MHT CET202613 April 2026Evening ShiftMathematicsApplication of DerivativesActual
If Rolle's theorem is applicable for the function f(x) = ( x^2+a x ) on [3, 4] with c (3, 4) such that f'(c) = 0 , then the value of f''(c) is
Options
- A1 48
- B1 36
- C1 24
- D1 12
Correct answer
D. 1 12
Step-by-step solution
Since Rolle's theorem is applicable for f(x) on [3, 4] , we have f(3) = f(4) . ( 9+a 3 ) = ( 16+a 4 ) 9+a 3 = 16+a 4 36 + 4a = 48 + 3a a = 12 The function becomes f(x) = (x^2+12) - x . Differentiating with respect to x : f'(x) = 2x x^2+12 - 1 x For f'(c) = 0 : 2c c^2+12 - 1 c = 0 2c^2 = c^2 + 12 c^2 = 12 Differentiating f'(x) again to find f''(x) : f''(x) = 2(x^2+12) - 2x(2x) (x^2+12)^2 + 1 x^2 = 24-2x^2 (x^2+12)^2 + 1 x^2 Substituting c^2 = 12 : f''(c) = 24 - 2(12) (12+12)^2 + 1 12 = 0 + 1 12 = 1 12 Answer: 1 12