Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Which one of the following function is continuous at ?Optio

Which one of the following function is continuous at \mathrm{x=3} ?

Option: 1

\mathrm{f(x)=\left\{\begin{array}{l} 2, \text { if } x=3 \\ x-1, \text { if } x>3 \\ \frac{x+3}{3}, \text { if } x<3 \end{array}\right.}


Option: 2

\mathrm{f(x)=\left\{\begin{array}{l} 4, \text { if }=3 \\ 8-x, \text { if } x \neq 3 \end{array}\right.}-


Option: 3

\mathrm{f(x)=\left\{\begin{array}{l} x+3, \text { if } x \leq 3 \\ x-4, \text { if } x>3 \end{array}\right.}


Option: 4

\mathrm{f(x)=\frac{1}{x^3-27} \text {, if } x \neq 3}


Answers (1)

best_answer

\mathrm{\text { If } L H L=R H L=f(a) \text {, i.e., } \underset{x \rightarrow a^{-}}{\operatorname{Lt}} f(x)=\operatorname{Lt}_{x \rightarrow a^{+}} f(x)=f(a) \text {, }}

then \mathrm{f(x)} is continuous at \mathrm{ x=a}

Let us take option (a)

                         \mathrm{\begin{aligned} & \text { LHL }=\operatorname{Lt}_{x \rightarrow 3^{-}} f(x)=\operatorname{Lt}_{x \rightarrow 3} \frac{x+3}{3}=2 \\ & \text { RHL }=\operatorname{Lt}_{x \rightarrow 3^{+}} f(x)=\operatorname{Lt}_{x \rightarrow 3}(x-1)=2 \end{aligned}}

and \mathrm{f(3)=2} so \mathrm{f(x)} is continuous at \mathrm{x=3}
\mathrm{\therefore} Option (a) is correct

Posted by

Rishi

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 2026 application correction begins on jeemain.nta.nic.in; edit details by December 2
anam-khan

JEE Mains 2026 LIVE: NTA to conduct session 1 from January 21 to 30; exam pattern, syllabus
anam-khan

JEE Main 2026 correction window opens tomorrow; NTA allows exam city change

Latest Question