Complex Numbers And Quadratic Equations question 690
Question: Consider $ f(x)=x^{2}-3x+a+\frac{1}{a}, $ $ a\in R-{0}, $ such that $ f(3)>0 $ and $ f(2)\le 0. $ If $ \alpha $ and $ \beta $ are the roots of equation $ f(x)=0 $ then the value of $ {{\alpha }^{2}}+{{\beta }^{2}} $ is equal to
Options:
A) greater than 11
B) less than 5
C) 5
D) depends upon a and a cannot be determined
Show Answer
Answer:
Correct Answer: C
Solution:
$ f(x)=x^{2}-3x+a+\frac{1}{a};f(3)=9-9+a+\frac{1}{a}>0 $
$ \Rightarrow a+\frac{1}{a}>0\Rightarrow a>0 $ $ f(2)=4-6+a+\frac{1}{a}\le 0\Rightarrow \frac{a^{2}-2a+1}{a}\le 0 $
$ \Rightarrow \frac{{{(a-1)}^{2}}}{a}\le 0\Rightarrow a=1 $ Therefore, $ f(x)=x^{2}-3x+2=0 $ has roots 1 and 2.
$ \therefore {{\alpha }^{2}}+{{\beta }^{2}}=5 $
BETA
