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.
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.
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.
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.