A) 2 and ? 3
B) 3 and 1
C) 3 and 2
D) 1 and 2
Correct Answer: C
Solution :
\[{{x}^{2}}-3x+2=\left( x-1 \right)\left( x-2 \right)\] \[\therefore \,\,{{x}^{2}}-3x+2\,\,divides\,\,{{x}^{4}}-{{x}^{3}}+a{{x}^{2}}+x+b\,\,exactly\] \[\Rightarrow \,\,\,\,\,\,x=1\] will satisfy \[{{x}^{4}}-{{x}^{3}}+a{{x}^{2}}+x+b\] \[\Rightarrow \,\,\,\,\,\,{{1}^{4}}-{{1}^{3}}+a\times {{1}^{2}}+1+b=0\] \[\Rightarrow \,\,\,\,\,\,a+b\,\,=-1\] ? (i) also \[x=2\] will satisfy \[{{x}^{4}}-{{x}^{3}}+a{{x}^{2}}+x+b\] \[\Rightarrow \,\,\,\,\,\,{{2}^{4}}-{{2}^{3}}+a\times {{2}^{2}}+2+b=0\] \[16-8+4a+b+2=0\] 4a+ b = - 10 .... (ii) On solving (i) and (ii), we have a = - 3 and b = [2]
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