Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

Let \mathrm{f(x)=1-|\cos x| } for all \mathrm{x \in R }. Then
 

Option: 1

\mathrm{ f^{\prime}\left(\frac{\pi}{2}\right)} does not exist
 


Option: 2

 \mathrm{f(x)} is not continuous everywhere
 


Option: 3

 \mathrm{f(x)} is not differentiable anywhere
 


Option: 4

\mathrm{ \lim _{x \rightarrow \pi / 2+0} f(x)=1}


Answers (1)

best_answer

1 as well as \mathrm{|\cos x|} are continuous everywhere. So, \mathrm{f(x)} is continuous everywhere.
1 is differentiable everywhere and \mathrm{|\cos x| } is not differentiable only at those \mathrm{x} where \mathrm{\cos x=0. So, f^{\prime}\left(\frac{\pi}{2}\right)} does not exist.

\mathrm{\lim _{x \rightarrow \pi / 2+0} f(x)=\lim _{h \rightarrow 0} f\left(\frac{\pi}{2}+h\right)=\lim _{h \rightarrow 0}\left\{1-\left|\cos \left(\frac{\pi}{2}+h\right)\right|\right\}=\lim _{h \rightarrow 0}\{1-\sin h\}=1 . }

Posted by

Suraj Bhandari

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