A) \[a,\frac{1}{a}\]
B) \[a,2a\]
C) \[a,\frac{1}{2a}\]
D) none of these
Correct Answer: A
Solution :
We have \[a({{x}^{2}}+1)-({{a}^{2}}+1)x=0\] \[\Rightarrow \]\[a{{x}^{2}}-({{a}^{2}}+1)x+a=0\] \[\Rightarrow \]\[(a{{x}^{2}}-{{a}^{2}}x)-(x-a)=0\] \[\Rightarrow \]\[ax(x-a)-1(x-a)=0\] \[\Rightarrow \]\[(x-a)(ax-1)=0\] \[\Rightarrow \] \[x=a,\frac{1}{a}\]You need to login to perform this action.
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