A) 2 and 8
B) 1 and 7
C) 5 and 3
D) 3 and 7
Correct Answer: B
Solution :
Put \[x=-1\]in \[{{x}^{3}}+(a+1)\,{{x}^{2}}-(b-2)x-6=0\] \[-1+(a+1)1-(b-2)(-1)-6=0\] \[-1+a+1+b-2-6=0\] \[a+b=8\] ?.. (i) Now put \[~x=2\] \[8+(a+1)\,4-(b-2)2-6=0\] \[8+4a+4-2b+4-6=0\] \[4a-2b=-10\Rightarrow 2a-b=-5\] ?? (ii) From (i) and (ii) \[\mathbf{a=1,}\,\,\,\mathbf{b=7}\]You need to login to perform this action.
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