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Quadratic Equation and Inequalities question

2020 · 8 Jan · Shift 2 · Q22
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  5. /2020 · 8 Jan · Shift 2 · Q22

Quadratic Equation and Inequalities question

2020 · 8 Jan · Shift 2 · Q22

JEE MainMathematicsQuadratic Equation and InequalitiesMCQ+4 / −1
Let α=−1+i32\alpha = {{ - 1 + i\sqrt 3 } \over 2}α=2−1+i3​​. If a=(1+α)∑k=0100α2ka = \left( {1 + \alpha } \right)\sum\limits_{k = 0}^{100} {{\alpha ^{2k}}}a=(1+α)k=0∑100​α2k and b=∑k=0100α3kb = \sum\limits_{k = 0}^{100} {{\alpha ^{3k}}}b=k=0∑100​α3k, then a and b are the roots of the quadratic equation :
  1. A
    x2 + 101x + 100 = 0
  2. B
    x2 + 102x + 101 = 0
  3. C
    x2 – 102x + 101 = 0
  4. D
    x2 – 101x + 100 = 0
View written solutionFree

Correct answer: C

  1. Identify α\alphaα

Given

α=−1+i32\alpha = \frac{-1+i\sqrt{3}}{2}α=2−1+i3​​

This is a cube root of unity:

α=ω,ω3=1,ω≠1\alpha = \omega, \quad \omega^3=1, \quad \omega\ne 1α=ω,ω3=1,ω=1

Also,

1+α+α2=01+\alpha+\alpha^2=01+α+α2=0

So,

1+α=−α21+\alpha=-\alpha^21+α=−α2
  1. Compute bbb
b=∑k=0100α3kb=\sum_{k=0}^{100}\alpha^{3k}b=k=0∑100​α3k

Since α3=1\alpha^3=1α3=1,

α3k=1∀k\alpha^{3k}=1 \quad \forall kα3k=1∀k

Hence,

b=∑k=01001=101b=\sum_{k=0}^{100}1=101b=k=0∑100​1=101
  1. Compute aaa

Given

a=(1+α)∑k=0100α2ka=(1+\alpha)\sum_{k=0}^{100}\alpha^{2k}a=(1+α)k=0∑100​α2k

Using 1+α=−α21+\alpha=-\alpha^21+α=−α2,

a=−α2∑k=0100α2ka=-\alpha^2\sum_{k=0}^{100}\alpha^{2k}a=−α2k=0∑100​α2k

Now simplify the sum:

∑k=0100α2k=1+α2+α4+⋯+α200\sum_{k=0}^{100}\alpha^{2k}=1+\alpha^2+\alpha^4+\cdots+\alpha^{200}k=0∑100​α2k=1+α2+α4+⋯+α200

Since α3=1\alpha^3=1α3=1, powers repeat modulo 333. Also α2=ω2\alpha^2=\omega^2α2=ω2, and

α2k=(α2)k\alpha^{2k}=(\alpha^2)^kα2k=(α2)k

This is a geometric series with ratio α2\alpha^2α2:

∑k=0100(α2)k=1−(α2)1011−α2\sum_{k=0}^{100}(\alpha^2)^k=\frac{1-(\alpha^2)^{101}}{1-\alpha^2}k=0∑100​(α2)k=1−α21−(α2)101​

But an easier way is to use periodicity. Since

α2k=1,α2,α,1,α2,α,…\alpha^{2k} = 1,\alpha^2,\alpha,1,\alpha^2,\alpha,\dotsα2k=1,α2,α,1,α2,α,…

The cycle length is 333, and

1+α+α2=01+\alpha+\alpha^2=01+α+α2=0

Now 101=99+2101=99+2101=99+2, so there are 333333 full cycles plus 222 extra terms:

∑k=0100α2k=33(1+α+α2)+(1+α2)=1+α2\sum_{k=0}^{100}\alpha^{2k}=33(1+\alpha+\alpha^2)+(1+\alpha^2)=1+\alpha^2k=0∑100​α2k=33(1+α+α2)+(1+α2)=1+α2

Therefore,

a=(1+α)(1+α2)a=(1+\alpha)(1+\alpha^2)a=(1+α)(1+α2)

Now expand:

(1+α)(1+α2)=1+α+α2+α3(1+\alpha)(1+\alpha^2)=1+\alpha+\alpha^2+\alpha^3(1+α)(1+α2)=1+α+α2+α3

Using 1+α+α2=01+\alpha+\alpha^2=01+α+α2=0 and α3=1\alpha^3=1α3=1,

a=0+1=1a=0+1=1a=0+1=1

So,

a=1,b=101a=1, \quad b=101a=1,b=101
  1. Form the quadratic equation whose roots are aaa and bbb

If roots are 111 and 101101101, then

  • Sum of roots:
1+101=1021+101=1021+101=102
  • Product of roots:
1⋅101=1011\cdot 101=1011⋅101=101

Hence the quadratic equation is

x2−(102)x+101=0x^2-(102)x+101=0x2−(102)x+101=0
  1. Match with options

This is

x2−102x+101=0\boxed{x^2-102x+101=0}x2−102x+101=0​

which is Option C.

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