Q41 / 90 · JEE Main 2012 (19 May Online) · Mathematics

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

  • A
    Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.
  • B
    Statement 1 is false, Statement 2 is true.
  • C
    Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.
  • Correct answer
    D
    Statement 1 is true, Statement 2 is false.

Solution

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

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

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