Let f be any function continuous on


Let $f$ be any function continuous on $[a, b]$ and twice differentiable on $(a, b)$. If for all $x \in(a, b), f^{\prime}(x)>0$ and

$f^{\prime \prime}(x)<0$, then for any $c \in(a, b), \frac{f(c)-f(a)}{f(b)-f(c)}$ is greater than:

  1. (1) $\frac{b+a}{b-a}$

  2. (2) 1

  3. (3) $\frac{b-c}{c-a}$

  4. (4) $\frac{c-a}{b-c}$

Correct Option: , 4


Since, function $f(x)$ is twice differentiable and continuous in $x \in[a, b]$. Then, by LMVT for $x \in[a, c]$

$\frac{f(c)-f(a)}{c-a}=f^{\prime}(\alpha), \alpha \in(a, c)$

Again by LMVT for $x \in[c, b]$

$\frac{f(b)-f(c)}{b-c}=f^{\prime}(\beta), \beta \in(c, b)$

$\because f^{\prime \prime}(x)<0 \Rightarrow f^{\prime}(x)$ is decreasing

$f^{\prime}(\alpha)>f^{\prime}(\beta) \Rightarrow \frac{f(c)-f(a)}{c-a}>\frac{f(b)-f(c)}{b-c}$

$\Rightarrow \frac{f(c)-f(a)}{f(b)-f(c)}>\frac{c-a}{b-c} \quad(\because f(x)$ is increasing $)$

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