MHT CET202618 April 2026Evening ShiftMathematicsApplication of DerivativesActual
Let g(x) = f(x) + f(1-x) and f''(x) < 0, 0 x 1 , then
Options
- Ag(x) increases on [ 1 2 , 1 ] and g(x) decreases on [0, 1 2 ]
- Bg(x) decreases on [0, 1]
- Cg(x) increases on [0, 1]
- Dg(x) decreases on [ 1 2 , 1 ] and g(x) increases on [0, 1 2 ]
Correct answer
D. g(x) decreases on [ 1 2 , 1 ] and g(x) increases on [0, 1 2 ]
Step-by-step solution
Given g(x) = f(x) + f(1-x) Differentiating with respect to x , we get: g'(x) = f'(x) - f'(1-x) It is given that f''(x) For g(x) to be increasing, g'(x) > 0 : f'(x) - f'(1-x) > 0 f'(x) > f'(1-x) Since f'(x) is strictly decreasing, the input values must satisfy the reverse inequality: x 2x Thus, g(x) is increasing on [0, 1 2 ] . For g(x) to be decreasing, g'(x) f'(x) - f'(1-x) f'(x) Again, since f'(x) is strictly decreasing, the input values must satisfy: x > 1 - x 2x > 1 x > 1 2 Thus, g(x) is decreasing on [ 1 2 , 1