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Let π6<θ<π12. Suppose α1 and β1 are the roots of the equation x22xsecθ+1=0 and α2 and β2 are the roots of the equation x2+2xtanθ1=0. If α1>β1 and α2>β2, then α1+β2 equals:
  1. 2secθ
  2. 2tanθ
  3. 2(secθtanθ)
  4. 0


Solution

The correct option is B 2tanθ
For x22xsecθ+1=0
x=2secθ ±4sec2θ42=secθ±tanθ

For x22xtanθ1=0
x=2tanθ ±4tan2θ+42=tanθ±secθ

π6<θ<π12 and α1>β2

α1=secθtanθ       and 
β2=tanθsecθ

α1+β2=2tanθ

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