Chapter 6: Applications of Derivatives — Online MCQ Test
MATHS · Grade 12 · CBSE(NCERT)
Practice Chapter 6: Applications of Derivatives with a free chapter-wise online MCQ test.
This chapter covers: 1. Rate of Change of QuantitiesCore Concept: The derivative \(\frac{dy}{dx}\) represents the rate of change of \(y\) with respect to \(x\).Chain Rule Application: If two variables....
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Chapter 6: Applications of Derivatives — Important Questions & Answers
The derivative dy/dx represents which of the following?
- A. The instantaneous rate of change of y with respect to x
- B. The average value of y over an interval
- C. The total change in y
- D. The slope of the function at infinity
Answer: A. The instantaneous rate of change of y with respect to x
The derivative dy/dx is defined as the instantaneous rate of change of y with respect to x at any given point.
The derivative dy/dx is defined as the instantaneous rate of change of y with respect to x at any given point.
If both x and y vary with respect to time t, then the relationship between dy/dx, dy/dt, and dx/dt is:
- A. dy/dx = (dy/dt) × (dx/dt)
- B. dy/dx = (dy/dt) ÷ (dx/dt), where dx/dt ≠ 0
- C. dy/dx = (dx/dt) ÷ (dy/dt)
- D. dy/dx = (dy/dt) + (dx/dt)
Answer: B. dy/dx = (dy/dt) ÷ (dx/dt), where dx/dt ≠ 0
By the chain rule, when two variables vary with respect to a third variable, dy/dx = (dy/dt)/(dx/dt), provided dx/dt ≠ 0.
By the chain rule, when two variables vary with respect to a third variable, dy/dx = (dy/dt)/(dx/dt), provided dx/dt ≠ 0.
The rate of change of area of a circle with respect to time is given by:
- A. dA/dt = πr dr/dt
- B. dA/dt = 2πr dr/dt
- C. dA/dt = 4πr dr/dt
- D. dA/dt = πr² dr/dt
Answer: B. dA/dt = 2πr dr/dt
Since A = πr², differentiating with respect to t gives dA/dt = 2πr(dr/dt).
Since A = πr², differentiating with respect to t gives dA/dt = 2πr(dr/dt).
The rate of change of volume of a sphere with respect to radius is:
- A. dV/dr = 4πr
- B. dV/dr = 4πr²
- C. dV/dr = (4/3)πr²
- D. dV/dr = 12πr²
Answer: B. dV/dr = 4πr²
Since V = (4/3)πr³, differentiating with respect to r gives dV/dr = 4πr².
Since V = (4/3)πr³, differentiating with respect to r gives dV/dr = 4πr².
A water tank is being filled such that the radius increases at 2 cm/s. The rate of change of volume when r = 5 cm is:
- A. 20π cm³/s
- B. 100π cm³/s
- C. 200π cm³/s
- D. 50π cm³/s
Answer: C. 200π cm³/s
Using dV/dt = 4πr²(dr/dt) = 4π(5)²(2) = 4π(25)(2) = 200π cm³/s.
Using dV/dt = 4πr²(dr/dt) = 4π(5)²(2) = 4π(25)(2) = 200π cm³/s.
Find the interval on which f(x) = x³ - 3x² - 9x + 5 is strictly increasing:
- A. (-∞, -1) and (3, ∞)
- B. (-1, 3)
- C. (-∞, ∞)
- D. (0, 3)
Answer: A. (-∞, -1) and (3, ∞)
f'(x) = 3x² - 6x - 9 = 3(x² - 2x - 3) = 3(x + 1)(x - 3). f'(x) > 0 when x < -1 or x > 3.
f'(x) = 3x² - 6x - 9 = 3(x² - 2x - 3) = 3(x + 1)(x - 3). f'(x) > 0 when x < -1 or x > 3.
A cylindrical can has volume V cm³. The material cost for the lateral surface area is Rs. 10/cm² and for the base is Rs. 20/cm². The total cost function is minimized when:
- A. h = 2r
- B. h = r
- C. h = r/2
- D. h = 4r
Answer: C. h = r/2
Cost = 20(2πr²) + 10(2πrh) = 40πr² + 20πrh. With constraint πr²h = V, substitute h = V/(πr²) and minimize with respect to r to find h = r/2.
Cost = 20(2πr²) + 10(2πrh) = 40πr² + 20πrh. With constraint πr²h = V, substitute h = V/(πr²) and minimize with respect to r to find h = r/2.
Two variables u and v are related by u = v² + 3v. If v is decreasing at 5 units per second, then du/dt when v = 2 is:
- A. -35 units/s
- B. -25 units/s
- C. 35 units/s
- D. 25 units/s
Answer: A. -35 units/s
du/dt = (du/dv)(dv/dt) = (2v + 3)(-5). When v = 2: du/dt = (4 + 3)(-5) = 7(-5) = -35 units/s.
du/dt = (du/dv)(dv/dt) = (2v + 3)(-5). When v = 2: du/dt = (4 + 3)(-5) = 7(-5) = -35 units/s.
The function f(x) = |x - 2| is NOT differentiable at x = 2 because:
- A. The function is not continuous at x = 2
- B. The left and right derivatives at x = 2 are not equal
- C. The function value at x = 2 is undefined
- D. The function is decreasing on both sides of x = 2
Answer: B. The left and right derivatives at x = 2 are not equal
At x = 2, the left derivative is -1 and the right derivative is +1. Since these are not equal, the function is not differentiable at x = 2, making it a critical point.
At x = 2, the left derivative is -1 and the right derivative is +1. Since these are not equal, the function is not differentiable at x = 2, making it a critical point.
A manufacturing company must minimize cost C(x) = x³ - 6x² + 12x + 50 for production quantity x. The production level that minimizes cost is:
- A. x = 1
- B. x = 2
- C. x = 3
- D. x = 4
Answer: B. x = 2
C'(x) = 3x² - 12x + 12 = 3(x² - 4x + 4) = 3(x - 2)² = 0 at x = 2. C''(x) = 6x - 12; C''(2) = 0 (test is inconclusive). Using the First Derivative Test or noting that (x - 2)² ≥ 0 means C'(x) ≥ 0, so x = 2 is a global minimum.
C'(x) = 3x² - 12x + 12 = 3(x² - 4x + 4) = 3(x - 2)² = 0 at x = 2. C''(x) = 6x - 12; C''(2) = 0 (test is inconclusive). Using the First Derivative Test or noting that (x - 2)² ≥ 0 means C'(x) ≥ 0, so x = 2 is a global minimum.