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MHT CET20243 May 2024Evening ShiftMathematicsQuadratic EquationActual

The equation ( cosp -1) x^2+( cosp ) x+ sinp =0 in the variable x , has real roots. Then p can take any value in the interval

Options

  1. A(0,2 )
  2. B(- , 0)
  3. C(- 2 , 2 )
  4. D(0, )

Correct answer

D. (0, )

Step-by-step solution

Given equation is ( p -1) x^2+( p ) x+ p =0 Comparing with a x^2+ bx + c =0 , we get a= p-1, b= p, c= p It has real roots. aligned & b^2-4 a c 0 & ^2 p -4( p -1)( p ) 0 & ^2 p -4 p p +4 p 0 & ^2 p -4 p p +4 ^2 p & +4 p -4 ^2 p 0 & ( p -2 p )^2+4 p (1- p ) 0 aligned ( p -2 p ) is always positive 1- p 0 for all values of p , p (0, )

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