Exams › JEE Advanced › Maths

Consider y(x) as a solution to the differential equation (1 + e^x) dy/dx + y e^x = 1, with the initial condition y(0) = 2. Which of the following assertions is/are accurate?

1. y(-4) equals 0
2. y(-2) equals 0
3. y(x) reaches a critical point within the range (-1, 0)
4. y(x) does not have a critical point within the range (-1, 0)

Correct answer: y(-4) equals 0

Solution

Solving the differential equation with the initial condition y(0) = 2 gives the solution y(x). Substituting x = -4 into the solution shows that y(-4) equals 0, confirming the correctness of this assertion.

Related JEE Advanced Maths questions

• Consider the differential equation associated with y = Σ (from i=1 to 3) C_i e^(m_i x), where C_i represents arbitrary constants and m₁, m₂, m₃ are the solutions of m³ - 7m + 6 = 0. If the equation is expressed as d³y/dx³ - 7 dy/dx + k y = 0, determine the value of k.
• Determine the order of the highest derivative raised to a power in the equation: (dy/dx)⁴ - 2x (d³y/dx³)² x² d²y/dx² d³y/dx³.
• The provided differential equation is expressed as a polynomial involving derivatives such as d²y/dx² and d³y/dx³. Among these, d³y/dx³ is the derivative of the highest order, and its greatest power in the equation is 2. What is the degree of this differential equation?
• When solving the differential equation d²y/dx² = 6x - 4, the integration gives dy/dx = 3x² - 4x + A. If dy/dx becomes zero at x = 1, what is the value of A?
• A curve passes through the point (1, π/6). Let the slope of the curve at each point (x, y) be y/x + sec(y/x), x > 0. Then the equation of the curve is -
• Let f: (0, ∞) → R be a differentiable function such that f'(x) = 2 - f(x)/x for all x ∈ (0, ∞) and f(1) ≠ 1. Then

⚔️ Practice JEE Advanced Maths free + battle 1v1 →