JEE MAIN - Mathematics (2019 - 8th April Morning Slot - No. 15)
If $$\alpha $$ and $$\beta $$ be the roots of the equation
x2 – 2x + 2 = 0, then the least value of n for which $${\left( {{\alpha \over \beta }} \right)^n} = 1$$ is :
2
5
4
3
Explanation
x2 – 2x + 2 = 0
$$ \therefore $$ x = $${{2 \pm \sqrt { - 4} } \over 2} = 1 \pm i$$
Now, $${\alpha \over \beta } = {{1 + i} \over {1 - i}} = {{{{\left( {1 + i} \right)}^2}} \over {1 - {i^2}}} = i$$
or $${\alpha \over \beta } = {{1 - i} \over {1 + i}} = {{{{\left( {1 - i} \right)}^2}} \over {1 + {i^2}}} = - i$$
$$ \therefore $$ $${\left( { i} \right)^n}$$ = 1
and $${\left( { - i} \right)^n}$$ = 1
So n must be a multiple of 4
$$ \therefore $$ Minimum value of n = 4
$$ \therefore $$ x = $${{2 \pm \sqrt { - 4} } \over 2} = 1 \pm i$$
Now, $${\alpha \over \beta } = {{1 + i} \over {1 - i}} = {{{{\left( {1 + i} \right)}^2}} \over {1 - {i^2}}} = i$$
or $${\alpha \over \beta } = {{1 - i} \over {1 + i}} = {{{{\left( {1 - i} \right)}^2}} \over {1 + {i^2}}} = - i$$
$$ \therefore $$ $${\left( { i} \right)^n}$$ = 1
and $${\left( { - i} \right)^n}$$ = 1
So n must be a multiple of 4
$$ \therefore $$ Minimum value of n = 4
Comments (0)
