Search any question & find its solution
Question:
Answered & Verified by Expert
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 $x=2$.
Options:
Statement 1: $\lim _{x \rightarrow 2^{-}} f(x)$ exists.
Statement 2: $f$ is continuous at $x=2$.
Solution:
1895 Upvotes
Verified Answer
The correct answer is:
Statement 1 is true, Statement 2 is false.
Statement 1 is true, Statement 2 is false.
$$
\begin{aligned}
& \text { Consider } \frac{x}{[x]} \leq f(x) \leq \sqrt{6-x} \\
& \Rightarrow \lim _{x \rightarrow 2^{-}} \frac{x}{[x]}=\frac{2}{1}=2
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \lim _{x \rightarrow 2^{-}} \sqrt{6-x}=2 \\
& \therefore \lim _{x \rightarrow 2^{-}} f(x)=2
\end{aligned}
$$
[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
\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
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.