If x is real, then the greatest and least values of the expression $\frac{x^{2} - 2 x + 2}{x^{2} + 3 x + 9}$

see full answer

Want to Fund your own JEE / NEET / Foundation preparation ??

Take the SCORE scholarship exam from home and compete for scholarships worth ₹1 crore!*

An Intiative by Sri Chaitanya

2 and $- \frac{2}{27}$

2 and $\frac{2}{27}$

2 and $\frac{1}{27}$

2 or 3

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$\begin{array}{l} L e t y = \frac{x^{2} - 2 x + 2}{x^{2} + 3 x + 9} \\ \Rightarrow x^{2} y + 3 x y + 9 y = x^{2} - 2 x + 2 \\ \Rightarrow (y - 1) x^{2} + (3 y + 2) x + 9 y - 2 = 0 \\ \forall x \in R, D \geq 0 \\ \Rightarrow {(3 y + 2)}^{2} - 4 (y - 1) (9 y - 2) \geq 0 \\ \Rightarrow 9 y^{2} + 4 + 12 y - 4 (9 y^{2} - 2 y - 9 y + 2) \geq 0 \\ \Rightarrow 9 y^{2} + 4 + 12 y - 36 y^{2} + 8 y + 36 - 8 \geq 0 \\ \Rightarrow - 27 y^{2} + 56 y - 4 \geq 0 \\ \Rightarrow 27 y^{2} - 56 y + 4 \leq 0 \\ \Rightarrow 27 y^{2} - 54 y - 2 y + 4 \leq 0 \\ \Rightarrow 27 y (y - 2) - 2 (y - 2) \leq 0 \\ \Rightarrow (y - 2) (27 y - 2) \leq 0 \\ \Rightarrow \frac{2}{27} \leq y \leq 2 \end{array}$