1. Problem: Verify if the statement sgn(−x) = −sgn(x) for all real numbers x is true. Step 1: Recall the definition of the sign function: $$\text{sgn}(x) = \begin{cases} -1 & x < 0 \\ 0 & x = 0 \\ 1 & x > 0 \end{cases}$$ Step 2: Evaluate sgn(−x) and −sgn(x) for different cases: - If $x > 0$, then $-x < 0$, so $\text{sgn}(-x) = -1$ and $-\text{sgn}(x) = -1$. - If $x < 0$, then $-x > 0$, so $\text{sgn}(-x) = 1$ and $-\text{sgn}(x) = 1$. - If $x = 0$, then $\text{sgn}(-0) = 0$ and $-\text{sgn}(0) = 0$. Step 3: Since $\text{sgn}(-x) = -\text{sgn}(x)$ holds for all $x \in \mathbb{R}$, the statement is TRUE. 2. Problem: Find the range of the relation $R = \{(x,y): y < x \text{ and } x + y < 2\}$. Step 1: The conditions are $y < x$ and $x + y < 2$. Step 2: From $x + y < 2$, we get $y < 2 - x$. Step 3: Since $y$ must satisfy both $y < x$ and $y < 2 - x$, the maximum $y$ for each $x$ is the minimum of $x$ and $2 - x$. Step 4: The function $\min(x, 2 - x)$ achieves its maximum at $x = 1$, where both are equal to 1. Step 5: Therefore, the range of $y$ is all values less than 1, i.e., $(-\infty, 1)$. Step 6: The statement that the range is $(-\infty, 1)$ is TRUE. 3. Problem: Check if the power function $f(x) = \frac{1}{\sqrt{x}}$ satisfies $f(xy) = f(x)f(y)$. Step 1: Compute $f(xy) = \frac{1}{\sqrt{xy}} = \frac{1}{\sqrt{x}\sqrt{y}}$. Step 2: Compute $f(x)f(y) = \frac{1}{\sqrt{x}} \cdot \frac{1}{\sqrt{y}} = \frac{1}{\sqrt{x}\sqrt{y}}$. Step 3: Since $f(xy) = f(x)f(y)$, the function satisfies the condition. Step 4: The statement says it does not satisfy the condition, so the statement is FALSE. 4. Problem: Find the domain of $f(x) = \sqrt{1 - \sqrt{x} - 3}$. Step 1: Simplify the expression inside the outer square root: $1 - \sqrt{x} - 3 = -2 - \sqrt{x}$. Step 2: For the square root to be defined, the inside must be $\geq 0$: $$-2 - \sqrt{x} \geq 0 \implies -\sqrt{x} \geq 2 \implies \sqrt{x} \leq -2$$ Step 3: Since $\sqrt{x} \geq 0$ for $x \geq 0$, $\sqrt{x} \leq -2$ is impossible. Step 4: Therefore, the domain is empty, so the given domain $[-4, -3] \cap [3,4]$ is FALSE. 5. Problem: Find the solution set of $\text{sgn}(x^3 + 125) = 0$. Step 1: $\text{sgn}(z) = 0$ if and only if $z = 0$. Step 2: Set $x^3 + 125 = 0 \implies x^3 = -125 \implies x = -5$. Step 3: The solution set is $\{-5\}$, so the statement is TRUE. 6. Problem: Given $f(x) = 3x - 2$ and $g(x) = 5x + 1$, find $(f \circ g \circ f)(2)$. Step 1: Compute $f(2) = 3(2) - 2 = 6 - 2 = 4$. Step 2: Compute $g(f(2)) = g(4) = 5(4) + 1 = 20 + 1 = 21$. Step 3: Compute $f(g(f(2))) = f(21) = 3(21) - 2 = 63 - 2 = 61$. Step 4: The value is 61, so the statement is TRUE. --- 1. Problem: Identify which pair of functions are NOT inverses. Step 1: Check each pair: - A: $f(x) = \sqrt{2x}$, $g(x) = 2x^2$. - $f(g(x)) = \sqrt{2(2x^2)} = \sqrt{4x^2} = 2|x|$ which is not $x$. - So, not inverses. - B: $f(x) = \log_3(2x)$, $g(x) = \frac{1}{2}3^x$. - $f(g(x)) = \log_3(2 \cdot \frac{1}{2}3^x) = \log_3(3^x) = x$. - $g(f(x)) = \frac{1}{2}3^{\log_3(2x)} = \frac{1}{2} \cdot 2x = x$. - So, inverses. - C: $f(x) = x^3$, $g(x) = x^2$. - $f(g(x)) = (x^2)^3 = x^6 \neq x$. - Not inverses. - D: $f(x) = 3 + \sqrt{x}$, $g(x) = (x-3)^2$. - $f(g(x)) = 3 + \sqrt{(x-3)^2} = 3 + |x-3|$ which is not $x$. - Not inverses. Step 2: Pairs A, C, and D are not inverses. The question asks for which pair is NOT inverse, so answer is C (since A and D are also not inverses, but C is the clearest mismatch). 2. Problem: Identify the power function. Step 1: A power function has the form $f(x) = kx^a$ for constants $k$ and $a$. Step 2: Check options: - A: $f(x) = \sqrt{\frac{1}{x}} = x^{-1/2}$, a power function. - B: $f(x) = 2^x$, exponential function. - C: $f(x) = 3x^2 + 4$, polynomial but not a single power term. - D: $f(x) = 3^{2x+1}$, exponential. Step 3: The power function is A. 3. Problem: Find the domain of $h(x) = \frac{x^2 + 6x + 5}{x^2 - 25}$. Step 1: Denominator $x^2 - 25 = (x-5)(x+5)$ must not be zero. Step 2: So, $x \neq 5$ and $x \neq -5$. Step 3: Domain is all real numbers except $\pm 5$, i.e., $(-\infty, -5) \cup (-5, 5) \cup (5, \infty)$. Step 4: The correct option is B. 4. Problem: Find the inverse of $f(x) = \sqrt{x + 5} + 3$. Step 1: Set $y = \sqrt{x + 5} + 3$. Step 2: Solve for $x$: $$y - 3 = \sqrt{x + 5}$$ $$ (y - 3)^2 = x + 5$$ $$x = (y - 3)^2 - 5$$ Step 3: So, $f^{-1}(x) = (x - 3)^2 - 5$. Step 4: The correct option is C. 5. Problem: Given $f(x) = ax + b$ and $f(f(x)) = 4x - 1$, find $b$. Step 1: Compute $f(f(x))$: $$f(f(x)) = a(ax + b) + b = a^2 x + ab + b$$ Step 2: Equate to $4x - 1$: $$a^2 x + ab + b = 4x - 1$$ Step 3: Equate coefficients: - $a^2 = 4 \implies a = \pm 2$ - $ab + b = -1 \implies b(a + 1) = -1$ Step 4: For $a = 2$, $b(2 + 1) = -1 \implies 3b = -1 \implies b = -\frac{1}{3}$. Step 5: For $a = -2$, $b(-2 + 1) = -1 \implies -b = -1 \implies b = 1$ (not in options). Step 6: The possible value of $b$ is $-\frac{1}{3}$. 6. Problem: Identify the function $f$ such that $f(-x) = -f(x)$ (odd function). Step 1: Check each option: - A: $f(x) = x \cdot \text{sgn}(x)$. - $f(-x) = -x \cdot \text{sgn}(-x) = -x \cdot (-\text{sgn}(x)) = x \cdot \text{sgn}(x) = f(x)$, so $f(-x) = f(x)$, even function. - B: $f(x) = x^5 + 3x^3 - 3$. - $f(-x) = (-x)^5 + 3(-x)^3 - 3 = -x^5 - 3x^3 - 3 \neq -f(x)$. - C: $f(x) = \frac{x^2 - 1}{x^3 + x}$. - Check parity: - Numerator is even, denominator is odd, so $f(-x) = \frac{x^2 - 1}{-x^3 - x} = -f(x)$. - So, $f$ is odd. - D: $f(x) = |x + 2|$, not odd. Step 2: The correct answer is C. 7. Problem: Find the inverse of $f(x) = \frac{4x}{3x + 4}$ with domain $\mathbb{R} \setminus \{-\frac{4}{3}\}$. Step 1: Set $y = \frac{4x}{3x + 4}$. Step 2: Solve for $x$: $$y(3x + 4) = 4x$$ $$3xy + 4y = 4x$$ $$3xy - 4x = -4y$$ $$x(3y - 4) = -4y$$ $$x = \frac{-4y}{3y - 4} = \frac{4y}{4 - 3y}$$ Step 3: So, inverse function is $g(x) = \frac{4x}{4 - 3x}$. Step 4: The correct option is B. --- Final answers: 1. TRUE 2. TRUE 3. FALSE 4. FALSE 5. TRUE 6. TRUE Multiple choice answers: 1. C 2. A 3. B 4. C 5. B 6. C 7. B