What is the domain and range of #y=sec^2x+1#?

1 Answer
Apr 3, 2017

Answer:

#"Domain="RR-{(2k+1)pi/2 | k in ZZ}.#

#"Range="{ x in RR | x >=2}, or, [2,oo).#

Explanation:

Recall that the Domain of #sec# fun. is #RR-{(2k+1)pi/2 | k in ZZ}#.

Clearly, so is the Domain of the given fun.

#because, |secx| >= 1 :. sec^2x >=1, &, :., y=sec^2x+1 >=2.#

This means that the Range of the fun. is,

#{ x in RR | x >=2}, or, [2,oo).#

Enjoy Maths.!