Exams › JEE Advanced › Maths

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?

1. 5
2. 7
3. 8
4. 9

Correct answer: 8

Solution

The absolute minimum value of y = f(x) on the interval x ∈ [0, 2] is 8, which can be determined by evaluating f(x) at the critical points and endpoints of the interval, considering the local minimum value of 5 at x = 1 and the properties of the function.

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?
• The function f(x) = 2 |x| + |x+2| - ||x+2|-2| - |2|x|| has a local minimum or a local maximum at x =
• If the function e^(f(x)) assumes its minimum in the interval [0,1] at x = 1/4, which of the following is true?

⚔️ Practice JEE Advanced Maths free + battle 1v1 →