A) 2
B) 1
C) 4
D) None of these
Correct Answer: B
Solution :
[b] \[{{x}^{2}}-ax-x+a-1=0\] \[{{x}^{2}}-ax-x+a=1\] \[\underset{\operatorname{int}eger}{\mathop{\underset{\downarrow }{\mathop{\left( x-a \right)}}\,}}\,\underset{\operatorname{int}eger}{\mathop{\underset{\downarrow }{\mathop{\left( x\text{ }-\text{ }1 \right)}}\,}}\,=\underset{\operatorname{int}eger}{\mathop{\underset{\downarrow }{\mathop{1}}\,}}\,\] \[x,a\text{ }\in \text{ }integer\] only possible if \[\Rightarrow \] \[\] is only single a for which it is possible
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