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

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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}]


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



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