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Mathematics

If a = x + 1x\dfrac{1}{\text{x}} and b = x - 1x\dfrac{1}{\text{x}}, then a2 - b2 is:

  1. x2 + 1x2\dfrac{1}{\text{x}^2}

  2. x2 - 1x2\dfrac{1}{\text{x}^2}

  3. 4

  4. 2(x2+1x2)2\Big(x^2 + \dfrac{1}{\text{x}^2}\Big)

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Answer

Given, a = x + 1x\dfrac{1}{\text{x}} and b = x - 1x\dfrac{1}{\text{x}}

⇒a2−b2=(x+1x)2−(x−1x)2=x2+(1x)2+2×x×1x−[x2+(1x)2−2×x×1x]=x2+1x2+2−[x2+1x2−2]=x2+1x2+2−x2−1x2+2=4\Rightarrow a^2 - b^2 = \Big(x + \dfrac{1}{x}\Big)^2 - \Big(x - \dfrac{1}{x}\Big)^2\\[1em] = x^2 + \Big(\dfrac{1}{x}\Big)^2 + 2 \times x \times \dfrac{1}{x} - \Big[x^2 + \Big(\dfrac{1}{x}\Big)^2 - 2 \times x \times \dfrac{1}{x}\Big]\\[1em] = x^2 + \dfrac{1}{x^2} + 2 - \Big[x^2 + \dfrac{1}{x^2} - 2\Big]\\[1em] = x^2 + \dfrac{1}{x^2} + 2 - x^2 - \dfrac{1}{x^2} + 2\\[1em] = 4

Hence, option 3 is the correct option.

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