Complex Numbers And Quadratic Equations question 23
Question: If the roots of the equation $ x^{2}-2ax+a^{2}+a-3=0 $ are real and less than 3, then [IIT 1999; MP PET 2000]
Options:
A) $ a<2 $
B) $ 2\le a\le 3 $
C) $ 3<a\le 4 $
D) $ a>4 $
Show Answer
Answer:
Correct Answer: A
Solution:
Given equation is $ x^{2}-2ax+a^{2}+a-3=0 $ If roots are real, then $ D\ge 0 $
Þ $ 4a^{2}-4(a^{2}+a-3)\ge 0\Rightarrow -a+3\ge 0 $
Þ $ a-3\le 0\Rightarrow a\le 3 $ As roots are less than 3, hence $ f(3)>0 $ $ 9-6a+a^{2}+a-3>0\Rightarrow a^{2}-5a+6>0 $
$ \Rightarrow (a-2)(a-3)>0\Rightarrow $ either $ a<2 $ or $ a>3 $ Hence $ a<2 $ satisfy all.
BETA
