#### Need solution for RD Sharma maths class 12 chapter 9 Differentiability exercise Fill in the blanks question  3

Answer: $z$

Hint: [r] is not continuous ⇒ not differentiable at every point.

If $\mathrm{f}(\mathrm{x})=\mathrm{f}_{1}(\mathrm{x})+\mathrm{f}_{2}(\mathrm{x})$ where $\mathrm{f}_{1}(\mathrm{x})$  is differentiable at $x=C$ and $\mathrm{f}_{2}(\mathrm{x})$ is not differentiable at $x=C$  then $\mathrm{f}(\mathrm{x})=\mathrm{f}_{1}(\mathrm{x})+\mathrm{f}_{2}(\mathrm{x})$ is also not differentiable at $x=c.$

Given: $\mathrm{f}(\mathrm{x})=\mathrm{x}-[\mathrm{x}]$

Solution:

We have $\mathrm{f}(\mathrm{x})=\mathrm{x}-[\mathrm{x}]$

Let $\mathrm{f}(\mathrm{x})=\mathrm{g}(\mathrm{x})+\mathrm{h}(\mathrm{x})$

Where $g(x)=x \text { and } h(x)=-[x]$

We know that [x] is not continuous at every point of of $\mathrm{x} \Rightarrow[\mathrm{x}]$ is not differentiable for all x

So , $h(x) = -[x]$ is also not continuous and this implies

$h(x) = --[x]$ is not differentiable for $x \in z$

Since $h(x)$ is not differentiable $\forall \mathrm{x} \in \mathbb{Z}$

$\Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{g}(\mathrm{x})+\mathrm{h}(\mathrm{x})$ is also not differentiable for all values of r ie when $x \in z$