Given that ' $x$ ' is real then the solution set of the equation $\sqrt{x - 1} + \sqrt{x + 1} = 1$ .

No solution

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2

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None of these

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Solution

The correct option is A No solution
$\sqrt{x - 1} + \sqrt{x + 1} = 1$ As $x > 1$
$∵ x - 1$ can't be negative.
$\Rightarrow \sqrt{x + 1} > 1$ $0 \sqrt{x - 1}$ is possitve
$∴$ there sum is $> 1$ always . hence no solution.

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