Chapter 4: Differential Equations — Online MCQ Test
BUSINESS MATHS AND STATISTICS · CLASS 12th · Tamil Nadu State Board
Practice Chapter 4: Differential Equations with a free chapter-wise online MCQ test.
This chapter covers: Exploring ordinary differential equations this chapter covers equation formation variable separation and first-order linear solutions. Students study second-order linear differenti....
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Chapter 4: Differential Equations — Important Questions & Answers
What are the order and degree of the differential equation: d^2y/dx^2 + 5(dy/dx) + y = 0?
- A. Order 1, Degree 2
- B. Order 2, Degree 1
- C. Order 2, Degree 2
- D. Order 1, Degree 1
Answer: B. Order 2, Degree 1
The order of a differential equation is the order of the highest derivative present (here, d^2y/dx^2, which is order 2). The degree is the power of this highest derivative, which is 1.
The order of a differential equation is the order of the highest derivative present (here, d^2y/dx^2, which is order 2). The degree is the power of this highest derivative, which is 1.
The integrating factor (I.F.) of the first-order linear differential equation dy/dx + Py = Q (where P and Q are functions of x only) is given by:
- A. e^(integral of P dx)
- B. e^(integral of Q dx)
- C. integral of (e^P dx)
- D. integral of P dx
Answer: A. e^(integral of P dx)
For a standard linear differential equation of the form dy/dx + Py = Q, the integrating factor is defined as I.F. = e^(integral of P dx).
For a standard linear differential equation of the form dy/dx + Py = Q, the integrating factor is defined as I.F. = e^(integral of P dx).
Find the degree of the differential equation: [1 + (dy/dx)^2]^(3/2) = d^2y/dx^2.
- A. 3
- B. 2
- C. 1
- D. Not defined
Answer: B. 2
To find the degree, we must express the equation in a rational form free of radical powers. Squaring both sides yields [1 + (dy/dx)^2]^3 = (d^2y/dx^2)^2. The power of the highest order derivative is 2.
To find the degree, we must express the equation in a rational form free of radical powers. Squaring both sides yields [1 + (dy/dx)^2]^3 = (d^2y/dx^2)^2. The power of the highest order derivative is 2.
The marginal cost (MC) of producing x units of a commodity is MC = dC/dx = 10 - 4x + 3x^2. If the fixed cost is ₹500, find the total cost function C(x).
- A. C(x) = x^3 - 2x^2 + 10x + 500
- B. C(x) = 3x^3 - 4x^2 + 10x + 500
- C. C(x) = x^3 - 2x^2 + 10x
- D. C(x) = x^3 - 4x^2 + 10x + 500
Answer: A. C(x) = x^3 - 2x^2 + 10x + 500
Integrating dC/dx yields C(x) = 10x - 2x^2 + x^3 + K. Since the fixed cost is ₹500, C(0) = 500, which means K = 500. Thus, C(x) = x^3 - 2x^2 + 10x + 500.
Integrating dC/dx yields C(x) = 10x - 2x^2 + x^3 + K. Since the fixed cost is ₹500, C(0) = 500, which means K = 500. Thus, C(x) = x^3 - 2x^2 + 10x + 500.
A firm's marginal revenue is dR/dx = 300 - 2x - 3x^2. If the total revenue R is zero when demand x is zero, find the demand function (price per unit p) where R = px.
- A. p = 300 - x - x^2
- B. p = 300 - 2x - 3x^2
- C. p = 300x - x^2 - x^3
- D. p = 300 - x/2 - x^2/3
Answer: A. p = 300 - x - x^2
Integrating the marginal revenue, we get R(x) = 300x - x^2 - x^3 + C. Since R(0) = 0, we have C = 0. Therefore, R(x) = 300x - x^2 - x^3. Since revenue R = px, the demand function p is R/x = 300 - x - x^2.
Integrating the marginal revenue, we get R(x) = 300x - x^2 - x^3 + C. Since R(0) = 0, we have C = 0. Therefore, R(x) = 300x - x^2 - x^3. Since revenue R = px, the demand function p is R/x = 300 - x - x^2.