Theory of Equations 1 Question 4
4. If $\alpha$ and $\beta$ are the roots of the equation $x^{2}-2 x+2=0$, then the least value of $n$ for which $\frac{\alpha}{\beta}{ }^{n}=1$ is
(a) 2
(b) 5
(c) 4
(d) 3
(2019 Main, 8 April I)
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Answer:
Correct Answer: 4. (c)
Solution:
- Given, $\alpha$ and $\beta$ are the roots of the quadratic equation,
$ \Rightarrow \quad \begin{array}{ll} x^{2}-2 x+2=0 \\ & (x-1)^{2}+1=0 \end{array} $
$ \begin{array}{lc} \Rightarrow & (x-1)^{2}=-1 \\ \Rightarrow & x-1= \pm i \\ \Rightarrow & x=(1+i) \text { or }(1-i) \end{array} $
Clearly, if $\alpha=1+i$, then $\beta=1-i$
According to the question $\frac{\alpha}{\beta}{ }^{n}=1$
$\Rightarrow \quad \frac{1+i^{n}}{1-i}=1$
$\Rightarrow \quad \frac{(1+i)(1+i)}{(1-i)(1+i)}^{n}=1 \quad$ [by rationalization]
$\Rightarrow \quad \frac{1+i^{2}+2 i}{1-i^{2}}{ }^{n}=1 \Rightarrow \frac{2 i}{2}^{n}=1 \Rightarrow i^{n}=1$
So, minimum value of $n$ is 4 .
$\left[\because i^{4}=1\right]$
