Verified Solution • 2025–26 Session moderate

QJEEM21Ma0147: JEE Main 2021 Mathematics - Class 11 JEEM21Ma0147 Solution: If , then p and q are roots of the equation :

If (3+i)100=299(p+iq){\left( {\sqrt 3 + i} \right)^{100}} = {2^{99}}(p + iq) , then p and q are roots of the equation :

Multiple Choice Options

  • A x2−(3−1)x−3=0{x^2} - \left( {\sqrt 3 - 1} \right)x - \sqrt 3 = 0
  • B x2+(3+1)x+3=0{x^2} + \left( {\sqrt 3 + 1} \right)x + \sqrt 3 = 0
  • C x2+(3−1)x−3=0{x^2} + \left( {\sqrt 3 - 1} \right)x - \sqrt 3 = 0
  • D x2−(3+1)x+3=0{x^2} - \left( {\sqrt 3 + 1} \right)x + \sqrt 3 = 0

Step-by-Step Verified Solution

(2eiπ/6)100=299(p+iq){\left( {2{e^{i\pi /6}}} \right)^{100}} = {2^{99}}(p + iq)

2100(cos⁡50π3+isin⁡50π3)=299(p+iq){2^{100}}\left( {\cos {{50\pi } \over 3} + i\sin {{50\pi } \over 3}} \right) = {2^{99}}(p + iq)

p+iq=2(cos⁡2π3+isin⁡2π3)p + iq = 2\left( {\cos {{2\pi } \over 3} + i\sin {{2\pi } \over 3}} \right)

p = −- 1, q = 3\sqrt 3

x2−(3−1)x−3=0{x^2} - (\sqrt 3 - 1)x - \sqrt 3 = 0

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