Let p, q ∈ R and (1-√(3)i )^200 = 2^199(p+iq), i = √(-1). Then p + q + q^2 and p - q + q^2 are roots of the equation.

Let p, q ∈ R and (1-√(3)i )^200 = 2^199(p+iq), i = √(-1).… — JEE Main 2023 Mathematics PYQ

JEE Main 2023 · 24 Jan · Shift 1 · Paper-1(B.E./B. Tech.)

Let p, q

\in \mathbb{R}

and

\left(1-\sqrt{3}i\right)^{200} = 2^{199}(p+iq)

i = \sqrt{-1}

. Then

p + q + q^2

and

p - q + q^2

are roots of the equation.

A.
$x^2 - 4x - 1 = 0$
B.
$x^2 + 4x - 1 = 0$
C.
$x^2 - 4x + 1 = 0$
D.
$x^2 + 4x + 1 = 0$

Show answer and solution

Correct option: C

Solution

Convert 1 - √3·i to polar: modulus = 2, argument = -π/3.

So (1 - √3·i)²⁰⁰ = 2²⁰⁰ · e^(-i·200π/3).

200π/3 = 66π + 2π/3, so e^(-i·200π/3) = e^(-i·2π/3) = cos(2π/3) - i·sin(2π/3) = -1/2 - i·(√3/2).

Thus (1 - √3·i)²⁰⁰ = 2²⁰⁰·(-1/2 - i·√3/2) = 2¹⁹⁹·(-1 - i√3).

Setting equal to 2¹⁹⁹(p + iq): p = -1, q = -√3.

Now compute the two expressions:
- p + q + q² = -1 + (-√3) + 3 = 2 - √3
- p - q + q² = -1 - (-√3) + 3 = 2 + √3

These two roots have sum = 4 and product = (2-√3)(2+√3) = 4 - 3 = 1.

The equation with sum 4 and product 1 is x² - 4x + 1 = 0.

[JEE Main 2023 · 24 Jan · Shift 1 · Paper-1(B.E./B. Tech.)]

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