1. Problem 11: Determine the properties of the function $f : [-3, \infty) \to [-8, \infty)$ defined by $f(x) = x^2 + 6x + 1$ and identify the correct statement among the given options. Step 1: Analyze $f(x) = x^2 + 6x + 1$. To understand the function's behavior, complete the square: $$f(x) = x^2 + 6x + 1 = (x^2 + 6x + 9) - 9 + 1 = (x + 3)^2 - 8$$ Step 2: Domain is $[-3, \infty)$, range is $[( -3 + 3)^2 - 8, \infty) = [-8, \infty)$ as given. Step 3: Check if $f$ is one-to-one (injective) on $[-3, \infty)$. Since $f(x) = (x+3)^2 -8$ is a parabola opening upwards with vertex at $x=-3$, over $[-3, \infty)$ the function is increasing and hence one-to-one. Step 4: Since the range matches $[-8, \infty)$ exactly on the domain $[-3, \infty)$, $f$ is onto (surjective) on the codomain. Step 5: Find the inverse. Solve $y = (x+3)^2 - 8$ for $x$: $$y + 8 = (x+3)^2 \implies x + 3 = \pm \sqrt{y + 8}$$ Because $x \geq -3$, $(x+3) \geq 0$, so: $$x + 3 = \sqrt{y + 8} \implies x = -3 + \sqrt{y + 8}$$ Step 6: Check the given inverse options against this result: - Option (E) states $f^{-1}(x) = -3 + \sqrt{8 + x}$ which matches our derived inverse. Hence, the correct answer is (E). 2. Problem 12: Let $p$ and $q$ be primes with $2 < p < q$, and let $M$ be the set of positive integers $n$ such that $n^5$ is divisible by $p^3$ and by $64q^{11}$. Find the least integer $n$ in $M$. Step 1: Factorize the divisibility conditions. $n^5$ divisible by $p^3$ means $p^3 | n^5$. Similarly, $64q^{11} = 2^6 q^{11}$ divides $n^5$. Step 2: Prime factorization condition: for each prime factor, the power in $n^5$ must be at least as large as that in the divisor. Let the prime factorization of $n$ include $p^{\alpha}$, $q^{\beta}$, and $2^{\gamma}$. From $p$: $5\alpha \geq 3 \implies \alpha \geq \dfrac{3}{5}$, so minimum integer $\alpha = 1$. From $q$: $5 \beta \geq 11 \implies \beta \geq \dfrac{11}{5} = 2.2$, so minimum $\beta = 3$. From $2$: $5 \gamma \geq 6 \implies \gamma \geq \dfrac{6}{5} = 1.2$, so minimum $\gamma = 2$. Step 3: Thus, $$n = 2^2 \times p^1 \times q^3 = 4 p q^3$$ Step 4: Compare this with the options: Option (C) is $4 p q^3$, which matches our result. So the answer is (C). 3. Problem 13: Graph $G$ has every two distinct vertices connected by exactly one edge (i.e., a complete graph), and $G$ has 190 edges. Find the number of vertices. Step 1: The number of edges in a complete graph with $v$ vertices is: $$\frac{v(v-1)}{2} = 190$$ Step 2: Multiply both sides by 2: $$v(v-1) = 380$$ Step 3: Solve quadratic: $$v^2 - v - 380 = 0$$ Using quadratic formula: $$v = \frac{1 \pm \sqrt{1 + 4 \times 380}}{2} = \frac{1 \pm \sqrt{1521}}{2} = \frac{1 \pm 39}{2}$$ Step 4: Two solutions: $$v = \frac{1 + 39}{2} = 20, \quad v = \frac{1 - 39}{2} = -19$$ Negative number of vertices is invalid, so $v=20$. Answer is (C).