\[\textbf{Question 1:}\]
\[ \text{Find the local maximum or minimum of } f(x) = x^3 – 3x^2 + 2. \]
\[ \text{(a) } x = 0 \text{ is a local maximum}, \quad \text{(b) } x = 0 \text{ is a local minimum}, \quad \text{(c) } x = 1 \text{ is a local minimum}, \quad \text{(d) } x = 1 \text{ is a local maximum} \]
\[ \text{Answer: C} \]
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
\[
\textbf{Solution:}
\]
\[
\text{To find inflection points, we compute the second derivative:}
\]
\[
f'(x) = \frac{d}{dx} (x^3 – 3x^2 + 4x + 1) = 3x^2 – 6x + 4
\]
\[
f”(x) = \frac{d}{dx} (3x^2 – 6x + 4) = 6x – 6
\]
\[
\text{Setting } f”(x) = 0 \text{ to find inflection points:}
\]
\[
6x – 6 = 0
\]
\[
x = 1
\]
\[
\text{Thus, the correct answer is } \boxed{x = 1}.
\]
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