JEE PYQ: Quadratic Equation Question 4
Question 4 - 2021 (18 Mar Shift 1)
The value of $3 + \dfrac{1}{4 + \dfrac{1}{3 + \dfrac{1}{4 + \cdots}}}$ is equal to
(1) $1.5 + \sqrt{3}$
(2) $2 + \sqrt{3}$
(3) $3 + 2\sqrt{3}$
(4) $4 + \sqrt{3}$
Show Answer
Answer: (1)
Solution
Let $x = 3 + \dfrac{1}{4 + \frac{1}{x}}$
So, $x = 3 + \frac{1}{4 + \frac{1}{x}} = 3 + \frac{x}{4x+1}$
$\Rightarrow (x - 3) = \frac{x}{(4x+1)}$
$\Rightarrow (4x + 1)(x - 3) = x$
$\Rightarrow 4x^2 - 12x + x - 3 = x$
$\Rightarrow 4x^2 - 12x - 3 = 0$
$x = \frac{12 \pm \sqrt{144 + 48}}{8} = \frac{12 \pm 4\sqrt{12}}{8} = \frac{3 \pm 2\sqrt{3}}{2}$
$x = \frac{3}{2} \pm \sqrt{3} = 1.5 \pm \sqrt{3}$
But only positive value is accepted.
So, $x = 1.5 + \sqrt{3}$
BETA
