Verified Solution • 2025–26 Session moderate

QJEEM20Ma6707: JEE Main 2020 Mathematics - Class 11 JEEM20Ma6707 Solution: Let . If and , then a and b are the roots of the quadratic…

Let α=−1+i32\alpha = {{ - 1 + i\sqrt 3 } \over 2} .
If a=(1+α)∑k=0100α2ka = \left( {1 + \alpha } \right)\sum\limits_{k = 0}^{100} {{\alpha ^{2k}}} and
b=∑k=0100α3kb = \sum\limits_{k = 0}^{100} {{\alpha ^{3k}}} , then a and b are the roots of the quadratic equation :

Multiple Choice Options

  • A x2 + 101x + 100 = 0
  • B x2 + 102x + 101 = 0
  • C x2 – 102x + 101 = 0
  • D x2 – 101x + 100 = 0

Step-by-Step Verified Solution

α=−1+i32\alpha = {{ - 1 + i\sqrt 3 } \over 2} = ω\omega

a=(1+α)∑k=0100α2ka = \left( {1 + \alpha } \right)\sum\limits_{k = 0}^{100} {{\alpha ^{2k}}}

= (1+ω)∑k=0100ω2k\left( {1 + \omega } \right)\sum\limits_{k = 0}^{100} {{\omega ^{2k}}}

= (1+ω)1(1−(ω2)101)1−ω2\left( {1 + \omega } \right){{1\left( {1 - {{\left( {{\omega ^2}} \right)}^{101}}} \right)} \over {1 - {\omega ^2}}}

= 1−ω2021−ω{{1 - {\omega ^{202}}} \over {1 - \omega }}

= 1−ω1−ω{{1 - \omega } \over {1 - \omega }} = 1

b=∑k=0100α3kb = \sum\limits_{k = 0}^{100} {{\alpha ^{3k}}}

= ∑k=0100ω3k\sum\limits_{k = 0}^{100} {{\omega ^{3k}}}

= 101 [as ω3{\omega ^3} = 1]

∴\therefore roots are 101 and 1

Then equation is = x2 – 102x + 101 = 0

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