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Mathematics

Statement 1: x > 0 and x2+1x2x^2 + \dfrac{1}{x^2} = 2, then x2−1x2x^2 - \dfrac{1}{x^2} = 0

Statement 2: (x+1x)2=x2+1x2+2(x + \dfrac{1}{x})^2 = x^2 + \dfrac{1}{x^2} + 2 = 2 + 2 = 4

(x−1x)2=x2+1x2−2(x - \dfrac{1}{x})^2 = x^2 + \dfrac{1}{x^2} - 2 = 2 - 2 = 0

and, (x2−1x2)=(x+1x)(x−1x)(x^2 - \dfrac{1}{x^2}) = (x + \dfrac{1}{x})(x - \dfrac{1}{x})

  1. Both the statements are true.

  2. Both the statements are false.

  3. Statement 1 is true, and statement 2 is false.

  4. Statement 1 is false, and statement 2 is true.

Expansions

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Answer

Given, x2 + 1x2\dfrac{1}{\text{x}^2} = 2

As we know,

⇒(x−1x)2=x2+1x2−2×x×1x⇒(x−1x)2=x2+1x2−2⇒(x−1x)2=2−2⇒(x−1x)2=0⇒x−1x=0⇒x−1x=0⇒(x+1x)2=x2+1x2+2×x×1x⇒(x+1x)2=x2+1x2+2⇒(x+1x)2=2+2⇒(x+1x)2=4⇒x+1x=4⇒x+1x=±2\Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} - 2 \times x \times \dfrac{1}{x}\\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} - 2\\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = 2 - 2\\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = 0\\[1em] \Rightarrow x - \dfrac{1}{x} = \sqrt{0}\\[1em] \Rightarrow x - \dfrac{1}{x} = 0\\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} + 2 \times x \times \dfrac{1}{x}\\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} + 2\\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = 2 + 2\\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = 4\\[1em] \Rightarrow x + \dfrac{1}{x} = \sqrt{4}\\[1em] \Rightarrow x + \dfrac{1}{x} = \pm 2\\[1em]

Now, we know that (x2−1x2)=(x+1x)(x−1x)(x^2 - \dfrac{1}{x^2}) = (x + \dfrac{1}{x})(x - \dfrac{1}{x})

= ±\pm 2 x 0

= 0.

∴ Both the statements are true.

Hence, option 1 is the correct option.

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