Question

Consider the function $y=\frac{x^2+x+1}{\left(x^2+1\right)}$, where $x=\tan \theta$ and $\theta \in R-(2 n+1) \frac{\pi}{2} ; n \in I$. Also $L$ is the least value of the function and $G$ is the greatest value of the function for all permissible values of $x$.
The value of $(L + G)$ is equal to

• A. 4
• B. 3
• C. 2
• D. 1

Answer 1

Answer: (C)

$y=\frac{\sec ^2 \theta+\tan \theta}{\sec ^2 \theta}=1+\sin \theta \cos \theta=1+\frac{1}{2} \sin 2 \theta$
$\Rightarrow y \in\left[\frac{1}{2}, \frac{3}{2}\right] $
$\therefore L =\frac{1}{2} ; G =\frac{3}{2}$
$\therefore L + G =2 \Rightarrow [ C ] \text { is correct. } $