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JEE Main 202117 MarchMorning Shift
Exam
JEE Main
Year
2021
Date
17 March
Shift
Morning Shift
Subject
Mathematics
Chapter
Quadratic Equation

JEE Main 2021 · 17 March · Morning Shift · Mathematics · Quadratic Equation

The value of 4 + (1)/(5 + (1)/(4 + (1)/(5 + (1)/(4 + ∞)))) is:

The value of 4+15+14+15+14+……∞4 + \dfrac{1}{5 + \dfrac{1}{4 + \dfrac{1}{5 + \dfrac{1}{4 + \ldots \ldots \infty}}}} is:

Options

  1. A2+25302 + \dfrac{2}{5}\sqrt{30}
  2. B2+45302 + \dfrac{4}{\sqrt{5}}\sqrt{30}
  3. C4+45304 + \dfrac{4}{\sqrt{5}}\sqrt{30}
  4. D5+25305 + \dfrac{2}{5}\sqrt{30}

Correct answer

A. 2+25302 + \dfrac{2}{5}\sqrt{30}

Step-by-step solution

Given y=4+15+14+15+14+……∞⏟yy = 4 + \dfrac{1}{5 + \mathop{\mathop{\dfrac{1}{4 + \dfrac{1}{5 + \dfrac{1}{4 + \ldots \ldots \infty}}}}_{⏟}}_{y}}

y=4+1(5+1y)y = 4 + \dfrac{1}{\left(5 + \dfrac{1}{y}\right)}

⇒y−4=y(5y+1)⇒ y - 4 = \dfrac{y}{( 5y + 1 )}

⇒5y2−20y−4=0⇒ 5{y}^{2} - 20y - 4 = 0

⇒y=20±(−20)2−4⋅5⋅(−4)2⋅5⇒ y = \dfrac{20 \pm \sqrt{{\left(- 20\right)}^{2} - 4 \cdot 5 \cdot \left(- 4\right)}}{2 \cdot 5}

⇒y=20±48010⇒ y = \dfrac{20 \pm \sqrt{480}}{10}

∴y=2−480100∴ y = 2 - \sqrt{\dfrac{480}{100}}→\torejected

So, y=2+480100=2+2530y = 2 + \sqrt{\dfrac{480}{100}} = 2 + \dfrac{2}{5}\sqrt{30}

Correct with Option (A).

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