A) - 6, - 4
B) - 6, 5
C) - 6, 4
D) 6, - 4
Correct Answer: D
Solution :
Let for real roots are \[\alpha ,\,\beta ,\,\gamma ,\,\delta \] then equation is \[(x-\alpha )\,(x-\beta )(x-\gamma )\,(x-\delta )=0\] \[{{x}^{4}}-(\alpha +\beta +\gamma +\delta ){{x}^{3}}+(\alpha \beta +\beta \gamma +\gamma \delta \]\[+\alpha \delta +\beta \delta +\alpha \gamma ){{x}^{2}}-(\alpha \beta \gamma +\beta \gamma \delta \]\[+\alpha \beta \delta +\alpha \gamma \delta )x+\alpha \beta \gamma \delta =0\] \[{{x}^{4}}-\sum \alpha .{{x}^{3}}+\sum \alpha \beta .{{x}^{2}}-\sum \alpha \beta \gamma .x+\alpha \beta \gamma \delta =0\] on comparing with \[{{x}^{4}}-4{{x}^{3}}+a{{x}^{2}}+bx+1=0\] \[\sum \alpha =4,\,\sum \alpha \beta \,=a\] \[\sum \alpha \beta \gamma =-b,\,\alpha \beta \gamma \delta =1\] For real roots, A.M. of roots \[\ge \]G.M. of roots \[\frac{1}{4}(\sum \alpha )\ge {{(\alpha \beta \gamma \delta )}^{1/4}}\]; \[\sum \alpha =4\] \[\therefore \] \[\frac{1}{4}\sum \alpha =\frac{1}{4}\times 4=1\] \[{{(\alpha \beta \gamma \delta )}^{1/4}}=1\]Þ \[\alpha \beta \gamma \delta =1\] \[\Sigma \alpha =4\] and \[\alpha \beta \gamma \delta =1\] \[\therefore \] \[\alpha =\beta =\gamma =\delta \] = 1 Now, \[\sum \alpha \beta =a\] \[\therefore a=\alpha \beta +\beta \gamma +\gamma \delta +\alpha \delta +\beta \delta +\alpha \gamma \] \[=1\times 1+1\times 1+1\times 1+1\times 1+1\times 1+1\times 1\] = 6 \[-b=\alpha \beta \gamma +\alpha \beta \delta +\alpha \gamma \delta +\beta \gamma \delta \] \[=1\times 1\times 1+1\times 1\times 1+1\times 1\times 1+1\times 1\times 1\] \[={{(1)}^{3}}+{{(1)}^{3}}+{{(1)}^{3}}+{{(1)}^{3}}=1+1+1+1=4\] \[\therefore \,b=-4\]; \[\therefore a=6\] and \[b=-4\].
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