The number of integral value(s) of x satisfying the inequality $\frac{\sqrt{x - 1}}{x - 2} \leq 0$ is

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Solution

$\frac{\sqrt{x - 1}}{x - 2} \leq 0 x - 1 \geq 0, for \sqrt{x - 1} to be defined \Rightarrow x \geq 1 As square root of a number is always non - negative . So, for \frac{\sqrt{x - 1}}{x - 2} to be less than 0 or negative, x - 2 < 0 \Rightarrow x < 2 ∴ 1 \leq x < 2 The number of integral values of x satisfying the given inequality is 1 .$

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