Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
If the range of the function fx=sin-1x+2tan-1x+x2+4x+1 is [p, q], then the value of p+q is
MathematicsFunctionsJEE Main
Solution:
1893 Upvotes Verified Answer
The correct answer is: 4
Domain of fx is -1,  1

fx=sin-1x+2tan-1x+x2+4x+1

fx=11-x2+21+x2+2x+4

=11-x2+21+x2+2x+2

  x-1,  1, fx>0

    fx is an increasing function

    p=minimum value of fx

=sin-1-1+2tan-1-1+-12+4-1+1

=-π2+2-π4-2=-π2-π2-2=-π-2

And q= maximum value of fx

=sin-11+2tan-11+1+4+1

=-π2+2π4+6=π+6

  So, the range of fx is -π-2,  π+6

p+q=π+6-π-2=4

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.