CAT 2020 Quant Question [Slot 1] with Solution 25

Let \(x + \frac{1}{x} = a\)

The given equation becomes, \({a^2} - 3a + 2 = 0\) \(a = 2\) or 1 i.e \(x + \frac{1}{x} = 2\) or \(x + \frac{1}{x} = 1\)

since \(x\) is real, \(x + \frac{1}{x} \ne 1;\therefore x + \frac{1}{x} = 2\)

\(\therefore \) The number of solutions = 1