Complex Numbers And Quadratic Equations question 18
Question: If $ x $ is real, then the maximum and minimum values of the expression $ \frac{x^{2}-3x+4}{x^{2}+3x+4} $ will be [IIT 1984]
Options:
A) 2, 1
B) $ 5,\frac{1}{5} $
C) $ 7,\frac{1}{7} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
Let $ y=\frac{x^{2}-3x+4}{x^{2}+3x+4} $
Þ $ (y-1)x^{2}+3(y+1)x+4(y-1)=0 $ For x is real $ D\ge 0 $
Þ $ 9{{(y+1)}^{2}}-16{{(y-1)}^{2}}\ge 0 $
Þ $ -7y^{2}+50y-7\ge 0 $
Þ $ 7y^{2}-50y+7\le 0 $
Þ $ (y-7)(7y-1)\le 0 $ Now, the product of two factors is negative if one in and one in . Case I : $ (y-7)\ge 0 $ and $ (7y-1)\le 0 $
Þ $ y\ge 7 $ and $ y\le \frac{1}{7} $ . But it is impossible Case II : $ (y-7)\le 0 $ and $ (7y-1)\ge 0 $
Þ $ y\le 7 $ and $ y\ge \frac{1}{7}\Rightarrow \frac{1}{7}\le y\le 7 $ Hence maximum value is 7 and minimum value is $ \frac{1}{7} $
BETA
