JEE PYQ: Quadratic Equation Question 33
Question 33 - 2019 (08 Apr Shift 1)
If $\alpha$ and $\beta$ be the roots of the equation $x^2 - 2x + 2 = 0$, then the least value of $n$ for which $\left(\frac{\alpha}{\beta}\right)^n = 1$ is:
(1) 2
(2) 5
(3) 4
(4) 3
Show Answer
Answer: (3)
Solution
The given quadratic equation is $x^2 - 2x + 2 = 0$.
Then, the roots of this equation are $\frac{2 \pm \sqrt{-4}}{2} = 1 \pm i$
Now, $\frac{\alpha}{\beta} = \frac{1-i}{1+i} = \frac{(1-i)^2}{1-i^2} = i$ (or $-i$)
So, $\frac{\alpha}{\beta} = \pm i$
Now, $\left(\frac{\alpha}{\beta}\right)^n = 1 \Rightarrow (\pm i)^n = 1$
$\Rightarrow n$ must be a multiple of 4.
Hence, the required least value of $n = 4$.
BETA
