A) \[ac{{x}^{2}}-bx+1=0\]
B) \[{{x}^{2}}-acx+bc+1=0\]
C) \[ac{{x}^{2}}+bx-1=0\]
D) \[{{x}^{2}}+acx-bc+11=0\]
E) \[ac{{x}^{2}}-bx-11=0\]
Correct Answer: A
Solution :
\[\because \] \[\alpha ,\beta \]are roots of equation, \[a{{x}^{2}}+bx+c=0\] \[(c\ne 0)\] \[\therefore \] \[\alpha +\beta =-\frac{b}{a},\alpha \beta =\frac{c}{a}\] \[\therefore \]Required equation is \[{{x}^{2}}-\left( \frac{1}{a\alpha +b}+\frac{1}{\alpha \beta +b} \right)x+\left( \frac{1}{a\alpha +b}.\frac{1}{a\beta +b} \right)\] \[\Rightarrow \] \[{{x}^{2}}-\left( \frac{a(\alpha +\beta )+2b}{{{a}^{2}}\alpha \beta +ab(\alpha +\beta )+{{b}^{2}}} \right)x\] \[+\left( \frac{1}{{{a}^{2}}\alpha \beta +ab(\alpha +\beta )+{{b}^{2}}} \right)=0\] \[\Rightarrow \] \[{{x}^{2}}-\left( \frac{a\left( -\frac{b}{a} \right)+2b}{{{a}^{2}}.\frac{c}{a}+ab\left( -\frac{b}{a} \right)+{{b}^{2}}} \right)x\] \[+\frac{1}{{{a}^{2}}.\frac{c}{a}+ab\left( -\frac{b}{a} \right)+{{b}^{2}}}=0\] \[\Rightarrow \] \[{{x}^{2}}-\frac{b}{ac}x+\frac{1}{ac}=0\] \[\Rightarrow \] \[ac{{x}^{2}}-bx+1=0\]
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