The number of roots A2x+B2x-1=1 is
Explanation:
A2x+B2x-1=1 (x≠1)
A2(x − 1) + B2(x) = x(x − 1) A2x − A2 + B2x = x2 − x x2 − x − A2x − B2x + A2 = 0 x2 − x[A2 + B2 + 1] + A2 = 0
Since it is a quadratic equation, the number of roots = 2 Now, ∆ = (A2 + B2 + 1)2 − 4A2 = A4 + B4 + 1 + 2A2B2 + 2B2 + 2A2 − 4A2 = A4 + B4 + 1 + 2A2B2 + 2B2 − 2A2
For roots to be real, A4 + B4 + 1 + 2A2B2 + 2B2 − 2A2 ≥ 0 ∴ B4 + 2B2(A2 + 1) + (A2 - 1)2 ≥ 0
The given equation will have one root if ∆ = 0. It can be clearly seen that for ∆ = 0, each of the three terms must be equal to 0 since each term is greater than or equal to 0. ∴ A = ±1 and B = 0. But for these values of A and B, x = 1 which is a contradiction.
So, ∆ ≠ 0
For ∆ > 0, two different roots are obtained. Therefore, the number of roots of the given equation will be 2.
Hence, option (b).
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