Question

The range of the function $f\left(x\right)=(co{s}^{2}x+4se{c}^{2}x)$ is

• A. $[4,\infty$)
• B. $[0,\infty$)
• C. $[5,\infty$)
• D. $[0,\infty$)

Answer 1

Answer: (C)

$[5,\infty$)

Given,

$f\left(x\right)={\cos }^{2}x+4{\sec }^{2}x$

$\Rightarrow f\left(x\right)={\cos }^{2}x+4\frac{1}{{\cos }^{2}x}$

$\Rightarrow f\left(x\right)=\frac{({\cos }^{2}x{)}^2+4}{{\cos }^{2}x}$

Since the minimum value of ${\cos }^{4}x$ is $0$ and the maximum value is $1$

Therefore, the minimum value of $f\left(x\right)$ will be when ${\cos }^{2}x$ is $1$ and the maximum value will be when ${\cos }^{2}x$ is $0$

when ${\cos }^{2}x$ is $1$ then the value of $f\left(x\right)=\frac{1+4}{1}=5$

And when ${\cos }^{2}x$ is $0$ $f\left(x\right)$ will be tending to infinity

Therefore, range of $f\left(x\right)=[5,\infty )$