Let $[P]$ denote the greatest integer $\leq P$, If $0 \leq a \leq 2$, then the number of integral values of…
Let $[P]$ denote the greatest integer $\leq P$, If $0 \leq a \leq 2$, then the number of integral values of ' $a$ ' such that $\lim _{x \rightarrow a}\left(\left[x^2\right]-[x]^2\right)$ does not exist is
3
2
1
0
Solution
$0 \leq a \leq 2$. Integral values of $a$ are $0,1,2$
$\begin{aligned} & \text { For } a=0 ; \lim _{x \rightarrow 0}\left(\left[x^2\right]-[x]^2\right)=\lim _{x \rightarrow 0}\left(\left[x^2\right]-[x]^2\right) \\ & \text { L.H.L. }=\lim _{h \rightarrow 0}\left(\left[(0-h)^2\right]-[0-h]^2\right)=0-(-1)^2=-1\end{aligned}$
$\text { R.H.L. }=\lim _{h \rightarrow 0}\left(\left[(0+h)^2\right]-[0+h]^2\right)=0$
$\therefore$ Limit doesn't exist for $a=0$
For $a=1:$ L.H.L. $=\lim _{h \rightarrow 0}\left(\left[(1-h)^2\right]-[1-h]^2\right)=0$
$\text { R.H.L. }=\lim _{h \rightarrow 0}\left(\left[(1+h)^2\right]-[1+h]^2\right)=1-1=0$
$\therefore$ Limit exist for $a=1$
For $a=2$ : L.H.L. $=\lim _{h \rightarrow 0}\left(\left[(2-h)^2\right]-[2-h]^2\right)=3-1=2$
$\text { R.H.L. }=\lim _{h \rightarrow 0}\left(\left[(2+h)^2\right]-[2+h]^2\right)=4-4=0$
$\therefore$ Limit doesn't exist for $a=2$
Hence, for 2 values of ' $a$ ' limit doesn't exist.