Exams › JEE Advanced › Maths

If phi(x) = f(x) + f(2a - x) where f''(x) > 0, a > 0 and 0 <= x <= 2a, then which statement is correct?

1. phi(x) increases on the interval (a, 2a)
2. phi(x) increases on the interval (0, a)
3. phi(x) decreases on the interval (a, 2a)
4. None of the above

Correct answer: phi(x) increases on the interval (a, 2a)

Solution

phi'(x) = f'(x) + f'(2a-x)*(-1) = f'(x) - f'(2a-x). Since f''(x) > 0 on [0,2a], f' is strictly increasing. For x in (a, 2a): 2a - x < a < x, so f'(2a-x) < f'(x) (since f' is increasing), therefore phi'(x) = f'(x) - f'(2a-x) > 0. So phi is increasing on (a, 2a). For x in (0, a): 2a-x > a > x, so f'(2a-x) > f'(x), so phi'(x) < 0 and phi decreases on (0, a).

Related JEE Advanced Maths questions

• What is the proportion between the height of the cone with the largest possible volume that can fit inside a sphere and the sphere's diameter?
• Identify the point on the curve 9y² = x³ where the normal to the curve intersects the coordinate axes at equal distances.
• The normal to the parametric curve defined by x = a(cos θ + sin θ) and y = a(sin θ − cos θ) at any given value of θ:
• How many critical points exist for the function y = f(x) within the interval x ∈ [0, 2], if the function satisfies d²y/dx² = 6x - 4 and has a local minimum value of 5 at x = 1?
• What is the absolute minimum value of y = f(x) on the interval x ∈ [0, 2], given that f(x) satisfies the equation d²y/dx² = 6x - 4 and has a local minimum value of 5 at x = 1?
• The function f(x) = 2 |x| + |x+2| - ||x+2|-2| - |2|x|| has a local minimum or a local maximum at x =

⚔️ Practice JEE Advanced Maths free + battle 1v1 →