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Mathematics

Assertion (A): x+1x=4 and 1x=2+3, then x =2−3x + \dfrac{1}{x} = 4 \text{ and } \dfrac{1}{x} = 2 + \sqrt{3}, \text{ then x } = 2 - \sqrt{3}

Reason (R):
x+1x=4⇒x+2+3=4⇒x=2−3x + \dfrac{1}{x} = 4 \\[1em] \Rightarrow x + 2 + \sqrt{3} = 4 \\[1em] \Rightarrow x = 2 - \sqrt{3}

  1. A is true, but R is false.

  2. A is false, but R is true.

  3. Both A and R are true, and R is the correct reason for A.

  4. Both A and R are true, and R is the incorrect reason for A.

Rational Irrational Nos

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Answer

Given, 1x=2+3 and x+1x=4\dfrac{1}{x} = 2 + \sqrt{3}\text{ and } x + \dfrac{1}{x} = 4

And,

⇒x+1x=4=x+2+3=4=x=4−(2+3)=x=4−2−3=x=2−3.\Rightarrow x + \dfrac{1}{x} = 4\\[1em] = x + 2 + \sqrt{3} = 4\\[1em] = x = 4 - (2 + \sqrt{3})\\[1em] = x = 4 - 2 - \sqrt{3}\\[1em] = x = 2 - \sqrt{3}.

So, assertion (A) is true.

If, 1x=2+3\dfrac{1}{x} = 2 + \sqrt{3}

⇒x=12+3=12+3×2−32−3 (On rationalizing)=2−3(2)2−(3)2=2−34−3=2−31=2−3.\Rightarrow x = \dfrac{1}{2 + \sqrt{3}}\\[1em] = \dfrac{1}{2 + \sqrt{3}} \times \dfrac{2 - \sqrt{3}}{2 - \sqrt{3}} \text{ (On rationalizing)} \\[1em] = \dfrac{2 - \sqrt{3}}{(2)^2 - (\sqrt{3})^2}\\[1em] = \dfrac{2 - \sqrt{3}}{4 - 3}\\[1em] = \dfrac{2 - \sqrt{3}}{1}\\[1em] = 2 - \sqrt{3}.

So, reason (R) is true.

∴ Both A and R are true, and R is the correct reason for A.

Hence, option 3 is correct option.

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