A) greater than 11
B) less than 5
C) 5
D) depends upon a and a cannot be determined
Correct Answer: C
Solution :
\[f(x)=\,\,{{x}^{2}}-3x+a+\frac{1}{a};\,\,f(3)=9-9+a+\frac{1}{a}>0\] \[\Rightarrow \,\,\,a+\frac{1}{a}>0\,\,\,\,\,\Rightarrow \,\,\,a>0\] \[\operatorname{f}(2)=4-6+a+\frac{1}{a}\,\,\le 0\,\,\,\Rightarrow \,\,\,\frac{{{a}^{2}}-2a+1}{a}\le 0\] \[\Rightarrow \,\,\,\,\frac{{{(a-1)}^{2}}}{a}\le 0\,\,\,\Rightarrow \,\,a=1\] Therefore, \[f(x)={{x}^{2}}-3x+2=0\] has roots 1 and 2. \[\therefore \,\,\,{{\alpha }^{2}}+{{\beta }^{2}}=5\]
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