Exams › JEE Advanced › Maths

Let f and g be differentiable functions such that g(x) = f(x) - x is a strictly increasing function. Consider F(x) = f(x) - x + x³. Then F is:

1. strictly increasing for all x in R
2. strictly decreasing for all x in R
3. strictly decreasing on (-inf, 1/sqrt(3)) and strictly increasing on (1/sqrt(3), +inf)
4. strictly increasing on (-inf, 1/sqrt(3)) and strictly decreasing on (1/sqrt(3), +inf)

Correct answer: strictly increasing for all x in R

Solution

Since g(x) = f(x) - x is strictly increasing, g'(x) > 0 for all x. Then F'(x) = f'(x) - 1 + 3x² = g'(x) + 3x² > 0 for all x, so F is strictly increasing on all of R.

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 →