Question

If x= 1/√2+1/2+1/2√2....., then the value of
x+1/x
(1) √2
(3) 3√2
(4) 4√2​

Answer 1

Answer:

2√2

Step-by-step explanation:

$x = \frac{1}{ \sqrt{2} } + \frac{1}{2} + \frac{1}{2 \sqrt{2} } + \frac{1}{4} ......... \infty \\ \\ taking \: \: \: \frac{1}{ \sqrt{2} } \: \: \: as \: common \\ \\ x = \frac{1}{ \sqrt{2} } (1 + \frac{1}{ \sqrt{2} } + \frac{1}{2} + \frac{1}{2 \sqrt{2} } + ......... \infty) \\ \\ putting \: \frac{1}{ \sqrt{2} } + \frac{1}{2} + \frac{1}{2 \sqrt{2} } + ......... \infty \: = x \\ \\ in \: eqation \\ \\ x = \frac{1}{ \sqrt{2} } (1 + x) \\ \\ \sqrt{2} x = 1 + x \\ \\ \sqrt{2} x - x = 1 \\ \\ x( \sqrt{2} - 1) = 1 \\ \\ x = \frac{1}{( \sqrt{2} - 1)} \\ \\ \frac{1}{x} = \sqrt{2} - 1 \: \: \: \: \: - (1)\\ \\ from \: x \: = \: \frac{1}{( \sqrt{2} - 1)} \\ \\ x = \frac{1}{( \sqrt{2} - 1)} \\ \\ x = \frac{1}{( \sqrt{2} - 1)} \times \frac{( \sqrt{2} + 1 )}{( \sqrt{2} + 1) } \\ \\ x = \frac{ \sqrt{2} + 1}{( { \sqrt{2}) }^{2} - {(1)}^{2} } \\ \\ x = \sqrt{2} + 1 \: \: \: \: - (2)$

Using equation (1) and (2)

$x + \frac{1}{x} = \sqrt{2} + 1 + \sqrt{2} - 1 \\ \\ x + \frac{1}{x} = 2 \sqrt{2}$

Hence the answer is 2√2

hope this will help you

(by the way i want to ask which class question is this ?)

Answer 2

Answer : 2√2

Here, x is the sum of terms of an infinite G. P.

We see first term, a = 1/√2

Common ratio = 1/√2

Sum of infinite terms of G. P

$= \frac{ a}{1 - r}$

For this G. P,

$x = \frac{ \frac{1}{ \sqrt{2} } }{1 - \frac{1}{ \sqrt{2} } } \\ \\ x = \frac{1}{ \sqrt{2} } \times \frac{ \sqrt{2} }{ \sqrt{2} - 1} \\ \\ x = \frac{1}{ \sqrt{2} - 1} - - (1)\\ \\ x = \frac{1}{ \sqrt{2} - 1} \times \frac{ \sqrt{2} + 1}{ \sqrt{2} + 1} \\ \\ x = \frac{ \sqrt{2} + 1}{2 - 1} \\ \\ x = \sqrt{2} + 1$

From (1)

$x = \frac{1}{ \sqrt{2} - 1} \\ \\ \frac{1}{x} = \sqrt{2} - 1$

Now,

$= x + \frac{1}{x} \\ \\ = \sqrt{2} + 1 + \sqrt{2} - 1 \\ \\ = 2 \sqrt{2}$

Required Answer is 2√2