The number of integral values of a for which (a2−3a+2)x2+(a2−5a+6)x+a2−4=0 is an identity in x is
A
0
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B
2
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C
1
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D
3
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Solution
The correct option is B1 If this given equation is identity than a2−3a+2=0, a=2,1 a2−5a+6=0, a=3,2 a2−4=0, a=2,−2 So, common value of a=2 Hence, option 'C' is correct.