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.!

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