If a and b are real numbers such that 2+α^4=a+bα, where α=(-1+i√3)/2, then a+b is equal to:

Correct answer: (A) 9. 2+α^4=a+bα 4+α2+4α^2=a+bα Now using, α=(-1+i√3)/2⇒2α+1^2=i3^2 ⇒α^2=-1+α So, a+bα=4-α-1+4α^2 a+bα=3+3α^2=91+2α+α^2 a+bα=91+2α+-1-α=9α ⇒a=0, b=9⇒a+b=9

Q40 / 75 · JEE Main 2020 (04 Sep Shift 2) · Mathematics

If $a$ and $b$ are real numbers such that ${(2 + α)}^{4} = a + b α$ , where $α = \frac{- 1 + i \sqrt{3}}{2}$ , then $a + b$ is equal to:

Correct answer
A
$9$
B
$24$
C
$33$
D
$57$

Solution

$(2 + α)^{4} = a + b α$
${(4 + α^{2} + 4 α)}^{2} = a + b α$

Now using, $α = \frac{- 1 + i \sqrt{3}}{2} \Rightarrow {(2 α + 1)}^{2} = {(i \sqrt{3})}^{2}$

$\Rightarrow α^{2} = - (1 + α)$

So, $a + b α = {(4 - α - 1 + 4 α)}^{2}$

$a + b α = {(3 + 3 α)}^{2} = 9 (1 + 2 α + α^{2})$
$a + b α = 9 (1 + 2 α + (- 1 - α)) = 9 α$

$\Rightarrow a = 0, b = 9 \Rightarrow a + b = 9$

Hence, the correct answer is option: 'A'.

If a and b are real numbers such that 2+α4=a+bα, where α=-1+i32, then a+b is equal to:

Solution

Relevant Questions

If $a$ and $b$ are real numbers such that ${(2 + α)}^{4} = a + b α$ , where $α = \frac{- 1 + i \sqrt{3}}{2}$ , then $a + b$ is equal to: