Q46 / 90 · JEE Main 2024 (27 Jan Shift 2) · Mathematics

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

  • A
    (-2,1)
  • Correct answer
    B
    (-3,0)
  • C
    -32,32
  • D
    (0,3)

Solution

Given: 132α+32113α+132α+33α+10=0

Applying, R1→R1-R2

⇒07676113α+132α+33α+10=0

Applying, C2→C2-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

Hence, the correct answer is option: 'B'.

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