Question

The range of function $f\left(x\right)=\frac{{\sec }^{2}x-\sec x+1}{{\sec }^{2}x+\sec x+1}$ is

• A. $[\frac{1}{3},3)$
• B. $[\frac{1}{3},1)\cup (1,3]$
• C. $R-(\frac{1}{3},3)$
• D. none of these

Answer 1

The correct option is B $[\frac{1}{3},1)\cup (1,3]$

$f\left(x\right)=\frac{{\sec }^{2}x-\sec x+1}{{\sec }^{2}x+\sec x+1}$

$f^{\prime }\left(x\right)=\frac{2\sec x\cdot \sec x\tan x-\sec x\tan x}{{\sec }^{2}x+\sec x+1}-\frac{({\sec }^{2}x-\sec x+1)}{({\sec }^{2}x+\sec x+1{)}^2}(2\sec x\sec x\tan x+\sec x\tan x)$

$=\frac{\sec x\tan x}{({\sec }^{2}x+\sec x+1)}[2\sec x-1-\frac{({\sec }^{2}x-\sec x+1}{{\sec }^{2}x+\sec x+1}(2\sec x+1\left)\right]$

$=\frac{\sec x\tan x}{({\sec }^{2}x+\sec x+1)}$

$\frac{[2{\sec }^{3}x+2{\sec }^{2}x+2\sec x-{\sec }^{2}x-\sec x-1-2{\sec }^{3}x+2{\sec }^{2}x-2\sec x-{\sec }^{2}x+\sec x-1]}{{\sec }^{2}x+\sec x+1}$

$=\frac{\sec x\tan x}{({\sec }^{2}x+\sec x+1{)}^2}[2{\sec }^{2}x-2]$

$=\frac{2\sec x\tan x{\tan }^{2}x}{({\sec }^{2}x+\sec x+1{)}^2}$

$f^{\prime }\left(x\right)=0$

$\Rightarrow \tan x=0$ or $\frac{1}{{\sec }^{2}x+\sec x+1}=0$

$\Rightarrow \sec x=\pm 1$ $\Rightarrow \sec x\to \infty$

$\Rightarrow \cos x=\pm 1$ $\Rightarrow \cos x\to 0$

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

$\cos x=1\Rightarrow f\left(x\right)=\frac{1}{3}\to$ minima

$\cos x\to 0\Rightarrow f\left(x\right)\to 1$ but $f\left(x\right)\ne 1$

$\cos x=-1\Rightarrow f\left(x\right)=3\to$ maxima

$\therefore$ Range $\left(f\right)=[\frac{1}{3},1)\cup (1,3]$.