Question

The function $f$ satisfying $\frac{f\left(b\right)-f\left(a\right)}{b-a}\ne f^{\prime }\left(x\right)$ for any $x\in (a,b)$ is :

• A. $f\left(x\right)=x^{1/3}$, $a=-1$, $b=1$
• B. $f\left(x\right)=1/x$; $a=1$, $b=4$
  
• C. $f\left(x\right)=x\left|x\right|$; $a=-1$,$b=1$
• D. None of these

Answer 1

The correct option is D None of these

$\frac{f\left(b\right)-f\left(a\right)}{b-a}\ne f^{\prime }\left(x\right)x\in (a,b)\left(i\right)f\left(x\right)=x^{1/3},a=-1,b=1\Rightarrow \frac{f\left(1\right)-f(-1)}{1-(-1)}=\frac{f\left(1\right)-f(-1)}{2}=\frac{(1{)}^{1/3}-(-1{)}^{1/3}}{2}=\frac{1-(-1)}{2}=1\Rightarrow f^{\prime }\left(x\right)=\frac{1}{3}{x}^{-2/3}\Rightarrow \frac{1}{3}{x}^{-2/3}=1\Rightarrow x^{-2/3}=3\Rightarrow {\left(x^{-2/3}\right)}^{-3/2}=3^{-3/2}\Rightarrow x={\left(\frac{1}{3}\right)}^{3/2}$

Clearly ${\left(\frac{1}{3}\right)}^{3/2}\in [-1,1]$. Hence, this does not hold true.

$\left(ii\right)f\left(x\right)=\frac{1}{x},a=1,b=4\Rightarrow \frac{f\left(b\right)-f\left(a\right)}{b-a}=\frac{(\frac{1}{4}-\frac{1}{1})}{4-1}=\frac{-3}{4}\times \frac{1}{3}=-\frac{1}{4}\Rightarrow f^{\prime }\left(x\right)=-\frac{1}{x^2}\Rightarrow -\frac{1}{x^2}=-\frac{1}{4}\Rightarrow x=\pm 2$

$x=2\in [1,4]$, So this also doees not hold true.

$\left(iii\right)f\left(x\right)=x\left|x\right|,a=-1,b=1\Rightarrow \frac{f\left(b\right)-f\left(a\right)}{b-a}=\frac{1\left|1\right|-(-1)|-1|}{1-(-1)}=\frac{1-(-1)}{2}=1\Rightarrow f^{\prime }\left(x\right)=\{\begin{array}{c}2xforx\ge 0\\ -2xforx<0\end{array}$

When, $2x=1\Rightarrow x=1/2$

and $-2x=1\Rightarrow x=-1/2$

So, this does not hold true either.

Therefore none of these will be the answer.