Question 1:
\[
\text{Which of the following is the domain of the function } f(x) = \frac{1}{x-2}?
\]
\[
\text{(a) } (-\infty, 2), \quad \text{(b) } (-\infty, 2) \cup (2, \infty), \quad \text{(c) } [2, \infty), \quad \text{(d) } (-\infty, \infty)
\]
Answer: B
Step by Step Solution
Solution:
\[
\text{The function } f(x) = \frac{1}{x-2} \text{ is undefined when } x – 2 = 0.
\]
\[
\Rightarrow x = 2
\]
\[
\text{Thus, the domain excludes } x = 2, \text{ meaning the correct domain is } (-\infty, 2) \cup (2, \infty).
\]
\[
\boxed{(-\infty, 2) \cup (2, \infty)}
\]
Question 2:
\[
\text{If } f(x) = x^2 + 3x – 4, \text{ what is } f(2)?
\]
\[
\text{(a) } 6, \quad \text{(b) } 8, \quad \text{(c) } 10, \quad \text{(d) } 4
\]
Answer: B
Step by Step Solution
Solution:
\[
f(x) = x^2 + 3x – 4
\]
\[
f(2) = (2)^2 + 3(2) – 4
\]
\[
= 4 + 6 – 4
\]
\[
= 8
\]
\[
\boxed{8}
\]
Question 3:
\[
\text{Which of the following functions is an example of an exponential function?}
\]
\[
\text{(a) } f(x) = 2^x, \quad \text{(b) } f(x) = x^2, \quad \text{(c) } f(x) = x + 1, \quad \text{(d) } f(x) = \sin(x)
\]
Answer: A
Step by Step Solution
Solution:
\[
\text{An exponential function has the form } f(x) = a^x, \text{ where } a > 0.
\]
\[
\text{The only function in the options that follows this form is } f(x) = 2^x.
\]
\[
\boxed{f(x) = 2^x}
\]
Question 4:
\[
\text{Which of the following is the inverse of } f(x) = 3x – 5?
\]
\[
\text{(a) } f^{-1}(x) = \frac{x+5}{3}, \quad \text{(b) } f^{-1}(x) = \frac{x-5}{3}, \quad \text{(c) } f^{-1}(x) = 3x + 5, \quad \text{(d) } f^{-1}(x) = \frac{x}{3} + 5
\]
Answer: A
Step by Step Solution
Solution:
\[
\text{To find the inverse, replace } f(x) \text{ with } y:
\]
\[
y = 3x – 5
\]
\[
\text{Swap } x \text{ and } y:
\]
\[
x = 3y – 5
\]
\[
\text{Solve for } y:
\]
\[
3y = x + 5
\]
\[
y = \frac{x+5}{3}
\]
\[
\boxed{f^{-1}(x) = \frac{x+5}{3}}
\]
Question 5:
\[
\text{The graph of the function } f(x) = \sin(x) \text{ is:}
\]
\[
\text{(a) } \text{Always increasing}, \quad \text{(b) } \text{Always decreasing}, \quad \text{(c) } \text{Periodic}, \quad \text{(d) } \text{Linear}
\]
Answer: C
Step by Step Solution
Solution:
\[
\text{The function } \sin(x) \text{ repeats its values in a regular interval.}
\]
\[
\text{This means it is periodic with period } 2\pi.
\]
\[
\boxed{\text{Periodic}}
\]
Question 6:
\[
\text{Which of the following is the derivative of the function } f(x) = x^3 + 2x^2 – 5x?
\]
\[
\text{(a) } 3x^2 + 4x – 5, \quad \text{(b) } 3x^2 – 4x + 5, \quad \text{(c) } 3x^2 + 2x – 5, \quad \text{(d) } 2x^3 + 4x – 5
\]
Answer: A
Step by Step Solution
Solution:
\[
\text{Using the power rule: } \frac{d}{dx} (x^n) = n x^{n-1}
\]
\[
\frac{d}{dx} (x^3) = 3x^2, \quad \frac{d}{dx} (2x^2) = 4x, \quad \frac{d}{dx} (-5x) = -5
\]
\[
\text{So, } f'(x) = 3x^2 + 4x – 5
\]
\[
\boxed{3x^2 + 4x – 5}
\]
Question 7:
\[
\text{The function } f(x) = \ln(x) \text{ is the inverse of which function?}
\]
\[
\text{(a) } e^x, \quad \text{(b) } 2^x, \quad \text{(c) } x^2, \quad \text{(d) } \sin(x)
\]
Answer: A
Step by Step Solution
Solution:
\[
\text{The natural logarithm function } \ln(x) \text{ is the inverse of } e^x.
\]
\[
\text{That means } f^{-1}(x) = e^x \Rightarrow f(x) = \ln(x).
\]
\[
\boxed{e^x}
\]
Question 9:
\[
\text{Which of the following functions is even?}
\]
\[
\text{(a) } f(x) = x^3, \quad \text{(b) } f(x) = x^2, \quad \text{(c) } f(x) = \sin(x), \quad \text{(d) } f(x) = x + 1
\]
Answer: B
Step by Step Solution
Solution:
\[
\text{A function is even if } f(-x) = f(x) \text{ for all } x \text{ in the domain.}
\]
– **Option (a):** \( f(x) = x^3 \)
\[
f(-x) = (-x)^3 = -x^3 \neq f(x)
\]
\( \Rightarrow \) Not even.
– **Option (b):** \( f(x) = x^2 \)
\[
f(-x) = (-x)^2 = x^2 = f(x)
\]
\( \Rightarrow \) Even function.
– **Option (c):** \( f(x) = \sin(x) \)
\[
f(-x) = \sin(-x) = -\sin(x) \neq f(x)
\]
\( \Rightarrow \) Not even.
– **Option (d):** \( f(x) = x + 1 \)
\[
f(-x) = -x + 1 \neq f(x)
\]
\( \Rightarrow \) Not even.
Thus, the correct answer is:
\[
\boxed{f(x) = x^2}
\]
Question 10:
\[
\text{The function } f(x) = 2x – 3 \text{ represents:}
\]
\[
\text{(a) } A linear model, \quad \text{(b) } An exponential model, \quad \text{(c) } A quadratic model, \quad \text{(d) } A cubic model
\]
Answer: A
Step by Step Solution
Solution:
\[
\text{A function is linear if it is in the form } f(x) = ax + b, \text{ where } a, b \text{ are constants}.
\]
\[
\text{The given function is } f(x) = 2x – 3.
\]
### Step 1: Check the highest power of \(x\)
– The highest power of \( x \) in \( f(x) = 2x – 3 \) is **1**.
– Linear functions always have the form \( ax + b \), where \( a \) is the slope.
### Step 2: Compare with given options
– **Option (a):** **Linear model** \( \checkmark \)
– **Option (b):** **Exponential model** \( (\text{Incorrect, as exponential functions have the form } a^x) \)
– **Option (c):** **Quadratic model** \( (\text{Incorrect, as quadratic functions have } x^2) \)
– **Option (d):** **Cubic model** \( (\text{Incorrect, as cubic functions have } x^3) \)
Since \( f(x) = 2x – 3 \) fits the form of a **linear function**, the correct answer is:
\[
\boxed{\text{A linear model}}
\]
Question 11:
\[
\text{Which of the following is the integral of } f(x) = 3x^2?
\]
\[
\text{(a) } x^3 + C, \quad \text{(b) } \frac{x^3}{3} + C, \quad \text{(c) } 3x^3 + C, \quad \text{(d) } 6x^2 + C
\]
Answer: A
Step by Step Solution
Solution:
\[
\text{The integral of a function } f(x) = ax^n \text{ follows the power rule:}
\]
\[
\int ax^n \,dx = \frac{a x^{n+1}}{n+1} + C, \quad \text{for } n \neq -1.
\]
### Step 1: Identify \( a \) and \( n \)
– Given function: \( f(x) = 3x^2 \)
– Here, \( a = 3 \) and \( n = 2 \).
### Step 2: Apply the Power Rule
\[
\int 3x^2 \,dx = \frac{3x^{2+1}}{2+1} + C
\]
\[
= \frac{3x^3}{3} + C
\]
### Step 3: Simplify
\[
= x^3 + C
\]
### Step 4: Compare with Given Options
– **Option (a):** \( x^3 + C \) **(Correct)**
– **Option (b):** \( \frac{x^3}{3} + C \) **(Incorrect)**
– **Option (c):** \( 3x^3 + C \) **(Incorrect)**
– **Option (d):** \( 6x^2 + C \) **(Incorrect)**
Thus, the correct answer is:
\[
\boxed{x^3 + C}
\]
Question 12:
\[
\text{The function } f(x) = e^x \text{ is:}
\]
\[
\text{(a) } \text{Always increasing}, \quad \text{(b) } \text{Always decreasing}, \quad \text{(c) } \text{Periodic}, \quad \text{(d) } \text{Linear}
\]
Answer: A
Step by Step Solution
Solution:
### Step 1: Understand the Given Function
– The function given is:
\[
f(x) = e^x
\]
where \( e \) is Euler’s number (\(\approx 2.718\)).
### Step 2: Find the Derivative
– The derivative of \( f(x) \) determines whether it is increasing or decreasing:
\[
f'(x) = \frac{d}{dx} e^x = e^x
\]
– Since \( e^x > 0 \) for all real numbers \( x \), the function is **always increasing**.
### Step 3: Analyze Other Options
– **Option (b) Always decreasing:** Incorrect, because \( e^x \) is never decreasing.
– **Option (c) Periodic:** Incorrect, because a periodic function repeats values at regular intervals, but \( e^x \) does not.
– **Option (d) Linear:** Incorrect, because \( e^x \) is an exponential function, not a linear one.
### Conclusion:
Since the function is **always increasing**, the correct answer is:
\[
\boxed{\text{(a) Always increasing}}
\]
Question 13:
\[
\text{What is the derivative of the function } f(x) = \cos(x)?
\]
\[
\text{(a) } \sin(x), \quad \text{(b) } -\sin(x), \quad \text{(c) } -\cos(x), \quad \text{(d) } \cos(x)
\]
Answer: B
Step by Step Solution
Solution:
### Step 1: Understand the Given Function
– The function given is:
\[
f(x) = \cos(x)
\]
### Step 2: Differentiate the Function
– Using the standard derivative rule for cosine:
\[
\frac{d}{dx} \cos(x) = -\sin(x)
\]
### Step 3: Analyze the Options
– **Option (a) \( \sin(x) \)**: Incorrect, since the derivative of \( \cos(x) \) is **negative** sine.
– **Option (b) \( -\sin(x) \)**: Correct, as calculated above.
– **Option (c) \( -\cos(x) \)**: Incorrect, as the derivative of \( \cos(x) \) is not \( -\cos(x) \).
– **Option (d) \( \cos(x) \)**: Incorrect, as differentiation of cosine does not return cosine.
### Conclusion:
Since the derivative of \( f(x) = \cos(x) \) is **\( -\sin(x) \)**, the correct answer is:
\[
\boxed{\text{(b) } -\sin(x)}
\]
Question 14:
\[
\text{Which of the following represents the equation of a rational function?}
\]
\[
\text{(a) } f(x) = \frac{x^2 – 1}{x + 2}, \quad \text{(b) } f(x) = x^2, \quad \text{(c) } f(x) = x + 1, \quad \text{(d) } f(x) = e^x
\]
Answer: A
Step by Step Solution
Solution:
### Step 1: Definition of a Rational Function
– A **rational function** is a function that can be expressed as the **quotient of two polynomials**:
\[
f(x) = \frac{P(x)}{Q(x)}
\]
where \( P(x) \) and \( Q(x) \) are polynomials and \( Q(x) \neq 0 \).
### Step 2: Analyzing the Given Options
– **Option (a) \( f(x) = \frac{x^2 – 1}{x + 2} \)**:
– The numerator \( x^2 – 1 \) and the denominator \( x + 2 \) are both polynomials.
– **This is a rational function.**
– **Option (b) \( f(x) = x^2 \)**:
– This is a polynomial function, **not a rational function**.
– **Option (c) \( f(x) = x + 1 \)**:
– This is also a polynomial function, **not a rational function**.
– **Option (d) \( f(x) = e^x \)**:
– This is an **exponential function**, **not a rational function**.
### Conclusion:
Since **only option (a)** is in the form of a rational function, the correct answer is:
\[
\boxed{\text{(a) } f(x) = \frac{x^2 – 1}{x + 2}}
\]
Question 15:
\[
\text{The limit of } \frac{1}{x} \text{ as } x \to 0 \text{ is:}
\]
\[
\text{(a) } \infty, \quad \text{(b) } 0, \quad \text{(c) } -\infty, \quad \text{(d) } \text{Undefined}
\]
Answer: D
Step by Step Solution
Solution:
### Step 1: Understanding the Limit
– The function given is:
\[
f(x) = \frac{1}{x}
\]
– We need to determine the limit of \( f(x) \) as \( x \to 0 \).
### Step 2: Evaluating the Left-Hand and Right-Hand Limits
#### Case 1: \( x \to 0^+ \) (Approaching 0 from the Right)
– As \( x \) gets **closer to 0 from the positive side** (i.e., small positive values of \( x \)),
\[
\frac{1}{x} \to +\infty
\]
– Example: \( f(0.1) = 10, \quad f(0.01) = 100 \), etc.
#### Case 2: \( x \to 0^- \) (Approaching 0 from the Left)
– As \( x \) gets **closer to 0 from the negative side** (i.e., small negative values of \( x \)),
\[
\frac{1}{x} \to -\infty
\]
– Example: \( f(-0.1) = -10, \quad f(-0.01) = -100 \), etc.
### Step 3: Conclusion
– Since the **left-hand limit** is \( -\infty \) and the **right-hand limit** is \( +\infty \), the two limits **do not match**.
– Therefore, the **overall limit does not exist**.
### Final Answer:
\[
\boxed{\text{Undefined}}
\]
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