Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The function  <img alt="\mathrm{f(x)=\left\{\begin{array}{l} 1, \quad|x| \geq 1 \\ \frac{1}{n^2}, \frac{1}{n}<|x|<\frac{1}{n-1}, n=2,3, \ldots \\ 0, \quad x=0 \end{array}\right

The function 

\mathrm{f(x)=\left\{\begin{array}{l} 1, \quad|x| \geq 1 \\ \frac{1}{n^2}, \frac{1}{n}<|x|<\frac{1}{n-1}, n=2,3, \ldots \\ 0, \quad x=0 \end{array}\right.}

Option: 1

is discontinuous at finitely many points


Option: 2

is continuous everywhere


Option: 3

is discontinuous only at \mathrm{x=\frac{1}{n} ; n \in Z-|0|}   and x= -0


Option: 4

none of these.


Answers (1)

best_answer

The function f is clearly continuous for |x|>1. We observe that
\mathrm{ \lim _{x \rightarrow-1^{+}} f(x)=1, \lim _{x \rightarrow-1^{-}} f(x)=\frac{1}{4} . }

Also, \mathrm{ \lim _{x \rightarrow \frac{1}{n}^{-}} f(x)=\frac{1}{n^2}~ and, ~\lim _{x \rightarrow \frac{1}{n}^{-}} f(x)=\frac{1}{(n+1)^2} }
Thus f is discontinuous for   \mathrm{ x= \pm \frac{1}{n}, n=1,2,3, }

Posted by

Shailly goel

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025 final answer key likely in February 2nd week; result date, marking scheme
anam-khan

JEE Mains 2025 Answer key (OUT) LIVE: Session 1 result by February 12 on jeemain.nta.nic.in; cut-off
anam-khan

JEE Main Marks vs Percentile 2025: How many marks do you need to score 95+ percentile?

Latest Question