If x - 1x - 2- x - 0 then value of x2 - 1x2 is equal to-

The correct option is B 2 Given: x+1x=2 To find: x^2+1x^2 Solution: x+1x=2 squaring both sides, we get (x+1x)^2=2^2 x^2+1x^2+2=4 ...

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If x + 1x =...

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If $x + \frac{1}{x} = 2, x \neq 0$ then value of $x^{2} + \frac{1}{x^{2}}$ is equal to:

1

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3

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x + \frac{1}{x} = 2

\neq

x^{2} + \frac{1}{x^{2}}

x^{2} + x + 1 = 0

(x + \frac{1}{x}) + (x^{2} + \frac{1}{x^{2}}) + . . . . + (x^{52} + \frac{1}{x^{52}})

t a n^{- 1} [\frac{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}]

| x | < \frac{1}{2}, x \neq 0

x^{2} - x + 1 = 0

{(x + \frac{1}{x})}^{2} + {(x^{2} + \frac{1}{x^{2}})}^{2} + {(x^{3} + \frac{1}{x^{3}})}^{2} + {(x^{5} + \frac{1}{x^{5}})}^{2}

x + \frac{1}{x} = 2

x^{2} + \frac{1}{x^{2}}

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