If a∈(−1,1) then roots of the quadratic equation (a−1)x2+ax+1−a2=0 are

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real

imaginary

both equal

none of these

answer is A.

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Detailed Solution

(a−1)x2+ax+1−a2=0∴ D=a2−4(a−1)1−a2 =a2−4a1−a2+41−a2 a−21−a22+41−a21−1−a2>0 for a∈(−1,1) Hence roots are real but not equal.

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