Q100 / 160 · AP EAMCET 2024 (23 May Shift 1) · Mathematics

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

  • A
    3
  • Correct answer
    B
    2
  • C
    1
  • D
    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.

Hence, the correct answer is option: 'B'.

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