Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

Range of  f(x) = sin-1x + tan-1x + cos-1x is

Option: 1

\left( {0\;,\;\frac{{3\pi }}{2}} \right)


Option: 2

\left( {\frac{\pi }{2}\;,\;\frac{{3\pi }}{2}} \right)


Option: 3

\left( { - \frac{\pi }{2}\;,\;\frac{{3\pi }}{2}} \right]


Option: 4

\left[ {\frac{\pi }{4}\;,\;\frac{{3\pi }}{4}} \right]


Answers (1)

best_answer

As we learnt

 

Domains and Ranges of Inverse Trigonometric Functions -

 

For \cot ^{-1}x

Domain \epsilon R

Range \epsilon \left [ 0, \pi \right ]

-

Clearly the domain of the function is [-1, 1]

                              We know that {\tan ^{ - 1}}x \in \left[ { - \frac{\pi }{4}\;,\;\frac{\pi }{4}} \right] for x \in [-1, 1].

                              Thus range is \left[ {\frac{\pi }{4}\;,\;\frac{{3\pi }}{4}} \right]$.

Posted by

shivangi.bhatnagar

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