Question

Let $f$ be any function continuous on $[a,b]$ and twice differentaible on $(a,b)$. If for all $x\in (a,b),f^{\prime }\left(x\right)>0$ and $f^{\prime \prime }\left(x\right)<0$, then for any $c\in (a,b),\frac{f\left(c\right)-f\left(a\right)}{f\left(b\right)-f\left(c\right)}$ is greater than :

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

Answer 1

The correct option is C $\frac{c-a}{b-c}$

$f$ is continuous on $[a,b]$ and twice differentaible on $(a,b)$.

$\therefore$ LMVT is applicable
For $p\in (a,c),f^{\prime }\left(p\right)=\frac{f\left(c\right)-f\left(a\right)}{c-a}$
For $q\in (c,b),f^{\prime }\left(q\right)=\frac{f\left(b\right)-f\left(c\right)}{b-c}$

$\therefore f^{\prime \prime }\left(x\right)<0\Rightarrow f^{\prime }\left(x\right)$ is decreasing

$\Rightarrow f^{\prime }\left(p\right)>f^{\prime }\left(q\right)$
$\Rightarrow \frac{f\left(c\right)-f\left(a\right)}{c-a}>\frac{f\left(b\right)-f\left(c\right)}{b-c}$

$\Rightarrow \frac{f\left(c\right)-f\left(a\right)}{f\left(b\right)-f\left(c\right)}>\frac{c-a}{b-c}$
(as $f^{\prime }\left(x\right)>0\Rightarrow f\left(x\right)$ is increasing)