First slide

Theory of equations

Question

Number of real solutions of the equation $\sqrt{x - \frac{1}{x}} + \sqrt{1 - \frac{1}{x}} = x is$

Difficult

A

1
B

2
C

0
D

infinite

Solution

$We have \sqrt{x - \frac{1}{x}} + \sqrt{1 - \frac{1}{x}} = x - - - (1)$
$\begin{array}{l} \Rightarrow \frac{(x - \frac{1}{x}) - (1 - \frac{1}{x})}{\sqrt{x - \frac{1}{x}} - \sqrt{1 - \frac{1}{x}}} = x \\ \Rightarrow \frac{x - \frac{1}{x} - 1 + \frac{1}{x}}{\sqrt{x - \frac{1}{x}} - \sqrt{1 - \frac{1}{x}}} = x \end{array}$
$\Rightarrow \frac{x - 1}{x} = \sqrt{x - \frac{1}{x}} - \sqrt{1 - \frac{1}{x}} - - - - (2)$
Adding (1) and (2), we get
$\begin{array}{l} 1 - \frac{1}{x} + x = 2 \sqrt{x - \frac{1}{x}} \\ \Rightarrow x - \frac{1}{x} - 2 \sqrt{x - \frac{1}{x}} + 1 = 0 \\ \Rightarrow {(\sqrt{x - \frac{1}{x}})}^{2} - 2 (1) \sqrt{x - \frac{1}{x}} + 1^{2} = 0 \\ \Rightarrow {(\sqrt{x - \frac{1}{x}} - 1)}^{2} = 0 \\ \Rightarrow \sqrt{x - \frac{1}{x}} = 1 \end{array}$
$\begin{array}{l} \Rightarrow x^{2} - x - 1 = 0 \\ \Rightarrow x = \frac{1 \pm \sqrt{5}}{2} \\ As x = \frac{1 - \sqrt{5}}{2} does not satisfy the given equation. \end{array}$

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