JEE Main202127 Aug 2021Morning ShiftMathematicsLimitsActual
If α , β are the distinct roots of x 2 + b x + c = 0 , then lim x → β e 2 x 2 + b x + c - 1 - 2 x 2 + b x + c ( x - β ) 2 is equal to
Options
- A2 b 2 + 4 c
- Bb 2 - 4 c
- C2 b 2 - 4 c
- Db 2 + 4 c
Correct answer
C. 2 b 2 - 4 c
Step-by-step solution
lim x → β e 2 x 2 + b x + c - 1 - 2 x 2 + b x + c x - β 2 lim x → β 1 1 + 2 x 2 + b x + c 1 ! + 2 2 x 2 + b x + c 2 2 ! + . . . - 1 - 2 x 2 + b x + c x - β 2 lim x → β 2 x 2 + b x + c 2 x - β 2 lim x → β 2 x - α 2 x - β 2 x - β 2 2 β - α 2 = 2 b 2 - 4 c