Applications of Differentiation MCQs in Calculus

By: Prof. Dr. Fazal Rehman Shamil | Last updated: February 12, 2025

\[\textbf{Question 1:}\]
\[ \text{Find the local maximum or minimum of } f(x) = x^3 – 3x^2 + 2. \]
\[ \text{(a) } x = 0 \text{ is a local maximum}, \quad \text{(b) } x = 0 \text{ is a local minimum}, \quad \text{(c) } x = 1 \text{ is a local minimum}, \quad \text{(d) } x = 1 \text{ is a local maximum} \]
\[ \text{Answer: C} \]

Step by Step Solution

\[ \textbf{Solution:} \]
\[ f'(x) = \frac{d}{dx} (x^3 – 3x^2 + 2) \]
\[ = 3x^2 – 6x \]
\[ \text{Set } f'(x) = 0 \text{ to find critical points:} \]
\[ 3x^2 – 6x = 0 \]
\[ 3x(x – 2) = 0 \Rightarrow x = 0, x = 2 \]
\[ \text{Using the second derivative test:} \]
\[ f”(x) = 6x – 6 \]
\[ f”(1) = 6(1) – 6 = 0 \text{ (inconclusive, so use first derivative test)} \]
\[ \text{Checking intervals around } x=1: \]
\[ f'(0) = 3(0)^2 – 6(0) = 0, \quad f'(2) = 3(2)^2 – 6(2) = 0 \]
\[ \text{Since the sign changes from negative to positive around } x=1, \text{ it is a local minimum.} \]
\[ \boxed{x = 1 \text{ is a local minimum.}} \]

\[\textbf{Question 2:}\]
\[ \text{The rate of change of the area of a circle with respect to its radius is given by:} \]
\[ \text{(a) } 2\pi r, \quad \text{(b) } \pi r^2, \quad \text{(c) } 4\pi r, \quad \text{(d) } \frac{2\pi r}{r} \]
\[ \text{Answer: A} \]

Step by Step Solution

\[ \textbf{Solution:} \]
\[ \text{The area of a circle is given by } A = \pi r^2. \]
\[ \text{Differentiating with respect to } r: \]
\[ \frac{dA}{dr} = \frac{d}{dr} (\pi r^2) \]
\[ = 2\pi r \]
\[ \boxed{2\pi r} \]

\[\textbf{Question 3:}\]
\[ \text{The derivative of the function } f(x) = x^2 + 3x \text{ gives:} \]
\[ \text{(a) } 2x + 3, \quad \text{(b) } x^2 + 3, \quad \text{(c) } 3x + 5, \quad \text{(d) } 2x + 6 \]
\[ \text{Answer: A} \]

Step by Step Solution

\[ \textbf{Solution:} \]
\[ \frac{d}{dx} (x^2 + 3x) \]
\[ = 2x + 3 \]
\[ \boxed{2x + 3} \]

\[\textbf{Question 4:}\]
\[
\text{What is the maximum or minimum of the function } f(x) = -x^2 + 4x + 1 \text{?}
\]
\[
\text{(a) } \text{Maximum at } x = 2, \quad \text{(b) } \text{Minimum at } x = 2, \quad \text{(c) } \text{Maximum at } x = 4, \quad \text{(d) } \text{Minimum at } x = 4
\]
Answer: A

Step by Step Solution

Solution:

\[
\text{The given function is } f(x) = -x^2 + 4x + 1.
\]
\[
\text{Since the coefficient of } x^2 \text{ is negative, the parabola opens downward, meaning there is a maximum point.}
\]
\[
\text{The vertex of a quadratic function } ax^2 + bx + c \text{ occurs at } x = \frac{-b}{2a}.
\]
\[
\text{For } f(x) = -x^2 + 4x + 1, a = -1 \text{ and } b = 4.
\]
\[
\Rightarrow x = \frac{-4}{2(-1)} = \frac{4}{2} = 2.
\]
\[
\text{Thus, the function has a maximum at } x = 2.
\]
\[
\boxed{\text{Maximum at } x = 2}
\]

\[\textbf{Question 5:}\]
\[
\text{The instantaneous rate of change of a function is given by:}
\]
\[
\text{(a) } \text{The first derivative}, \quad \text{(b) } \text{The second derivative}, \quad \text{(c) } \text{The slope of the curve}, \quad \text{(d) } \text{The area under the curve}
\]
Answer: A

Step by Step Solution

Solution:

\[
\text{The instantaneous rate of change of a function at a point is given by the derivative.}
\]
\[
\text{The first derivative } f'(x) \text{ represents the slope of the tangent line at any point on the curve.}
\]
\[
\text{Thus, the correct answer is }
\]
\[
\boxed{\text{The first derivative}}
\]

\[\textbf{Question 6:}\]
\[
\text{What is the meaning of the second derivative of a function?}
\]
\[
\text{(a) } \text{Concavity of the graph}, \quad \text{(b) } \text{Slope of the graph}, \quad \text{(c) } \text{Rate of change of the rate of change}, \quad \text{(d) } \text{All of the above}
\]
Answer: D

Step by Step Solution

Solution:

\[
\text{The second derivative } f”(x) \text{ provides important information about a function.}
\]
\[
\text{1. It determines the concavity of the graph. If } f”(x) > 0, \text{ the function is concave up; if } f”(x) < 0, \text{ it is concave down.} \] \[ \text{2. It describes the rate of change of the first derivative, meaning it represents the rate of change of the rate of change.} \] \[ \text{3. It is related to the slope of the function indirectly, as it determines how the slope is changing.} \] \[ \text{Thus, the correct answer is } \] \[ \boxed{\text{All of the above}} \]

\[
\textbf{Question 7:}
\]

\[
\text{The critical points of the function } f(x) = x^3 – 6x^2 + 9x \text{ are at:}
\]

\[
\text{(a) } x = 0, \quad x = 1, \quad x = 3, \quad \text{(b) } x = 1, \quad x = 3, \quad \text{(c) } x = 0, \quad x = 2, \quad \text{(d) } x = 2
\]

\[
\text{Answer: B}
\]

Step by Step Solution

\[
\textbf{Solution:}
\]

\[
\text{To find the critical points, we first compute the derivative:}
\]

\[
f'(x) = \frac{d}{dx} (x^3 – 6x^2 + 9x)
\]

\[
= 3x^2 – 12x + 9
\]

\[
\text{Setting } f'(x) = 0 \text{ to find critical points:}
\]

\[
3x^2 – 12x + 9 = 0
\]

\[
\text{Solving } 3(x^2 – 4x + 3) = 0
\]

\[
3(x-1)(x-3) = 0
\]

\[
\Rightarrow x – 1 = 0 \quad \text{or} \quad x – 3 = 0
\]

\[
\Rightarrow x = 1, \quad x = 3
\]

\[
\text{Thus, the correct answer is } \boxed{x = 1, x = 3}.
\]

\[
\textbf{Question 8:}
\]

\[
\text{The marginal cost in economics is the derivative of:}
\]

\[
\text{(a) } \text{Total cost function}, \quad \text{(b) } \text{Revenue function}, \quad \text{(c) } \text{Profit function}, \quad \text{(d) } \text{Demand function}
\]

\[
\text{Answer: A}
\]

Step by Step Solution

\[
\textbf{Solution:}
\]

\[
\text{The marginal cost is defined as the derivative of the total cost function } C(x).
\]

\[
\text{Mathematically, it is represented as:}
\]

\[
MC = C'(x)
\]

\[
\text{Thus, the correct answer is } \boxed{\text{Total cost function}}.
\]

\[
\textbf{Question 9:}
\]

\[
\text{In optimization problems, the value of } x \text{ where } f'(x) = 0 \text{ is called:}
\]

\[
\text{(a) } \text{Critical point}, \quad \text{(b) } \text{Inflection point}, \quad \text{(c) } \text{Turning point}, \quad \text{(d) } \text{Point of inflection}
\]

\[
\text{Answer: A}
\]

Step by Step Solution

\[
\textbf{Solution:}
\]

\[
\text{A critical point occurs where } f'(x) = 0 \text{ or } f'(x) \text{ is undefined.}
\]

\[
\text{Thus, the correct answer is } \boxed{\text{Critical point}}.
\]

\[
\textbf{Question 10:}
\]

\[
\text{The slope of the tangent to the curve } y = f(x) \text{ at a point } (x_0, y_0) \text{ is given by:}
\]

\[
\text{(a) } f'(x_0), \quad \text{(b) } f(x_0), \quad \text{(c) } f”(x_0), \quad \text{(d) } x_0
\]

\[
\text{Answer: A}
\]

Step by Step Solution

\[
\textbf{Solution:}
\]

\[
\text{The derivative } f'(x) \text{ represents the slope of the tangent to the curve } y = f(x).
\]

\[
\text{At } x = x_0, \text{ the slope is given by } f'(x_0).
\]

\[
\text{Thus, the correct answer is } \boxed{f'(x_0)}.
\]

\[
\textbf{Question 11:}
\]

\[
\text{The rate of change of the volume of a sphere with respect to its radius is given by:}
\]

\[
\text{(a) } 4\pi r^2, \quad \text{(b) } 3\pi r^2, \quad \text{(c) } 4\pi r^3, \quad \text{(d) } \frac{4\pi r^3}{3}
\]

\[
\text{Answer: A}
\]

Step by Step Solution

\[
\textbf{Solution:}
\]

\[
\text{The volume of a sphere is given by:}
\]

\[
V = \frac{4}{3} \pi r^3
\]

\[
\text{Taking the derivative with respect to } r:
\]

\[
\frac{dV}{dr} = 4\pi r^2
\]

\[
\text{Thus, the correct answer is } \boxed{4\pi r^2}.
\]

\[
\textbf{Question 12:}
\]

\[
\text{If } f(x) = \ln(x^2 + 1), \text{ find the derivative of } f(x).
\]

\[
\text{(a) } \frac{2x}{x^2 + 1}, \quad \text{(b) } \frac{1}{x^2 + 1}, \quad \text{(c) } \frac{2x}{x^2 + 2}, \quad \text{(d) } \frac{1}{2x + 1}
\]

\[
\text{Answer: A}
\]

Step by Step Solution

\[
\textbf{Solution:}
\]

\[
\text{Using the chain rule:}
\]

\[
f'(x) = \frac{d}{dx} \ln(x^2 + 1)
\]

\[
= \frac{1}{x^2 + 1} \cdot \frac{d}{dx} (x^2 + 1)
\]

\[
= \frac{1}{x^2 + 1} \cdot (2x)
\]

\[
= \frac{2x}{x^2 + 1}
\]

\[
\text{Thus, the correct answer is } \boxed{\frac{2x}{x^2 + 1}}.
\]

\[
\textbf{Question 13:}
\]

\[
\text{If the velocity of a particle is given by } v(t) = 3t^2 – 5t + 2, \text{ what is the acceleration at } t = 2?
\]

\[
\text{(a) } 12, \quad \text{(b) } 10, \quad \text{(c) } 8, \quad \text{(d) } 6
\]

\[
\text{Answer: A}
\]

Step by Step Solution

\[
\textbf{Solution:}
\]

\[
\text{Acceleration is given by the derivative of velocity:}
\]

\[
a(t) = v'(t) = \frac{d}{dt} (3t^2 – 5t + 2)
\]

\[
= 6t – 5
\]

\[
\text{At } t = 2:
\]

\[
a(2) = 6(2) – 5 = 12 – 5 = 12
\]

\[
\text{Thus, the correct answer is } \boxed{12}.
\]

\[
\textbf{Question 14:}
\]

\[
\text{The second derivative test can be used to determine:}
\]

\[
\text{(a) } \text{Concavity of a function}, \quad \text{(b) } \text{Increasing or decreasing function}, \quad \text{(c) } \text{Local maximum or minimum}, \quad \text{(d) } \text{All of the above}
\]

\[
\text{Answer: D}
\]

Step by Step Solution

\[
\textbf{Solution:}
\]

\[
\text{The second derivative } f”(x) \text{ provides information about concavity:}
\]

\[
f”(x) > 0 \Rightarrow \text{Concave up (local minimum)}
\]

\[
f”(x) < 0 \Rightarrow \text{Concave down (local maximum)} \] \[ \text{Since it helps determine concavity and local extrema, the correct answer is } \boxed{\text{All of the above}}. \]

\[
\textbf{Question 15:}
\]

\[
\text{Find the points of inflection of the function } f(x) = x^3 – 3x^2 + 4x + 1.
\]

\[
\text{(a) } x = 0, \quad \text{(b) } x = 1, \quad \text{(c) } x = 2, \quad \text{(d) } \text{None of these}
\]

\[
\text{Answer: B}
\]

Step by Step Solution

\[
\textbf{Solution:}
\]

\[
\text{To find inflection points, we compute the second derivative:}
\]

\[
f'(x) = \frac{d}{dx} (x^3 – 3x^2 + 4x + 1) = 3x^2 – 6x + 4
\]

\[
f”(x) = \frac{d}{dx} (3x^2 – 6x + 4) = 6x – 6
\]

\[
\text{Setting } f”(x) = 0 \text{ to find inflection points:}
\]

\[
6x – 6 = 0
\]

\[
x = 1
\]

\[
\text{Thus, the correct answer is } \boxed{x = 1}.
\]

More MCQs on Calculus

  1. Functions and Models MCQs in Calculus
  2. Limits and Derivatives MCQs in Calculus
  3. Differentiation Rules MCQs in Calculus
  4. Applications of Differentiation MCQs in Calculus
  5. Integrals MCQs in Calculus
  6. Applications of Integration MCQs in Calculus
  7. Techniques of Integration MCQs in Calculus
  8. Differential Equations MCQs in Calculus
  9. Parametric Equations and Polar Coordinates MCQs in Calculus
  10. Infinite Sequences and Series MCQs in Calculus
  11. Vectors and the Geometry of Space MCQs in Calculus
  12. Vector Functions MCQs in Calculus
  13. Partial Derivatives MCQs in Calculus
  14. Multiple Integrals MCQs in Calculus
  15. Vector Calculus MCQs in Calculus
  16. Second-Order Differential Equations MCQs in Calculus

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