Let $f(x)=\sec^{-1}[1+\cos^2x]$, where [.] denotes the greatest integer function. Then the range of f (x) is
Answer & explanation
Correct answer: option 3
Clearly, f(x) is defined for all x ∈ R.
For any x ∈ R, we have $0≤\cos^2x≤1⇒1≤1+\cos^2x≤2$
$⇒[1+\cos^2x]=\left\{\begin{matrix}1,&for\,x∈ R-\{nπ:n∈ R\}\\2,&for\,x=nπ,n∈ R\end{matrix}\right.$
Hence, range $(f)=\{\sec^{-1}1, \sec^{-1}2\}$