If $y + \dfrac{1}{y} = 2$, then $y^5 + \dfrac{1}{y^5}$ =

1. 5

2. 3

3. 2

4. 1

Given,

$y + \dfrac{1}{y} = 2$

$\Rightarrow \dfrac{y^2 + 1}{y} = 2 \\[1em] \Rightarrow y^2 + 1 = 2y \\[1em] \Rightarrow y^2 - 2y + 1 = 0 \\[1em] \Rightarrow (y - 1)^2 = 0 \\[1em] \Rightarrow y-1 = 0 \\[1em] \Rightarrow y = 1$

Substituting value of y, we get :

$\Rightarrow y^5 + \dfrac{1}{y^5} = (1)^5 + \dfrac{1}{(1)^5} = 1 + 1 = 2$.

Hence, option 3 is correct option.