Basics & LogarithmsmediumJEE Main 2021Free

Basics & Logarithms — JEE Maths practice question (JEE Main 2021)

JEE Maths question with a full step-by-step solution.

Question
The value of 3+14+13+14+13+…∞3+\dfrac{1}{4+\dfrac{1}{3+\dfrac{1}{4+\dfrac{1}{3+\ldots\infty}}}} is equal to
A1.5+31.5+\sqrt{3}correct
B2+32+\sqrt{3}
C3+233+2\sqrt{3}
D4+34+\sqrt{3}
Solution
Step 1: Let x=3+14+13+14+⋯x=3+\dfrac{1}{4+\dfrac{1}{3+\dfrac{1}{4+\cdots}}}. Step 2: The pattern inside repeats as 3+14+1x3+\dfrac{1}{4+\dfrac{1}{x}}. So:
x=3+14+1x=3+14x+1x=3+x4x+1x=3+\dfrac{1}{4+\dfrac{1}{x}}=3+\dfrac{1}{\dfrac{4x+1}{x}}=3+\dfrac{x}{4x+1}
 Step 3: Simplify:
x−3=x4x+1  ⇒  (x−3)(4x+1)=xx-3=\dfrac{x}{4x+1}\;\Rightarrow\;(x-3)(4x+1)=x
4x2+x−12x−3=x4x^{2}+x-12x-3=x
4x2−12x−3=04x^{2}-12x-3=0
 Step 4: Solve:
x=12±144+488=12±1928=12±838=3±232x=\dfrac{12\pm\sqrt{144+48}}{8}=\dfrac{12\pm\sqrt{192}}{8}=\dfrac{12\pm 8\sqrt{3}}{8}=\dfrac{3\pm 2\sqrt{3}}{2}
 Step 5: Since x=3+…>3x=3+\ldots>3, take the positive root:
x=3+232=1.5+3x=\dfrac{3+2\sqrt{3}}{2}=1.5+\sqrt{3}
 Answer: (1) 1.5+31.5+\sqrt{3}
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