ is decreasing

Also by the defintion of definition increasing function we get,

$f\text{'}\left(p\right)>f\text{'}\left(q\right)$

$$\begin{gathered} \frac{\left(f\left(c\right)-f\left(a\right)\right)}{\left(c-a\right)}>\frac{\left(f\left(b\right)-f\left(c\right)\right)}{\left(b-c\right)} \\ \frac{\left(f\left(c\right)-f\left(a\right)\right)}{\left(f\left(b\right)-f\left(c\right)\right)}>\frac{\left(c-a\right)}{\left(b-c\right)} \end{gathered}$$

$f\text{'}\left(x\right)>0$, $f\left(x\right)$ is increasing.

Therefore, the value of the $\left(f\left(c\right)-f\left(a\right)]\right)÷\left(f\left(b\right)-f\left(c\right)]\right)$ is $\frac{\left(c-a\right)}{\left(b-c\right)}$.

Hence, the correct answer is option (C).