The total number of integral values of a so that x2−(a+1)x+a−1=0 has integral roots is equal to

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none of these

answer is .

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Detailed Solution

x2−(a+1)x+a−1=0⇒ (x−a)(x−1)=1Now, a∈I and we want x to be an integer. Hence,x−a=1,x−1=1 or x−a=−1,x−1=−1⇒ a=1 in both cases

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