Question

Let $f$ be a function defined as $f\left(x\right)={\sec }^{-1}(1+{\cos }^{2}x),$ then

• A. the domain of $f$ is $R$
• B. the domain of $f$ is $[0,\pi ]$
• C. the range of $f$ is $[0,\frac{\pi }{2}]$
• D. the range of $f$ is $[0,\frac{\pi }{3}]$

Answer 1

Answer: (A), (D)

the range of $f$ is $[0,\frac{\pi }{3}]$

$f\left(x\right)={\sec }^{-1}(1+{\cos }^{2}x)$
We know, $-1\le \cos x\le 1$
$\Rightarrow 0\le {\cos }^{2}x\le 1$
$\Rightarrow 1\le {\cos }^{2}x+1\le 2$
Since, $\cos x$ is defined in $R$ and domain of ${\sec }^{-1}y$ is $(-\infty ,-1]\cup [1,\infty ),$ hence domain of $f$ is $R.$

Hence, the range of the function is
$[{\sec }^{-1}1,{\sec }^{-1}2]=[0,\frac{\pi }{3}]$