Let [P] denote the greatest integer leq P. If | vedclass

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(B) Given $0 \leq a \leq 2$. The integral values of $a$ are $0, 1, 2$.Let $f(x) = [x^2] - [x]^2$.For $a = 0$:$	ext{L.H.L.} = \lim_{h ightarrow 0} (

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$2$

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(B) Given $0 \leq a \leq 2$. The integral values of $a$ are $0, 1, 2$.Let $f(x) = [x^2] - [x]^2$.For $a = 0$:$	ext{L.H.L.} = \lim_{h ightarrow 0} ([(0-h)^2] - [0-h]^2) = [h^2] - [-h]^2 = 0 - (-1)^2 = -1$.$	ext{R.H.L.} = \lim_{h ightarrow 0} ([(0+h)^2] - [0+h]^2) = [h^2] - [h]^2 = 0 - 0 = 0$.Since $	ext{L.H.L.} 
eq 	ext{R.H.L.}$,the limit does not exist at $a = 0$.For $a = 1$:$	ext{L.H.L.} = \lim_{h ightarrow 0} ([(1-h)^2] - [1-h]^2) = [1-2h+h^2] - [1-h]^2 = 0 - 0^2 = 0$.$	ext{R.H.L.} = \lim_{h ightarrow 0} ([(1+h)^2] - [1+h]^2) = [1+2h+h^2] - [1+h]^2 = 1 - 1^2 = 0$.Since $	ext{L.H.L.} = 	ext{R.H.L.}$,the limit exists at $a = 1$.For $a = 2$:$	ext{L.H.L.} = \lim_{h ightarrow 0} ([(2-h)^2] - [2-h]^2) = [4-4h+h^2] - [2-h]^2 = 3 - 1^2 = 2$.$	ext{R.H.L.} = \lim_{h ightarrow 0} ([(2+h)^2] - [2+h]^2) = [4+4h+h^2] - [2+h]^2 = 4 - 2^2 = 0$.Since $	ext{L.H.L.} 
eq 	ext{R.H.L.}$,the limit does not exist at $a = 2$.Thus,the limit does not exist for $2$ values of $a$ ($a=0$ and $a=2$).

Let $[P]$ denote the greatest integer $\leq P$. If $0 \leq a \leq 2$,then the number of integral values of $a$ such that $\lim _{x \rightarrow a}([x^2]-[x]^2)$ does not exist is:

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