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Mathematics

Assertion (A): If x ≥ 5 and x+1xx +\dfrac{1}{x} = m, then x−1xx - \dfrac{1}{x} = m - 2.

Reason (R): (x−1x)2=(x+1x)2−4(x -\dfrac{1}{x})^2 = (x +\dfrac{1}{x})^2 - 4

⇒ x−1x=(x+1x)2−4x - \dfrac{1}{x} = \sqrt{(x+\dfrac{1}{x})^2-4}

= m2−4=m−2\sqrt{m^2-4}= m - 2

  1. A is true, R is false.
  2. A is false, R is true.
  3. Both A and R are true.
  4. Both A and R are false.

Expansions

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Answer

Both A and R are false.

Explanation

Given,

x+1x=mx +\dfrac{1}{x} = m

Squaring both sides, we get,

⇒(x+1x)2=m2⇒x2+(1x)2+2×x×1x=m2⇒x2+(1x)2+2=m2⇒x2+(1x)2=m2−2⇒ \Big(x +\dfrac{1}{x}\Big)^2 = m^2\\[1em] ⇒ x^2 + \Big(\dfrac{1}{x}\Big)^2 + 2 \times x \times \dfrac{1}{x} = m^2\\[1em] ⇒ x^2 + \Big(\dfrac{1}{x}\Big)^2 + 2 = m^2\\[1em] ⇒ x^2 + \Big(\dfrac{1}{x}\Big)^2 = m^2 - 2\\[1em]

Subtracting 2 from both sides, we get,

⇒x2+(1x)2−2=m2−2−2⇒x2+(1x)2−2×x×1x=m2−4⇒(x−1x)2=m2−4⇒x−1x=m2−4⇒ x^2 + \Big(\dfrac{1}{x}\Big)^2 - 2 = m^2 - 2 - 2\\[1em] ⇒ x^2 + \Big(\dfrac{1}{x}\Big)^2 - 2 \times x \times \dfrac{1}{x} = m^2 - 4\\[1em] ⇒ \Big(x - \dfrac{1}{x}\Big)^2 = m^2 - 4\\[1em] ⇒ x - \dfrac{1}{x} = \sqrt{m^2 - 4}

∴ Assertion (A) is false.

(x−1x)2=(x+1x)2−4⇒x−1x=(x+1x)2−4⇒m2−4=(m2−4)\Big(x - \dfrac{1}{x}\Big)^2 = \Big(x +\dfrac{1}{x}\Big)^2 - 4\\[1em] ⇒ x - \dfrac{1}{x} = \sqrt{\Big(x +\dfrac{1}{x}\Big)^2 - 4}\\[1em] ⇒ \sqrt{m^2 - 4} = \sqrt{\Big(m^2 - 4)}\\[1em]

∴ Reason (R) is false.

Hence, both Assertion (A) and Reason (R) are false.

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