Question

Consider the function $y=\frac{x^2+x+1}{\left(x^2+1\right)}$, where $x=\tan \theta$ and $\theta \in R-(2n+1)\frac{\pi }{2};n\in I$. 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)

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