JEE Main20238 Apr 2023Evening ShiftMathematicsContinuity and DifferentiabilityActual
Let k and m be positive real numbers such that the function f x = 3 x 2 + k x + 1 , 0 < x < 1 m x 2 + k 2 , x ≥ 1 is differentiable for all x > 0 . Then 8 f ' ( 8 ) f ' 1 8 is equal to
Correct answer
0
Step-by-step solution
Since, f x = 3 x 2 + k x + 1 ;   0 < x < 1 m x 2 + k 2 ;       x ≥ 1 is differentiable at x = 1 , so function must be continuous at x = 1 , hence LHL = RHL 3 + k 2 = m + k 2 ⇒ k 2 - k 2 + m - 3 = 0       . . . . 1 And, f ' x = 6 x + k 2 x + 1 ;   0 < x < 1 2 m x ;       x > 1 So, f ' 1 - = f ' 1 + ⇒ 6 + k 2 2 = 2 m ⇒ m = 3 + k 4 2       . . . 2 Putting in 1 , we get k 2 - k 2 + 3 + k 4 2 - 3