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NDA 2019
Exam
NDA
Year
2019
Subject
Mathematics
Chapter
Quadratic Equation

NDA 2019 · Mathematics · Quadratic Equation

If A=\ x ∈ ℤ: x^3-1=0\ and B=\ x ∈ ℤ: x^2+x+.1=0\ , where Z is set...

If A={x∈Z:x3−1=0}A=\left\{x \in \mathbb{Z}: x^3-1=0\right\} and B={x∈Z:x2+x+1=0}B=\left\{x \in \mathbb{Z}: x^2+x+\right.1=0\}, where Z is set of complex numbers, then what is A∩B\mathrm{A} \cap \mathrm{B} equal to ?

Options

  1. ANull set
  2. B{−1+3i2,−1−3i2}\left\{\frac{-1+\sqrt{3} i}{2}, \frac{-1-\sqrt{3} i}{2}\right\}
  3. C{−1+3i4,−1−3i4}\left\{\frac{-1+\sqrt{3} i}{4}, \frac{-1-\sqrt{3} i}{4}\right\}
  4. D{1+3i2,1−3i2}\left\{\frac{1+\sqrt{3} i}{2}, \frac{1-\sqrt{3} i}{2}\right\}

Correct answer

B. {−1+3i2,−1−3i2}\left\{\frac{-1+\sqrt{3} i}{2}, \frac{-1-\sqrt{3} i}{2}\right\}

Step-by-step solution

Given, A={x∈z:x3−1=0}B={x∈z:x2+x+1=0}\mathrm{A}=\left\{\mathrm{x} \in \mathrm{z}: \mathrm{x}^3-1=0\right\}B=\left\{x \in z: x^2+x+1=0\right\} The roots of x3−1=0x^3-1=0 are 1,w,w21, w, w^2 The roots of x2+x+1=0x^2+x+1=0 are ω,ω2∴A∩B={1,ω,ω2}∩{ω,ω2}\omega, \omega^2\therefore A \cap B=\left\{1, \omega, \omega^2\right\} \cap\left\{\omega, \omega^2\right\}={ω,ω2}={−1+3i2,−1−3i2}\begin{aligned} & =\left\{\omega, \omega^2\right\} \\ & =\left\{\frac{-1+\sqrt{3} i}{2}, \frac{-1-\sqrt{3} i}{2}\right\}\end{aligned}$$
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