Complex Numbers And Quadratic Equations question 12
Question: If x is real, then the maximum and minimum values of expression $ \frac{x^{2}+14x+9}{x^{2}+2x+3} $ will be [Dhanbad Engg. 1968]
Options:
A) 4, - 5
B) 5, - 4
C) - 4, 5
D) - 4, - 5
Show Answer
Answer:
Correct Answer: A
Solution:
Let $ y=\frac{x^{2}+14x+9}{x^{2}+2x+3} $
Þ $ y(x^{2}+2x+3)-x^{2}-14x-9=0 $
$ \Rightarrow $ $ (y-1)x^{2}+(2y-14)x+3y-9=0 $ For real x, its discriminant $ \ge 0 $ i.e. $ 4{{(y-7)}^{2}}-4(y-1)3(y-3)\ge 0 $
Þ $ y^{2}+y-20\le $ 0 or $ (y-4)(y+5)\le 0 $ Now, the product of two factors is negative if these are of opposite signs. So following two cases arise: Case I: $ y-4\ge 0 $ or $ y\ge 4 $ and $ y+5\le 0 $ or $ y\le -5 $ This is not possible. Case II: $ y-4\le 0 $ or $ y\le 4 $ and $ y+5\ge 0 $ or $ y\ge -5 $ Both of these are satisfied if $ -5\le y\le 4 $ Hence maximum value of y is 4 and minimum value is - 5.
BETA
