Question

Namaste! Hello! Annyeong!

$\sqrt{\frac{1}{3}\sqrt{\frac{1}{3} \sqrt{\frac{1}{3}... \infty}}} \: \: \: \:$

a) 1/2
b) 2/3
c) 1/3
d) 3/4

Full solution needed!!!! ​

Answer 1

Answer -

$\sf{\green{Option\:c)}\:\dfrac{1}{3}}$

$\rule{200}2$

Explanation -

Let $\implies\:\sf\sqrt{\dfrac{1}{3}\sqrt{\dfrac{1}{3} \sqrt{\dfrac{1}{3}... \infty}}}\:\:\: \: \: \: =\:x$

$\implies\: \sf\sqrt{\dfrac{1}{3}\sqrt{\dfrac{1}{3} \sqrt{\dfrac{1}{3}... \infty}}} \: \: \: =\:x$

$\implies \sf\sqrt{\dfrac{1}{3}\bigg(\sqrt{\dfrac{1}{3} \sqrt{\dfrac{1}{3}... \infty}}}\bigg) \: \: \: \: =\:x$

$\implies\:\sf\sqrt{ \dfrac{1}{3} \: x \: } \: = \: x$

Now, do squaring on both sides

$\implies\:\sf \bigg(\sqrt{ \dfrac{1}{3} \: x \: } \bigg)^{2} \: = \: x ^{2}$

$\implies\:\sf{\dfrac{x}{3}\:=\:x^2}$

$\implies\:\sf{\dfrac{1}{3}\:=\:\dfrac{x^2}{x}}$

$\implies\:\sf{\dfrac{1}{3}\:=\:x}$

$\implies\:\sf{\bold{x\:=\:\dfrac{1}{3}}}$

•°• $\sf \sqrt{\dfrac{1}{3}\sqrt{\dfrac{1}{3} \sqrt{\dfrac{1}{3}... \infty}}} \: \: \: \: =\:\dfrac{1}{3}$

Answer 2

Answer:

$\large \bold\red {(c)x = \frac{1}{3} }$

Step-by-step explanation:

We have been given that,

$\sqrt{\frac{1}{3}\sqrt{\frac{1}{3} \sqrt{\frac{1}{3}... \infty}}}$

Now,

To find its value,

Let's assume that,

$\sqrt{\frac{1}{3}\sqrt{\frac{1}{3} \sqrt{\frac{1}{3}... \infty}}} \: \: \: \: = x$

Clearly,

The series is defined up to Infinity,

Therefore,

We can take this as,

$= > \sqrt{\frac{1}{3}(\sqrt{\frac{1}{3} \sqrt{\frac{1}{3}... \infty}})} \: \: \: \: = x \\ \\ = > \sqrt{ \frac{1}{3} x} = x$

Now,

Squaring both the sides,

We get,

$= > {( \sqrt{ \frac{x}{3} }) }^{2} = {x}^{2} \\ \\ = > \frac{x}{3} = {x}^{2} \\ \\ = > 3 {x}^{2} = x \\ \\ = > 3 {x}^{2} - x = 0 \\ \\ = > x(3x - 1) = 0$

Therefore,

We have,

$= > x = 0$

And,

$= > 3x - 1 = 0 \\ = > 3x = 1 \\ = > x = \frac{1}{3}$

Thus,

We have,

$\large \bold{x = 0} \: \: and \: \: \large \bold{x = \frac{1}{3} }$

But,

In the options we have only,

$\large \bold {(c)x = \frac{1}{3} }$

Hence,

(c) is the correct option.