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

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Answer 1 · 2017-04-03T12:10:45

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

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

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.!