Complex Numbers And Quadratic Equations question 19
Question: If $ x $ is real, then the value of $ \frac{x^{2}+34x-71}{x^{2}+2x-7} $ does not lie between [Roorkee 1983]
Options:
A) -9 and -5
B) -5 and 9
C) 0 and 9
D) 5 and 9
Show Answer
Answer:
Correct Answer: D
Solution:
Let $ y=\frac{x^{2}+34x-71}{x^{2}+2x-7} $
Þ $ x^{2}(y-1)+2(y-17)x+(71-7y)=0 $ For real values of x, its discriminant $ D\ge 0 $
$ \Rightarrow 4{{(y-17)}^{2}}-4(y-1)(71-7y)\ge 0 $
$ \Rightarrow (y^{2}-3+y+289)-(71y-7y^{2}-71+7y)\ge 0 $
$ \Rightarrow y^{2}-14y+45\ge 0\Rightarrow (y-5)(y-9)\ge 0 $ It is possible when both $ y-5 $ and $ y-9 $ are negative or both positive. Let $ y-5\le 0\Rightarrow y\le 5 $ and $ y-9\le 0\Rightarrow y\le 9 $ . Hence $ y\le 5 $ …..(i) If $ y-5\ge 0\Rightarrow $ $ y\ge 5 $ and $ y-9\ge 0\Rightarrow y\ge 9 $ Hence $ y\ge 9 $ . …..(ii) Therefore y does not lie between 5 and 9.
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