Exams › JEE Advanced › Maths

Let f(x) be a thrice-differentiable polynomial such that f(1) = 1, f(2) = 8, f(3) = 27, and f(4) = 64. Which of the following is/are always true?

1. f'''(x) = 6 for all x in R
2. There always exists at least one x in (1, 4) such that f'''(x) = 6
3. There always exists at least one x in (2, 3) such that f'(x) = 19 = f'''(x) = f(x)
4. There always exists at least one x in (1, 2) such that f'(x) = 7 = f'''(x) = f(x)

Correct answer: There always exists at least one x in (1, 4) such that f'''(x) = 6

Solution

Let g(x) = f(x) - x³; then g vanishes at 1,2,3,4. By applying Rolle's theorem three times, g''' must vanish at some point in (1,4), which means f'''(x) = 6 there. Option A is not guaranteed because f could have higher-degree terms.

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 →