CAT 2004 | Question: 50

Question

Let $y=\dfrac{1}{2+ \dfrac{1}{3+ \dfrac{1}{2+ \dfrac{1}{3+ \dots } } } }$ what is the value of $y?$

• A. $\frac{\sqrt{13} +3} {2}$
• B. $\frac{\sqrt{13} -3} {2}$
• C. $\frac{\sqrt{15} +3} {2}$
• D. $\frac{\sqrt{15} -3} {2}$

Answer 1

$y=\frac{1}{2+ \frac{1}{3+ \frac{1}{2+ \frac{1}{3+ \dots } } } }$

$y=\frac{1}{2+ \frac{1}{3+y} }$

$y=\frac{1}{ \frac{2y+7}{3+y} }$

$2y^{2}$ + 7y = 3 + y

$2y^{2}$ + 6y – 3 = 0

=$\frac{-6± \sqrt{36-4(-3) 2}} {4}$

=$ \frac {±\sqrt{15}-3} {2}$

y=$\frac {\sqrt{15}-3} {2}$

 

Hence, Option(4) $\frac {\sqrt{15}-3} {2}$ is the correct choice.