Question

Mark the correct alternative in each of the following:
Let $f\left(x\right)=x-\left[x\right],xϵR,$ then $f^{\prime }\left(\frac{1}{2}\right)$ is

• A. $\frac{3}{2}$
  
• B. 1
  
• C. 0
  
• D. -1

Answer 1

The correct option is B 1


Given: $f\left(x\right)=x-\left[x\right],xϵR,$
Now,
For $0\le x<1,\left[x\right]=0$
$\therefore f\left(x\right)=x-0=x,\forall xϵ[0,1]$
Differentiating both sides with respect to x, we get
$f^{\prime }\left(x\right)=1,\forall xϵ[0,1]$
$\therefore f^{\prime }\left(\frac{1}{2}\right)=1$
Hence, the correct answer is option (b).