MHT CET20243 May 2024Evening ShiftMathematicsQuadratic EquationActual
The equation ( cosp -1) x^2+( cosp ) x+ p =0 in the variable x , has real roots. Then p can take any value in the interval
Options
- A(0,2 )
- B(- , 0)
- C(- 2 , 2 )
- D(0, )
Correct answer
D. (0, )
Step-by-step solution
Given equation is ( p -1) x^2+( p ) x+ p =0 Comparing with ax ^2+ bx + c =0 , we get a= p-1, b= p, c= p It has real roots. b^2-4 a c 0 aligned & ^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 aligned array ll & ( p-2 p)^2+4 p(1- p) 0 & ( p-2 p) is always positive & 1- p 0 for all values of p, p (0, ) array