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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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} (

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(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:

Similar Questions

If $a, b, c$ and $d$ are positive,then $\lim_{x \to \infty} \left(1 + \frac{1}{a+bx}\right)^{c+dx}$ is equal to:

If $l, m$ $(l < m)$ are roots of $ax^2 + bx + c = 0$,then $\lim_{x \rightarrow \alpha} \frac{|ax^2 + bx + c|}{ax^2 + bx + c} = $

$\lim _{n}$ ${\rightarrow \infty} \sqrt{2} \left[ \frac{(2+\sqrt{2})^n + (2-\sqrt{2})^n}{(2+\sqrt{2})^n - (2-\sqrt{2})^n} \right] =$

$\lim _{n \rightarrow \infty} \frac{n !}{(n+1) !-n !} = $

$\lim _{x \rightarrow 2}\left(\sum_{n=1}^{9} \frac{x}{n(n+1) x^{2}+2(2 n+1) x+4}\right)$ is equal to :

Vedclass Test Series

Exam Paper Generator

Online Exam Module