Let α and β be two real roots of the equation k + 1 t a n 2 x - 2 ⋅ λ tan ⁡ x = 1…

Let α and β be two real roots of the equation k+1tan2x-2λtanx=1-k, where k-1 and λ are real numbers. If tan2α+β=50, then a value of λ is
  1. 102
  2. 10
  3. 5
  4. 52

Solution

k+1tan2x-2λtanx+k-1=0
tanα+tanβ=2λk+1
tanαtanβ=k-1k+1
tanα+β=2λk+11-k-1k+1=2λ2=λ2
tan2α+β=λ22=50
λ=±10

Asked in: JEE Main 2020 (07 Jan Shift 1)

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