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

Apr 3, 2017

$\text{Domain=} \mathbb{R} - \left\{\left(2 k + 1\right) \frac{\pi}{2} | k \in \mathbb{Z}\right\} .$

$\text{Range=} \left\{x \in \mathbb{R} | x \ge 2\right\} , \mathmr{and} , \left[2 , \infty\right) .$

#### Explanation:

Recall that the Domain of $\sec$ fun. is $\mathbb{R} - \left\{\left(2 k + 1\right) \frac{\pi}{2} | k \in \mathbb{Z}\right\}$.

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,

$\left\{x \in \mathbb{R} | x \ge 2\right\} , \mathmr{and} , \left[2 , \infty\right) .$

Enjoy Maths.!