CA Foundation P3 · Chapter 8

Basic Applications of Differential and Integral Calculus in Business and Economics MCQs with Answers

9 multiple-choice questions on Basic Applications of Differential and Integral Calculus in Business and Economics for CA Foundation P3 Quantitative Aptitude. Try each one before revealing the answer and explanation.

1. Question 1

   The derivative of y = x⁵ + 3x² − 7 with respect to x is:

   • A) 5x⁴ + 6x − 7
   • B) x⁴ + 6x
   • C) 5x⁴ + 3x
   • D) 5x⁴ + 6x
   Show answer & explanation

   Answer: D) 5x⁴ + 6x

   Using d/dx(xⁿ) = nxⁿ⁻¹ and d/dx(constant) = 0: dy/dx = 5x⁴ + 3(2x) − 0 = 5x⁴ + 6x. Keeping the constant −7 is a common error.

2. Question 2

   The derivative of x²eˣ with respect to x is:

   • A) 2xeˣ
   • B) eˣ(x² + 2x)
   • C) x²eˣ
   • D) eˣ(x² + 2)
   Show answer & explanation

   Answer: B) eˣ(x² + 2x)

   By the product rule, d/dx(uv) = u'v + uv'. With u = x², v = eˣ: 2x·eˣ + x²·eˣ = eˣ(x² + 2x). Differentiating only one factor gives 2xeˣ.

3. Question 3

   The total cost function of a firm is C = 0.5x² + 20x + 400 (in ₹). The marginal cost when x = 30 units is:

   • A) ₹50
   • B) ₹1,450
   • C) ₹48.33
   • D) ₹35
   Show answer & explanation

   Answer: A) ₹50

   MC = dC/dx = x + 20. At x = 30, MC = 30 + 20 = ₹50. ₹1,450 is the total cost (450 + 600 + 400), ₹48.33 is the average cost (1,450/30), and ₹35 forgets to differentiate 0.5x² correctly.

4. Question 4

   The derivative of y = log(3x² + 1) with respect to x is:

   • A) 1/(3x² + 1)
   • B) 6x
   • C) 3x/(3x² + 1)
   • D) 6x/(3x² + 1)
   Show answer & explanation

   Answer: D) 6x/(3x² + 1)

   By the chain rule, d/dx[log u] = (1/u)(du/dx). Here u = 3x² + 1 and du/dx = 6x, so dy/dx = 6x/(3x² + 1). Omitting du/dx gives 1/(3x² + 1).

5. Question 5

   The profit function of a firm is P = −2x² + 120x − 500 (in ₹). The maximum profit is:

   • A) ₹1,800
   • B) ₹4,900
   • C) ₹1,300
   • D) ₹3,100
   Show answer & explanation

   Answer: C) ₹1,300

   dP/dx = −4x + 120 = 0 gives x = 30, and d²P/dx² = −4 < 0 confirms a maximum. P(30) = −2(900) + 3,600 − 500 = −1,800 + 3,600 − 500 = ₹1,300. Ignoring the sign of −2x² gives ₹4,900.

6. Question 6

   ∫(6x² − 4x + 5) dx is equal to:

   • A) 12x − 4
   • B) 2x³ − 2x² + 5x + C
   • C) 6x³ − 4x² + 5x + C
   • D) 2x³ − 4x² + 5x + C
   Show answer & explanation

   Answer: B) 2x³ − 2x² + 5x + C

   Using ∫xⁿ dx = xⁿ⁺¹/(n + 1): ∫6x² dx = 2x³, ∫4x dx = 2x², ∫5 dx = 5x. Hence the integral is 2x³ − 2x² + 5x + C. 12x − 4 is the derivative, not the integral.

7. Question 7

   The value of the definite integral ∫ from 1 to 3 of (2x + 1) dx is:

   • A) 12
   • B) 8
   • C) 6
   • D) 10
   Show answer & explanation

   Answer: D) 10

   ∫(2x + 1) dx = x² + x. Evaluating: (9 + 3) − (1 + 1) = 12 − 2 = 10. Taking only the upper-limit value gives 12.

8. Question 8

   The demand function for a product is q = 100 − 2p. The price elasticity of demand (in absolute value) at p = 20 is:

   • A) 0.67
   • B) 2.00
   • C) 1.50
   • D) 0.40
   Show answer & explanation

   Answer: A) 0.67

   At p = 20, q = 100 − 40 = 60 and dq/dp = −2. Elasticity = −(p/q)(dq/dp) = (20/60) x 2 = 0.667 ≈ 0.67 (two decimals). Using q/p instead of p/q gives 1.50, and using 100 in place of q gives 0.40.

9. Question 9

   The marginal revenue function of a firm is MR = 50 − 4x. The corresponding demand (average revenue) function is:

   • A) p = 50 − 4x
   • B) p = 50x − 2x²
   • C) p = 50 − 2x
   • D) p = 50 − x
   Show answer & explanation

   Answer: C) p = 50 − 2x

   Total revenue R = ∫(50 − 4x) dx = 50x − 2x² + C. Since revenue is zero when nothing is sold, C = 0. Demand price p = R/x = 50 − 2x. 50x − 2x² is total revenue, not the demand function.