CAT 2002 — QA Question 14

$\frac{A^2}{x}+\frac{B^2}{x-1}=1 (x\ne 1)$

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).