The values of α , for which 1 3 2 α + 3 2 1 1 3 α + 1 3 2 α + 3 3 α + 1 0 = 0 , lie in the interval

The values of α, for which 132α+32113α+132α+33α+10=0, lie in the interval

  1. (-2,1)
  2. (-3,0)
  3. -32,32
  4. (0,3)

Solution

Given: 132α+32113α+132α+33α+10=0

Applying, R1R1-R2

07676113α+132α+33α+10=0

Applying, C2C2-C3

00761-αα+132α+33α+10=0

0-0+763α+1+2α2+3α=0

2α2+6α+1=0

α=-6±36-4212×2

α=-6±274

α=-3±72

α-3±2.62

α-0.42, -5.62

α-0.2, -2.8

α-3,0

Asked in: JEE Main 2024 (27 Jan Shift 2)

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