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JEE Main 201219 MayOnline
Exam
JEE Main
Year
2012
Date
19 May
Mode
Online
Subject
Mathematics
Chapter
Continuity and Differentiability

JEE Main 2012 · 19 May · Online · Mathematics · Continuity and Differentiability

Let f:[1,3] arrow R be a function satisfying (x)/([x]) ≤ f(x) ≤ 6-x...

Let f:[1,3]→Rf:[1,3] \rightarrow R be a function satisfying x[x]≤f(x)≤6−x\frac{x}{[x]} \leq f(x) \leq \sqrt{6-x}, for all x≠2x \neq 2 and f(2)=1f(2)=1, where RR is the set of all real numbers and [x][x] denotes the largest integer less than or equal to xx. Statement 1: lim⁡x→2−f(x)\lim _{x \rightarrow 2^{-}} f(x) exists. Statement 2: ff is continuous at x=2x=2.

Options

  1. AStatement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.
  2. BStatement 1 is false, Statement 2 is true.
  3. CStatement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.
  4. DStatement 1 is true, Statement 2 is false.

Correct answer

D. Statement 1 is true, Statement 2 is false.

Step-by-step solution

 Consider x[x]\leqf(x)≤6−x⇒lim⁡x→2−x[x]=21=2\begin{aligned}& \text{ Consider } \frac{x}{[x]} \leqf(x) \leq \sqrt{6-x}\\ & \Rightarrow \lim _{x \rightarrow 2^{-}} \frac{x}{[x]}=\frac{2}{1}=2 \end{aligned}⇒lim⁡x→2−6−x=2∴lim⁡x→2−f(x)=2\begin{aligned}& \Rightarrow \lim _{x \rightarrow 2^{-}} \sqrt{6-x}=2 \\ & \therefore \lim _{x \rightarrow 2^{-}}f(x)=2 \end{aligned}[By Sandwich theorem] Now lim⁡x→2+x[x]=1,lim⁡x→2+6−x=2\lim _{x \rightarrow 2^{+}} \frac{x}{[x]}=1, \lim _{x \rightarrow 2^{+}} \sqrt{6-x}=2 Hence by Sandwich theorem lim⁡x→2+f(x)\lim _{x \rightarrow 2^{+}} f(x) does not exists. Therefore ff is not continuous at x=2x=2. Thus statement-1 is true but statement- 2 is not true
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